Although the words "he," "him," and "his"
are used sparingly in this manual to enhance com-
munication, they are not intended to be gender
driven nor to affront or discriminate against
anyone reading Mathematics, Volume 1,
NAVEDTRA 10069-D1.
PREFACE
The purpose of this Rate Training Manual is to aid those enlisted
men and women who need a basic knowledge of mathematics 'to carry
out their Navy duties. Obviously, to serve the wide variety of ratings
needing basic mathematics, the text must be general in nature and is
not directed, therefore, toward any one specific rating.
The early chapters that contain basic arithmetic have been designed
to give an insight into the theory behind computational processes. Even
students who have mastered basic arithmetic rules should find these
chapters interesting and useful.
Beginning with chapters on number systems and positive and
negative whole numbers, the course continues with discussions of
fractions, decimals and percents, exponents, and radicals. Following these
topics are chapters concerning common logarithms and the shde rule,
algebraic fundamentals, and factoring of polynomials.
Linear equations in one variable and in two variables are discussed in
separate chapters, followed by a chapter on ratio, proportion, and
variation. Following this are discussions of dependence, functions, and
formulas; complex numbers; and quadratic equations. The topics covered
in the last three chapters are plane figures, geometric constructions and
solid figures, and numerical trigonometry.
This training module was prepared by the Naval Education and
Training Program Development Center, Pensacola, Florida for the Chief
of Naval Education and Training.
1980 Edition
Reprinted 1985
Stock Ordering No.
0502-LP-050-3460
Published by
NAVAL EDUCATION AND TRAINING PROGRAM
DEVELOPMENT CENTER
UNITED STATES
GOVERNMENT PRINTING OFFICE
WASHINGTON, D.C.: 1980
THE UNITED STATES NAVY
GUARDIAN OF OUR COUNTRY
The United States Navy is responsible for maintaining control of the sea
and is a ready force on watch at home and overseas, capable of strong
action to preserve the peace or of instant offensive action to win in war.
It is upon the maintenance of this control that our country's glorious
future depends; the United States Navy exists to make it so.
WE SERVE WITH HONOR
Tradition, valor, and victory are the Navy's heritage from the past. To
these may be added dedication, discipline, and vigilance as the watchwords
of the present and the future.
At home or on distant stations we serve with pride, confident in the respect
of our country, our shipmates, and our families.
Our responsibilities sober us; our adversities strengthen us.
Service to God and Country is our special privilege. We serve with honor.
THE FUTURE OF THE NAVY
The Navy will always employ new weapons, new techniques, and
greater power to protect and defend the United States on the sea, under
the sea, and in the air.
Now and in the future, control of the sea gives the United States her
greatest advantage for the maintenance of peace and for victory in war.
Mobility, surprise, dispersal, and offensive power are the keynotes of
the new Navy. The roots of the Navy lie in a strong belief in the
future, in continued dedication to our tasks, and in reflection on our
heritage from the past.
Never have our opportunities and our responsibilities been greater.
ii
CONTENTS
Chapter Page
1. Number systems and sets 1
2. Positive integers 7
3. Signed numbers 19
4. Common fractions 28
5. Decimals 45
6. Percentage and measurement 55
7. Exponents and radicals 65
8. Logarithms and the slide rule 80
9. Fundamentals of algebra 98
10. Factoring polynomials Ill
11. Linear equations in one variable 120
12. Linear equations in two variables 130
13. Ratio, proportion, and variation 141
14. Dependence, functions, and formulas 151
15. Complex numbers 158
16. Quadratic equations in one variable 167
17. Plane figures 181
18. Geometric constructions and solid figures 190
19. Numerical trigonometry 199
Appendix
I. Squares, cubes, square roots, cube roots, logarithms,
and reciprocals of numbers 210
n. Natural sines, cosines, and tangents of angles
from to 90 213
in. Mathematical symbols 219
IV. Weights and measures 220
V. Formulas 221
Index 222
ACTIVE DUTY ADVANCEMENT REQUIREMENTS
REQUIREMENTS *
El to E2
E2 to E3
#tE3
to E4
#E4
to E5
tE5
toE6
|E6toE7
t E7 to E8
t E8 to E9
SERVICE
4 mos.
service
or
comple-
tion of
recruit
training.
6 mos.
as E-2.
6 mos.
as E-3.
12 mos.
as E-4.
24 mos.
as E-5.
36 mos.
as E-6.
8 years
total
enlisted
service.
36 mos.
as E-7.
8 of 11
years
total
service
must be
enlisted.
24 mos.
as E-8.
10 of 13
years
total
service
must be
enlisted.
SCHOOL
Recruit
Training.
Class A
for PR3,
DT3, PT3.
AME3,
HM 3
Class B
for AGC
MUC,
MNC.
PRACTICAL
FACTORS
Locally
prepared
check-
offs.
Records of Practical Factors, NavPers 1414/1, must be
completed for E-3 and all PO advancements.
PERFORMANCE
TEST
Specified ratings must complete
applicable performance tests be-
fore taking examinations.
ENLISTED
PERFORMANCE
EVALUATION
As used by CO
when approving
advancement.
Counts toward performance factor credit in ad-
vancement multiple.
EXAMINATIONS**
Locally
prepared
tests.
See
below.
Navy-wide examinations required
for all PO advancements.
Navy-wide,
selection board.
NAVY TRAINING
COURSE (INCLUD-
ING MILITARY
REQUIREMENTS)
Required for E-3 and all PO advancements
unless waived because of school comple-
tion, but need not be repeated if identical
course has already been completed. See
NavPers 10052 (current edition).
Correspondence
courses and
recommended
reading. See
NavPers 10052
(current edition).
AUTHORIZATION
Commanding
Officer
U.S. Naval Examining
Center
Bureau of Naval Personnel
* All advancements require commanding officer's recommendation.
t 1 year obligated service required for E-5 and E-6; 2 years for E-6, E-7, E-8 and E-9.
# Military leadership exam required for E-4 and E-5.
** For E-2 to E-3, NAVEXAMCEN exams or locally prepared tests may be used.
INACTIVE DUTY ADVANCEMENT REQUIREMENTS
REQUIREMENTS*
El to E2 E2 to E3
E3to
E4 E4 to E5
E5 to E6 E
6toE7
E8
E9
TOTAL
15 moi. 18 mos.
TIME
4 mot. 6 moi.
24 mos. 3
6 mos.
36 mos.
24 mos.
IN
GRADE
TOTAL
1 4 days 1 4 days
TRAINING
14 days 14 days
28 days 4
2 days
42 days
28 days
DUTY IN
GRADE t
PERFORMANCE
Specified ratings must complete applicable
performance tests before taking exami-
TESTS
nation.
DRILL
PARTICIPATION
Satisfactory participation as a
member of a drill unit.
PRACTICAL FACTORS
Record of Practical Factors, NavPers 1414/1, must be completed
for all advancements.
(INCLUDING MILITARY
REQUIREMENTS)
NAVY TRAINING
COURSE (INCLUDING
Completion of applicable course or courses must be entered
MILITARY REQUIRE-
in service record.
MENTS)
Standard
Standard Exam,
Exom
Standard
Exam
Selection Board.
EXAMINATION
Standard
c or
required for all PO
Also pass Mil.
Exam
Rating
Advancements.
Leadership
Training.
Exam for E-4
and E-5.
AUTHORIZATION
Commanding
Officer
U.S. Naval Examining
Center
Bureau of Naval
Personnel
* Recommendation by commanding officer required for all advancements,
t Active duty periods may be substituted for training duty.
CHAPTER 1
NUMBER SYSTEMS AND SETS
Mathematics is a basic tool. Some use of
mathematics is found in every rating in the
Navy, from the simple arithmetic of counting
for inventory purposes to the complicated equa-
tions encountered in computer and engineering
work. Storekeepers need mathematical compu-
tation in their bookkeeping. Damage Control-
men need mathematics to compute stress, cen-
ters of gravity, and maximum permissible roll.
Electronics principles are frequently stated by
means of mathematical formulas. Navigation
and engineer ing also use mathematics to a great
extent. As maritime warfare becomes more
and more complex, mathematics achieves ever
increasing importance as an essential tool.
From the point of view of the individual there
are many incentives for learning the subject.
Mathematics better equips him to do his pres-
ent job. It will help him in attaining promotions
and the corresponding pay increases. Statisti-
cally it has been found that one of the best indi-
cators of a man's potential success as a naval
officer is his understanding of mathematics.
This training course begins with the basic
facts of arithmetic and continues through some
of the early stages of algebra. An attempt is
made throughout to give an understanding of
why the rules of mathematics are true. This is
done because it is felt that rules are easier to
learn and remember if the ideas that led to
their development are understood.
Many of us have areas in our mathematics
background that are hazy, barely understood, or
troublesome. Thus, while it may at first seem
beneath your dignity to read chapters on funda-
mental arithmetic, these basic concepts may be
just the spots where your difficulties lie. These
chapters attempt to treat the subject on an adult
level that will be interesting and informative.
COUNTING
Counting is such a basic and natural process
that we rarely stop to think about it. The proc-
ess is based on the idea of ONE-TO-ONE COR-
RESPONDENCE , which is easily demonstrated
by using the fingers. When children count on
their fingers, they are placing each finger in
one-to-one correspondence with one of the ob-
jects being counted. Having outgrown finger
counting, we use numerals.
NUMERALS
Numerals are number symbols. One of the
simplest numeral systems is the Roman nu-
meral system, in which tally marks are used to
represent the objects being counted. Roman
numerals appear to be a refinement of the tally
method still in use today. By this method, one
makes short vertical marks until a total of four
is reached; when the fifth tally is counted, a
diagonal mark is drawn through the first four
marks. Grouping by fives in this way is remi-
niscent of the Roman numeral system, in which
the multiples of five are represented by special
symbols.
A number may have many "names." For
example, the number 6 may be indicated by any
of the following symbols: 9-3, 12/2, 5 + 1, or
2x3. The important thing to remember is that
a number is an idea; various symbols used to
indicate a number are merely different ways of
expressing the same idea.
POSITIVE WHOLE NUMBERS
The numbers which are used for counting in
our number system are sometimes called natu-
ral numbers. They are the positive whole num-
bers, or to use the more precise mathematical
term, positive INTEGERS. The Arabic nu-
merals from through 9 are called digits, and
an integer may have any number of digits. For
example, 5, 32, and 7,049 are all integers. The
number of digits in an integer indicates its
rank; that is, whether it is "in the hundreds,"
"in the thousands," etc. The idea of ranking
numbers in terms of tens, hundreds, thousands,
etc., is based on the PLACE VALUE concept.
PLACE VALUE
Although a system such as the Roman nu-
meral system is adequate for recording the
results of counting, it is too cumbersome for
purposes of calculation. Before arithmetic
could develop as we know it today, the following
two important concepts were needed as addi-
tions to the counting process:
1. The idea of as a number.
2. Positional notation (place value).
Positional notation is a form of coding in
which the value of each digit of a number de-
pends upon its position in relation to the other
digits of the number. The convention used in
our number system is that each digit has a
higher place value than those digits to the right
of it.
The place value which corresponds to a given
position in a number is determined by the BASE
of the number system. The base which is most
commonly used is ten, and the system with ten
as a base is called the decimal system (decem
is the Latin word for ten). Any number is as-
sumed to be a base-ten number, unless some
other base is indicated. One exception to this
rule occurs when the subject of an entire dis-
cussion is some base other than ten. For ex-
ample, in the discussion of binary (base two)
numbers later in this chapter, all numbers are
assumed to be binary numbers unless some
other base is indicated.
DECIMAL SYSTEM
In the decimal system, each digit position in
a number has ten times the value of the position
adjacent to it on the right. For example, in the
number 11, the 1 on the left is said to be in the
"tens place," and its value is 10 times as great
as that of the 1 on the right. The 1 on the right
is said to be in the "units place," with the un-
derstanding that the term "unit" in our system
refers to the numeral 1. Thus the number 11
is actually a coded symbol which means "one
ten plus one unit." Since ten plus one is eleven,
the symbol 11 represents the number eleven.
Figure 1-1 shows the names of several digit
positions in the decimal system. If we apply
this nomenclature to the digits of the integer
235, then this number symbol means "two hun-
dreds plus three tens plus five units." This
number may be expressed in mathematical
symbols as follows:
2x lOx 10 + 3x 10 + 5x 1
Notice that this bears out our earlier statement:
each digit position has 10 times the value of the
position adjacent to it on the right.
W99
UNITS
TENS
HUNDREDS
THOUSANDS
Figure 1-1. Names
of digit positions .
The integer 4,372 is a number symbol whose
meaning is "four thousands plus three hundreds
plus seven tens plus two units." Expressed in
mathematical symbols, this number is as fol-
lows:
4 x 1000 +3x100+7x10+2x1
This presentation may be broken down further,
in order to show that each digit position as 10
times the place value of the position on its
right, as follows:
4 x 10 x 100 + 3 x 10 x 10 + 7 x 10 x 1 + 2x1
The comma which appears in a number sym-
bol such as 4,372 is used for "pointing off" the
digits into groups of three beginning at the
right-hand side. The first group of three digits
on the right is the units group; the second group
is the thousands group; the third group is the
millions group; etc. Some of these groups are
shown in table 1-1.
Table 1-1. Place values and grouping.
Billions
group
Millions
group
Thousands
group
Units
group
CO
w
o
CQ
O
i-H
<0
f-t
pwJ
^H CO
3 T3
21
i-^ C
al
" CO m
CO
"C 3 w
'O ^ w
O 3 a
o
<u -; a
ij) qj
cu c3
S-sJ
"03
"i
"32-2
C 23
cu S
HH
B C is
3 a) c
KL* *S
" 1=1
W H H
53 H P
By reference to table 1-1, we can verify that
5,432,786 is read as follows: five million, four
hundred thirty-two thousand, seven hundred
eighty-six. Notice that the word "and" is not
necessary when reading numbers of this kind.
Practice problems:
1. Write the number symbol for seven thousand
two hundred eighty-one.
2. Write the meaning, in words, of the symbol
23,469.
3. If a number is in the millions, it must have
at least how many digits ?
4. If a number has 10 digits, to what number
group (thousands, millions, etc.) does it
belong?
Answers:
1. 7,281
2. Twenty-three thousand, four hundred sixty-
nine.
3. 7
4. Billions
BINARY SYSTEM
The binary number system is constructed in
the same manner as the decimal system. How-
ever, since the base in this system is two, only
two digit symbols are needed for writing num-
bers. These two digits are 1 and 0. In order
to understand why only two digit symbols are
needed in the binary system, we may make
some observations about the decimal system
and then generalize from these.
One of the most striking observations about
number systems which utilize the concept of
place value is that there is no single-digit sym-
bol for the base. For example, in the decimal
system the symbol for ten, the base, is 10. This
symbol is compounded from two digit symbols,
and its meaning may be interpreted as "one
base plus no units." Notice the implication of
this where other bases are concerned: Every
system uses the same symbol for the base,
namely 10. Furthermore, the symbol 10 is not
called "ten" except in the decimal system.
Suppose that a- number system were con-
structed with five as a base. Then the only
digit symbols needed would be 0, 1, 2, 3, and 4.
No single-digit symbol for five is needed, since
the symbol 10 in a base-five system with place
value means "one five plus no units." In gen-
eral, in a number system using base N, the
largest number for which a single -digit symbol
is needed is N minus 1. Therefore, when the
base is two the only digit symbols needed are
1 and 0.
An example of a binary number is the sym-
bol 101. We can discover the meaning of this
symbol by relating it to the decimal system.
Figure 1-2 shows that the place value of each
digit position in the binary system is two times
the place value of the position adjacent to it on
the right. Compare this with figure 1-1, in
which the base is ten rather than two.
Figure 1-2. Digit positions
in the binary system.
Placing the digits of the number 101 in their
respective blocks on figure 1-2, we find that
101 means "one four plus no twos plus one unit."
Thus 101 is the binary equivalent of decimal 5.
If we wish to convert a decimal number, such
as 7, to its binary equivalent, we must break it
into parts which are multiples of 2. Since 7 is
equal to 4 plus 2 plus 1, we say that it "con-
tains" one 4, one 2, and one unit. Therefore
the binary symbol for decimal 7 is 111.
The most common use of the binary number
system is in electronic digital computers. All
data fed to a typical electronic digital computer
is converted to binary form and the computer
performs its calculations using binary arith-
metic rather than decimal arithmetic. One of
the reasons for this is the fact that electrical
and electronic equipment utilizes many switch-
ing circuits in which there are only two operat-
ing conditions. Either the circuit is "on" or it
is "off," and a two-digit number system is
ideally suited for symbolizing such a situation.
Details concerning binary arithmetic are be-
yond the scope of this volume, but are available
in Mathematics, Volume 3, NavPers 10073, and
in Basic Electonics, NavPers 10087 -A.
Practice problems:
1. Write the decimal equivalents of the binary
numbers 1101, 1010, 1001, and 1111.
2. Write the binary equivalents of the decimal
numbers 12, 7, 14, and 3.
Answers:
1. 13, 10, 9, and 15
2. 1100, 111, 1110, and 11
SETS
Any serious study of mathematics leads the
student to investigate more than one text and
more than one way of approaching each new
topic. At the time of printing of this course,
much emphasis is being placed on so-called
modern math in the public schools. Conse-
quently, the trainee who uses this course is
likely to find considerable material, in his par-
allel reading, which uses the ideas and termi-
nology of the "new" math.
In the following paragraphs, a very brief in-
troduction to some of the set theory of modern
math is presented. Although the remainder of
this course is not based on set theory, this brief
introduction should help in making the transi-
tion from traditional methods to newer, experi-
mental methods.
DEFINITIONS AND SYMBOLS
The word "set" implies a collection or group-
ing of similar objects or symbols. The objects
in a set have at least one characteristic in com-
mon, such as similarity of appearance or pur-
pose. A set of tools would be an example of a
group of objects not necessarily similar in ap-
pearance but similar in purpose. The objects
or symbols in a set are called members or
ELEMENTS of the set.
The elements of a mathematical set are usu-
ally symbols, such as numerals, lines, or points.
For example, the integers greater thanzero and
less than 5 form a set, as follows:
{1, 2, 3, 4}
Notice that braces are used to indicate sets.
This is of ten done where the elements of the set
are not too numerous.
Since the elements of the set {2, 4, 6} are
the same as the elements of {4, 2, 6}, these two
sets are said to be equal. In other words, equal-
ity between sets has nothing to do with the order
in which the elements are arranged. Further-
more, repeated elements are not necessary.
That is, the elements of {2, 2, 3, 4} are simply
2, 3, and 4. Therefore the sets (2, 3, 4} and
{2, 2, 3, 4} are equal.
Practice problems:
1. Use the correct symbols to designate the set
of odd positive integers greater than and
less than 10.
2. Use the correct symbols to designate the set
of names of days of the week which do not
contain the letter "s".
3. List the elements of the set of natural num-
bers greater than 15 and less than 20.
4. Suppose that we have sets as follows:
A ={1,2,3} C ={1,2,3,4}
B ={1,2,2,3} D = {l, 1,2,3}
Which of these sets are equal ?
Answers:
1. jl, 3,5,7, 9}
2. {Monday, Friday}
3. 16, 17, 18, and 19
4. A = B = D
SUBSETS
Since it is inconvenient to enumerate all of
the elements of a set each time the set is men-
tioned, sets are often designated by a letter.
For example, we could let S represent the set
of all integers greater than and less than 10.
In symbols, this relationship could be stated
as follows:
S ={1,2,3,4,5,6,7,8,9}
Now suppose that we have another set, T,
which comprises all positive even integers less
than 10. This set is then defined as follows:
T ={2,4,6, 8}
Notice that every element of T is also an ele-
ment of S. This establishes the SUBSET rela-
tionship; T is said to be a subset of S.
POSITIVE INTEGERS
The most fundamental set of numbers is the
set of positive integers. This set comprises
the counting numbers (natural numbers) and in-
cludes, as subsets, all of the sets of numbers
which we have discussed. The set of natural
numbers has an outstanding characteristic: it
is infinite. This means that the successive
elements of the set continue to increase in size
without limit, each number being larger by 1
than the number preceding it. Therefore there
is no "largest" number; any number that we
might choose as larger than all others could be
increased to a larger number simply by adding
1 to it.
One way to represent the set of natural num-
bers symbolically would be as follows:
{1,2,3,4,5,6,...}
The three dots, called ellipsis, indicate that the
pattern established by the numbers shown con-
tinues without limit. In other words, the next
number in the set is understood to be 7, the
next after that is 8, etc.
POINTS AND LINES
In addition to the many sets which can be
formed with number symbols, we frequently
find it necessary in mathematics to work with
sets composed of points or lines.
A point is an idea, rather than a tangible ob-
ject, just as a number is. The mark which is
made on a piece of paper is merely a symbol
representing the point. In strict mathematical
terms, a point has no dimensions (physical size)
at all. Thus a pencil dot is only a rough picture
of a point, useful for indicating the location of
the point but certainly not to be confused with
the idea.
Now suppose that a large number of points
are placed side by side to form a "string."
Picturing this arrangement by drawing dots on
paper, we would have a "dotted line." If more
dots were placed between the dots already in
the string, with the number of dots increasing
until we could not see between them, we would
have a rough picture of a line. Once again, it
is important to emphasize that the picture is
only a symbol which represents an ideal line.
The ideal line would have length but no width or
thickness.
The foregoing discussion leads to the con-
clusion that a line is actually a set of points.
The number of elements in the set is infinite,
since the line extends in both directions without
limit.
The idea of arranging points together to
form a line may be extended to the formation of
planes (flat surfaces). A mathematical plane
is determined by three points which do not lie
on the same line. It is also determinedby two
intersecting lines.
Line Segments and Rays
When we draw a "line," label its end points
A and B, and call it "line AB," we really mean
LINE SEGMENT AB. A line segment is a sub-
set of the set of points comprising a line.
When a line is considered to have a starting
point but no stopping point (that is, it extends
without limit in one direction), it is called a
RAY. A ray is not a line segment, because it
does not terminate at both ends; it may be ap-
propriate to refer to a ray as a "half-line."
THE NUMBER LINE
As in the case of a line segment, a ray is a
subset of the set of points comprising a line.
All three lines, line segments, and rays are
subsets of the set of points comprising a plane.
Among the many devices used for represent-
ing a set of numbers, one of the most useful is
the number line. To illustrate the construction
of a number line, let us place the elements of
the set of natural numbers in one-to-one cor-
respondence with points on a line. Since the
natural numbers are equally spaced, we select
points such that the distances between them are
equal. The starting point is labeled 0, the next
point is labeled 1, the next 2, etc., using the
natural numbers in normal counting order. (See
fig. 1-3.) Such an arrangement is often referred
to as a scale, a familiar example being the
scale on a thermometer.
Thus far in our discussion, we have not men-
tioned any numbers other than integers. The
number line is an ideal device for picturing the
Figure 1-3. A number line.
5
relationship between integers and other num-
bers such as fractions and decimals. It is clear
that many points, other than those representing
integers, exist on the number line. Examples
are the points representing the numbers 1/2
(located halfway between and 1) and 2.5 (lo-
cated halfway between 2 and 3).
An interesting question arises, concerning
the "in-between" points on the number line:
How many points (numbers) exist between any
two integers? To answer this question, suppose
that we first locate the point halfway between
and 1, which corresponds to the number 1/2.
Then let us locate the point halfway between
and 1/2, which corresponds to the number 1/4.
The result of the next such halving operation
would be 1/8, the next 1/16, etc. If we need
more space to continue our halving operations
on the number line, we can enlarge our "pic-
ture" and then continue.
It soon becomes apparent that the halving
process could continue indefinitely; that is,
without limit. In other words, the number of
points between and 1 is infinite. The same is
true of any other interval on the number line.
Thus, between any two integers there is an infi-
nite set of numbers other than integers. If this
seems physically impossible, considering that
even the sharpest pencil point has some width,
remember that we are working with ideal points,
which have no physical dimensions whatsoever.
Although it is beyond the scope of this course
to discuss such topics as orders of infinity, it
is interesting to note that the set of integers
contains many subsets which are themselves
infinite. Not only are the many subsets of num-
bers other than integers infinite, but also such
subsets as the set of all odd integers and the
set of all even integers. By intuition we see
that these two subsets are infinite, as follows:
If we select a particular odd or even integer
which we think is the largest possible, a larger
one can be formed immediately by merely
adding 2.
Perhaps the most practical use for the num-
ber line is in explaining the meaning of nega-
tive numbers. Negative numbers are discussed
in detail in chapter 3 of this course.
CHAPTER 2
POSITIVE INTEGERS
The purpose of this chapter is to review the
methods of combining integers. We have al-
ready used one combination process in our dis-
cussion of counting. We will extend the idea of
counting, which is nothing more than simple ad-
dition, to develop a systematic method for add-
ing numbers of any size. We will also learn
the meaning of subtraction, multiplication, and
division.
ADDITION AND SUBTRACTION
In the following discussion, it is assumed
that the reader knows the basic addition and
subtraction tables, which present such facts as
the following: 2 + 3 = 5, 9 + 8 = 17, 8-3 = 5,
etc.
The operation of addition is indicated by a
plus sign (+) as in 8 + 4 = 12. The numbers 8
and 4 are ADDENDS and the answer (12) is their
SUM. The operation of subtraction is indicated
by a minus sign (-) as in 9 - 3 = 6. The number
9 is the MINUEND, 3 is the SUBTRAHEND, and
the answer (6) is their DIFFERENCE.
REGROUPING
Addition may be performed with the addends
arranged horizontally, if they are small enough
and not too numerous. However, the most com-
mon method of arranging the addends is to place
them in vertical columns. In this arrangement,
the units digits of all the addends are alined
vertically, as are the tens digits, the hundreds
digits, etc. The following example shows three
addends arranged properly for addition:
357
1,845
22
It is customary to draw a line below the last
addend, placing the answer below this line. Sub-
traction problems are arranged in columns in
the same manner as for addition, with a line at
the bottom and the answer below this line.
Carry and Borrow
Problems involving several addends, with
two or more digits each, usually produce sums
in one or more of the columns which are greater
than 9. For example, suppose that we perform
the following addition:
357
845
22
1,224
The answer was found by a process called
"carrying." In this process extra digits, gen-
erated when a column sum exceeds 9, are car-
ried to the next column to the left and treated
as addends in that column. Carrying may be
explained by grouping the original addends.
For example, 357 actually means 3 hundreds
plus 5 tens plus 7 units. Rewriting the problem
with each addend grouped in terms of units,
tens, etc., we would have the following:
300 +50+7
800 +40+5
20+2
1,100 + 110 + 14
The "extra" digit in the units column of the
answer represents 1 ten. We regroup the col-
umns of the answer so that the units column has
no digits representing tens, the tens column has
no digits representing hundreds, etc., as follows:
1,100 + 110 + 14 = 1,100 + 110+10 + 4
= 1,100 + 120 + 4
= 1,100 + 100 + 20 + 4
= 1,200 + 20 + 4
= 1,000 + 200 + 20 + 4
= 1,224
When we carry the 10 from the expression
10 + 4 to the tens column and place it with the
110 to make 120, the result is the same as if
we had added 1 to the digits 5, 4, and 2 in the
tens column of the original problem. There-
fore, the thought process in addition is as fol-
lows: Add the 7, 5, and 2 in the units column,
getting a sum of 14. Write down the 4 in the
units column of the answer and carry the 1 to
the tens column. Mentally add the 1 along with
the other digits in the tens column, getting a
sum of 12. Write down the 2 in the tens column
of the answer and carry the 1 to the hundreds
column. Mentally add the 1 along with the other
digits in the hundreds column, getting a sum of
12. Write down the 2 in the hundreds column of
the answer and carry the 1 to the thousands
column. If there were other digits in the thou-
sands column to which the 1 could be added, the
process would continue as before. Since there
are no digits in the thousands column of the
original problem, this final 1 is not added to
anything, but is simply written in the thousands
place in the answer.
The borrow process is the reverse of carry-
ing and is used in subtraction. Borrowing is
not necessary in such problems as 46 - 5 and
58 - 53. In the first problem, the thought proc-
ess may be "5 from 6 is 1 and bring down the 4
to get the difference, 41." In the second prob-
lem, the thought process is "3 from 8 is 5" and
"5 from 5 is zero," and the answer is 5. More
explicitly, the subtraction process in these ex-
amples is as follows:
40+ 1 =41
50 + 8
50 + 3
0+5 = 5
This illustrates that we are subtracting units
from units and tens from tens.
Now consider the following problem where
borrowing is involved:
43
8
If the student uses the borrowing method, he
may think "8 from 13 is 5 and bring down 3 to
get the difference, 35." In this case what actu-
ally was done is as follows:
30 +
13
8
30 + 5 = 35
A 10 has been borrowed from the tens column
and combined with the 3 in the units column to
make a number large enough for subtraction of
the 8. Notice that borrowing to increase the
value of the digit in the units column reduces
the value of the digit in the tens column by 1.
Sometimes it is necessary to borrow in more
than one column. For example, suppose that we
wish to subtract 2,345 from 5,234. Grouping
the minuend and subtrahend in units, tens, hun-
dreds, etc., we have the following:
5,000 + 200+30 + 4
2,000 + 300 + 40+5
Borrowing a 10 from the 30 in the tens column,
we regroup as follows:
5,000 + 200 + 20+14
2,000 + 300 + 40 + 5
The units column is now ready for subtrac-
tion. By borrowing from the hundreds column,
we can regroup so that subtraction is possible
in the tens column, as follows:
5,000 + 100 + 120 + 14
2,000 + 300 + 40 + .5
In the final regrouping, we borrow from the
thousands column to make subtraction possible
in the hundreds column, with the foil owing result:
4,000 + 1,100 + 120 + 14
2,000 + 300+40+5
2,000+ 800 + 80+ 9 = 2,889
In actual practice, the borrowing and re-
grouping are done mentally. The numbers are
written in the normal manner, as follows:
5,234
-2,345
2,889
The following thought process is used: Borrow
from the tens column, making the 4 become 14.
Subtracting in the units column, 5 from 14 is 9.
In the tens column, we now have a 2 in the min-
uend as a result of the first borrowing opera-
tion. Some students find it helpful at first to
cancel any digits that are reduced as a result
of borrowing, jotting down the digit of next lower
value just above the canceled digit.
been done in the following example:
4 12
This has
-2,345
2,889
After canceling the 3, we proceed with the
subtraction, one column at a time. We borrow
from the hundreds column to change the 2 that
we now have in the tens column into 12. Sub-
tracting in the tens column, 4 from 12 is 8.
Proceeding in the same way for the hundreds
column, 3 from 11 is 8. Finally, in the thou-
sands column, 2 from 4 is 2.
Practice problems. In problems 1 through
4, add the indicated numbers. In problems 5
through 8, subtract the lower number from the
upper.
1. Add 23, 468, 7, and 9,045.
2. 129 3.
5.
129
958
787
436
709
594
9,497
6,364
4,269
9,785
6. 8,700
5,008
Answers:
1.
5.
9,543
115
2.
6.
2,310
3,692
4. 67,856
22,851
44,238
97,156
7. 7,928
5,349
3. 29,915
7. 2,579
75,168
28,089
4. 232,101
8. 47,079
Denominate Numbers
Numbers that have a unit of measure asso-
ciated with them, such as yard, kilowatt, pound,
pint, etc., are called DENOMINATE NUMBERS.
The word "denominate" means the numbers
have been given a name; they are not just ab-
stract symbols. To add denominate numbers,
add all units of the same kind. Simplify the re-
sult, if possible. The following example illus-
trates the addition of 6 ft 8 in. to 4 ft 5 in.:
6ft
4ft
8 in.
5 in.
10 ft 13 in.
Since 13 in. is the equivalent of 1 ft 1 in., we
regroup the answer as 11 ft 1 in.
A similar problem would be to add 20 de-
grees 44 minutes 6 seconds to 13 degrees 22
minutes 5 seconds. This is illustrated as fol-
lows:
20 deg 44 min 6 sec
13 deg 22 min 5 sec
33 deg 66 min 11 sec
min
This answer is regrouped as 34 deg 6
11 sec.
Numbers must be expressed in units of the
same kind, in order to be combined. For in-
stance, the sum of 6 kilowatts plus 1 watt is not
7 kilowatts nor is it 7 watts. The sum can only
be indicated (rather than performing the opera-
tion) unless some method is used to write these
numbers in units of the same value.
Subtraction of denominate numbers also in-
volves the regrouping idea. If we wish to sub-
tract 16 deg 8 min 2 sec from 28 deg 4 min
3 sec, for example, we would have the following
arrangement:
28 deg 4 min 3 sec
-16 deg 8 min 2 sec
In order to subtract 8 min from 4 min we re-
group as follows:
27 deg 64 min 3 sec
-16 deg 8 min 2 sec
11 deg 56 min 1 sec
Practice problems. In problems 1, 2, and 3 add. In
problems 4, 5, and 6 subtract the lower number from
the upper.
1. 6yd
2yd
7 in.
9 in.
10 in.
2. 9hr 47 min 51 sec
3 hr 36 min 23 sec
5 hr 15 min 23 sec
4. 15 hr 25 min 10 sec
6 hr 50 min 35 sec
5. 125 deg
47 deg 9 min 14 sec
3. 10 wks 5 days 7 hrs 6. 20 wks 2 days 10 hrs
22 wks 3 days 10 hrs 7 wks 6 days 15 hrs
3 wks 4 days 12 hrs
Answers:
1. 9 yd 2 ft 2 in.
2. 18 hr 39 min 37 sec
3. 36 wks 6 days 5 hr
4. 8 hr 34 min 35 sec
5. 77 deg 50 min 46 sec
6. 12 wks 2 days 19 hr
Mental Calculation
Mental regrouping can be used to avoid the
necessity of writing down some of the steps, or
of rewriting in columns, when groups of one-
digit or two-digit numbers are to be added or
subtracted.
One of the most common devices for rapid
addition is recognition of groups of digits whose
"am is 10. For example, in the following prob-
(m two "ten groups" have been marked with
races:
10
To add this column as grouped, you would say
to yourself, "7, 17, 22, 32." The thought should
be just the successive totals as shown above
and not such cumbersome steps as "7 + 10, 17,
+ 5, 22, + 10, 32."
When successive digits appear in a column
and their sum is less than 10, it is often con-
venient to think of them, too, as a sum rather
than separately. Thus, if adding a column in
which the sum of two successive digits is 10 or
less, group them as follows:
10
The thought process here might be, as shown
by the grouping, "5, 14, 24."
Practice problems. Add the following col-
umns from the top down, as in the preceding
example:
1. 2
7
3
6
4
1
2. 4
6
7
8
1
8
3. 88
36
59
82
28
57
4. 57
32
64
97
79
44
Answers, showing successive mental steps:
1. 2, 12, 22, 23 - - Final answer, 23
2. 10, 17, 26, 34 - - Final answer, 34
3. Units column: 14, 23, 33, 40 - - Write down
0, carry 4.
Tens column: 12, 20, 30, 35 - - Final an-
swer, 350.
4. Units column: 9, 20, 29, 33 - - Write down
3, carry 3.
Tens column: 8, 17, 26, 37 - - Final an-
swer, 373.
SUBTRACTION. In an example such as
73 - 46, the conventional approach is to place
46 under 73 and subtract units from units and
tens from tens, and write only the difference
without the intermediate steps. To do this, the
best method is to begin at the left. Thus, in the
example 73 - 46, we take 40 from 73 and then
take 6 from the result. This is done mentally,
however, and the thought would be "73, 33, 27,"
or "33, 27." In the example 84 - 21 the thought
is "64, 63" and in the example 64 - 39 the thought
is "34, 25."
Practice problems. Mentally subtract and
write only the difference:
1. 47 - 24
2. 69 - 38
3. 87 - 58
4. 86 - 73
5. 82 - 41
6. 30 - 12
Answers, showing successive mental steps:
1. 27, 23 - - Final answer, 23
2. 39, 31 - - Final answer, 31
3. 37, 29 - - Final answer, 29
4. 16, 13 - - Final answer, 13
5. 42, 41 - - Final answer, 41
6. 20, 18 - - Final answer, 18
MULTIPLICATION AND DIVISION
Multiplication may be indicated by a multi-
plication sign (x) between two numbers, a dot
10
Between two numoers, or parenmeses arouna
one or both of the numbers to be multiplied. The
following examples illustrate these methods:
6x 8 = 48
6 8 = 48
6(8) = 48
(6)(8) = 48
Notice that when a dot is used to indicate
multiplication, it is distinguished from a deci-
mal point or a period by being placed above the
line of writing, as in example 2, whereas a
period or decimal point appears on the line.
Notice also that when parentheses are used to
indicate multiplication, the numbers to be mul-
tiplied are spaced closer together than they are
when the dot or x is used.
In each of the four examples just given, 6 is
the MULTIPLIER and 8 is the MULTIPLICAND.
Both the 6 and the 8 are FACTORS, and the
more modern texts refer to them this way. The
"answer" in a multiplication problem is the
PRODUCT; in the examples just given, the
product is 48.
Division usually is indicated either by a
division sign (+) or by placing one number over
another number with a line between the num-
bers, as in the following examples:
1. 8-4 = 2
2. |=2
The number 8 is the DIVIDEND, 4 is the DIVI-
SOR, and 2 is the QUOTIENT.
MULTIPLICATION METHODS
The multiplication of whole numbers may be
thought of as a short process of adding equal
numbers. For example, 6(5) and 6x5 are read
as six 5's. Of course we could write 5 six times
and add, but if we learn that the result is 30 we
can save time. Although the concept of adding
equal numbers is quite adequate in explaining
multiplication of whole numbers, it is only a
special case of a more general definition, which
will be explained later in multiplication involv-
ing fractions.
Grouping
Let us examine the process involved in mul-
tiplying 6 times 27 to get the product 162. We
first arrange the factors in the following manner:
x6
162
The thought process is as follows:
1. 6 times 7 is 42. Write down the 2 and
carry the 4.
2. 6 times 2 is 12. Add the 4 that was car-
ried over from step 1 and write the result, 16,
beside the 2 that was written in step 1.
3. The final answer is 162.
Table 2-1 shows that the factors were grouped
in units, tens, etc. The multiplication was done
in three steps: Six times 7 units is 42 units (or
4 tens and 2 units) and six times 2 tens is 12
tens (or 1 hundred and 2 tens). Then the tens
were added and the product was written as 162.
Table 2-1. Multiplying by a
one-digit number.
W
T:
o>
T3
CO
CO
c
r-t
2
c
S3
H
P
2
7
6(27) = 162
6
4
2
1
2
1
6
2
In preparing numbers for multiplication as
in table 2-1, it is important to place the digits
of the factors in the proper columns; that is,
units must be placed in the units column, tens
in tens column, and hundreds in hundreds col-
umn. Notice that it is not necessary to write
the zero in the case of 12 tens (120) since the 1
and 2 are written in the proper columns. In
practice, the addition is done mentally, and just
the product is written without the intervening
steps.
Multiplying a number with more than two
digits by a one-digit number, as shown in table
2-2, involves no new ideas. Three times 6 units
is 18 units (1 ten and 8 units), 3 times tens is
0, and 3 times 4 hundreds is 12 hundreds (1
11
3(406) = 1,218
Thousands
Hundreds
a
o>
H
a
i-H
!
4
6
3
1
2
1
8
1
2
1
8
thousand and 2 hundreds). Notice that it is not
necessary to write the O's resulting from the
step "3 times tens is 0." The two terminal
O's of the number 1,200 are also omitted, since
the 1 and the 2 are placed in their correct col-
umns by the position of the 4.
Partial Products
In the example, 6(8) = 48, notice that the
multiplying could be done another way to get
the correct product as follows:
6(3 + 5) = 6 x 3 + 6 x 5
That is, we can break 8 into 3 and 5, multiply
each of these by the other factor, and add the
partial products. This idea is employed in
multiplying by a two-digit number. Consider
the following example:
43
x27
1,161
Breaking the 27 into 20 + 7, we have 7 units
times 43 plus 2 tens times 43, as follows:
43(20 + 7) = (43)(7) + (43)(20)
Since 7 units times 43 is 301 units, and 2 tens
times 43 is 86 tens, we have the following:
86 = 8 hundreds, 6 tens
1,161
As long as the partial products are written
in the correct columns, we can multiply begin-
ning from either the left or the right of the
multiplier. Thus, multiplying from the left, we
have
43
x27
86
301
1,161
Multiplication by a number having more places
involves no new ideas.
End Zeros
The placement of partial products must be
kept in mind when multiplying in problems in-
volving end zeros, as in the following example:
1,080
We have units times 27 plus 4 tens times 27,
as follows:
27
x40
108
1,080
The zero in the units place plays an important
part in the reading of the final product. End
zeros are often called "place holders" since
their only function in the problem is to hold the
digit positions which they occupy, thus helping to
place the other digits in the problem correctly.
The end zero in the foregoing problem can
be accounted for very nicely, while at the same
time placing the other digits correctly, by means
of a shortcut. This consists of offsetting the 40
one place to the right and then simply bringing
12
down the 0, without using it as a multiplier at
all. The problem would appear as follows:
27
x40
1,080
If the problem involves a multiplier with
more than one end 0, the multiplier is offset as
many places to the right as there are end O's.
For example, consider the following multipli-
cation in which the multiplier, 300, has two
end O's:
220
x300
66,000
Notice that there are as many place -holding
zeros at the end in the product as there are
place-holding zeros in the multiplier and the
multiplicand combined.
Placement of Decimal Points
In any whole number in the decimal system,
there is understood to be a terminating mark,
called a decimal point, at the right-hand end of
the number. Although the decimal point is sel-
dom shown except in numbers involving decimal
fractions (covered in chapter 5 of this course),
its location must be known. The placement of
the decimal point is automatically taken care of
when the end O's are correctly placed.
Practice problems. Multiply in each of the
following problems:
1. 287 x 8
2. 67x49
3. 940 x 20
Answers:
1. 2,296
2. 3,283
3. 18,800
DIVISION METHODS
4. 807 x 28
5. 694 x 80
6. 9,241 x 7,800
4. 22,596
5. 55,520
6. 72,079,800
Just as multiplication can be considered as
repeated addition, division can be considered as
repeated subtraction. For example, if we wish
to divide 12 by 4 we may subtract 4 from 12 in
successive steps and tally the number of times
that the subtraction is performed, as follows:
12
_ *
8
4 *
As indicated by the asterisks used as tally
marks, 4 has been subtracted 3 times. This
result is sometimes described by saying that
"4 is contained in 12 three times."
Since successive subtraction is too cumber-
some for rapid, concise calculation, methods
which treat division as the inverse of multipli-
cation are more useful. Knowledge of the mul-
tiplication tables should lead us to an answer
for a problem such as 12 * 4 immediately, since
we know that 3 x 4 is 12. However, a problem
such as 84 + 4 is not so easy to solve by direct
reference to the multiplication table.
One way to divide 84 by 4 is to note that 84
is the same as 80 plus 4. Thus 84 + 4 is the
same as 80 * 4 plus 4 + 4. In symbols, this can
be indicated as follows:
(When this type of division symbol is used, the
quotient is written above the vinculum as shown.)
Thus, 84 divided by 4 is 21.
From the foregoing example, it can be seen
that the regrouping is useful in division as well
as in multiplication. However, the mechanical
procedure used in division does not include
writing down the regrouped form of the divi-
dend. After becoming familiar with the proc-
ess, we find that the division can be performed
directly, one digit at a time, with the regrouping
taking place mentally. The following example
illustrates this:
14
4/56
4_
16
16
The thought process is as follows: "4 is con-
tained in 5 once" (write 1 in tens place over
the 5); "one times 4 is 4" (write 4 in tens place
13
MATHEMATICS, VOLUME 1
under 5, take the difference, and bring down 6);
and "4 is contained in 16 four times" (write 4
in units place over the 6). After a little prac-
tice, many people can do the work shown under
the dividend mentally and write only the quo-
tient, if the divisor has only 1 digit.
The divisor is sometimes too large to be
contained in the first digit of the dividend. The
following example illustrates a problem of this
kind:
Since 2 is not large enough to contain 7, we
divide 7 into the number formed by the first two
digits, 25. Seven is contained 3 times in 25; we
write 3 above the 5 of the dividend. Multiplying,
3 times 7 is 21; we write 21 below the first two
digits of the dividend. Subtracting, 25 minus 21
is 4; we write down the 4 and bring down the 2
in the units place of the dividend. We have now
formed a new dividend, 42. Seven is contained
6 times in 42; we write 6 above the 2 of the
dividend. Multiplying as before, 6 times 7 is 42;
we write this product below the dividend 42.
Subtracting, we have nothing left and the divi-
sion is complete.
Estimation
When there are two or more digits in the
divisor, it is not always easy to determine the
first digit of the quotient. An estimate must be
made, and the resulting trial quotient may be
too large or too small. For example, if 1,862
is to be divided by 38, we might estimate that
38 is contained 5 times in 186 and the first digit
of our trial divisor would be 5 . However, mul-
tiplication reveals that the product of 5 and 38
is larger than 186. Thus we would change the 5
in our quotient to 4, and the problem would then
appear as follows:
On the other hand, suppose that we had esti-
mated that 38 is contained in 186 only 3 times.
We would then have the following:
38
3
T862
114
72
Now, before we make any further moves in the
division process, it should be obvious that some-
thing is wrong. If our new dividend is large
enough to contain the divisor before bringing
down a digit from the original dividend, then the
trial quotient should have been larger . In other
words, our estimate is too small.
Proficiency in estimating trial quotients is
gained through practice and familiarity with
number combinations. For example, after a
little experience we realize that a close esti-
mate can be made in the foregoing problem by
thinking of 38 as "almost 40." It is easy to see
that 40 is contained 4 times in 186, since 4
times 40 is 160. Also, since 5 times 40 is 200,
we are reasonably certain that 5 is too large
for our trial divisor.
Uneven Division
In some division problems such as 7*3,
there is no other whole number that, when mul-
tiplied by the divisor, will give the dividend.
We use the distributive idea to show how divi-
sion is done in such a case. For example, 7-^3
could be written as follows:
Thus, we see that the quotient also carries one
unit that is to be divided by 3. It should now be
clear that 3/37 = 3/30 + 7, and that this can be
further reduced as follows:
30
3
- + - = 10
33
2 +i= 12 4-
In elementary arithmetic the part of the divi-
dend that cannot be divided evenly by the divisor
is often called a REMAINDER and is placed
next to the quotient with the prefix R. Thus, in
the foregoing example where the quotient was
method of indicating uneven division is useful
in examples such as the following:
Suppose that $13 is available for the pur-
chase of spare parts, and the parts needed cost
$3 each. Four parts can be bought with the
available money, and $1 will be left over. Since
it is not possible to buy 1/3 of a part, express-
ing the result as 4 R 1 gives a more meaningful
answer than 4 1/3.
Placement of Decimal Points
In division, as in multiplication, the place-
ment of the decimal point is important. Deter-
mining the location of the decimal point and the
number of places in the quotient can be rela-
tively simple if the work is kept in the proper
columns. For example, notice the vertical
alinement in the following problem:
We notice that the first two places in the divi-
dend are used to obtain the first place in the
quotient. Since 3 is in the hundreds column
there are two more places in the quotient (tens
place and units place). The decimal point in the
quotient is understood to be directly above the
position of the decimal point in the dividend. In
the example shown here, the decimal point is
not shown but is understood to be immediately
after the second 1.
Checking Accuracy
The accuracy of a division of numbers can
be checked by multiplying the quotient by the
divisor and adding the remainder, if any. The
result should equal the dividend. Consider the
following example:
5203
42/218541
210
85
84
Check:
141
126
15
5203
x 42
10406
20812
218526
+ 15
218541
DENOMINATE NUMBERS
We have learned that denominate numbers
are not difficult to add and subtract, provided
that units, tens, hundreds, etc., are retained in
their respective columns. Multiplication and
division of denominate numbers may also be
performed with comparative ease, by using the
experience gained in addition and subtraction.
Multiplication
In multiplying denominate numbers by inte-
gers, no new ideas are needed. If in the prob-
lem 3(5 yd 2 ft 6 in.) we remember that we can
multiply each part separately to get the correct
product (as in the example, 6(8) = 6(3) + 6(5)),
we can easily find the product, as follows:
5 yd 2 ft 6 in.
x 3
15 yd 6 ft 18 in.
Simplifying, this is
17 yd 1 ft 6 in.
When one denominate number is multiplied
by another, a question arises concerning the
products of the units of measurement. The
product of one unit times another of the same
kind is one square unit. For example, 1 ft
times 1 ft is 1 square foot, abbreviated sq ft;
2 in. times 3 in. is 6 sq in.; etc. If it becomes
necessary to multiply such numbers as 2 yd 1 ft
times 6 yd 2 ft, the foot units may be converted
to fractions of a yard, as follows:
(2 yd 1 ft)(6 yd 2 ft) = (2 1/3 yd)(6 2/3 yd)
In order to complete the multiplication, a
knowledge of fractions is needed. Fractions
are discussed in chapter 4 of this training
course.
Division
The division of denominate numbers requires
division of the highest units first; and if there
is a remainder, conversion to the next lower
unit, and repeated division until all units have
been divided.
In the example (24 gal 1 qt 1 pt) -s- 5, we per-
form the following steps:
15
MATHEMATICS, VOLUME 1
Step 1: 4 gal
5/24 gal
20
4 gal (left over)
Step 2: Convert the 4 gal left over to 16 qt and
add to the 1 qt.
Step 3:
3qt
5/TTqt
15
2 qt (left over)
Step 4: Convert the 2 qt left over to 4 pt and
add to the 1 pt.
Step 5:
Therefore, 24 gal 1 qt 1 pt divided by 5 is
4 gal 3 qt 1 pt.
Practice problems. In problems 1 through 4,
divide as indicated. In problems 5 through 8,
multiply or divide as indicated.
1. 549 + 9
2. 470/63
3. 25/^300
4. 64/74,816
Answers:
1. 61
2. 7 R 29
3. 92
4. 1,169
5. 4 hr 26 min 16 sec
x 5
6. 3(4 gal 3 qt 1 pt)
7. 67 deg 43 min 12 sec
8. 5/63 Ib 11 oz
5. 22 hr 11 min 20 sec
6. 14 gal 2 qt 1 pt
7. 33 deg 51 min 36 sec
8. 12 Ib 11 4/5 oz
ORDER OF OPERATIONS
When a series of operations involving addi-
tion, subtraction, multiplication, or division is
indicated, the order in which the operations are
performed is important only if division is in-
volved or if the operations are mixed. A se-
ries of individual additions, subtractions, or
multiplications may be performed in any order.
Thus, in
4 + 2 + 7 + 5 = 18
or
or
100 - 20 - 10 - 3 = 67
4x2x7x5= 280
the numbers may be combined in any order de-
sired. For example, they may be grouped easily
to give
and
and
6 + 12 = 18
97 - 30 = 67
40 x 7 = 280
A series of divisions should be taken in the
order written.
Thus,
100 H- 10 * 2 = 10 + 2 = 5
In a series of mixed operations, perform multi-
plications and divisions in order from left to right, then
perform additions and subtractions in order from left
to right.
For example
100 -* 4 x 5 = 25 x 5 = 125
and
60 - 25 + 5 = 60 - 5 = 55
Now consider
60 - 25 - 5 + 15 - 100 + 4 x 10
= 60 - 5 + 15 - 100 + 4 x 10
= 60 - 5 + 15 - 100 4- 40
= 115 - 105
= 10
Practice problems. Evaluate each of the
following expressions:
1. 9 4 3 + 2
2. 18 - 2 x 5 + 4
3. 90 + 2 + 9
4. 75 + 5 x 3 + 5
5. 7+1-8x4+16
Answers:
1. 5
2. 12
3. 5
4. 9
5. 6
MULTIPLES AND FACTORS
Any number that is exactly divisible by a
given number is a MULTIPLE of the given
number. For example, 24 is a multiple of 2, 3,
4, 6, 8, and 12, since it is divisible by each of
these numbers. Saying that 24 is a multiple of
3, for instance, is equivalent to saying that 3
multiplied by some whole number will give 24.
Any number is a multiple of itself and also of 1 .
Any number that is a multiple of 2 is an
EVEN NUMBER. The even numbers begin with
2 and progress by 2's as follows:
2,4,6,8, 10, 12, ...
Any number that is not a multiple of 2 is an
ODD NUMBER. The odd numbers begin with 1
and progress by 2's, as follows:
1, 3, 5, 7, 9, 11, 13, ...
Any number that can be divided into a given
number without a remainder is a FACTOR of
the given number. The given number is a mul-
tiple of any number that is one of its factors.
For example, 2, 3, 4, 6, 8, and 12 are factors
of 24. The following four equalities show vari-
ous combinations of the factors of 24:
24 = 24
24 = 12
1
2
24 = 8 3
24 = 6 4
K the number 24 is factored as completely as
possible, it assumes the form
24 = 2 2 2 3
ZERO AS A FACTOR
If any number is multiplied by zero, the
product is zero. For example, 5 times zero
equals zero and may be written 5(0) = 0. The
zero factor law tells us that, if the product of
two or more factors is zero, at least one of the
factors must be zero.
PRIME FACTORS
A number that has factors other than itself
and 1 is a COMPOSITE NUMBER. For exam-
ple, the number 15 is composite. It has the
factors 5 and 3.
A number that has no factors except itself
and 1 is a PRIME NUMBER. Since it is some-
times advantageous to separate a composite
number into prime factors, it is helpful to be
able to recognize a few prime numbers quickly.
The following series shows all the prime num-
bers up to 60:
2, 3, 5, 1, 11, 13, 17, 19, 23, 29, 31, 37,41,
43,47,53,59.
Notice that 2 is the only even prime number.
All other even numbers are divisible by 2.
Notice also that 51, for example, does not ap-
pear in the series, since it is a composite num-
ber equal to 3 x 17.
K a factor of a number is prime, it is called
a PRIME FACTOR. To separate a number into
prime factors, begin by taking out the smallest
factor. If the number is even, take out all the
2's first, then try 3 as a factor, etc. Thus, we
have the following example:
540 = 2 270
=2-2-135
=2-2-3-45
= 2 2 3 3 15
= 2-2- 3-3-3-5
Since 1 is an understood factor of every num-,
ber, we do not waste space recording it as one
of the factors in a presentation of this kind.
A convenient way of keeping track of the
prime factors is in the short division process
as follows:
2/54Q
2/2IQ
3/45
3/15
5/5 _
1
17
If a number is odd, its factors will be odd
numbers. To separate an odd number into
prime factors, take out the 3's first, if there
are any. Then try 5 as a factor, etc. As an
example,
5,775 = 3 1,925
=3-5-385
=3-5-5-77
= 3 5 5 7 11
Practice problems:
1. Which of the following are prime numbers
and which are composite numbers?
25, 7, 18, 29, 51
2. What prime numbers are factors of 36?
3. Which of the following are multiples of 3?
45, 53, 51, 39, 47
4. Find the prime factors of 27.
Answers:
1. Prime: 7, 29
Composite: 25, 18, 51
2. 36 = 2 2 3 3
3. 45, 51, 39
4. 27 = 3 3 3
Tests for Divisibility
It is often useful to be able to tell by inspec-
tion whether a number is exactly divisible by
one or more of the digits from 2 through 9. An
expression which is frequently used, although it
is sometimes misleading, is "evenly divisible."
This expression has nothing to do with the con-
cept of even and odd numbers, and it probably
should be avoided in favor of the more descrip-
tive expression, "exactly divisible." For the re-
mainder of this discussion, the word "divisible"
has the same meaning as "exactly divisible."
Several tests for divisibility are listed in the
following paragraphs:
1. A number is divisible by 2 if its right-
hand digit is even.
2. A number is divisible by 3 if the sum of
its digits is divisible by 3. For example, the
digits of the number 6,561 add to produce the
sum 18. Since 18 is divisible by 3, we know
that 6,561 is divisible by 3.
3. A number is divisible by 4 if the number
formed by the two right-hand digits is divisible
by 4. For example, the two right-hand digits of
the number 3,524 form the number 24. Since
24 is divisible by 4, we know that 3,524 is di-
visible by 4.
4. A number is divisible by 5 if its right-
hand digit is or 5.
5. A number is divisible by 6 if it is even
and the sum of its digits is divisible by 3. For
example, the sum of the digits of 64,236 is 21,
which is divisible by 3. Since 64,236 is also an
even number, we know that it is divisible by 6.
6. No short method has been found for de-
termining whether a number is divisible by 7.
7. A number is divisible by 8 if the number
formed by the three right-hand digits is divisi-
ble by 8. For example, the three right-hand
digits of the number 54,272 form the number
272, which is divisible by 8. Therefore, we
know that 54,272 is divisible by 8.
8. A number is divisible by 9 if the sum of
its digits is divisible by 9. For example, the
sum of the digits of 546,372 is 27, which is di-
visible by 9. Therefore we know that 546,372
is divisible by 9.
Practice problems. Check each of the fol-
lowing numbers for divisibility by all of the
digits except 7:
1. 242,431,231,320
2. 844,624,221,840
3. 988,446,662,640
4. 207,634,542,480
Answers: All of these numbers are divisible
by 2, 3, 4, 5, 6, 8, and 9.
18
CHAPTER 3
SIGNED NUMBERS
The positive numbers with which we have
worked in previous chapters are not sufficient
for every situation which may arise. For ex-
ample, a negative number results in the opera-
tion of subtraction when the subtrahend is larger
than the minuend.
NEGATIVE NUMBERS
When the subtrahend happens to be larger
than the minuend, this fact is indicated by plac-
ing a minus sign in front of the difference, as
in the following:
12 - 20 = -8
The difference, -8, is said to be NEGATIVE. A
number preceded by a minus sign is a NEGA-
TIVE NUMBER. The number -8 is read "minus
eight." Such a number might arise when we
speak of temperature changes. If the tempera-
ture was 12 degrees yesterday and dropped 20
degrees today, the reading today would be
12 - 20, or -8 degrees.
Numbers that show either a plus or minus
sign are called SIGNED NUMBERS. An un-
signed number is understood to be positive and
is treated as though there were a plus sign
preceding it.
If it is desired to emphasize the fact that a
number is positive, a plus sign is placed in
front of the number, as in +5, which is read
"plus five." Therefore, either +5 or 5 indi-
cates that the number 5 is positive. If a num-
ber is negative, a minus sign must appear in
front of it, as in -9.
In dealing with signed numbers it should be
emphasized that the plus and minus signs have
two separate and distinct functions. They may
indicate whether a number is positive or nega-
tive, or they may indicate the operation of ad-
dition or subtraction.
When operating entirely with positive num-
bers, it is not necessary to be concerned with
this distinction since plus or minus signs indi-
cate only addition or subtraction. However,
when negative numbers are also involved in a
computation, it is important to distinguish be-
tween a sign of operation and the sign of a
number.
DIRECTION OF MEASUREMENT
Signed numbers provide a convenient way of
indicating opposite directions with a minimum
of words. For example, an altitude of 20ft
above sea level could be designated as +20 ft.
The same distance below sea level would then
be designated as -20 ft. One of the most com-
mon devices utilizing signed numbers to indicate
direction of measurement is the thermometer.
Thermometer
The Celsius (centigrade) thermometer shown
in figure 3-1 illustrates the use of positive and
negative numbers to indicate direction of travel
above and below 0. The mark is the change-
over point, at which the signs of the scale num-
bers change from - to +.
When the thermometer is heated by the sur-
rounding air or by a hot liquid in which it is
placed, the mercury expands and travels up the
tube. After the expanding mercury passes 0,
the mark at which it comes to rest is read as a
positive temperature. If the thermometer is
allowed to cool, the mercury contracts. After
passing in its downward movement, any mark
at which it comes to rest is read as a negative
temperature.
Rectangular Coordinate System
As a matter of convenience, mathematicians
have agreed to follow certain conventions as to
the use of signed numbers in directional meas-
urement. For example, in figure 3-2, a direc-
tion to the right along the horizontal line is
positive, while the opposite direction (toward
the left) is negative. On the vertical line, di-
rection upward is positive, while direction
downward is negative. A distance of -3 units
along the horizontal line indicates a measure-
ment of 3 units to the left of starting point 0, A
distance of -3 units on the vertical line indicates
19
n
I30C
I2O
no
WATER BEGINS BOILING
STEAM BEGINS
CONDENSING
80
70
60
50
40
3O
20
+IO
/
1
10
DEGF
'
i
?EES
'
-
ICE BEGINS MELTING
WATER BEGINS
FREEZING
MERCU
-10
20
;
s
\
r
RY
" _
-30
-40;
O
WATER
O
y.
^rSrfl
o
BOILING
Figure 3-1. Celsius (centigrade)
temperature scale.
a measurement of 3 units below the starting
point.
The two lines of the rectangular coordinate
system which pass through the position are
the vertical axis and horizontal axis. Other
vertical and horizontal lines may be included,
forming a grid. When such a grid is used for
the location of points and lines, the resulting
"picture" containing points and lines is called a
GRAPH.
STARTING POINr
I 2. 3
Figure 3-2. Rectangular
coordinate system.
The Number Line
Sometimes it is important to know the rela-
tive greatness (magnitude) of positive and nega-
tive numbers. To determine whether a partic-
ular number is greater or less than another
number, think of all the numbers both positive
and negative as being arranged along a hori-
zontal line. (See fig. 3-3.)
-5 -4 -3 -2 -I
+ 1
+E +3 +4 +5
Figure 3-3. Number line showing both
positive and negative numbers.
Place zero at the middle of. the line. Let the
positive numbers extend from zero toward the
right. Let the negative numbers extend from
zero toward the left. With this arrangement,
positive and negative numbers are so located
that they progress from smaller to larger num-
bers as we move from left to right along the
line. Any number that lies to the right of a
given number is greater than the given number.
A number that lies to the left of a given number
is less than the given number. This arrange-
ment shows that any negative number is smaller
than any positive number.
The symbol for "greater than" is >. The
symbol for "less than" is <. It is easy to dis-
tinguish between these symbols because the
symbol used always opens toward the larger
number. For example, "7 is greater than 4"
can be written 7 > 4 and "-5 is less than -1"
can be written -5 < -1.
20
Absolute Value
The ABSOLUTE VALUE of a number is its
numerical value when the sign is dropped. The
absolute value of either +5 or -5 is 5. Thus,
two numbers that differ only in sign have the
same absolute value.
The symbol for absolute value consists of
two vertical bars placed one on each side of the
number, as in I -5 I = 5. Consider also the
following:
I 4 - 20 I = 16
|+7| = |-7| = 7
The expression | -7 | is read "absolute value of
minus seven."
When positive and negative numbers are
used to indicate direction of measurement, we
are concerned only with absolute value, if we
wish to know only the distance covered. For
example, in figure 3-2, if an object moves to
the left from the starting point to the point in-
dicated by -2, the actual distance covered is 2
units. We are concerned only with the fact that
1-2 | = 2, if our only interest is in the distance
and not the direction.
OPERATING WITH SIGNED NUMBERS
The number line can be used to demonstrate
addition of signed numbers. Two cases must
be considered; namely, adding numbers with
like signs and adding numbers with unlike signs.
ADDING WITH LIKE SIGNS
As an example of addition with like signs,
suppose that we use the number line (fig. 3-4)
to add 2 + 3. Since these are signed numbers,
we indicate this addition as (+2) + (+3). This
emphasizes that, among the three + signs shown,
two are number signs and one is a sign of
-*H 1
p*-
1
1
.,
- f M
1
bl
C
1
1
i "l
1 1
1 1 1 1 1 1 I
operation. Line a (fig. 3-4) above the number
line shows this addition. Find 2 on the number
line. To add 3 to it, go three units more in a
positive direction and get 5.
To add two negative numbers on the number
line, such as -2 and -3, find -2 on the number
line and then go three units more in the nega-
tive direction to get -5, as in b (fig. 3-4) above
the number line.
Observation of the results of the foregoing
operations on the number line leads us to the
following conclusion, which may be stated as a
law: To add numbers with like signs, add the
absolute values and prefix the common sign.
ADDING WITH UNLIKE SIGNS
To add a positive and a negative number,
such as (-4) + (+5), find +5 on the number line
and go four units in a negative direction, as in
line c above the number line in figure 3-4.
Notice that this addition could be performed in
the other direction. That is, we could start at
-4 and move 5 units in the positive direction.
(See line d, fig. 3-4.)
The results of our operations with mixed
signs on the number line lead to the following
conclusion, which maybe stated as a law: To
add numbers with unlike signs, find the differ-
ence between their absolute values and prefix
the sign of the numerically greater number.
The following examples show the addition of
the numbers 3 and 5 with the four possible com-
binations of signs:
8
-3
-5
-8
3
-5
-3
5
-12-11-10-9 -8 -7 -6 -S-4 -3 -2-1 I 2 3 A S 6 7 fl 9 10 1 1 12
Figure 3-4. Using the number line to add.
In the first example, 3 and 5 have like signs
and the common sign is understood to be posi-
tive. The sum of the absolute values is 8 and no
sign is prefixed to this sum, thus signifying that
the sign of the 8 is understood to be positive.
In the second example, the 3 and 5 again have
like signs, but their common sign is negative.
The sum of the absolute values is 8, and this
time the common sign is prefixed to the sum.
The answer is thus -8.
In the third example, the 3 and 5 have unlike
signs. The difference between their absolute
values is 2, and the sign of the larger addend is
negative. Therefore, the answer is -2.
In the fourth example , the 3 and 5 again have
unlike signs. The difference of the absolute
21
values is still 2, but this time the sign of the
larger addend is positive. Therefore, the sign
prefixed to the 2 is positive (understood) and
the final answer is simply 2.
These four examples could be written in a
different form, emphasizing the distinction be-
tween the sign of a number and an operational
sign, as follows:
(+3) + (+5) = +8
(-3) + (-5)
(+3) + (-5)
(-3) + (+5)
= -8
= -2
= +2
Practice problems. Add as indicated:
-10 + 5 = (-10) + (+5) = ?
Add -9, -16, and 25
- 7 - 1 - 3 = (-7) + (-1) + (-3) = ?
4. Add -22 and -13
Answers:
1. -5
2.
SUBTRACTION
3. -11
4. -35
Subtraction is the inverse of addition. When
subtraction is performed, we "take away" the
subtrahend. This means that whatever the value
of the subtrahend, its effect is to be reversed
when subtraction is indicated. In addition, the
sum of 5 and -2 is 3. In subtraction, however,
to take away the effect of the -2, the quantity +2
must be added. Thus the difference between
+5 and -2 is +7.
Keeping this idea in mind, we may now pro-
ceed to examine the various combinations of
subtraction involving signed numbers. Let us
first consider the four possibilities where the
minuend is numerically greater than the sub-
trahend, as in the following examples:
8
5
8
-5
-8
-5
-13
We may show how each of these results is
obtained by use of the number line, as shown in
figure 3-5.
In the first example, we find +8 on the num-
ber line, then subtract 5 by making a movement
that reverses its sign. Thus, we move to the
left 5 units. The result (difference) is +3. (See
line a, fig. 3-5.)
In the second example, we find +8 on the
number line, then subtract (-5) by making a
movement that will reverse its sign. Thus we
move to the right 5 units. The result in this
case is +13. (See line b, fig. 3-5.)
In the third example, we find -8 on the num-
ber line, then subtract 5 by making a movement
that reverses its sign. Thus we move to the
left 5 units. The result is -13. (See line c,
fig. 3-5.)
In the fourth example, we find -8 on the
number line, then reverse the sign of -5 by
moving 5 units to the right. The result is -3.
(See line d, fig. 3-5.)
Next, let us consider the four possibilities
that arise when the subtrahend is numerically
greater than the minuend, as in the following
examples:
5
8
-3
5
-8
13
-5
8
-13
-5
-8
In the first example, we find +5 on the num-
ber line, then subtract 8 by making a movement
(t
o
w \
10
P
1
^1
^ 1
i
i
i
IQ '
r i
-13 -12 -I MO -9 -fl -7-6-5-4-3-2-1 I 2345 6 7 8 9 10
12 13
Figure 3-5. -Subtraction by use of the number line.
22
that reverses its sign. Thus we move to the
left 8 units. The result is -3. (See line e,
fig. 3-5.)
In the second example, we find +5 on the
number line, then subtract -8 by makingamove-
ment to the right that reverses its sign. The
result is 13. (See line f, fig. 3-5.)
In the third example, we find -5 on the num-
ber line, then reverse the sign of 8 by a move-
ment to the left. The result is -13. (See line g,
fig. 3-5.)
In the fourth example, we find -5 on the num-
ber line, then reverse the sign of -8 by a move-
ment to the right. The result is 3. (See line h,
fig. 3-5.)
Careful study of the preceding examples
leads to the following conclusion, which is
stated as a law for subtraction of signed num-
bers: In any subtraction problem, mentally
change the sign of the subtrahend and proceed
as in addition.
Practice problems. In problems 1 through 4,
subtract the lower number from the upper. In
5 through 8, subtract as indicated.
1. 17
-10
2. -12
8
3. -9
-13
4. 7
16
5. 1 -(-5) = ?
6. -6 -(-8) = ?
7. 14 - 7 -(-3) =
8. -9 - 2 = ?
Answers:
1. 27
5. 6
2. -20
6. 2
3. 4
7. 10
4. -9
8. -11
MULTIPLICATION
To explain the rules for multiplication of
signed numbers, we recall that multiplication
of whole numbers may be thought of as short-
ened addition. Two types of multiplication
problems must be examined; the first type in-
volves numbers with unlike signs, and the sec-
ond involves numbers with like signs.
Unlike Signs
Consider the example
multiplicand is negative.
3(-4), in which the
This means we are
to add -4 three times; that is, 3(-4) is equal to
(-4) + (-4) + (-4), which is equal to -12. For
example, if we have three 4-dollar debts, we
owe 12 dollars in all.
When the multiplier is negative, as in -3(7),
we are to take away 7 three times. Thus, -3(7)
is equal to -(7) - (7) - (7) which is equal to -21.
For example, if 7 shells were expended in one
firing, 7 the next, and 7 the next, there would
be a loss of 21 shells in all. Thus, the rule is
as follows: The product of two numbers with
unlike signs is negative.
The law of signs for unlike signs is some-
times stated as follows: Minus times plus is
minus; plus times minus is minus. Thus a
problem such as 3(-4) can be reduced to the
following two steps:
1. Multiply the signs and write down the
sign of the answer before working with the
numbers themselves.
2. Multiply the numbers as if they were un-
signed numbers.
Using the suggested procedure, the sign of
the answer for 3(-4) is found to be minus. The
product of 3 and 4 is 12, and the final answer
is -12. When there are more than two numbers
to be multiplied, the signs are taken in pairs
until the final sign is determined.
Like Signs
When both factors are positive, as in 4(5),
the s-ign of the product is positive. We are to
add +5 four times, as follows:
4(5) = 5 ..+ 5 + 5 + 5 = 20
When both factors are negative, as in -4(-5),
the sign of the product is positive. We are to
take away -5 four times.
-4(-5) = -(-5) - (-5) - (-5) - (-5)
= +5 +5 +5 +5
= 20
Remember that taking away a negative 5 is the
same as adding a positive 5. For example,
suppose someone owes a man 20 dollars and
pays him back (or diminishes the debt) 5 dollars
at a time. He takes away a debt of 20 dollars
by giving him four positive 5 -dollar bills, or a
total of 20 positive dollars in all.
The rule developed by the foregoing example
is as follows: The product of two numbers with
like signs is positive.
Knowing that the product of two positive num-
bers or two negative numbers is positive, we
can conclude that the product of any even num-
ber of negative numbers is positive. Similarly,
23
the product of any odd number of negative num-
bers is negative.
The laws of signs may be combined as fol-
lows: Minus times plus is minus; plus times
minus is minus; minus times minus is plus;
plus times plus is plus. Use of this combined
rule may be illustrated as follows:
4(-2) (-5) (6) (-3) = -720
Taking the signs in pairs, the understood plus
on the 4 times the minus on the 2 produces a
minus. This minus times the minus on the 5
produces a plus. This plus times the under-
stood plus on the 6 produces a plus. This plus
times the minus on the 3 produces a minus, so
we know that the final answer is negative. The
product of the numbers, disregarding their
signs, is 720; therefore, the final answer is
-720.
Practice problems. Multiply as indicated:
1. 5(-8) = ?
2. -7(3) (2) = ?
3. 6(-l)(-4) = ?
4. -2(3) (-4) (5) (-6) = ?
Answers:
1. -40
2. -42
DIVISION
3. 24
4. -720
Because division is the inverse of multipli-
cation, we can quickly develop the rules for
division of signed numbers by comparison with
the corresponding multiplication rules, as in
the following examples:
1. Division involving two numbers with un-
like signs is related to multiplication with un-
like signs, as follows:
3(-4) = -12
Therefore,
-12
= -4
Thus, the rule for division with unlike signs is:
The quotient of two numbers with unlike signs
is negative.
2. Division involving two numbers with like
signs is related to multiplication with like signs,
as follows:
3(-4) = -12
Therefore, ^p = 3
Thus the rule for division with like signs is:
The quotient of two numbers with like signs is
positive.
The following examples show the application
of the rules for dividing signed numbers:
3
-12
-3
= 4
-12 = 4
3
12 _ A
-3
Practice problems. Multiply and divide as
indicated:
1. 15 * -5
2. -2(-3)/-6
Answers:
1. -3
2. -1
SPECIAL CASES
3 (-3) (4)
4. -81/9
3. 2
4. -9
Two special cases arise frequently in which
the laws of signs may be used to advantage.
The first such usage is in simplifying subtrac-
tion; the second is in changing the signs of the
numerator and denominator when division is
indicated in the form of a fraction.
Subtraction
The rules for subtraction may be simplified
by use of the laws of signs, if each expression
to be subtracted is considered as being multi-
plied by a negative sign. For example, 4 -(-5)
is the same as 4 + 5, since minus times minus
is plus. This result also establishes a basis
for the rule governing removal of parentheses.
The parentheses rule, as usually stated, is:
Parentheses preceded by a minus sign may be
removed, if the signs of all terms within the
parentheses are changed. This is illustrated
as follows:
12 -(3 - 2 + 4) = 12 - 3 + 2 - 4
24
The reason for the changes of sign is clear
when the negative sign preceding the parenthe-
ses is considered to be a multiplier for the
whole parenthetical expression.
Division in Fractional Form
Division is often indicated by writing the
dividend as the numerator, and the divisor as
the denominator, of a fraction. In algebra,
everyfraction is considered to have three signs.
The numerator has a sign, the denominator has
a sign, and the fraction itself, taken as a whole,
has a sign. In many cases, one or more of
these signs will be positive, and thus will not be
shown. For example, in the following fraction
the sign of the numerator and the sign of the
denominator are both positive (understood) and
the sign of the fraction itself is negative:
Fractions with more than one negative sign
are always reducible to a simpler form with at
most one negative sign. For example, the sign
of the numerator and the sign of the denomina-
tor may be both negative. We note that minus
divided by minus gives the same result as plus
divided by plus. Therefore, we may change to
the less complicated form having plus signs
(understood) for both numerator and denomina-
tor, as follows:
-15 +15 15
-5
+5
Since -15 divided by -5 is 3, and 15 divided
by 5 is also 3, we conclude that the change of
sign does not alter the final answer. The same
reasoning may be applied in the following ex-
ample, in which the sign of the fraction itself is
negative:
-15
-5
+15
+5
!.
5
When the fraction itself has a negative sign, as
in this example, the fraction may be enclosed
in parentheses temporarily, for the purpose of
working with the numerator and denominator
only. Then the sign of the fraction is applied
separately to the result, as follows:
If a fraction has a negative sign in one of the
three sign positions, this sign may be moved to
another position. Such an adjustment is an ad-
vantage in some types of complicated expres-
sions involving fractions. Examples of this
type of sign change follow:
!5_ = -15 _ 15
"5 5-5
In the first expression of the foregoing ex-
ample, the sign of the numerator is positive
(understood) and the sign of the fraction is neg-
ative. Changing both of these signs, we obtain
the second expression. To obtain the third ex-
pression from the second, we change the sign
of the numerator and the sign of the denomina-
tor. Observe that the sign changes in each case
involve a pair of signs. This leads to the law
of signs for fractions: Any two of the three
signs of a fraction may be changed without al-
tering the value of the fraction.
AXIOMS AND LAWS
An axiom is a self-evident truth. It is a
truth that is so universally accepted that it does
not require proof. For example, the statement
that "a straight line is the shortest distance
between two points" is an axiom from plane
geometry. One tends to accept the truth of an
axiom without proof, because anything which is
axiomatic is, by its very nature, obviously true.
On the other hand, a law (in the mathematical
sense) is the result of defining certain quanti-
ties and relationships and then developing logi-
cal conclusions from the definitions.
AXIOMS OF EQUALITY
The four axioms of equality with which we
are concerned in arithmetic and algebra are
stated as follows:
1. If the same quantity is added to each of
two equal quantities, the resulting quantities
are equal. This is sometimes stated as follows:
If equals are added to equals, the results are
equal. For example, by adding the same quan-
tity (3) to both sides of the following equation,
we obtain two sums which are equal:
-2
All of this can be done mentally.
-2 = -3
+ 3 = -3
1 = 1
+ 1
+ 1 +
25
111C Ji CDUXLD AJ. C Cl^UO..!.. JL UJL C.ft.a.lJll^/XC;, WJ 0UM~
tracting 2 from both sides of the following equa-
tion we obtain results which are equal:
5 = 2 + 3
5-2 = 2 + 3-2
3 = 3
3. If two equal quantities are multiplied by
the same quantity, the resulting products are
equal. This is sometimes stated as follows: If
equals are multiplied by equals, the products
are equal. For example, both sides of the fol-
lowing equation are multiplied by -3 and equal
results are obtained:
e duueuuo 10 uic
L cgai
5 = 2 + 3
(-3) (5)= (-3) (2 +
-15 = -15
3)
4. If two equal quantities are divided by the
same quantity, the resulting quotients are equal.
This is sometimes stated as follows: If equals
are divided by equals, the results are equal.
For example, both sides of the following equa-
tion are divided by 3, and the resulting quotients
are equal:
12 + 3 = 15
12 + 3 _ JL5
3 3
4 + 1 = 5
These axioms are especially useful when
letters are used to represent numbers. If we
know that 5x = -30, for instance, then dividing
both 5x and -30 by 5 leads to the conclusion
that x = -6.
LAWS FOR COMBINING NUMBERS
Numbers are combined in accordance with
the following basic laws:
1. The associative laws of addition and mul-
tiplication.
2. The commutative laws of addition and
multiplication.
3. The distributive law.
of the manner in which they are grouped. For
example, 6 + 3 + 1 is the same as 6 + (3 + 1) or
(6 +3) +1.
This law can be applied to subtraction by
changing signs in such a way that all negative
signs are treated as number signs rather than
operational signs. That is, some of the ad-
dends can be negative numbers. For example,
6-4-2 can be rewritten as 6 + (-4) + (-2).
By the associative law, this is the same as
6 + [(-4) + (-2)] or [6 + (-4)] + (-2).
However, 6-4-2 is not the same as 6 - (4 - 2);
the terms must be expressed as addends before
applying the associative law of addition.
Associative Law of Multiplication
This law states that the product of three or
more factors is the same regardless of the
manner in which they are grouped. For ex-
ample, 6 3 2 is the same as (6 3) 2 or
6 (3 2). Negative signs require no special
treatment in the application of this law.
For example, 6 (-4) (-2) is the same
as [6- (-4)J -(-2) or 6 -[(-4) -(-2)].
Commutative Law of Addition
The word "commute" means to change, sub-
stitute or move from place to place. The com-
mutative law of addition states that the sum of
two or more addends is the same regardless of
the order in which they are arranged. For ex-
ample, 4 + 3 + 2 is the same as 4 + 2 + 3 or
2+4 + 3.
This law can be applied to subtraction by
changing signs so that all negative signs be-
come number signs and all signs of operation
are positive. For example, 5 - 3 - 2 is changed
to 5 + (-3) + (-2), which is the same as 5 + (-2)
+ (-3) or (-3) +5 + (-2).
Commutative Law of Multiplication
This law states that the product of two or
more factors is the same regardless of the
order in which the factors are arranged. For
example, 3 4 5 is the same as 5 3 4 or
26
4-3-5. Negative signs require no special 2(3 + 4 + 5) = 2-3 + 2'4 + 2'5
treatment in the application of this law. For =6 + 8 + 10
example, 2 (-4) (-3) is the same as (-4)
(-3) 2 or (-3) 2 (-4). To verify the distributive law, we note that
2(3 + 4 + 5) is the same as 2(12) or 24. Also,
Distributive Law f 6 .t ?. + "> is 2 , 4 ' For ^cation of the dis-
tributive law where negative signs appear, the
This law combines the operations of addition following procedure is recommended:
and multiplication. The word "distributive" re- 3(4 - 2) = 3 [4 + (-2)]
fers to the distribution of a common multiplier = 3(4) + 3(-2)
among the terms of an additive expression. =12-6
For example, = 6
27
CHAPTER 4
COMMON FRACTIONS
The emphasis in previous chapters of this
course has been on integers (whole numbers).
In this chapter, we turn our attention to num-
bers which are not integers. The simplest type
of number other than an integer is a COMMON
FRACTION. Common fractions and integers
together comprise a set of numbers called the
RATIONAL NUMBERS; this set is a subset of
the set of real numbers.
The number line may be used to show the
relationship between integers and fractions.
For example, if the interval between and 1 is
marked off to form three equal spaces (thirds),
then each space so formed is one-third of the
total interval. If we move along the number line
from toward 1, we will have covered two of
the three "thirds" when we reach the second
mark. Thus the position of the second mark
represents the number 2/3. (See fig. 4-1.)
-2
>3
I
Figure 4-1. Integers and fractions on the
number line.
The numerals 2 and 3 in the fraction 2/3 are
named so that we may distinguish between them;
2 is the NUMERATOR and 3 is the DENOMINA-
TOR. In general, the numeral above the di-
viding line in a fraction is the numerator and
the numeral below the line is the denominator.
The numerator and denominator are the TERMS
of the fraction. The word "numerator" is re-
lated to the word "enumerate." To enumerate
means to "tell how many"; thus the numerator
tells us how many fractional parts we have in
the indicated fraction. To denominate means to
"give a name" or "tell what kind"; thus the de-
nominator tells us what kind of parts we have
(halves, thirds, fourths, etc.).
Attempts to define the word "fraction" in
mathematics usually result in a statement sim-
ilar to the following: A fraction is an indicated
division. Any division maybe indicated by plac-
ing the dividend over the divisor and drawing a
line between them. By this definition, any num-
ber which can be written as the ratio of two in-
tegers (one integer over the other) can be con-
sidered as a fraction. This leads to a further
definition: Any number which can be expressed
as the ratio of two integers is a RATIONAL
number. Notice that every integer is a rational
number, because we can write any integer as
the numerator of a fraction having 1 as its de-
nominator. For example, 5 is the same as 5/1.
It should be obvious from the definition that
every common fraction is also a rational
number.
TYPES OF FRACTIONS
Fractions are often classified as proper or
improper. A proper fraction is one in which the
numerator is numerically smaller than the de-
nominator. An improper fraction has a nu-
merator which is larger than its denominator.
MIXED NUMBERS
When the denominator of an improper frac-
tion is divided into its numerator, a remainder
is produced along with the quotient, unless the
numerator happens to be an exact multiple of
the denominator. For example, 7/5 is equal to
1 plus a remainder of 2. This remainder may
be shown as a dividend with 5 as its divisor, as
follows:
T_
5
The expression 1 + 2/5 is a MIXED NUM-
BER. Mixed numbers are usually written with-
out showing the plus sign; that is, 1 + 2/5 is
O
the same as 1 4 or 1 2/5. When a mixed num-
ber is written as 1 2/5, care must be taken to
insure that there is a space between the 1 and
the 2; otherwise, 1 2/5 might be taken to mean
12/5.
28
MEASUREMENT FRACTIONS
Measurement fractions occur in problems
such as the following:
If $2 were spent for a stateroom rug at $3
per yard, how many yards were bought? If $6
had been spent we could find the number of yards
by simply dividing the cost per yard into the
amount spent. Since 6/3 is 2, two yards could
be bought for $6. The same reasoning applies
when $2 are spent, but in this case we can only
indicate the amount purchased as the indicated
division 2/3. Figure 4-2 shows a diagram for
both the $6 purchase and the $2 purchase.
- 2YDS -
1 YD
1 $2 $3 $4 $5 $6
2 YARDS PURCHASED FOR $ 6 AT $ 3 PER YARD
-IYD-
$4 $Z p $7
% YARD PURCHASED FOR $2 AT $3 PER YARD
Figure 4-2. Measurement fractions.
PARTITIVE FRACTIONS
The difference between measurement frac-
tions and partitive fractions is explained as
follows: Measurement fractions result when we
determine how many pieces of a given size can
be cut from a larger piece. Partitive fractions
result when we cut a number of pieces of equal
size from a larger piece and then determine the
size of each smaller piece. For example, if 4
equal lengths of pipe are to be cut from a 3-foot
pipe, what is the size of each piece? If the
problem had read that 3 equal lengths were to
be cut from a 6-foot pipe, we could find the size
of each pipe by dividing the number of equal
lengths into the overall length. Thus, since 6/3
is 2, each piece would be 2 feet long. By this
same reasoning in the example, we divide the
overall length by the number of equal parts to
get the size of the individual pieces; that is,
3/4 foot. The partitioned 6-foot and 3-foot
pipes are shown in figure 4-3.
1
1 PART
1 PART _,
1 PART
1
1
1
1
EFT
1
2 FT
1
2 FT
1
|
IFT 2
FT 3FT 4
FT
5 FT
6
6 FEET DIVIDED INTO 3 EQUAL PARTS OF Z FEET EACH
I PARTI I PARTi I PARTi I PART_J
~ 3, CT~n~ ^, rr" 3^ cr~n 3, CT t
I I
2FT
3FT
3 FEET DIVIDED INTO 4 EQUAL PARTS OF ^ OFA FOOT EACH
Figure 4-3. Partitive fractions
EXPRESSING RELATIONSHIPS
When a fraction is used to express a rela-
tionship, the numerator and denominator take
on individual significance. In this frame of
reference, 3/4 means 3 out of 4, or 3 parts in
4, or the ratio of 3 to 4. For example, if 1 out
of 3 of the men in a division are on liberty, then
it would be correct to state that 1/3 of the
division are on liberty. Observe that neither of
these ways of expressing the relationship tells
us the actual number of men; the relationship
itself is the important thing.
Practice problems.
1. What fraction of 1 foot is 11 inches?
2. Represent 3 out of 8 as a fraction.
3. Write the fractions that indicate the rela-
tionship of 2 to 3; 8 divided by 9; and 6 out of
7 equal parts.
4. The number 6-r- means 6-
Answers:
1. 11/12
2. 3/8
3. 2/3; 8/9; 6/7
4. plus
EQUIVALENT FRACTIONS
It will be recalled that any number divided
by itself is 1. For example, 1/1, 2/2, 3/3, 4/4,
and all other numbers formed in this way, have
the value 1. Furthermore, any number multi-
plied by 1 is equivalent to the number itself.
For example, 1 times 2 is 2, 1 times 3 is 3,
1 times 1/2 is 1/2, etc.
These facts are used in changing the form
of a fraction to an equivalent form which is
more convenient for use in a particular problem.
29
will be in a different form, as follows:
2J
2-5
__
10
Figure 4-4 shows that -? of line a is equal to
c
y- of line b where line a equals line b. Line a
is marked off in fifths and line b is marked off
fi *?
in tenths. It can readily be seen that TQ and g-
measure distances of equal length.
1 J
3
1
1
1
1
1
6
1 |
1
1
1 i. J
5 5
) 4
5 5
T
J..LJL.1.JL.L.Z.JL.LJ2.
10 10 10 10 10 10 10 10 10 10
Figure 4-4. Equivalent fractions.
The markings on a ruler show equivalent
fractions. The major division of an inch divides
it into two equal parts. One of these parts
represents -jr. The next smaller markings divide
the inch into four equal parts. It will be noted that
two of these parts represent the same distance as
1 21
i; that is, 4 equals |-. Also, the next smaller
Lt Tt
markings break the inch into 8 equal parts. How
many of these parts are equivalent to \ inch?
Ci
The answer is found by noting that- equals i.
Practice problems. Using the divisions on a
ruler for reference, complete the following
exercise:
3"
~A ~ 1C
?_
16
o 1- -L
*' 8 ~ 16
4
-1
16
A review of the foregoing exercise will re-
veal that in each case the right-hand fraction
could be formed by multiplying both the nu-
merator and the denominator of the left-hand
fraction by the same number. In each case the
number may be determined by dividing the de-
nominator of the right-hand fraction by the de-
nominator of the left-hand fraction. Thus in
problem 1, both terms of -jwere multiplied by 2.
In problem 3, both terms were multiplied by 4.
It is seen that multiplying both terms of a frac-
tion by the same number does not change the
value of the fraction.
1 2
Since -^ equals -T, the reverse must also be
t *
2 1
true; that is -7 must be equal to ^ This can
likewise be verified on a ruler. We have al-
M 119 *?
ready seen that % is the same as -=, jg equals -^,
2 1
and |r equals -r. We see that dividing both terms
8 4
of a fraction by the same number does not
change the value of the fraction.
FUNDAMENTAL RULE OF FRACTIONS
The foregoing results are combined to form
the fundamental rule of fractions, which is
stated as follows: Multiplying or dividing both
terms of a fraction by the same number does
not change the value of the fraction. This is
one of the most important rules used in dealing
with fractions.
The following examples show how the funda-
mental rule is used:
1 Change 1/4 to twelfths. This problem is set
up as follows:
I - 2-
4 ' 12
The first step is to determine how many 4's
are contained in 12. The answer is 3, so we
know that the multiplier for both terms of the
fraction is 3, as follows:
3 1
3*4
__
12
30
Chapter 4-COMMON FRACTIONS
2. What fraction with a numerator of 6 is equal
to 3/4?
SOLUTION:
6.
7
3_
4
We note that 6 contains 3 twice; therefore we
need to double the numerator of the right-hand
fraction to make it equivalent to the numerator
of the fraction we seek. We multiply both terms
of 3/4 by 2, obtaining 8 as the denominator of
the new fraction, as follows:
1 1
4 ' 2
3. Change 6/16 to eighths.
SOLUTION:
16
both terms of the preceding example by 6 re-
duces the fraction to lowest terms. In computa-
tion, fractions should usually be reduced to
lowest terms where possible.
If the greatest common factor cannot readily
be found, any common factor may be removed
and the process repeated until the fraction is in
18
lowest terms: Thus, 75- could first be divided
48
by 2 and then by 3.
18 * 2 _ _9_
48 - 2 ~ 24
9 + 3 3
24 + 3 ~ 8
Practice problems. Reduce the following
fractions to lowest terms:
We note that the denominator of the fraction
which we seek is 1/2 as large as the denomina-
tor of the original fraction. Therefore the new
fraction may be formed by dividing both terms
of the original fraction by 2, as follows:
-
- 48
4
*' 60
2
2 ' 20
5.
18
24
6.
35
56
9
144
Practice
ber in each
3 30
11 8 ~ ?
44 _?
*' 48 ~ 12
6*2
3
the missing
7
1
num-
12
= 72
= 25
.nuowci o.
'!
-i
IMPROPER
Although
**
..f
FRACTIONS
the "impropei
16 * 2 ~ 8
problems. Supply
of the following:
3 -2- - -2-
J% 90 10
4 I^fi.
*' 6 ?
3.1
Answers:
1. 80
2. 11
3. 27
4. 36
5. 6
6. 15
REDUCTION TO LOWEST TERMS
It is frequently desirable to change a frac-
tion to an equivalent fraction with the smallest
possible terms; that is, with the smallest pos-
sible numerator and denominator. This process
n
is called REDUCTION. Thus, ^ reduced to
lowest terms is v-. Reduction can be accom-
plished by finding the largest factor that is
common to both the numerator and denominator
and dividing- hnth nf these terms bv it. Dividing
quite "proper" mathematically, it is usually
customary to change it to a mixed number. A
recipe may call for VJT cups of milk, but would
o
not call for = cups of milk.
Since a fraction is an indicated division, a
method is already known for reduction of im-
proper fractions to mixed numbers. The im-
Q
proper fraction -s- may be considered as the di-
vision of 8, by 3. This division is carried out
as follows:
2 R 2 = 2f
O
3/5
6
2
MATHEMATICS, VOLUME 1
The truth of this can be verified another way:
o C
If 1 equals -r> then 2 equals -5-. Thus ,
These examples lead to the following con-
clusion, which is stated as a rule: To change
an improper fraction to a mixed number, divide
the numerator by the denominator and write the
fractional part of the quotient in lowest terms.
Practice problems. Change the following
fractions to mixed numbers:
1. 31/20
2. 33/9
Answers:
o o.
2. 3
3. 65/20
4. 45/8
3 3
J. 6
4. 5
OPERATING WITH MIXED NUMBERS
In computation, mixed numbers are often un-
wieldy. As it is possible to change any im-
proper fraction to a mixed number, it is like-
wise possible to change any mixed number to an
improper fraction. The problem can be reduced
to the finding of an equivalent fraction and a
simple addition.
EXAMPLE: Change 2-= to an improper fraction.
SOLUTION:
Step 1: Write 2-^ as a whole number plus a
fraction, 2 + -r-.
Step 2: Change 2 to an equivalent fraction
with a denominator of 5, as follows:
2(5) _ 10
W) ~ 5
Step 3: Add + - =
Thus, 2 = ^
o D
2
EXAMPLE: Write 5^ as an improper fraction.
SOLUTION:
,2 , 2
5 9 = 5 + "9
_ 1
1 " 9
5(9) .. 45
1(9) 9
45
47
Thus,
47
9
In each of these examples, notice that the
multiplier used in step 2 is the same number as
the denominator of the fractional part of the
original mixed number. This leads to the fol-
lowing conclusion, which is stated as a rule:
To change a mixed number to an improper frac-
tion, multiply the whole -number part by the
denominator of the fractional part and add the
numerator to this product. The result is the
numerator of the improper fraction; its denom-
inator is the same as the denominator of the
fractional part of the original mixed number.
Practice problems. Change the following
mixed numbers to improper fractions:
'4
3
3.
4. 4
10
Answers:
'f
*.*
2.
20
NEGATIVE FRACTIONS
A fraction preceded by a minus sign is nega-
tive. Any negative fraction is equivalent to a
positive fraction multiplied by -1. For example,
Chapter 4-COMMON FRACTIONS
1
5
The number - -= is read "minus two-fifths."
o
We know that the quotient of two numbers
with unlike signs is negative. Therefore,
anri __ .
I and ^5 = - 7
This indicates that a negative fraction is equiv-
alent to a fraction with either a negative nu-
merator or a negative denominator.
2
The fraction -r is read "two over minus
o
-2
five." The fraction -;r- is read "minus two
over five."
A minus sign in a fraction can be moved
about at will. It can be placed before the nu-
merator, before the denominator, or before the
fraction itself. Thus,
JL
-5
2_
5
Moving the minus sign from numerator to
denominator, or vice versa, is equivalent to
multiplying the terms of the fraction by -1.
This is shown in the following examples:
and
A fraction may be regarded as having three
signs associated with it the sign of the numer-
ator, the sign of the denominator, and the sign
preceding the fraction. Any two of these signs
may be changed without changing the value of
the fraction. Thus,
_3
4
-3
4
-4
-4
OPERATIONS WITH FRACTIONS
It will be recalled from the discussion of
denominate numbers that numbers must be of
the same denomination to be added. We can add
pounds to pounds, pints to pints, but not ounces
to pints. If we think of fractions loosely as de-
nominate numbers, it will be seen that the rule
of likeness applies also to fractions. We can
add eighths to eighths, fourths to fourths, but
1 2
not eighths to fourths. To add-jr- inch to ? inch
we simply add the numerators and retain the
denominator unchanged. The denomination is
fifths; as with denominate numbers, we add 1
fifth to 2 fifths to get 3 fifths, or |.
LIKE AND UNLIKE FRACTIONS
We have shown that like fractions are added
by simply adding the numerators and keeping the
denominator. Thus,
3 + 2 _ 5_
8 "8
or
16 16 ~ 16
Similarly we can subtract like fractions by
subtracting the numerators.
8
7-2
_5
8
The following examples will show that like
fractions may be divided by dividing the nu-
merator of the dividend by the numerator of
the divisor.
SOLUTION: We may state the problem as a
1 1
question: "How many times does - appear in-,
8 o
1 3
or how many times may -g- be taken from-g-?"
3/8 - 1/8 = 2/8
2/8 - 1/8 = 1/8
1/8 - 1/8 =0/8=0
(1)
(2)
(3)
We see that 1/8 can be subtracted from 3/8
three times. Therefore,
3/8 +1/8=3
When the denominators of fractions are un-
equal, the fractions are said to be unlike. Ad-
dition, subtraction, or division cannot be per-
formed directly on unlike fractions. The
proper application of the fundamental rule,
however, can change their form so that they
become like fractions; then all the rules for
like fractions apply.
LOWEST COMMON DENOMINATOR
To change unlike fractions to like fractions,
it is necessary to find a COMMON DENOMINA-
TOR and it is usually advantageous to find the
LOWEST COMMON DENOMINATOR (LCD).
This is nothing more than the least common
multiple of the denominators.
Least Common Multiple
H a number is a multiple of two or more
different numbers, it is called a COMMON
MULTIPLE. Thus, 24 is a common multiple of
6 and 2. There are many common multiples of
these numbers. The numbers 36, 48, and 54, to
name a few, are also common multiples of 6
and 2.
The smallest of the common multiples of a
set of numbers is called the LEAST COMMON
MULTIPLE. It is abbreviated LCM. The least
common multiple of 6 and 2 is 6. To find the
least common multiple of a set of numbers,
first separate each of the numbers into prime
factors.
Suppose that we wish to find the LCM of 14,
24, and 30. Separating these numbers into
prime factors we have
14 = 2 7
24 = 2 3 3
30 = 2 3 5
The LCM will contain each of the various prime
factors shown. Each prime factor is used the
greatest number of times that it occurs in any
one of the numbers. Notice that 3, 5, and Teach
occur only once in any one number. On the
other hand, 2 occurs three times in one number.
We get the following result:
LCM = 2 3 3 5 7
= 840
Thus, 840 is the least common multiple of 14,
24, and 30.
Greatest Common Divisor
The largest number that can be divided into
each of two or more given numbers without a
remainder is called the GREATEST COMMON
DIVISOR of the givennumbers. It is abbreviated
GCD. It is also sometimes called the HIGHEST
COMMON FACTOR.
In finding the GCD of a set of numbers, se-
parate the numbers into prime factors just as
for LCM. The GCD is the product of only those
factors that appear in all of the numbers. Notice
in the example of the previous section that 2 is
the greatest common divisor of 14, 24, and 30.
Find the GCD of 650, 900, and 700. The pro-
cedure is as follows:
650 = 2
900 = 2 2
700 = 2 2
GCD = 2
5* 13
- 3 2 5 2
5 2 7
5 2 = 50
Notice that 2 and5 2 are factors of each num-
ber. The greatest common divisor is 2 x 25 = 50.
USING THE LCD
Consider the example
1 . 1
"2 + 3"
The numbers 2 and 3 are both prime; so the
LCD is 6.
Therefore
and
1-1
3 " 6
Thus, the addition of - and - is performed as
2 3
follows :
I
2
In the example
10 is the LCD.
i
6
I JL
5 + 10
34
Therefore,
.1
10
_2_
10
To
_3_
10
_1
2
EXAMPLE:
Practice problems. Change the fractions in
each of the following groups to like fractions
with least common denominators:
a - 3' 6
2- 12' "3
Answers:
3 I I 2
- 2' 4' 3
I JL i
6' 10' 5
O 1
Here we change 7? to the mixed number 1 y. Then
10^ = 10 + 1 + i
i 1 i
' 6' 6
o _ _
' 12' 12
ADDITION
4.
JL J. JL
12' 12' 12
A JL _i
30' 30' 30
It has been shown that in adding like frac-
tions we add the numerators. In adding unlike
fractions , the fractions must first be changed so
that they have common denominators. We apply
these same rules in adding mixed numbers. It
will be remembered that a mixed number is an
indicated sum.
Thus, 2 -5- is really 2 +.
o o
Add-
ing can be done in any order. The following
examples will show the application of these
rules:
EXAMPLE:
EXAMPLE:
Add
*i
We first change the fractions so that they are
like and have the least common denominator
and then proceed as before.
4 12
~3 = 2 12
EXAMPLE:
Add
12
8
This could have been written as follows:
2 - 2
2 2 " 2 8
1 _ 2_
4 ~ 8
c 2 ; 2
5 + = 5
11 3
Since -^ equals 1 -^, the final answer is found
o o
as follows:
35
"I
Practice problems. Add, and reduce the
sums to simplest terms:
2
y <J. ?
3. 6f
2 3
li
5. = 4
8 &> *
Jr- 1
Answers:
1. 3
28
2. 2
3. 9
13
"20
4 2 31
4 - 2 40
5. 5
8
The following example demonstrates a prac-
tical application of addition of fractions:
EXAMPLE: Find the total length of the piece
of metal shown in figure 4-5 (A).
SOLUTION: First indicate the sum as follows:
(B)
Figure 4-5. Adding fractions to obtain
total length or spacing.
EXAMPLE: Subtract li from 5 -|
o o
-
1648416
Changing to like fractions and adding numerators,
9 .
12
, 14
12
9
56
16 +
16 H
16 "
H I6 H
" 16 ~
16
3 A
16
We see that whole numbers are subtracted from
whole numbers; fractions from fractions.
The total length is 3 i inches.
EXAMPLE: Subtract from
o 5
Practice problem. Find the distance from
the center of the first hole to the center of the
last hole in the metal plate shown in figure
4-5 (B).
Answer:
SUBTRACTION-
2 -^ inches
lo
The rule of likeness applies in the sub-
traction of fractions as well as in addition.
Some examples will show that cases likely to
arise may be solved by use of ideas previously
developed.
5
1
Changing to like fractions with an LCD, we have
32
40
_5_
40
_27
40
36
EXAMPLE-. Subtract from 3 -
J.A
o 2 _ , _8_
6 3 ~ J 12
U 11
12 12
Regrouping 3 ^ we have
12
12 12
Then
20
12
11 - li
12 " 12
_
12
Practice problems. Subtract the lower num-
ber from the upper number and reduce the
difference to simplest terms:
70 K 1
, 9-2. <JK_2- 4 is S 2
J. . ^T" A. TT- O. O T^T T. J J. 6 "7^
l
Answers:
" 18
> 3
3. 2}
The following problem demonstrates sub-
traction of fractions in a practical situation.
EXAMPLE: What is the length of the dimen-
sion marked X on the machine bolt shown in
figure 4-6 (A)?
SOLUTION: Total the lengths of the known
parts.
*_ .- _ .--
4 84 2 ~ 64 64 64 ~ 64
Subtract this sum from the overall length.
o _ 1
2 ~ l
49
64
64
64
ii
64
1 15
64
15
The answer is 1 j^j inch.
(A)
(B)
Figure 4-6. Finding unknown dimensions
by subtracting fractions.
Practice problem. Find the length of the
dimension marked Y on the machine bolt in
figure 4-6 (B).
Answer: 2 inches
MULTIPLICATION
The fact that multiplication by a fraction does
not increase the value of the product may con-
fuse those who remember the definition of mul-
tiplication presented earlier for whole numbers.
It was stated that 4(5) means 5 is taken as an
addend 4 times.
How is it then that -(4) is 2, a
Ct
number less than 4 ? Obviously our idea of
multiplication must be broadened.
Consider the following products:
4(4) = 16
3(4) = 12
2(4) = 8
1(4) = 4
37
|(4) = 2
{(4) = 1
Notice that as the multiplier decreases, the
product decreases, until, when the multiplier
is a fraction, the product is less than 4 and
continues to decrease as the fraction decreases.
The fraction introduces the "part of" idea:
-5(4) means ~ of 4; j(4) means -j of 4.
A 4 4
The definition of multiplication stated for
whole numbers may be extended to include frac-
tions. Since 4(5) means that 5 is to be used 4
times as an addend, we can say that with frac-
tions the numerator of the multiplier tells how
many times the numerator of the multiplicand
is to be used as an addend. By the same rea-
soning, the denominator of the multiplier tells
how many times the denominator of the mul-
tiplicand is to be used as an addend. The fol-
lowing examples illustrate the use of this idea:
1. The fraction A is multiplied by the whole
\&
number 4 as follows:
From these examples a general rule is
developed: To find the product of two or more
fractions multiply their numerators together
and write the result as the numerator of the
product; multiply their denominators and write
the result as the denominator of the product;
reduce the answer to lowest terms.
In using this rule with whole numbers, write
each whole number as a fraction with 1 as the
denominator. For example, multiply 4 times
1/12 as follows:
4 x
_
12
__
12
JL
12
1
= 3
In using this rule with mixed numbers, re-
write all mixed numbers as improper frac-
tions before applying the rule, as follows:
2 1 X 1-1 X 1
2 3 X 2 ~ 3 X 2
_! _l x JL
12 " 1 X 12
.1 + 1 + 1 + 1
__
12
12
1
T
This example shows that 4 (1/12) is the same as
12 '
Another way of thinking about the multiplica-
tion of 1/12 by 4 is as follows:
12
2. The fraction 2/3 is multiplied by 1/2 as
follows:
2
3"
2
6"
1
3
A second method of multiplying mixed num-
bers makes use of the distributive law. This
law states that a multiplier applied to a two-part
expression is distributed over both parts. For
example, to multiply 6 -^ by 4 we may rewrite
6 -i as 6 + 1/3. Then the problem can be written
3
as 4(6 + 1/3) and the multiplication proceeds as
follows:
4(6 + 1/3) = 24 + 4/3
= 25 + 1/3
- 25!
- 25 3
Cancellation
Computation can be considerably reduced by
dividing out (CANCELLING) factors common to
both the numerator and the denominator. We
recognize a fraction as an indicated division.
Thinking of ~ as an indicated division, we re-
9
member that we can simplify division by show-
ing both dividend and divisor as the indicated
38
Chapter 4- COMMON FRACTIONS
products of their factors and then dividing like
2x 3-1= ?
factors, or canceling. Thus,
*4 3
6 2x3
9 ~ 3 x 3
3 5
9 ... 10 _ 6 _ 1& _ 15
4 X 3 ~ A X T~T
Dividing the factor 3 in the numerator by 3 in
2 1
the denominator gives the following simplified
7 *
result:
2
1
Practice problems. Determine the following
2 x % _ 2_
products, using the general rule and canceling
3 x $ 3
where possible:
1
1. x 12 3. 5 x 5 x -
This method is most advantageous when done
8 9 33
before any other computation. Consider the
112 3 41
example,
2> 2" X "3 X ~5~ 4 ' T X 6 6> ~3 X 6"
100
XX
Answers:
325
The product in factored form is
1 7 1 Q o 2 e; 2
1. 7- 6. Z- b. -
1x3x2
o 1 4 4 1 6 2
3x2x5
15 2 9
Rather than doing the multiplying and then
The following problem illustrates the mul-
/
reducing the result , it is simpler to cancel
uU
tiplication of fractions in a practical situation.
like factors first, as follows:
EXAMPLE: Find the distance between the cen-
ter lines of the first and fifth rivets connecting
1 1
the two metal plates shown in figure 4-7 (A).
1 x t x I 1
% x I x 5 5
SOLUTION: The distance between two adjacent
1 1
rivets, centerline to centerline, is 4 1/2 times
the diameter of one of them.
Likewise,
Thus,
1
1 I
1 space = 4 -i- x -|
t X it X ~9 = 1f
9 5
1 I
= "2 X ~8
1
45
Here we mentally factor 6 to the form 3x2,
16
and 4 to the form 2x2. Cancellation is a
valuable tool in shortening operations with
There are 4 such spaces between the first and
fractions.
fifth rivets. Therefore, the total distance, D,
The general rule may be applied to mixed
is found as follows:
numbers by simply changing them to improper
J- AC A P 1
fractions.
, 45 45 1 1 1
D = Ax B~ 4~ U 4
ThllC!
4
MATHEMATICS, VOLUME 1
1
J II 1
1 1
1 1
1
\ 5 ,.. DIAMETER
_- . 9 ft.
(B)
=DIAMETER
-RIVET SPACING =-
5k DIAMETERS
Figure 4-7. Application of multiplication of fractions
in determining rivet spacing.
The distance is 11 - inches
12-4
Practice problem. Find the distance between
the centers of the two rivets shown in figure
4-7 (B).
Answer: 4 inches
Ib
DIVISION
There are two methods commonly used for
performing division with fractions. One is the
common denominator method and the other is
the reciprocal method.
Common Denominator Method
The common denominator method is an adap-
tation of the method of like fractions. The rule
is as follows: Change the dividend and divisor
to like fractions and divide the numerator of
the dividend by the numerator of the divisor.
This method can be demonstrated with whole
numbers, first changing them, to fractions with
1 as the denominator. For example, 12 * 4 can
be written as follows:
If the dividend and divisor are both fractions,
as in 1/3 divided by 1/4, we proceed as follows:
1 -1 - A J_
3 4 ~ 12 ' 12
= 4 + 3_
12 + 12
Reciprocal Method
The word "reciprocal" denotes an inter-
changeable relationship. It is used in mathe-
matics to describe a specific relationship be-
tween two numbers. We say that two numbers
are reciprocals of each other if their product
is one. In the example 4 x A = 1, the fractions
and -1 are reciprocals. Notice the interchange-
ability: 4 is the reciprocal of -J and I is the re-
4 4
ciprocal of 4.
What is the reciprocal of -^-7 It must be a
3
number which, when multiplied by =-, produces
the product, 1. Therefore,
SOLUTION:
i .1 JL _ 2. ^ jL
2 22
= 2-5
1 1
Check:
4. Wliat is the reciprocal of 3 ~?
Q
We see that -r- is the only number that could ful-
O
fill the requirement. Notice that the numerator
q
and denominator of were simply interchanged
to get its reciprocal. If we know a number, we
can always find its reciprocal by dividing 1 by
the number. Notice this principle in the follow-
ing examples:
1. What is the reciprocal of 7?
SOLUTION:
Check:
1 _ 8 .25
~~
= 8-25
_
25
= 1
Check:
Notice that the cancellation process in this ex-
ample does not show the usual 1's which result
when dividing a number into itself. For ex-
ample, when 7 cancels 7, the quotient 1 could be
shown beside each of the 7's. However, since 1
as a factor has the same effect whether it is
written in or simply understood, the 1's need
not be written.
2. What is the reciprocal of ?
o
i - -
" '
A
8
.8*S,orf
Check:
The foregoing examples lead to the rule for
finding the reciprocal of any number: The re-
ciprocal of a number is the fraction formed
when 1 is divided by the number . (If the final
result is a whole number, it can be considered
as a fraction whose denominator is 1.) A short-
cut rule which is purely mechanical and does
not involve reasoning may be stated as follows:
To find the reciprocal of a number, express
the number as a fraction and then invert the
fraction.
When the numerator of a fraction is 1, the
reciprocal is a whole number. The smaller the
fraction, the greater is the reciprocal. For ex-
ample, the reciprocal of * is 1,000.
Also, the reciprocal of any whole number is a
proper fraction. Thus the reciprocal of 50 is
1
50'
Practice problems. Write the reciprocal of
each of the following numbers:
1. 4 2. -|- 3. 2^- 4. 17 5. 4 6. -7
O Lt &
3. What is the reciprocal of ^
Answers
'"4 2>
3
3.
5
4.
17
5. J- 6.
1
5
41
The reciprocal method of division makes use
of the close association of multiplication and
division. In any division problem, we must find
the answer to the following question: What
number multiplied by the divisor yields the
dividend? For example, if the problem is to
divide 24 by 6, we must find the factor which,
when multiplied by 6, yields 24. Experience
tells us that the number we seek is 1/6 of 24.
Thus, we may rewrite the problem as follows:
24 + 6 =
24
Check:
= 4
6 x 4 = 24
speed and the possibility of cancellation of like
factors, which simplifies the computation. It
is the suggested method once the principles be-
come familiar.
EXAMPLE:
f.4.7
Common Denominator
Method
1-4=1 -20
5 55
= 2+20
.1
20
_L
10
Reciprocal Method
1, 4 -l x !
5 ' 4 " 5 X 4
In the example 1 -5- * 3, we could write 3 x ? =
It
\. The number we seek must be one-third of EXAMPLE:
10
1 -i. Thus we can do the division by taking one-
third of 1-i; that is, we multiply !-- by the re-
2
ciprocal of 3.
1 . Q _ i 1 v 1
2 ' 3 ~ * 2 X "3
= 1x1
Common Denominator
Method
2 ~3 T 3 = I T I
= 8*9
1
" 9
Reciprocal Method
2 f^
Check:
<**-_ i
3 x 2 ~ 2 - 1 2
The rule for division by the reciprocal
method is: Multiply the dividend by the recipro-
cal of the divisor. This is sometimes stated in
short form as follows: Invert the divisor and
multiply.
The following examples of cases that arise
in division with fractions will be solved by both
the reciprocal method and the common denom-
inator method. The common denominator
method more clearly shows the division proc-
ess and is easier for the beginner to grasp.
The reciprocal method is more obscure as to
the reason for its use but has the advantage of
EXAMPLE: 9 -y =
9 ,1 = 63 ^ 1
7 7 7
= 63-2
- 83 .. 1
~ 2 ~ 31 2
Method
9x7
1x2
63
EXAMPLE:
10 + 5 = ?
42
Chapter 4-COMMON FRACTIONS
Common Denominator
Method
Reciprocal Method
fractions, the resulting expression is called
a complex fraction. The following expression
is a complex fraction:
m - ?! - 40 23
1U ^ 7^ T^
10 + 5 1 = 10 x gl
444
3/5
37T
=40-23
10 x 4
1 x 23
This should be read "three-fifths over three-
_ 40 ,17
fourths" or "three-fifths divided by
three-
23 " 23
_ 40 _ t 17
fourths." Any complex fraction may
be sim-
" 23 " 23
plified by writing it as a division problem, as
follows:
EXAMPLE: -| +
1 = ?
3
4
3/5 3 3
374 ~ 5 4
Common Denominator
Method
Reciprocal Method
_ # 4
2183
2124
3 * 4 = 12 : 12
_**_ . ,, _ ^__ v ~ r
3 ' 4 " 3 1
= 4/5
= 8+3
8 ,> 2
Similarly,
33
8 9 2
- T ~ ''T
, 1 2
3 10 . 5 X0 2 4 , :
1
w w
, 1 3 ' 2 " 3 "JS 3 X 3
EXAMPLE: - +
-1 = ?
2
16
10
Complex fractions may also contain an in-
Common Denominator
Method
Reciprocal Method
dicated operation in the numerator or
inator or both. Thus,
denom-
3 45 24
9 3 9 10
1 1
16 T 10 " 80 " 80
16 T 10 ' 16 x 3
I + I
=45 +24
3 5
1 + 1
AK 1 C
>- *" -^ *^
A* 1 ?
jlp J5
55
" ![ 8~
8
is a complex fraction. To simplify
such a
fraction we simplify the numerator and
denom-
= il
15 , 7
inator and proceed as follows:
8
"8 8
Practice problems
Perform the following
2" + lf_6" t "6"_6"
division by the reciprocal method:
9 1 ~ 10 2
5 + 5 T
1 3 . 2 , o 1 . ,
1 5 ._5 4 1 ,4.
_ 5 . 2
f ' 8 " 3 *' *3
2 8 ' 16 36
~ 6 T 1
Answers:
-l x l
Q R
1 _JL 9 1-2.
J> .d A i JL
3. 2 4. 1
6 2
16 9
2
e
COMPLEX FRACTIONS
When the numerator or denominator, or both,
in a fraction are themselves composed of
- 12
Mixed numbers appearing in complex fractions
usually show the plus sign.
MATHEMATICS, VOLUME 1
Thus,
might be written
-?-
7 +
Practice problems,
complex fractions:
Simplify the following
24-
L T
8
3.
4.
1.
Answers:
* 2
3. 1
19
36
16 " 32
4. 18
Complex fractions may arise in electronics
when it is necessary to find the total resistance
of several resistances in parallel as shown in
figure 4-8. The rule is: The total resistance
of a parallel circuit is 1 divided by the sum of
the reciprocals of the separate resistances.
Written as a formula, this produces the follow-
ing expression:
(B)
I
Rl"
=- 5fl<
T
R2S R3<
2n< 10CX<
Figure 4-8. Application of complex fractions
in calculating electrical resistance.
346
The LCD of the fractions -i, 4, and! is 12.
Tt U
Thus,
R =
__
12 12 12
1 JL
12
R,
R
EXAMPLE: Find the total resistance of the
parallel circuit in figure 4-8 (A). Substituting
the values 3, 4, and 6 for the letters Rj, R
and R 3 , we have the following:
2>
= 1 ohms (measure of resistance).
3
Practice problem: Find the total resistance
of the parallel circuit in figure 4-8 (B).
Answer: 1-7- ohms.
CHAPTER 5
DECIMALS
The origin and meaning of the word "decimal"
were discussed in chapter 1 of this course. Also
discussed in chapter 1 were the concept of place
value and the use of the number ten as the base
for our number system. Another term which is
frequently used to denote the base of a number
system is RADIX. For example, two is the
radix of the binary system and ten is the radix
of the decimal system. The radix of a number
system is always equal to the number of differ-
ent digits used in the system. For example, the
decimal system, with radix ten, has ten digits:
through 9.
DECIMAL FRACTIONS
A decimal fraction is a fraction whose de-
nominator is 10 or some power of 10, such as
100, 1,000, or 10,000.
Thus, ,
and
215
100' 1000
are decimal fractions. Decimal fractions have
special characteristics that make computation
much simpler than with other fractions.
Decimal fractions complete our decimal
system of numbers. In the study of whole num-
bers, we found that we could proceed to the left
from the units place, tens, hundreds, thousands,
and on indefinitely to any larger place value,
but the development stopped with the units place.
Decimal fractions complete the development so
that we can proceed to the right of the units
place to any smaller number indefinitely.
Figure 5-1 (A) shows how decimal fractions
complete the system. It should be noted that as
we proceed from left to right, the value of each
place is one-tenth the value of the preceding
place, and that the system continues uninter-
rupted with the decimal fractions.
Figure 5-1 (B) shows the system again, this
time using numbers. Notice in (A) and (B) that
the units place is the center of the system and
that the place values proceed to the right or
left of it by powers of ten. Ten on the left is
balanced by tenths on the right, hundreds by
hundredths, thousands by thousandths, etc.
Notice that 1/10 is one place to the right of
the units riicrit 1/1(10 is two nlanes to thp
etc. (See fig. 5-1.) If a marker is placed after
the units digit, we can decide whether a decimal
digit is in the tenths, hundredths, or thousandths
position by counting places to the right of the
marker. In some European countries, the
marker is a comma; but in the English-speaking
countries, the marker is the DECIMAL POINT.
Thus, ~ is written 0.3. To write ^ it is
necessary to show that 3 is in the second place
to the right of the decimal point, so a zero is
inserted in the first place. Thus, ~ is written
0.03. Similarly,
can be written by insert-
ing zeros in the first two places to the right of
the decimal point. Thus, r~r is written 0.003.
In the number 0.3, we say that 3 is in the first
decimal place; in 0.03, 3 is in the second deci-
mal place; and in 0.003, 3 is in the third deci-
mal place. Quite frequently decimal fractions
are simply called decimals when written in this
shortened form.
WRITING DECIMALS
Any decimal fraction may be written in the
shortened form by a simple mechanical process.
Simply begin at the right-hand digit of the nu-
merator and count off to the left as many places
as there are zeros in the denominator. Place
the decimal point to the left of the last digit
counted. The denominator may then be dis-
regarded. If there are not enough digits, as
many place-holding zeros as are necessary are
added to the left of the left-hand digit in the
numerator.
Thus, in
beginning with the digit 3,
we count off four places to the left, adding two
O's as we count, and place the decimal point to
the extreme left. (See fig. 5-2.) Either form
is read "twenty-three ten-thousandths."
When a decimal fraction is written in the
shortened form, there will always be as many
ripnimal nlarps In thp shnrtpnpd fnrtn afi
Figure 5-1. Place values including decimals.
10000
/ -*-*-*-
-i, A no 7
'
PLACE HOLDING
ZEROS AOOED
Figure 5-2. Conversion
of a decimal fraction
to shortened form.
are zeros in the denominator of the fractional
form.
Figure 5-3 shows the fraction rllttl and
what is meant when it is changed to the short-
ened form. This figure is presented to show
further that each digit of a decimal fraction
holds a certain position in the digit sequence
and has a particular value.
By the fundamental rule of fractions, it
should be clear that - = ^- = -^-. Writing
10 100 1000
the same values in the shortened way, we have
0.5 = 0.50 = 0.500. In other words, the value of
a decimal is not changed by annexing zeros at
the right-hand end of the number. This is not
100000
OF
\Z TENTHS OR .2
I4HUNDREDTHS OR.O4
THOUSANDTHS OR. 003
TEN . T HOUSANDTHS OR .0005
I 8 HUNDRED-THOUSANDTHS OR .QOOOfl
.14358
Figure 5-3. Steps in the conversion of a
decimal fraction to shortened form.
true of whole numbers. Thus, 0.3, 0.30, and
0.300 are equal but 3, 30, and 300 are not equal.
Also notice that zeros directly after the deci-
mal point do change values. Thus 0.3 is not
equal to either 0.03 or 0.003.
Decimals such as 0.125 are frequently seen.
Although the on the left of the decimal point
is not required, it is often helpful. This is par-
ticularly true in an expression such as 32 * 0.1.
In this expression, the lower dot of the division
symbol must not be crowded against the decimal
point; the serves as an effective spacer. If
any doubt exists concerning the clarity of an
expression such as .125, it should be written as
0.125.
46
Practice problems. In problems 1 through 4,
change the fractions to decimals. In problems
5 through 8, write the given numbers as deci-
mals:
1. 8/100 5. Four hundredths
2. 5/1000 6. Four thousandths
3. 43/1000 7. Five hundred one ten-
4. 32/10000 thousandths
8. Ninety- seven thousandths
Answers:
1. 0.08 5. 0.04
2. 0.005 6. 0.004
3. 0.043 7. 0.0501
4. 0.0032 8. 0.097
READING DECIMALS
To read a decimal fraction in full, we read
both its numerator and denominator, as in read-
ing common fractions. To read 0.305, we read
"three hundred five thousandths." The denomi-
nator is always 1 with as many zeros as deci-
mal places. Thus the denominator for 0.14 is
1 with two zeros, or 100. For 0.003 it is 1,000;
for 0.101 it is 1,000; and for 0.3 it is 10. The
denominator may also be determined by count-
ing off place values of the decimal. For 0.13
we may think "tenths, hundredths "and the frac-
tion is in hundredths. In the example 0.1276 we
may think "tenths, hundredths, thousandths,
ten- thousandths." We see that the denominator
is 10,000 and we read the fraction "one thou-
sand two hundred seventy- six ten- thousandths."
A whole number with a fraction in the form
of a decimal is called a MIXED DECIMAL.
Mixed decimals are read in the same manner
as mixed numbers. We read the whole number
in the usual way followed by the word "and" and
then read the decimal. Thus, 160.32 is read
"one hundred sixty and thirty-two hundredths."
The word "and" in this case, as with mixed
numbers, means plus. The number 3.2 means
three plus two tenths.
It is also possible to have a complex deci-
mal. A COMPLEX DECIMAL contains a com-
mon fraction. The number 0.3^ is a complex
decimal and is read "three and one-third tenths."
The number 0.87^ means 87^ hundredths. The
A i
common fraction in each case forms a part of
the last or right-hand place.
In actual practice when numbers are called
out for recording, the above procedure is not
used. Instead, the digits are merely called out
in order with the proper placing of the decimal
point. For example, the number 216.003 is
read, "two one six point zero zero three." The
number 0.05 is read, "zero point zero five."
EQUIVALENT DECIMALS
Decimal fractions may be changed to equiv-
alent fractions of higher or lower terms, as is
the case with common fractions. If each deci-
mal fraction is rewritten in its common frac-
tion form, changing to higher terms is accom-
plished by multiplying both numerator and
denominator by 10, or 100, or some higher
power of 10. For example, if we desire to
change y- to hundredths, we may do so by mul-
tiplying both numerator and denominator by 10.
Thus,
5 = 50
10 100
In the decimal form, the same thing may be ac-
complished by simply annexing a zero. Thus,
0.5 = 0.50
Annexing a on a decimal has the same ef-
fect as multiplying the common fraction form
of the decimal by 10/10. This is an application
of the fundamental rule of fractions. Annexing
two O's has the same effect as multiplying the
common fraction form of the decimal by 100/100;
annexing three O's has the same effect as mul-
tiplying by 1000/1000; etc.
REDUCTION TO LOWER TERMS
Reducing to lower terms is known as ROUND-
OFF, or simply ROUNDING, when dealing with
decimal fractions. If it is desired to reduce
6.3000 to lower terms, we may simply drop as
many end zeros as necessary since this is
equivalent to dividing both terms of the fraction
by some power of ten. Thus, we see that 6.3000
is the same as 6.300, 6.30, or 6.3.
It is frequently necessary to reduce a num-
ber such as 6.427 to some lesser degree of
precision. For example, suppose that 6.427 is
to be rounded to the nearest hundredth. The
question to be decided is whether 6.427 is closer
47
to e.42 or b.43. The oest way to decide mis
question is to compare the fractions 420/1000,
427/1000, and 430/1000. It is obvious that
427/1000 is closer to 430/1000, and 430/1000
is equivalent to 43/100; therefore we say that
6.427, correct to the nearest hundredth, is 6.43.
A mechanical rule for rounding off can be
developed from the foregoing analysis. Since
the digit in the tenths place is not affected when
we round 6.427 to hundredths, we may limit our
attention to the digits in the hundredths and
thousandths places. Thus the decision reduces
to the question whether 27 is closer to 20 or 30.
Noting that 25 is halfway between 20 and 30, it
is clear that anything greater than 25 is closer
to 30 than it is to 20.
In any number between 20 and 30, if the digit
in the thousandths place is greater than 5, then
the number formed by the hundredths and thou-
sandths digits is greater than 25. Thus we
would round the 27 in our original problem to
30, as far as the hundredths and thousandths
digits are concerned. This result could be sum-
marized as follows: When rounding to hun-
dredths, if the digit in the thousandths place is
greater than 5, increase the digit in the hun-
dredths place by 1 and drop the digit in the
thousandths place.
The digit in the thousandths place may be
any one of the ten digits, through 9. If these
ten digits are split into two groups, one com-
posed of the five smaller digits (0 through 4)
and the other composed of the five larger digits,
then 5 is counted as one of the larger digits.
Therefore, the general rule for rounding off is
stated as follows: If the digit in the decimal
place to be eliminated is 5 or greater, increase
the digit in the next decimal place to the left
by 1. K the digit to be eliminated is less than 5,
leave the retained digits unchanged.
The following examples illustrate the rule
for rounding off:
1. 0.1414 rounded to thousandths is 0.141.
2. 3.147 rounded to tenths is 3.1.
3. 475 rounded to the nearest hundred is 500.
Observe carefully that the answer to exam-
ple 2 is not 3.2. Some trainees make the error
of treating the rounding process as a kind of
chain reaction, in which one first rounds 3.147
to 3.15 and then rounds 3.15 to 3.2. The error
of this method is apparent when we note that
147/1000 is closer to 100/1000 than it is to
200/1000.
Problems of the following type are some-
times confusing: Reduce 2.998 to the nearest
nunareatn. TO arop tne ena ngure we musi in-
crease the next figure by 1. The final result is
3.00. We retain the zeros to show that the an-
swer is carried to the nearest hundredth.
Practice problems. Round off as indicated:
1. 0.5862 to hundredths
2. 0.345 to tenths
3. 2346 to hundreds
4. 3.999 to hundredths
1.
2.
Answers:
0.59
0.3
3. 2300
4. 4.00
CHANGING DECIMALS
TO COMMON FRACTIONS
Any decimal may be reduced to a common
fraction. To do this we simply write out the
numerator and denominator in full and reduce
to lowest terms. For example, to change 0.12
to a common fraction, we simply write out the
fraction in full,
_
100
and reduce to lowest terms,
3
it
= 25
25
Likewise, 0.77 is written
77_
100
but this is in lowest terms so the fraction can-
not be further reduced.
Oneway of checking to see if a decimal frac-
tion can be reduced to lower terms is to con-
sider the makeup of the decimal denominator.
The denominator is always 10 or a power of 10.
Inspection shows that the prime factors of 10
are 5 and 2. Thus, the numerator must be di-
visible by 5 or 2 or both, or the fraction cannot
be reduced.
EXAMPLE: Change the decimal 0.0625 to a
common fraction and reduce to lowest terms.
48
SOLUTION: 0.0625 =
625
10000
- 625 * 25
10000 + 25
_
16
25
400
Complex decimals are changed to common
fractions by first writing out the numerator and
denominator in full and then reducing the re-
sulting complex fraction in the usual way. For
example, to reduce 0.12|, we first write
Li
100
Writing the numerator as an improper fraction
we have
25
2
100
and applying the reciprocal method of division,
we have
1
ii x .I. - 1
2 100 " 8
4
Practice problems. Change the following
decimals to common fractions in lowest terms:
1. 0.25
2. 0.375
Answers:
1. 1/4
2. 3/8
3. 0.6*
4. 0.031
5
3. 5/8
4. 4/125
CHANGING COMMON
FRACTIONS TO DECIMALS
The only difference between a decimal frac-
tion and a common fraction is that the decimal
fraction has 1 with a certain number of zeros
(in other words, a power of 10) for a denomina-
tor. Thus, a common fraction can be changed
to a decimal if it can be reduced to a fraction
having a power of 10 for a denominator.
If the denominator of the common fraction in
its lowest terms is made up of the prime fac-
tors 2 or 5 or both, the fraction can be con-
verted to an exact decimal. If some other prime
factor is present, the fraction cannot be con-
verted exactly. The truth of this is evident
when we consider the denominator of the new
fraction. It must always be 10 or a power of 10,
and we know the factors of such a number are
always 2's and 5's.
The method of converting a common fraction
to a decimal is illustrated as follows:
EXAMPLE: Convert 3/4 to a decimal.
SOLUTION:
3
4
300
400
300
4
= 75 x
= 0.75
J_
100
1
100
Notice that the original fraction could have been
rewritten as 3000/4000, in which case the re-
sult would have been 0.750. On the other hand,
if the original fraction had been rewritten as
30/40, the resulting division of 4 into 30 would
not have been possible without a remainder.
When the denominator in the original fraction
has only 2's and 5's as factors, so that we know
a remainder is not necessary, the fraction
should be rewritten with enough O's to complete
the division with no remainder.
Observation of the results in the foregoing
example leads to a shortcut in the conversion
method. Noting that the factor 1/100 ultimately
enters the answer in the form of a decimal, we
could introduce the decimal point as the final
step without ever writing the fraction 1/100.
Thus the rule for changing fractions to deci-
mals is as follows:
1. Annex enough O's to the numerator of the
original fraction so that the division will be
exact (no remainder).
2. Divide the original denominator into the
new numerator formed by annexing the O's.
3. Place the decimal point in the answer so
that the number of decimal places in the answer
is the same as the number of O's annexed to the
original numerator.
49
If a mixed number in common fraction form
is to be converted, convert only the fractional
part and then write the two parts together. This
is illustrated as follows:
2 + .75 = 2.75
Practice problems. Convert the following
common fractions and mixed numbers to deci-
mal form:
3.
4. 2:
32 16
Answers:
1. 0.25 2. 0.375 3. 0.15625 4. 2.3125
When a common fraction generates such a
repeating decimal, it becomes necessary to
arbitrarily select a point at which to cease the
repetition. This may be done in two ways. We
may write the decimal fraction by rounding off
at the desired point. For example, to round off
the decimal generated by i to hundredths, we
carry the division to thousandths, see that this
figure is less than 5, and drop it. Thus, 5-
o
rounded to hundredths is 0.33. The other method
is to carry the division to the desired number
of decimal places and carry the remaining in-
complete division as a common fraction that
is, we write the result of a complex decimal.
For example, carried to thousandths would be
Nonterminating Decimals
As stated previously, if the denominator of a
common fraction contains some prime factor
other than 2 or 5, the fraction cannot be con-
verted completely to a decimal. When such
fractions are converted according to the fore-
going rule, the decimal resulting will never
terminate. Consider the fraction 1/3. Apply-
ing the rule, we have
The division will continue indefinitely. Any
common fraction that cannot be converted ex-
actly yields a decimal that will never terminate
and in which the digits sooner or later recur.
In the previous example, the recurring digit
was 3. In the fraction 5/11, we have
.4545
11/5.0000
4 4
60
5JL
50
44
60
55
The recurring digits are 4 and 5.
Practice problems. Change the following
common fractions to decimals with three places
and carry the incomplete division as a common
fraction:
J7
13
Answers:
1. 0.538^
1 J
2. 0.555^
y
3 -
3 ' 15
3. 0.266f
4. 0.416|
o
'r!
OPERATION WITH DECIMALS
In the study of addition of whole numbers, it
was established that units must be added to
units, tens to tens, hundreds to hundreds, etc.
For convenience, in adding several numbers,
units were written under units, tens under tens,
etc. The addition of decimals is accomplished
in the same manner.
50
Chapter 5 -DECIMALS
ADDITION
In adding decimals, tenths are written under
tenths, hundredths under hundredths, etc. When
this is done, the decimal points fall in a straight
line. The addition is the same as in adding
whole numbers. Consider the following example:
2.18
34.35
0.14
4.90
41.57
Adding the first column on the right gives 17
hundredths or 1 tenth and 7 hundredths. As
with whole numbers, we write the 7 under the
hundredths column and add the 1 tenth in the
tenths column that is, the column of the next
higher order. The sum of the tenths column is
15 tenths or 1 unit and 5 tenths. The 5 is writ-
ten under the tenths column and the 1 is added
in the units column.
It is evident that if the decimal points are
kept in a straight line that is, if the place
values are kept in the proper columns addition
with decimals may be accomplished in the ordi-
nary manner of addition of whole numbers. It
should also be noted that the decimal point of
the sum falls directly under the decimal points
of the addends.
SUBTRACTION
Subtraction of decimals likewise involves no
new principles. Notice that the place values of
the subtrahend in the following example are
fixed directly under the corresponding place
values in the minuend. Notice also that this
causes the decimal points to be alined and that
the figures in the difference (answer) also re-
tain the correct columnar alinement.
45.76
-31.87
13.89
We subtract column by column, as with whole
numbers, beginning at the right.
Practice problems. Add or subtract as
indicated:
1. 12.3 + 2.13 + 4 + 1.234
2. 0.5 + 0.04 + 12.001 + 10
3. 237.5 - 217.9
4. 9.04 - 7.156
Answers:
1. 19.664
2. 22.541
MULTIPLICATION
3.
4.
19.6
1.884
Multiplication of a decimal by a whole num-
ber may be explained by expressing the decimal
as a fraction.
EXAMPLE: Multiply 6.12 by 4.
SOLUTION:
1 1UU
= 24.48
4 612 = 2448
1 100 100
When we perform the multiplication keeping
the decimal form, we have
6.12
4
24.48
By common sense, it is apparent that the whole
number 4 times the whole number 6, with some
fraction, will yield a number in the neighbor-
hood of 24. Hence, the placing of the decimal
point is reasonable.
An examination of several examples will re-
veal that the product of a decimal and a whole
number has the same number of decimal places
as the factor containing the decimal. Zeros, if
any, at the end of the decimal should be rejected.
Multiplication of Two Decimals
To show the rule for multiplying two deci-
mals together, we multiply the decimal in frac-
tional form first and then in the conventional
way, as in the following example:
0.4 x 0.37
Writing these decimals as common fractions,
we have
4 37
10 100
_ 4 x 37
10 x 100
_ 148
1000
= 0.148
51
0.4
3.
0.148
The placing of the decimal point is reasonable,
since 4 tenths of 37 hundredths is a little less
than half of 37 hundredths, or about 15 hun-
dredths.
Consider the following example:
4.316 x 3.4
In the common fraction form, we have
4316 34 _ 4316 x 34
1000 10 " 1000 x 10
146744
10000
= 14.6744
In the decimal form the problem is
4.316
3.4
17264
12948
14.6744
We note that 4 and a fraction times 3 and a
fraction yields a product in the neighborhood of
12. Thus, the decimal point is in the logical
place.
In the above examples it should be noted in
each case that when we multiply the decimals
together we are multiplying the numerators.
When we place the decimal point by adding the
number of decimal places in the multiplier and
multiplicand, we are in effect multiplying the
denominators.
When the numbers multiplied together are
thought of as the numerators, the decimal points
may be temporarily disregarded and the num-
bers may be considered whole. This justifies
the apparent disregard for place value in the
multiplication of decimals. We see that the
rule for multiplying decimals is only a modifi-
cation of the rule for multiplying fractions.
To multiply numbers in which one or more
of the factors contain a decimal, multiply as
though the numbers were whole numbers. Mark
off as many decimal places in the product as
there are decimal places in the factors together.
6.5
xO.Ol
Answers:
1. 0.074
3. 0.065
4.
0.0073
x5.4
2. 0.315
4. 0.03942
Multiplying by Powers of 10
Multiplying by a power of 10 (10, 100, 1,000,
etc.) is done mechanically by simply moving
the decimal point to the right as many places
as there are zeros in the multiplier. For ex-
ample, 0.00687 is multiplied by 1,000 by mov-
ing the decimal point three places to the right
as follows:
1,000 x 0.00687 = 6.87
Multiplying a number by 0.1, 0.01, 0.001,
etc., is done mechanically by simply moving
the decimal point to the left as many places as
there are decimal places in the multiplier. For
example, 348.2 is multiplied by 0.001 by moving
the decimal point three places to the left as
follows:
348.2 x 0.001 = 0.3482
DIVISION
When the dividend is a whole number,
we
recognize the problem of division as that of
converting a common fraction to a decimal.
Thus in the example 5 + 8, we recall that the
problem could be written
5000
1000
o = 5000 * 8
1000
625
1000
= .625
This same problem may be worked by the
following, more direct method:
52
how many decimal places it Is desired to carry
the quotient. If it is decided to terminate a
quotient at the third decimal place, the division
should be carried to the fourth place so that the
correct rounding off to the third place may be
determined.
When the dividend contains a decimal, the
same procedure applies as when the dividend is
whole. Notice the following examples (rounded
to three decimal places):
1. 6.31 + 8
.789
2. 0.0288 * 32
0.0009 = 0.001
32/0.0288
288
Observe in each case (including the case
where the dividend is whole), that the quotient
contains the same number of decimal places as
the number used in the dividend. Notice also
that the place values are rigid; that is, tenths
in the quotient appear over tenths in the divi-
dend, hundredths over hundredths, etc.
Practice problems. In the following division
problems, round off each quotient correct to
three decimal places.
1. 10 * 6
2. 23.5 * 16
Answers:
1. 1.667
2. 1.469
Decimal Divisors
3. 2.743 * 77
4. 1.00 + 3
3. 0.036
4. 0.333
In the foregoing examples, the divisor in
each case was an integer. Division with divi-
sors which are decimals may be accomplished
by changing the divisor and dividend so that the
divisor becomes a whole number.
the division problem as a fraction. Multiply
the numerator (dividend) and denominator (divi-
sor) by 10, 100, or some higher power of 10;
the power of 10 must be large enough to change
the divisor to a whole number. This rule is
illustrated as follows:
2.568 * 0.24 =
2.568
0.24
2.568
0.24
256.8
24
100
100
Thus 2.568 divided by 0.24 is the same as 256.8
divided by 24.
From the mechanical standpoint, the fore-
going rule has the effect of moving the decimal
point to the right, as many places as necessary
to change the divisor to an integer. Therefore
the rule is sometimes stated as follows: When
the divisor is a decimal, change it to a whole
number by moving the decimal point to the
right. Balance the change in the divisor by
moving the decimal point in the dividend an
equal number of places to the right.
The following example illustrates this ver-
sion of the rule:
9 1.1
0.9A/81.9A9
The inverted v, called a caret, is used as a
marker to indicate the new position of the deci-
mal point. Notice that the decimal point in the
quotient is placed immediately above the caret
in the dividend. Alinement of the first quotient
digit immediately above the 1 in the dividend,
and the second quotient digit above the 9, as-
sures that these digits are placed properly with
respect to the decimal point.
Practice problems. In the following division
problems, round off each quotient to three dec-
imal places:
1. 0.02958 + 0.12
2. 30.625 * 3.5
Answers:
1. 0.247
2. 8.750
3. 4610 + 0.875
4. 0.000576 * 0.008
3.. 5268.571
4. 0.072
53
Dividing by Powers of 10
Division of any number by 10, 100, 1,000,
etc., is really just an exercise in placing the
decimal point of a decimal fraction. Thus,
5,031 * 100 may be thought of as the decimal
fraction -TQQ-; to remove the denominator, we
simply count off two places from the right.
Thus,
5031
100
= 50.31
The following three examples serve to illus-
trate this procedure further:
401 + 10 = 40.1
2 + 1,000 = .002
11,431 v 100 = 114.31
If the dividend already contains a decimal
part, begin counting with the first number to
the left of the decimal point. Thus, 243.6 * 100 =
2.436. When the decimal point is not shown in
a number, it is always considered to be to the
right of the right-hand digit.
Dividing by 0.1, 0.01, 0.001, etc., may also
be accomplished by a simple mechanical rule.
We simply begin at the position of the decimal
point in the dividend and count off as many
places to the right as there are decimal places
in the divisor. The decimal point is then placed
to the right of the last digit counted. E there
are not enough digits, zeros may be added.
The foregoing rule is based on the fact that
0.1 is really JQ, 0.01 is ^, 0.001 is ~^, etc.
For example,
23 - 0.1 = 23 + ji
= 230
Notice that dividing by 0.1 is the same as
multiplying by 10. Likewise,
1
234.1 + 0.001 = 234.1 *
= 234.1 x
= 234,100
1000
1000
and
24 * 0.01 =
-
= 2,400
Practice problems. Divide by relocation of
the decimal point.
1. 276 + 100
2. 2,845 * 1,000
Answers:
1. 2.76
2. 2.845
3. 276 * 0.01
4. 2,845 * 0.001
3. 27,600
4. 2,845,000
54
CHAPTER 6
PERCENTAGE AND MEASUREMENT
In the discussion of decimal fractions, it was
shown that for convenience in writing fractions
whose denominators are 10 or some power of
10, the decimal point could be employed and the
denominators could be dropped. Thus, this spe-
cial group of fractions could be written in a
much simpler way. As early as the 15th cen-
tury, businessmen made use of certain decimal
fractions so much that they gave them the spe-
cial designation PERCENT.
MEANING OF PERCENT
The word "percent" is derived from Latin.
It was originally "per centum," which means
"by the hundred." Thus the statement is often
made that "percent means hundredths."
Percentage deals with the group of decimal
fractions whose denominators are 100 that is,
fractions of two decimal places. Since hun-
dredths were used so frequently, the decimal
point was dropped and the symbol % was placed
after the number and read "percent" (per 100).
Thus, 0.15 and 15% represent the same value,
15/100. The first is read "15 hundredths," and
the second is read "15 percent." Both mean 15
parts out of 100.
Ordinarily, percent is used in discussing
relative values. For example, 25 percent may
convey an idea of relative value or relationship.
To say "25 percent of the crew is ashore" gives
an idea of what part of the crew is gone, but it
does not tell how many. For example, 25 per-
cent of the crew would represent vastly different
numbers if the comparison were made between
an LSM and a cruiser. When it is necessary
to use a percent in computation, the number is
written in its decimal form to avoid confusion.
By converting all decimal fractions so that
they had the common denominator 100, men
found that they could mentally visualize the
relative size of the part of the whole that was
being considered.
CHANGING DECIMALS TO PERCENT
Since per cent means hundredths, any decimal
may be changed to percent by first expressing
it as a fraction with 100 as the denominator.
The numerator of the fraction thus formed in-
dicates how many hundredths we have, and
therefore it indicates "how many percent" we
have. For example, 0.36 is the same as 36/100.
Therefore, 0.36 expressed as a percentage
would be 36 percent. By the same reasoning,
since 0.052 is equal to 5.2/100, 0.052 is the
same as 5.2 percent.
In actual practice, the step in which the de-
nominator 100 occurs is seldom written down.
The expression in terms of hundredths is con-
verted mentally to percent. This results in the
following rule: To change a decimal to percent,
multiply the decimal by 100 and annex the per-
cent sign (%). Since multiplying by 100 has the
effect of moving the decimal point two places to
the right, the rule is sometimes stated as fol-
lows: To change a decimal to percent, move
the decimal point two places to the right and
annex the percent sign.
Changing Common Fractions and
Whole Numbers To Percent
Common fractions are changed to percent by
first expressing them as decimals. For exam-
ple, the fraction 1/4 is equivalent to the deci-
mal 0.25. Thus 1/4 is the same as 25 percent.
Whole numbers may be considered as special
types of decimals (for example, 4 may be writ-
tenas 4.00) and thus may be expressed in terms
of percentage. The meaning of an expression
such as 400 percent is vague unless we keep in
mind that percentage is a form of comparison.
For example, a question which often arises is
"How can I have more than 100 percent of some-
thing, if 100 percent means all of it?"
This question seems reasonable, if we limit
our attention to such quantities as test scores.
However, it is also reasonable to use percent-
age in comparing a current set of data with a
previous set. For example, if the amount of
electrical power used by a Navy facility this
year is double the amount used last year, then
this year's power usage is 200 percent of last
year's usage.
55
The meaning of a phrase such as "200 per-
cent of last year's usage" is often misinter-
preted. A total amount that is 200 percent of
the previous amount is not the same as an in-
crease of 200 percent. The increase in this
case is only 100 percent, for a total of 200. If
the increase had been 200 percent, then the
new usage figure would be 300 percent of the
previous figure.
Baseball batting averages comprise a spe-
cial case in which percentage is used with only
occasional reference to the word "percent."
The percentages in batting averages are ex-
pressed in their decimal form, with the figure
1.000 representing 100 percent. Although a
batting average of 0.300 is referred to as "bat-
ting 300," this is actually erroneous nomencla-
ture from the strictly mathematical standpoint.
The correct statement, mathematically, would
be "batting point three zero zero" or "batting
30 percent."
Practice problems. Change each of the fol-
lowing numbers to percent:
1. 0.0065
2. 1.25
Answers:
1. 0.65%
2. 125%
3. 0.363
4. 3/4
3. 36.3%
4. 75%
5. 7
6. 1/2
5. 700%
6. 50%
CHANGING A PERCENT
TO A DECIMAL
Since we do not compute with numbers in the
percent form, it is often necessary to change a
percent back to the decimal form. The proce-
dure is just opposite to that used in changing
decimals to percents: To change a percent to a
decimal, drop the percent sign and divide the
number by 100. Mechanically, the decimal
point is simply shifted two places to the left
and the percent sign is dropped. For example,
25 percent is the same as the decimal 0.25.
Percents larger than 100 percent are changed
to decimals by the same procedure as ordinary
percents. For example, 125 percent is equiva-
lent to 1.25.
Practice problems. Change the following
percents to decimals:
0.63%
3. 125%
4. 25%
5. 5;
6.
Answers:
1. 0.025
2. 0.0063
3. 1.25 5. 5.75% = 0.0575
4. 0.25 6. 9.50% = 0.095
THE THREE PERCENTAGE CASES
To explain the cases that arise in problems
involving percents, it is necessary to define the
terms that will be used. Rate (r) is the number
of hundredths parts taken. This is the number
followed by the percent sign. The base (b) is
the whole on which the rate operates. Percent-
age (p) is the part of the base determined by
the rate. In the example
5% of 40 = 2
5% is the rate, 40 is the base, and 2 is the
percentage.
There are three cases that usually arise in
dealing with percentage, as follows:
Case I To find the percentage when the
base and rate are known.
EXAMPLE: What number is 6% of 50 ?
Case n-To find the rate when the base and
percentage are known.
EXAMPLE: 20 is what percent of 60 ?
Case IE To find the base when the percent-
age and rate are known.
EXAMPLE: The number 5 is 25% of what
number ?
Case I
In the example
6% of 50 = ?
the "of" has the same meaning as it does in
fractional examples, such as
of 16 =
In other words, "of" means to multiply. Thus,
to find the percentage, multiply the base by the
rate. Of course the rate must be changed from
a percent to a decimal before multiplying can
56
0.06 x 50 = 3
The number that is 6% of 50 is 3.
FRACTIONAL PERCENTS.-Af r ac tional
percent represents a part of 1 percent. In a
case such as this, it is sometimes easier to
find 1 percent of the number and then find the
fractional part. For example, we would find
1/4 percent of 840 as follows:
1% of 840 = 0.01 x 840
= 8.40
Therefore, \% of 840 = 8.40 x ^
= 2.10
Case II
To explain case II and case III, we notice in
the foregoing example that the base corresponds
to the multiplicand, the rate corresponds to the
multiplier, and the percentage corresponds to
the product.
50 (base or multiplicand)
.06 (rate or multiplier)
3.00 (percentage or product)
Recalling that the product divided by one of its
factors gives the other factor, we can solve the
following problem:
?% of 60 = 20
We are given the base (60) and percentage (20).
60 (base)
__?_ (rate)
20 (percentage)
We then divide the product (percentage) by the
multiplicand (base) to get the other factor (rate) .
Percentage divided by base equals rate. The
rate is found as follows:
60
% (rate)
quotient in the decimal form first, and finally
as a percent.
Case in
The unknown factor in case III is the base,
and the rate and percentage are known.
EXAMPLE: 25% of 7 = 5
? (base)
.25 (rate)
5.00 (percentage)
We divide the product by its known factor to
find the other factor. Percentage divided by
rate equals base. Thus,
.25
= 20 (base)
The rule for case III may be stated as follows:
To find the base when the rate and percentage
are known, divide the percentage by the rate.
Practice problems. In each of the following
problems, first determine which case is in-
volved; then find the answer.
1. What is
of 740?
2. 7.5% of 2.75 = ?
3. 8 is 2% of what number?
4. ?% of 18 = 15.
5. 12% of ? = 12.
6. 8 is what percent of 32?
Answers:
1. Case I; 5.55
2. Case I; 0.20625
3. Case III; 400
4. Case II; 83^%
5. Case HI; 100
6. Case II; 25%
57
places, For example, we may oe asked to add
such numbers as 4.1 and 32.31582. How should
they be added? Should zeros be annexed to 4.1
until it is of the same order as the other deci-
mal (to the same number of places)? Or, should
.31582 be rounded off to tenths? Would the sum
be accurate to tenths or hundred-thousandths?
The answers to these questions depend on how
the numbers orignially arise.
Some decimals are finite or are considered
as such because of their use. For instance, the
decimal that represents TT, that is 0.5, is as
u
accurate at 0.5 as it is at 0.5000. Likewise,
the decimal that represents ? has the value
o
0.125 and could be written just as accurately
with additional end zeros. Such numbers are
said to be finite. Counting numbers are finite.
Dollars and cents are examples of finite values.
Thus, $10.25 and $5.00 are finite values.
To add the decimals that represent $ and ^,
O J-t
it is not necessary to round off 0.125 to tenths.
Thus, 0.5 + 0.125 is added as follows:
0.500
0.125
0.625
Notice that the end zeros were added to 0.5 to
carry it out the same number of places as 0.12 5.
It is not necessary to write such place-holding
zeros if the figures are kept in the correct col-
umns and decimal points are alined. Decimals
that have a definite fixed value may be added or
subtracted although they are of different order.
On the other hand, if the numbers result
from measurement of some kind, then the ques-
tion of how much to round off must be decided
in terms of the precision and accuracy of the
measurements.
ESTIMATION
Suppose that two numbers to be added re-
sulted from measurement. Let us say that one
number was measured with a ruler marked off
in tenths of an inch and was found, to the near-
est tenth of an inch, to be 2.3 inches. The other
tion between marks on any measuring instru-
ment is subject to human error. Experience
has shown that the best the average person can
do with consistency is to decide whether a
measurement is more or less than halfway be-
tween marks. The correct way to state this
fact mathematically is to say that a measure-
ment made with an instrument marked off in
tenths of an inch involves a maximum probable
error of 0.05 inch (five hundredths is one-half
of one tenth). By the same reasoning, the prob-
able error in a measurement made with an in-
strument marked in thousandths of an inch is
0.0005 inch.
PRECISION
In general, the probable error in any meas-
urement is one-half the size of the smallest
division on the measuring instrument. Thus
the precision of a measurement depends upon
how precisely the instrument is marked. It is
important to realize that precision refers to
the size of the smallest division on the scale; it
has nothing to do with the correctness of the
markings. In other words, to say that one in-
strument is more precise than another
does not imply that the less p r e c i s e in-
strument is poorly manufactured. In fact, it
would be possible to make an instrument with
very high apparent precision, and yet mark it
carelessly so that measurements taken with it
would be inaccurate.
From the mathematical standpoint, the pre-
cision of a number resulting from measurement
depends upon the number of decimal places;
that is, a larger number of decimal places
means a smaller probable error. In 2.3 inches
the probable error is 0.05 inch, since 2.3 actu-
ally lies somewhere between 2.25 and 2.35. In
1.426 inches there is a much smaller probable
error of 0.0005 inch. If we add 2.300 + 1.426
and get an answer in thousandths, the answer,
3.726 inches, would appear to be precise to
thousandths; but this is not true since there
was a probable error of .05 in one of the ad-
dends. Also 2.300 appears to be precise to
thousandths but in this example it is precise
only to tenths. It is evident that the precision
of a sum is no greater than the precision of the
58
least precise addend. It can also be shown that
the precision of a difference is no greater than
the less precise number compared.
To add or subtract numbers of different or-
ders, all numbers should first be rounded off to
the order of the least precise number. In the
foregoing example, 1.426 should be rounded to
tenths that is, 1.4.
This rule also applies to repeating decimals.
Since it is possible to round off a repeating
decimal at any desired point, the degree of pre-
cision desired should be determined and all re-
peating decimals to be added should be rounded
to this level. Thus, to add the decimals gener-
12 5
ated by j, -z, and rz correct to thousandths,
first round off each decimal to thousandths, and
then add, as follows:
.333
.667
.417
1.417
When a common fraction is used in recording
the results of measurement, the denominator of
the fraction indicates the degree of precision.
For example, a ruler marked in sixty-fourths
of an inch has smaller divisions than one
marked in sixteenths of an inch. Therefore a
4
measurement of 3gj inches is more precise
64
1
than a measure of 3 inches, even though the
16
two fractions are numerically equal. Remember
that a measurement of 3~ inches
64
contains a
probable error of only one -half of one sixty -
fourth of an inch. On the other hand, if the
smallest division on the ruler is one- sixteenth of
an inch, then a measurement of 3 inches con-
16
tains a probable error of one thirty-second of
an inch.
ACCURACY
Even though a number may be very precise,
which indicates that it was measured with an
instrument having closely spaced divisions, it
may not be very accurate. The accuracy of a
measurement depends upon the relative size of
the probable error when compared with the
quantity being measured. For example, a dis-
tance of 25 yards on a pistol range may be
measured carefully enough to be correct to the
nearest inch. Since there are 900 inches in 25
yards, this measurement is between 899.5
inches and 900.5 inches. When compared with
the total of 900 inches, the 0.5-inch probable
error is not very great.
On the other hand, a length of pipe may be
measured rather precisely and found to be 3.2
inches long. The probable error here is 0.05
inch, and this measurement is thus more pre-
cise than that of the pistol range mentioned be-
fore. To compare the accuracy of the two meas-
urements, we note that 0.05 inch out of a total
of 3.2 inches is the same as 0.5 inch out of 32
inches. Comparing this with the figure obtained
in the other example (0.5 inch out of 900), we
conclude that the more precise measurement is
actually the less accurate of the two measure-
ments considered.
It is important to realize that the location of
the decimal point has no bearing on the accu-
racy of the number. For example, 1.25 dollars
represents exactly the same amount of money
as 125 cents. These are equally accurate ways
of representing the same quantity, despite the
fact that the decimal point is placed differently.
Practice problems. In each of the following
problems, determine which number of each pair
is more accurate and which is more precise:
1. 3.72 inches or 2,417 feet
2. 2.5 inches or 17.5 inches
3 7
3. 5j inches or 12^- inches
4. 34.2 seconds or 13 seconds
Answers:
1. 3.72 inches is more precise.
2,417 feet is more accurate.
2. The numbers are equally precise.
17.5 inches is more accurate.
7
3. 12^- inches is more precise and more accu-
o
rate.
4. 34.2 seconds is more precise and more ac-
curate .
Percent of Error
The accuracy of a measurement is deter-
mined by the RELATIVE ERROR. The relative
59
MA i ntL IVIH. J. iL. 13 , VULiUiVilli J.
error is the ratio between the probable error
and the quantity being measured. This ratio is
simply the fraction formed by using the prob-
able error as the numerator and the measure-
ment itself as the denominator. For example,
suppose that a metal plate is found to be 5.4
inches long, correct to the nearest tenth of an
inch. The maximum probable error is five
hundredths of an inch {one-half of one tenth of
an inch) and the relative error is found as
follows:
probable error _ 0.05
measured value ~ 5i4
5
540
Thus the relative error is 5 parts out of 540.
Relative error is usually expressed as PER-
CENT OF ERROR. When the denominator of
the fraction expressing the error ratio is di-
vided into the numerator, a decimal is obtained.
This decimal, converted to percent, gives the
percent of error. For example, the error in
the foregoing problem could be stated as 0.93
percent, since the ratio 5/540 reduces to 0.0093
(rounded off) in decimal form.
Significant Digits
The accuracy of a measurement is often de-
scribed in terms of the number of significant
digits used in expressing it. If the digits of a
number resulting from measurement are exam-
ined one by one, beginning with the left-hand
digit, the first digit that is not is the first
significant digit. For example, 2345 has four
significant digits and 0.023 has only two sig-
nificant digits.
The digits 2 and 3 in a measurement such as
0.023 inch signify how many thousandths of an
inch comprise the measurement. The O's are
of no significance in specifying the number of
thousandths in the measurement; their presence
is required only as "place holders" in placing
the decimal point.
A rule that is often used states that the sig-
nificant digits in a number begin with the first
nonzero digit (counting from left to right) and
end with the last digit. This implies that can
be a significant digit if it is not the first digit
in the number. For example, 0.205 inch is a
measurement having three significant digits.
The between the 2 and the 5 is significant
because it is a part of the number specifying
how many hundredths are in the measurement.
The rule stated in the foregoing paragraph
fails to classify final O's on the right. For ex-
ample, in a number such as 4,700, the number
of significant digits might be two, three, or
four. If the O's merely locate the decimal point
(that is, if they show the number to be approxi-
mately forty -seven hundred rather than forty
seven), then the number of significant digits is
two. However, if the number 4,700 represents
a number such as 4,730 rounded off to the near-
est hundred, there are three significant digits.
The last merely locates the decimal point. If
the number 4,700 represents a number such as
4,700.4 rounded off, then the number of signifi-
cant digits is four.
Unless we know how a particular number
was measured, it is sometimes impossible to
determine whether right-hand O's are the result
of rounding off. However, in a practical situa-
tion it is normally possiWe to obtain informa-
tion concerning the instruments used and the
degree of precision of the original data before
any rounding was done.
In a number such as 49.30 inches, it is rea-
sonable to assume that the in the hundredths
place would not have been recorded at all if it
were not significant. In other words, the in-
strument used for the measurement can be read
to the nearest hundredth of an inch. The on
the right is thus significant. This conclusion
can be reached another way by observing that
the in 49.30 is not needed as a place holder in
placing the decimal point. Therefore its pres-
ence must have some other significance.
The facts concerning significant digits may
be summarized as follows:
1. Digits other than Oare always significant.
2. Zero is significant when it falls between
significant digits.
3. Any final to the right of the decimal
point is significant.
4. When a is present only as a place
holder for locating the decimal point, it is not
significant.
5. The following categories comprise the
significant digits of any measurement number:
a. The first nonzero left-hand digit is
significant.
b. The digit which indicates the precision
of the number is significant. This is the digit
farthest to the right, except when the right-hand
digit is 0. If it is 0, it may be only a place
holder when the number is an integer.
60
c. All digits between significant digits
are significant.
Practice problems. Determine the percent
of error and the number of significant digits in
each of the following measurements:
1. 5.4 feet
2. 0.00042 inch
Answers:
3. 4.17 sec
4. 147.50 miles
1. Percent of error: 0.93%
Significant digits: 2
2. Percent of error: 1.19%
Significant digits: 2
3. Percent of error: 0.12%
Significant digits: 3
4. Percent of error: 0.0034%
Significant digits: 5
CALCULATING WITH
APPROXIMATE NUMBERS
The concepts of precision and accuracy form
the basis for the rules which govern calculation
with approximate numbers (numbers resulting
from measurement).
Addition and Subtraction
A sum or difference can never be more pre-
cise than the least precise number in the cal-
culation. Therefore, before adding or sub-
tracting approximate numbers, they should be
rounded to the same degree of precision. The
more precise numbers are all rounded to the
precision of the least precise number in the
group to be combined. For example, the num-
bers 2.95, 32.7, and 1.414 would be rounded to
tenths before adding as follows:
3.0
32.7
1.4
Multiplication and Division
When two numbers are multiplied, the result
often has several more digits than either of the
original factors. Division also frequently pro-
duces more digits in the quotient than the orig-
inal data possessed, if the division is "carried
out" to several decimal places. Results such
as these appear to have more significant digits
than the original measurements from which they
came, giving the false impression of greater
accuracy than is justified. In order to correct
this situation, the following rule is used:
In order to multiply or divide two approxi-
mate numbers having an equal number of sig-
nificant digits, round the answer to the same
number of significant digits as are shown in one
of the original numbers. If one of the original
factors has more significant digits than the
other, round the more accurate number before
multiplying. It should be rounded to one more
significant digit than appears in the less accurate
number; the extra digit protects the answer from
the effects of multiple rounding. After perform-
ing the multiplication or division, round the
result to the same number of significant digits
as are shown in the less accurate of the original
factors.
Practice problems:
1. Find the sum of the sides of a triangle in
which the lengths of the three sides are as
follows: 2.5 inches, 3.72 inches, and 4.996
inches.
2. Find the product of the length and width of a
rectangle which is 2.95 feet long and 0.9046
foot wide.
Answers:
1. 11.2 inches
2. 2.67 square feet
MICROMETERS AND VERNIERS
Closely associated with the study of deci-
mals is a measuring instrument known as a
micrometer. The ordinary micrometer is ca-
pable of measuring accurately to one -thousandth
of an inch. One -thousandth of an inch is about
the thickness of a human hair or a thin sheet of
paper. The parts of a micrometer are shown
in figure 6-1.
MICROMETER SCALES
The spindle and the thimble move together.
The end of the spindle (hidden from view in
figure 6-1) is a screw with 40 threads per inch.
Consequently, one complete turn of the thimble
moves the spindle one -fortieth of an inch or
61
MATHEMATICS, VOLUME 1
THIMBLE RATCHET STOP
LOCKNUT SLEEVE
(A)
(B)
Figure 6-1. (A) Parts of a micrometer;
(B) micrometer scales.
0.025 inch since JQ is equal to 0.025. The
sleeve has 40 markings to the inch. Thus each
space between the markings on the sleeve is
also 0.025 inch. Since 4 such spaces are 0.1
inch (that is, 4 x 0.025), every fourth mark is
labeled in tenths of an inch for convenience in
reading. Thus, 4 marks equal 0.1 inch, 8 marks
equal 0.2 inch, 12 marks equal 0.3 inch, etc.
To enable measurement of a partial turn,
the beveled edge of the thimble is divided into
25 equal parts. Thus each marking on the
thimble is ^r of a complete turn, or ^ of TT
25
25
40
of an inch. Multiplying ^F times 0.025 inch, we
find that each marking on the thimble repre-
sents 0.001 inch.
READING THE MICROMETER
It is sometimes convenient when learning to
read a micrometer to writedown the component
parts of the measurement as read on the scales
and then to add them. For example, in figure
6-1 (B) there are two major divisions visible
(0.2 inch). One minor division is showing
clearly (0.025 inch). The marking on the thimble
nearest the horizontal or index line of the sleeve
is the second marking (0.002 inch). Adding
these parts, we have
0.200
0.025
0.002
0.227
Thus, the reading is 0.227 inch. As explained
previously, this is read verbally as "two hun-
dred twenty-seven thousandths." A more skill-
ful method of reading the scales is to read all
digits as thousandths directly and to do any
adding mentally. Thus, we read the major divi-
sion on the scale as "two hundred thousandths "
and the minor division is added on mentally.
The mental process for the above setting then
would be "two hundred twenty -five; two hundred
twenty-seven thousandths."
Practice problems:
1. Read each of the micrometer settings shown
in figure 6-2.
(A)
/ 1 I 1 t 5 T
(E)
(G)
1 2 } 4 9 >
(I)
Fieure 6-2. Micrometer settings.
Answers:
1. (A) 0.750 (F) 0.009
(B) 0.201 (G) 0.662
(C) 0.655 (H) 0.048
(D) 0.075 (I) 0.526
(E) 0.527
VERNIER
Sometimes the marking on the thimble of the
micrometer does not fall directly on the index
line of the sleeve. To make possible readings
even smaller than thousandths, an ingenious
device is introduced in the form of an additional
scale. This scale, called a VERNIER, was
named after its inventor, Pierre Vernier. The
vernier makes possible accurate readings to
the ten -thousandth of an inch.
Principle of the Vernier
Suppose a ruler has markings every tenth of
an inch but it is desired to read accurately to
hundredths. A separate, freely sliding vernier
scale (fig. 6-3) is added to the ruler. It has 10
markings on it that take up the same distance
as 9 markings on the ruler scale . Thus , each
19 9
space on the vernier is TTrof r~- inch, or
inch.
nier
Each vernier space is inch smaller than a
~-
How much smaller is a space on the ver-
than a space on the ruler? The ruler
space is rr- inch, or r^ inch, and the vernier
10
100
space is r^r inch. The vernier space is smaller
by the difference between these two numbers,
as follows:
10
100
9
100
1
100
DECIMAL RULER (ENLARGED)
1 1 1
.ir
1 1 1 1
10
)0
1 1 1 1
*
1 1
1 1 1 1
51834
till
> 6 7 8 9
1 L-
--2- /
100 /
VERMER
Figure 6-3. Vernier scale.
ruler space.
As an example of the use of the vernier
scale, suppose that we are measuring the steel
bar shown in figure 6-4. The end of the bar
almost reaches the 3 -inch mark on the ruler,
and we estimate that it is about halfway between
2.9 inches and 3.0 inches. The vernier marks
help us to decide whether the exact measure-
ment is 2.94 inches, 2.95 inches, or 2. 96 inches.
DECIMAL RULER IENLAROED)
3 4
I i i i i I r f i i. I I i ,i-,i .1 i
VEHNIEH
-STEEL BAH
BEING MEASURED
Figure 6-4. Measuring with a vernier.
The on the vernier scale is spaced the
distance of exactly one ruler mark (in this case,
one tenth of an inch) from the left hand end of
the vernier. Therefore the is at a position
between ruler marks which is comparable to
the position of the end of the bar. In other
words, the on the vernier is about halfway
between two adjacent marks on the ruler, just
as the end of the bar is about halfway between
two adjacent marks. The 1 on the vernier scale
is a little closer to alinement with an adjacent
ruler mark; in fact, it is one hundredth of an
inch closer to alinement than the 0. This is
because each space on the vernier is one hun-
dredth of an inch shorter than each space on
the ruler.
Each successive mark on the vernier scale
is one hundredth of an inch closer to alinement
than the preceding mark, until finally alinement
is achieved at the 5 mark. This means that the
on the vernier must be five hundredths of an
inch from the nearest ruler mark, since five
increments, each one hundredth of an inch in
size, were used before a mark was found in
alinement.
We conclude that the end of the bar is five
hundredths of an inch from the 2.9 mark on the
ruler, since its position between marks is ex-
actly comparable to that of the on the vernier
scale. Thus the value of our measurement is
2.95 inches.
63
through for any distance between markings.
Suppose the mark fell seven tenths of the dis-
tance between ruler markings. It would take
seven vernier markings , a loss of one-hundredth
of an inch each time, to bring the marks in line
at 7 on the vernier .
The vernier principle may be used to get
fine linear readings, angular readings, etc.
The principle is always the same. The vernier
has one more marking than the number of mark-
ings on an equal space of the conventional scale
of the measuring instrument. For example, the
vernier caliper (fig. 6-5) has 25 markings on
the vernier for 24 on the caliper scale. The
caliper is marked off to read to fortieths (0.025)
of an inch, and the vernier extends the accuracy
to a thousandth of an inch.
Figure 6-5. A vernier caliper.
Vernier Micrometer
By adding a vernier to the micrometer, it is
possible to read accurately to one ten-thousandth
of an inch. The vernier markings are on the
sleeve of the micrometer and are parallel to
the thimble markings. There are 10 divisions
on the vernier that occupy the same space as 9
divisions on the thimble. Since a thimble space
is one thousandth of an inch, a vernier space is
10 of iooo inch > or 16600 inch - rt is 10606 inch
less than a thimble space. Thus, as in the pre-
ceding- explanation of verniers, it is possible to
read the nearest ten- thousandth of an inch by
reading the vernier digit whose marking coin-
cides with a thimble marking.
In figure 6-6 (A), the last major division
showing fully on the sleeve index is 3. The
third minor division is the last mark clearly
and below the index is the 8 (0.008). The ver-
nier marking that matches a thimble marking
is the fourth (0.0004). Adding them all together,
we have,
0.3000
0.0750
0.0080
0.0004
0.3834
The reading is 0.3834 inch. With practice these
readings can be made directly from the microm-
eter, without writing the partial readings.
-20
0123' r
"liiilmlinlni :
-15
-5
(A)
116
-10
I Z
Imtmfi"
01234!
-0
20
it;
(O
C\J
F-I5
10
IE)
IF)
Figure 6-6. Vernier micrometer settings.
Practice problems:
1. Read the micrometer settings in figure 6-6.
Answers:
1. (A) See the foregoing example.
(B) 0.1539 (E) 0.4690
(C) 0.2507 (F) 0.0552
(D) 0.2500
64
CHAPTER 7
EXPONENTS AND RADICALS
The operation of raising a number to a power
is a special case of multiplication in which the
factors are all equal. In examples such as
4 2 = 4 x 4 = 16 and 5 3 = 5x5x5 = 125, the
number 16 is the second power of 4 and the
number 125 is the third power of 5. The ex-
pression 5 3 means that three 5's are to be mul-
tiplied together. Similarly, 4 2 means 4x4.
The first power of any number is the number
itself. The power is the number of times the
number itself is to be taken as a factor.
The process of finding a root is the inverse
of raising a number to a power. A root is a
special factor of a number, such as 4 in the
expression 4 2 = 16. When a number is taken
as a factor two times, as in the expression
4 x 4 = 16, it is called a square root. Thus, 4
is a square root of 16. By the same reasoning,
2 is a cube root of 8, since 2x2x2 is equal
to 8. This relationship is usually written as
2 3 = 8.
POWERS AND ROOTS
A power of a number is indicated by an EX-
PONENT, which is a number in small print
placed to the right and toward the top of the
number. Thus, in 4 3 = 64, the number 3 is the
EXPONENT of the number 4. The exponent 3
indicates that the number 4, called the BASE,
is to be raised to its third power. The expres-
sion is read "4 to the third power (or 4 cubed)
equals 64." Similarly, 5 2 = 25 is read "5 to the
second power (or 5 squared) equals 25." Higher
powers are read according to the degree indi-
cated; for example, "fourth power," "fifth
power," etc.
When an exponent occurs, it must always be
written unless its value is 1. The exponent 1
usually is not written, but is understood. For
example, the number 5 is actually 5 1 . When we
work with exponents, it is important to remem-
ber that any number that has no written expo-
nent really has an exponent equal to 1.
A root of a number can be indicated by plac-
ing a radical sign, -vT, over the number and
within the notch of the radical sign. Thus, N/154
indicates the cube root of 64, and \H32 indicates
the fifth root of 32. The number that indicates
the root is called the INDEX of the root. In the
case of the square root, the index, 2, usually is
not shown. When a radical has no index, the
square root is understood to be the one desired.
For example, ^T36 indicates the square root of
36. The line above the number whose root is to
be found is a symbol of grouping called the vin-
culum. When the radical symbol is used, a vin-
culum, long enough to extend over the entire
expression whose root is to be found, should be
attached.
Practice problems. Raise to the indicated
power or find the root indicated.
3. 4 3 4. 25 3
7. ^Tl25 8. Vl2
1.
5.
1.
5.
2 3
Answers
8
4
2. 6 2
6. 3 J1
2. 36
6. 2
3. 64
7. 5
4. 15,625
8. 2
NEGATIVE INTEGERS
Raising to a power is multiplication in which
all the numbers being multiplied together are
equal. The sign of the product is determined,
as in ordinary multiplication, by the number of
minus signs. The number of minus signs is odd
or even, depending on whether the exponent of
the base is odd or even. For example, in the
problem
(-2) 3 = (-2) (-2) (-2) = -8
there are three minus signs. The result is
negative. In
(-2) 6 = 64
there are six minus signs. The result is posi-
Thus, when the exponent of a negative num-
ber is odd, the power is negative; when the ex-
ponent is even, the power is positive.
As other examples, consider the following:
(-3) 4 = 81
2\ 3 _ _8_
5/ " 125
(-2) 8 = 256
(-1)5 = -1
Positive and negative numbers belong to the
class called REAL NUMBERS. The square of a
realnumber ispositive. For example, (-7) 2 = 49
and 7 2 = 49. The expression (-7) 2 is read
"minus seven squared." Note that either seven
squared or minus seven squared gives us +49.
We cannot obtain -49 or any other negative
number by squaring any real number, positive
or negative.
Since there is no real number whose square
is a negative number, it is sometimes said that
the square root of a negative number does not
exist. However, an expression under a square
root sign may take on negative values. While
the square root of a negative number cannot
actually be found, it can be indicated.
The indicated square root of a negative num-
ber is called an IMAGINARY NUMBER. The
number J^T, for example, is said to be imagi-
nary. It is read "square root of minus seven."
Imaginary numbers are discussed in chapter 15
of this course.
FRACTIONS
We recall that the exponent of a number tells
the number of times that the number is to be
taken as a factor. A fraction is raised to a
power by raising the numerator and the denom-
inator separately to the power indicated. The
expression (=J means is used twice as a
factor. Thus,
_ 3 3 _ 3f
-7X7-72
__
49
Similarly,
Since a minus sign can occupy any one of
three locations in a fraction, notice that evalu-
/ 1\ 2
ating (--s) is equivalent to
The process of taking a root of a number is
the inverse of the process of raising the num-
ber to a power, and the method of taking the
root of a fraction is similar. We may simply
take the root of each term separately and write
the result as a fraction. Consider the following
examples:
Practice problems. Find the values for. the
indicated operations:
2. T
Answers:
1. 1/9
5. 4/6
DECIMALS
2. 9/16
6. 4/5
3. 36/25
7. 2/3
4. 8/27
8. 3/7
25
When a decimal is raised to a power, the
number of decimal places in the result is equal
to the number of places in the decimal multi-
plied by the exponent. For example, consider
(0.12) 3 . There are two decimal places in 0.12
and 3 is the exponent. Therefore, the number
of places in the power will be 3(2) = 6. The re-
sult is as follows:
(0.12) 3 = 0.001728
The truth of this rule is evident when we re-
call the rule for multiplying decimals. Part of
the rule states: Mark off as many decimal
places in the product as there are decimal
places in the factors together. If we carry out
66
for any decimal raised to any power by simply
carrying out the multiplication indicated by the
exponent.
Consider these examples:
(1.4) 2 = 1.96
(0.12) 2 = 0.0144
(0.4) 3 = 0.064
(0.02) 2 = 0.0004
(0.2) 2 = 0.04
Finding a root of a number is the inverse of
raising a number to a power. To determine the
number of decimal places in the root of a per-
fect power, we divide the number of decimal
places in the radicand by the index of the root.
Notice that this is just the opposite of what was
done in raising a number to a power.
Consider *1 0.0625. The square root of 625
is 25. There are four decimal places in the
radicand, 0.0625, and the index of the root is 2.
Therefore, 4*2 = 2 is the number of decimal
places in the root. We have
= 0.25
4 x 4, we see that 4 is used as a factor five
times. Therefore 4 3 x 4 2 is the same as 4 s .
This result could be written as follows:
4 3 x4 2 =4x4x4x4x4
Notice that three of the five 4's came from
the expression 4 3 , and the other two 4's came
from the expression 4 2 . Thus we may rewrite
the problem as follows:
4 3 x 4 2 = 4< 3+2 >
= 4 s
The law of exponents for multiplication may
be stated as follows: To multiply two or more
powers having the same base, add the exponents
and raise the common base to the sum of the
exponents. This law is further illustrated by
the following examples:
2 3 x 2 4 = 2 7
3 x 3 2 = 3 3
15 4 x 15 2 = 15 6
10 2 x 10' s = 10 2 ' 5
Similarly,
LAWS OF EXPONENTS
All of the laws of exponents may be devel-
oped directly from the definition of exponents.
Separate laws are stated for the following five
cases:
1. Multiplication.
2. Division.
3. Power of a power.
4. Power of a product.
5. Power of quotient.
MULTIPLICATION
To illustrate the law of multiplication, we
examine the following problem:
Common Errors
It is important to realize that the base must
be the same for each factor, in order to apply
the laws of exponents. For example, 2 3 x 3 is
neither 2 5 nor 3 5 . There is no way to apply the
law of exponents to a problem of this kind. An-
other common mistake is to multiply the bases
together. For example, this kind of error in
the foregoing problem would imply that 2 3 x 3 2
is equivalent to 6 s , or 7776. The error of this
may be proved as follows:
8x9
2 3 x
3 2 =
= 72
DIVISION
The law of exponents for division may be
developed from the following example:
x
= 6'
67
Cancellation of the five 6's in the divisor with
five of the 6's in the dividend leaves only two
6's, the product of which is 6 2 .
This result can be reached directly by noting
that 6 2 is equivalent to 6 ( - 7 ~ 5 \ In other words,
we have the following:
6 7 + 6 s = 6 (7 - 5)
= 6 2
Therefore the law of exponents for division is
as follows: To divide one power into another
having the same base, subtract the exponent of
the divisor from the exponent of the dividend.
Use the number resulting from this subtraction
as the exponent of the base in the quotient.
Use of this rule sometimes produces a neg-
ative exponent or an exponent whose value is 0.
These two special types of exponents are dis-
cussed later in this chapter.
POWER OF A POWER
Consider the example (3 2 ) 4 . Remembering
that an exponent shows the number of times the
base is to be taken as a factor and noting in
this case that 3 2 is considered the base, we
have
(3 2 ) 4 = 3 2 3 2 3 2 3 2
Also in multiplication we add exponents. Thus,
3 2 - 3 2 - 3 2 3 2 = 3 (2+2+2 +2) = 3 8
Therefore,
(3 2 ) 4 = 3 (4X2)
= 3 s
The laws of exponents for the power of a
power may be stated as follows: To find the
power of a power, multiply the exponents. It
should be noted that this case is the only one in
which multiplication of exponents is performed,
POWER OF A PRODUCT
Consider the example (3 2 5) 3 . We know
that
Thus 3, 2, and 5 appear three times each as
factors, and we can show this with exponents as
3 3 , 2 3 , and 5 3 . Therefore,
(3 2 5) 3 = 3 3 2 3 5 3
The law of exponents for the power of a
product is as follows: The power of a product
is equal to the product obtained when each of
the original factors is raised to the indicated
power and the resulting powers are multiplied
together.
POWER OF A QUOTIENT
The law of exponents for a power of an indi-
cated quotient may be developed from the fol-
lowing example:
2\ 3 _ 2 2 2
3/ ~ 3 ' 3 ' 3
_ 2-2-2
"3-3-3
2!
~ 3 3
Therefore,
The law is stated as follows: The power of
a quotient is equal to the quotient obtained when
the dividend and divisor are each raised to the
indicated power separately, before the division
is performed.
Practice problems. Raise each of the fol-
lowing expressions to the indicated power:
3.
3 - 2
5
2. 3" + 3"
Answers:
4. (-3 2 ) 3
5
& ' 5
6. (3 2 7) 2
1. 3 4 x 2 6 = 5,184
2. 27
3.
1
125
(3
5) 3 = (3 - 2 - 5)(3 - 2 - 5) (3 - 2 - 5)
4. [(-3) 2 ] 3 = 729
5. 25
6. 9 4 - 49 = 1,764
SPECIAL EXPONENTS
ONE AS AN EXPONENT
Thus far in this discussion of exponents, the
emphasis has been on exponents which are posi-
tive integers. There are two types of exponents
which are not positive integers, and two which
are treated as special cases even though they
may be considered as positive integers,
ZERO AS AN EXPONENT
Zero occurs as an exponent in the answer to
a problem such as 4 3 + 4 3 . The law of expo-
nents for division states that the exponents are
to be subtracted. This is illustrated as follows:
=
The number 1 arises as an exponent some-
times as a result of division. In the example
5 3
2 we subtract the exponents to get
= 5
This problem may be worked another way as
follows:
Therefore,
5 1 = 5
= 5
Another way of expressing the result of
dividing 4 3 by 4 3 is to use the fundamental
axiom which states that any number divided by
itself is 1. In order for the laws of exponents
to hold true in all cases, this must also be true
when any number raised to a power is divided
by itself. Thus, 4 3 /4 3 must equal 1.
Since 4 3 /4 3 has been shown to be equal to
both 4 and 1, we are forced to the conclusion
that 4 = 1.
By the same reasoning,
We conclude that any number raised to the
first power is the number itself. The exponent
1 usually is not written but is understood to
exist.
NEGATIVE EXPONENTS
If the law of exponents for division is ex-
tended to include cases where the exponent of
the denominator is larger, negative exponents
arise. Thus,
5 _ s
- 5
_ =
- 5
|2-5
= 3
-3
Also,
Therefore,
*-'
5 = I
Thus we see that any number divided by itself
results in a exponent and has a value of 1.
By definition then, any number (other than zero)
raised to the zero power equals 1. This is fur-
ther illustrated in the following examples:
3 = 1
400 = 1
0.02 = 1
v5
K5) = 1
Another way of expressing this problem is as
follows:
3-3-3 3
Therefore,
r 3 * Jr
We conclude that a number N with a negative
exponent is equivalent to a fraction having the
following form: Its numerator is 1; its denomi-
nator is N with a positive exponent whose abso-
lute value is the same as the absolute value of
the original exponent. In symbols, this rule
may be stated as follows:
N a =
N
69
Also,
N
NJ
The following examples further illustrate
the rule:
' 12
Notice that the sign of an exponent may be
changed by merely moving the expression which
contains the exponent to the other position in the
fraction. The sign of the exponent is changed
as this move is made. For example,
1
nr 2
....
'
= i x
10 2
i?
1
Therefore,
By using the foregoing relationship, a prob-
lem such as 3 -5- 5" 4 may be simplified as fol-
lows:
= 3 x 5 4
FRACTIONAL EXPONENTS
Fractional exponents obey the same laws as
do integral exponents. For example,
4 l/2 x 4 1/2 _ 4 ( 1/2 + 1/2)
= 4 2 / 2
= 4 1 = 4
Another way of expressing this would be
4 1/2 x 4 l/2 = (4 1/2) 2
_ 4 (l/-2 x 2)
= 4 1 = 4
Observe that the number 4 1/2 , when squared
in the foregoing example, produced the number
4 as an answer. Recalling that a square root of
a number N is a number x such that x 2 = N, we
conclude that 4 1/2 is equivalent to "sTi". Thus
we have a definition, as follows: A fractional
exponent of the form 1/r indicates a root, the
index of which is r. This is further illustrated
in the following examples:
2 1/2
1/3
Also,
6 2/3 = (6 l/ 3) 2 =
2/3
3 2/3
Notice that in an expression such as 8" 1 ' we
can either find the cube root of 8 first or square
8 first, as shown by the following example:
(8
1/32
2 2 = 4 and (8 2 ) 1/3
= 4
All the numbers in the evaluation of 8 2/3
remain small if the cube root is found before
raising the number to the second power. This
order of operation is particularly desirable in
evaluating a number like 64 5/6 . If 64 were first
raised to the fifth power, a large number would
result. It would require a great deal of unnec-
essary effort to find the sixth root of 64 s . The
result is obtained easily, if we write
64
5/6
= (64 1/6 ) 5 =
2 = 32
If an improper fraction occurs in an expo-
nent, such as 7/3 in the expression 2 7/3 , it is
customary to keep the fraction in that form
rather than express it as a mixed number. In
fraction form an exponent shows immediately
what power is intended and what root is in-
tended. However, 2 7/3 can be expressed in
another form and simplified by changing the
improper fraction to a mixed number and writ-
ing the fractional part in the radical form as
follows:
7/3
,1/3
= 4
70
The law of exponents for multiplication may
be combined with the rule for fractional expo-
nents to solve problems of the following type:
PROBLEM: Evaluate the expression 4 2 - 5 .
The laws of exponents form the basis for
calculation using powers of 10. The following
list includes several decimals and whole num-
bers expressed as powers of 10:
SOLUTION:
2
1/3
2.5 = 4 2 X 4 O.S
= 16 x 4 1/2
= 16 x 2
= 32
Practice problems:
1. Perform the indicated division:
2. Find the product: 7 2/5 x 7 1/10 x 7 3/1
3. Rewrite with a positive exponent and sim-
plify: 9~ 1/2
4. Evaluate 100 3/2
5. Evaluate (8) 5
Answers:
i. 2 3/3 + 2 l/3 = VT
9 78/10
& I
Q 1 1
3. gTTT = 3
4. 1,000
5. 1
SCIENTIFIC NOTATION
AND POWERS OF 10
Technicians, engineers, and others engaged
in scientific work are often required to solve
problems involving very large and very small
numbers. Problems such as
22,684 x 0.00189
0.0713 x 83 x 7
are not uncommon. Solving such problems by
the rules of ordinary arithmetic is laborious
and time consuming. Moreover, the tedious
arithmetic process lends itself to operational
errors. Also there is difficulty in locating the
decimal point in the result. These difficulties
can be greatly reduced by a knowledge of the
powers of 10 and their use.
10,000
=
10 4
1,000
=
10 3
100
=
10 2
10
=
10 1
1
=
10
1 __
KT 1
.01 =
10~ 2
.001 =
10~ 3
.0001 =
10~ 4
The concept of scientific notation may be
demonstrated as follows:
60,000 = 6.0000 x 10,000
= 6 x 10 4
538 = 5.38 x 100
= 5.38 x 10 2
Notice that the final expression in each of
the foregoing examples involves a number be-
tween 1 and 10, multiplied by a power of 10.
Furthermore, in each case the exponent of the
power of 10 is a number equal to the number of
digits between the new position of the decimal
point and the original position (understood) of
the decimal point.
We apply this reasoning to write any number
in scientific notation; that is, as a number be-
tween 1 and 10 multiplied by the appropriate
power of 10. The appropriate power of 10 is
found by the following mechanical steps:
1. Shift the decimal point to standard posi-
tion, which is the position immediately to the
right of the first nonzero digit.
2. Count the number of digits between the
new position of the decimal point and its origi-
nal position. This number indicates the value
of the exponent for the power of 10.
3. If the decimal point is shifted to the. left,
the sign of the exponent of 10 is positive; if the
decimal point is shifted to the right, the sign of
the exponent is negative.
The validity of this rule, for those cases in
which the exponent of 10 is negative, is demon-
strated as follows:
71
0.00657 = 6.57 x 0.001
= 6.57 x 10~ 3
0.348 = 3.48 x 0.1
= 3.48 x 10~ J
Further examples of the use of scientific
notation are given as follows:
543,000,000 = 5.43 x 10 8
186 = 1.86 x 10 2
243.01 = 2.4301 x 10 2
0.0000007 = 7 x 10~ 7
0.00023 = 2.3 x 10~ 4
Multiplication Using Powers of 10
From the law of exponents for multiplication
we recall that to multiply two or more powers
to the same base we add their exponents. Thus,
10 4 x 10 2 = 10 6
We see that multiplying powers of 10 together
is an application of the general rule. This is
demonstrated in the following examples:
1. 10,000 x 100 = 10* x 10 2
= 10 4 + 2
= 10 5
2. 0.0000001 x 0.001 = 10~ 7 x 10~ 3
3.
4.
= 10' 10
10,000 x 0.001 = 10 4 x 10~ 3
= 10 4 ~ 3
= 10
23,000 x 500 = ?
23,000 = 2.3 x 10 4
500 = 5 x 10 2
Therefore,
23,000 x 500 = 2.3 x 10 4 x 5 x 10 2
= 2.3 x 5 x 10 4 x 10 2
= 11.5 x 10 6
= 1.15 x JO 7
5. 62,000 x 0.0003 x 4,600 = ?
62,000 = 6.2 x 10 4
0.0003 = 3 x 10~ 4
4,600 = 4.6 x 10 3
Therefore,
62,000 x 0.0003 x 4,600 = 6.2 x 3
x 4.6 x 10 4 x 10~ 4 x 10 3
= 85.56 x 10 3
= 8.556x 10 4
Practice problems. Multiply, using powers
of 10. For the purposes of this exercise, treat
all numbers as exact numbers:
1. 10,000 x 0.001 x 100
2. 0.000350 x 5,000,000 x 0.0004
3. 3,875 x 0.000032 x 3,000,000
4. 7,000 x 0.015 x 1.78
Answers:
1. 1.0 x 10 3
2. 7.0 x lO" 1
3. 3.72 x 10 s
4. 1.869 x 10 2
Division Using Powers of 10
The rule of exponents for division states
that, for powers of the same base, the exponent
of the denominator is subtracted from the ex-
ponent of the numerator. Thus,
= 10 7 '
10;
10 ;
It should be remembered that powers may
be transferred from numerator to denominator
or from denominator to numerator by simply
changing the sign of the exponent. The follow-
ing examples illustrate the use of this rule for
powers of 10:
1.
72,000
0.0012
7.2 x 10 4
1.2 x 10~ 3
7.2
1.2
= 6 x 10 7
x 10 4 x 10 3
72
Chapter 7-EXPONENTS AND RADICALS
2.
44 x 10 " 4 44
x 10~ 4 x 10 5
Answers:
11 x 10~ 5 11
41 r\
1.
2.4 x 10~ 6
x 10
2.
3.6 x 10
Combined Multiplication
and Division
3.
9.8 x 10~ 2
Using the rules already shown, multiplica-
tion and division involving powers of 10 may be
combined. The usual method of solving such
problems is to multiply and divide alternately
until the problem is completed. For example,
36,000 x 1.1 x 0.06
0.012 x 2,200
Rewriting this problem in scientific notation,
we have
3.6x 10 4 x 1.1 x 6x 10 2
1.2x 10~ 2 x 2.2 x 10 3
_ 3.6 x 1.1 x 6
1.2 x 2.2
= 9 x 10
= 90
x 10
Notice that the elimination of O's, wherever
possible, simplifies the computation and makes
it an easy matter to place the decimal point.
SIGNIFICANT DIGITS. -One of the most im-
portant advantages of scientific notation is the
fact that it simplifies the task of determining
the number of significant digits in a number.
For example, the fact that the number 0.00045
has two significant digits is sometimes ob-
scured by the presence of the O's. The confu-
sion can be avoided by writing the number in
scientific notation, as follows:
0.00045 = 4.5 x 10~ 4
Practice problems. Express the numbers in
the following problems in scientific notation
and round off before performing the calculation.
In each problem, round off calculation numbers
to one more digit than the number of significant
digits in the least accurate number; round the
answer to the number of significant digits in.
the least accurate number:
1. 0.000063 x 50.4 x 0.007213
2.
780 x 0.682 x 0.018
0.015 x 216 x 1.78
72 x 0.0624 x 0.0353
3. 0.000079 x 0^00036
Other Applications
The applications of powers of 10 may be
broadened to include problems involving recip-
rocals and powers of products.
RECIPROCALS. The following example il-
lustrates the use of powers of 10 in the forma-
tion of a reciprocal:
250,000 x 300 x 0.02
_ 1
~ 2.5 x 10 s x 3 x 10 2 x 2 x 10~ 2
= 10~ 5
2.5 x 3 x 2
_ 10~ 5
~ 15
Rather than write the numerator as 0.00001,
write it as the product of two factors, one of
which may be easily divided, as follows:
10~ s 10 2 x 10~ 7
15
15
n''
x in'
x 10
= 6.67 x 10~ 7
= 0.000000667
POWER OF A PRODUCT. -The following
example illustrates the use of powers of 10 in
finding the power of a product:
(80,000 x 2 x 10 s ) 2 = (8 x 10 4 x 2 x 10 s ) 2
= 8 2 x 2 2 x (10 4+5 ) 2
= 64 x 4 x 10 18
= 256 x 10 18
= 2.56 x 10 20
RADICALS
An expression such as \T2, /5~, or Va + b
in tne radical torm. me wora radical is ae-
rived from the Latin word "radix, "which means
"root." The word "radix" itself is more often
used in modern mathematics to refer to the
base of a number system, such as the base 2 in
the binary system. However, the word "radical"
is retained with its original meaning of "root."
The radical symbol (\T) appears to be a dis-
tortion of the initial letter "r" from the word
"radix." With long usage, the r gradually lost
its significance as a letter and became dis-
torted into the symbol as we use it. The vin-
culum helps to specify exactly which of the
letters and numbers following the radical sign
actually belong to the radical expression.
The number under a radical sign is theRAD-
ICAND. The index of the root (except in the
case of a square root) appears in the trough of
the radical sign. The index tells what root of
the radicand is intended. For example, in V 32,
the radicand is 32 and the index of the root is 5.
The fifth root of 32 is intended. In \/~50, the
square root of 50 is intended. When the index
is 2, it is not written, but is understood.
If we can find one square root of a number
we can always find two of them. Remember
(3) 2 is 9 and (-3) 2 is also 9. Likewise (4) 2 and
(-4) 2 both equal 16 and (5) 2 and (-5) 2 both equal
25. Conversely, 4^ is +3 or -3, ^/T6 is +4 or
-4, and ^25 is +5 or -5. When we wish to show
a number that may be either positive or nega-
tive, we may use the symbol which is read
"plus or minus." Thus 3 means "plus or
minus 3." Usually when a number is placed
under the radical sign, only its positive root is
desired and, unless otherwise specified, it is
the only root that need be found.
COMBINING RADICALS
A number written in front of another number
and intended as a multiplier is called a COEF-
FICIENT. The expression 5x means 5 times x;
aymeans a times y; and 7 N/T means 7 times
v2. In these examples, 5 is the coefficient of
x, a is the coefficient of y, and 7 is. the coeffi-
cient of A /2~.
Radicals having the same index and the same
radicand are SIMILAR. Similar radicals may
have different coefficients in front of the radi-
cal sign. For example, 3 *J~2, \/T, and |- \T2
cais is tne same as mat stated lor aaaing de-
nominate numbers: Add only units of the same
kind. For example, we could add 2 N/T and
4 N/T because the "unit" in each of these num-
bers is the same (N/~3)_. By the same reasoning,
we could not add 2 */3 and 4 "/~5 because these
are not similar radicals.
Addition and Subtraction
When addition or subtraction of similar rad-
icals is indicated, the radicals are combined by
adding or subtracting their coefficients and
placing the result in front of the radical. Add-
ing 3 N/T and 5 \T~2~ is similar to adding 3 bolts
and 5 bolts. The following examples illustrate
the addition and subtraction of similar radical
expressions:
= 8 /2"
+ 1/3 (NTS) = 5/6 (\T3)
1. 3 \T2~
2. 1/2
3. N/Tr
4. -5 ITT -2 N/T+ ? Vr= o
Example 4 illustrates a case that is some-
times troublesome. The sum of the coefficients,
-5, -2, and 7, is 0. Therefore, the coefficient
of the answer would be 0, as follows:
O(\TT) = o x
Thus the final answer is 0, since multiplied
by any quantity is still 0.
Practice problems. Perform the indicated
operations:
1. 4
2.
- \T3~+ 5
3. -\T5 - 6
4. -2 N/TJO - 7 "xTIS"
Answers
1. 8
2. 1
3. -5
4. -9
74
otner radical, multiplication is intended. Some-
times a dot is placed between the radicals, but
not always. Thus, either N/T- vTT or -/T-sTTT
means multiplication.
When multiplication or division of radicals
is indicated, several radicals having the same
index can be combined into one radical, if de-
sired. Radicals having the same index are said
to be of the SAME ORDER. For example, "/"2
is a radical of the second order. The radicals
^/and "'/IT are of the same order.
If radicals are of the same order, the radi-
cands can be multiplied or divided and placed
under one radical symbol. For example , "/IT
multiplied by N/~3~is the same as V 5 x 3. Also,
N/1T divided by */3" is the same as V6 - 3. If
coefficients appear before the radicals, they
also must be included in the multiplication or
division. This is illustrated in the following
examples:
1.
!".3 N/T
= 2.N/~2.3.
/F
= 2-3 N/T.
vTT
= 2-3 \TJT1
>
= 6 \/10
15 VF
_15 /F
2.
= 5x\T2"
= 5 \T2
It is important to note that what we have
said about multiplication and division does not
apply to addition. A typical error is to treat
the expression N/9 + 4 as if it were equivalent
to N/1F + \T4T These expressions cannot be
equivalent, since 3 + 2 is not equivalent to N! 13.
FACTORING RADICALS.- A radical can be
split into two or more radicals of the same or-
der if the radicand can be factored. This is
illustrated in the following examples:
lent torm that is easier to use. A radical is in
its simplest form when no factor can be re-
moved from the radical, when there is no frac-
tion under the radical sign, and when the index
of the root cannot be reduced. A factor can be
removed from the radical if it occurs a number
of times equal to the index of the root. The fol-
lowing examples illustrate this:
Removing a factor that occurs a number of
times equal to the index of the root is equiva-
lent to separating a radical into two radicals so
that one radicand is a perfect power. The rad-
ical sign can be removed from the number that
is a perfect square, cube, fourth power, etc.
The root taken becomes the coefficient of the
remaining radical.
In order to simplify radicals easily, it is
convenient to know the squares of whole num-
bers up to about 25 and a few of the smaller
powers of the numbers 2, 3, 4, 5, and 6. Table
7-1 shows some frequently used powers of
numbers.
Table 7-1. Powers of numbers.
= vT= 2
I 2
= 1
14 2
= 196
2 2
= 4
15 2
= 225
3 2
= 9
16 2
= 256
4 2
= 16
1?2
= 289
5 2
= 25
18 2
= 324
6 2
= 36
19 2
= 361
7 2
= 49
20 2
= 400
8 2
= 64
21 2
= 441
9 2
= 81
22 2
= 484
10 2
= 100
23 2
= 529
II 2
= 121
24 2
= 576
12 2
= 144
25 2
= 625
13 2
= 169
(A)
75
Table 7-1. Powers of numbers Continued.
2 1 = 2
3 1 = 3
2 2 = 4
3 2 = 9
2 3 = 8
3 3 = 27
2 4 = 16
3 4 = 81
2 s = 32
3 5 = 243
2 = 64
(C)
2 7 = 128
2 R = 256
(B)
4 1 = 4
5 1 = 5
4 2 = 16
5 2 = 25
4 3 = 64
5 3 = 125
4 4 = 256
5 4 = 625
(D)
(E)
6 1 = 6
6 2 = 36
6 3 = 216
(F)
Referring to table 7-1 (A), we see that the
series of numbers
1, 4, 9, 16, 25, 36, 49, 64, 81, 100
comprises all the perfect squares from 1 to 100
inclusive. If any one of these numbers appears
under a square root symbol, the radical sign
can be removed immediately. This is illus-
trated as follows:
\f2lT = 5
-JUT = 9
A radicand such as 75, which has a perfect
square (25) as a factor, can be simplified as
follows:
3
-V/T
This procedure is further illustrated in the fol-
lowing problems:
1. N/1T =
By reference to the perfect fourth powers in
table 7-1, we may simplify a radical such as
v~405. Noting that 405 has the perfect fourth
power 81 as a factor, we have the following:
105 =
= 3
As was shown with fractional exponents,
taking a root is equivalent to dividing the expo-
nent of a power by the index of the root. If a
factor of the radicand has an exponent that is
not a multiple of the index of the root, the fac-
tor may be separated so that one exponent is
divisible by the index, as in
</37 = N/^TS = 36/2 . 31/2 = 33 . ,/y = 27 N/T
Consider also
V2 3 3 7 5 = N/2 2 2 3 6 3 5
= 2 3 3 (V2 3 5)
= 54 JM)
If the radicand is a large number, the per-
fect powers that are factors are not always ob-
vious. In such a case the radicand can be sepa-
rated into prime factors. For example,
/ 8,820 = N/2 2 3 2 5 7 2
= 2 3 7 \nr
= 42
Practice problems. Simplify the radicals
and reduce to lowest terms:
= 5
3(^10)
76
Answers:
1. 3
2.
3. 6(</~3)
4. 7
RATIONAL AND
IRRATIONAL NUMBERS
Real and imaginary numbers make up the
number system of algebra. Imaginary numbers
are discussed in chapter 15 of this course.
Real numbers are either rational or irrational.
The word RATIONAL comes from the word
"ratio." A number is rational if it can be ex-
pressed as the quotient, or ratio, of two whole
numbers. Rational numbers include fractions
like 2/7, whole numbers, and radicals if the
radical sign is removable.
Any whole number is rational. Its denomi-
o
nator is 1. For instance, 8 equals -r, which is
the quotient of two integers. A number like
*fl6 is rational, since it can be expressed as
4
the quotient of two integers in the form y. The
following are also examples of rational numbers:
denominators are cnangea immediately to deci-
mals, as in
7 _ 7
sfZ " 1.4142
the process of evaluating a fraction becomes an
exercise in long division. Such a fraction can
be evaluated quickly by first changing the de-
nominator to a rational number. Converting a
fraction with an irrational number in its de-
nominator to an equivalent fraction with a ra-
tional number in the denominator is called
RATIONALIZING THE DENOMINATOR.
Multiplying a fraction by 1 leaves the value
of the fraction unchanged. Since any number
divided by itself equals 1, it follows, for exam-
ple, that
= 1
If the numerator and denominator of -^ are
each multiplied by \f2~, another fraction having
the same value is obtained. The result is
1- l/o~> which equals -g-
i y o
-6
2, -6, which equals -=-
3.
=, which equals -=-
Any rational number can be expressed as the
quotient of two integers in many ways. For
example,
7 -1- M _ 11
1 ~ 2 ~ 3 "
An IRRATIONAL number is a real number
that cannot be expressed as the ratio of two in-
o
tegers. The numbers N/T, 5 *T2, N/T, -g NT20",
2
and -Ff are examples of irrational numbers.
The denominator of the new equivalent frac-
tion is 2, which is rational. The decimal value
of the fraction is
= 7(0.7071) = 4.9497
To rationalize the denominator in _
multiply the numerator and denominator by
We get
We
5(3)
, or _L_
15 15
Practice problems. Rationalize the denomi-
nator in each of the following:
Rationalizing Denominators
1.
3.
Expressions such as 77 and
77^ an p. r?
rational numbers in the denominator.
have ir-
If the
2.
4 -
77
Answers:
1. 3 -T2
2.
3.
4.
N/T
EVALUATING RADICALS
Any radical expression has a decimal equiv-
alent which may be exact if the radicand is a
rational number. If the radicand is not rational,
the root may be expressed as a decimal ap-
proximation, but it can never be exact. A pro-
cedure similar to long division may be used for
calculating square root and cube root, and
higher roots may be calculated by means of
methods based on logarithms and higher math-
ematics. Tables of powers and roots have been
calculated for use in those scientific fields in
which it is frequently necessary to work with
roots.
SQUARE ROOT PROCESS
The arithmetic process for calculation of
square root is outlined in the following para-
graphs:
1. Begin at the decimal point and mark the
number off into groups of two digits each, mov-
ing both to the right and to the left from the
decimal point. This may leave an odd digit at
the right-hand or left-hand end of the number,
or both. For example, suppose that the number
whose square root we seek is 9025. The num-
ber marked off as specified would be as follows:
2. Find the greatest number whose square
is contained in the left-hand group (90). This
number is 9, since the square of 9 is 81. Write
9 above the first group. Square this number (9),
place its square below the left-hand group, and
subtract, as follows:
Bring down the next group (25) and place it be-
side the 9, as shown. This is the new dividend
(925).
3. Multiply the first digit in the root (9) by
20, obtaining 180 as a trial divisor. This trial
divisor is contained in the new dividend (925)
five times; thus the second digit of the root ap-
pears to be 5. However, this number must be
added to the trial divisor to obtain a "true
divisor." If the true divisor is then too large
to use with the second quotient digit, this digit
must be reduced by 1. The procedure for step 3
is illustrated as follows:
5.
189
185
II
9 25
9 25
00
The number 180, resulting from the multi-
plication of 9 by 20, is written as a trial divisor
beside the new dividend (925), as shown. The
quotient digit (5) is then recorded and the trial
divisor is adjusted, becoming 185. The trial
quotient (180) is crossed out.
4. The true divisor (185) is multiplied by
the second digit (5) and the product is placed
below the new dividend (925). This step is
shown in the illustration for step 3. When the
product in step 4 is subtracted from the new
dividend, the difference is 0; thus, in this ex-
ample, the root is exact.
5. In some problems, the difference is not
after all of the digits of the original number
have been used to form new dividends. Such
problems may be carried further by adding O's
on the right-hand end of the original number,
just as in normal long division. However, in
the square root process the O's must be added
and used in groups of 2.
Practice problems. Find the square root of
each of the following numbers:
1. 9.61
2. 123.21
Answers:
1. 3.1 2. 11.1
TABLES OF ROOTS
3. 0.0025
3. 0.05
The decimal values of square roots and cube
roots of numbers with as many as 3 or 4 digits
can be found from tables. The table in appen-
dix I of this course gives the square roots and
cube roots of numbers from 1 to 100. Most of
the values given in such tables are approximate
numbers which have been rounded off.
78
For example , the fourth column in appendix I
shows that N/ 72 = 8.4853, to 4 decimal places.
By shifting the decimal point we can obtain
other square roots. A shift of two places in the
decimal point in the radicand corresponds to a
shift of one place in the same direction in the
square root.
The following examples show the effect, as
reflected in the square root, of shifting the
location of the decimal point in the number
whose square root we seek:
\T72 = 8.4853
N/~0772 = 0.84853
N/0.0072 = 0.084853
V7,200 = 84.853
Cube Root
The fifth column in appendix I shows that the
cube root of 72 is 4.1602. By shifting the deci-
mal point we immediately have the cube roots
of certain other numbers involving the same
digits. A shift of three places in the decimal
point in the radicand corresponds to a shift of
one place in the same direction in the cube
root.
Compare the following examples:
^72 = 4.1602
\/ 0.072 = 0.41602
\/ 72,000 = 41.602
Many irrational numbers in their simplified
forms involve */2" and \^3~. Since these radicals
occur often, it is convenient to remember their
decimal equivalents as follows:
"/2~ = 1.4142 and \/T= 1.7321
Thus any irrational numbers that do not contain
any radicals other than "/or '\T3~can be con-
verted to decimal forms quickly without re-
ferring to tables.
For example consider
NT72 = 6 \T= 6(1.4142) = 8.485
NT2T = 3 \T3"= 3(1.7321) = 5.196
Keep in mind that the decimal equivalents of
NHTand N/~3"as used in the foregoing examples
are not exact numbers and the results obtained
with them are approximate in the fourth deci-
mal place.
79
CHAPTER 8
LOGARITHMS AND THE SLIDE RULE
Logarithms represent a specialized use of
exponents. By means of logarithms, computa-
tion with large masses of data can be greatly
simplified. For example, when logarithms are
used, the process of multiplication is replaced
by simple addition and division is replaced by
subtraction. Raising to a power by means of
logarithms is done in a single multiplication,
and extracting a root reduces to simple division.
DEFINITIONS
In the expression 2 3 = 8, the number 2 is
the base (not to be confused with the base of the
number system), and 3 is the exponent which
must be used with the base to produce the num-
ber 8. The exponent 3 is the logarithm of 8
when the base is 2. This relationship is usually
stated as follows: The logarithm of 8 to the
base 2 is 3. In general, the logarithm of a
number N with respect to a given base is the
exponent which must be used with the base to
produce N. Table 8-1 illustrates this.
Table 8-1. Logarithms with various bases.
Exponential
form
Logarithmic form
2 3 =
8
Iog 2 8 = 3
4 2 =
16
Iog 4 16 = 2
5 =
1
Iog 5 l =
27 2/3 =
9
Iog 27 9 = 2/3
Table 8-1 shows that the logarithmic rela-
tionship may be expressed equally well in either
of two forms; these are the exponential form
and the logarithmic form. Observe, in table
8-1, that the base of a logarithmic expression
is indicated by placing a subscript just below
and to the right of the abbreviation "log." Ob-
serve also that the word "logarithm" is abbre-
viated without using a period.
The equivalency of the logarithmic and ex-
ponential forms may be used to restate the fun-
damental definition of logarithms in its most
useful form, as follows:
b" = N implies that log b N = x
In words, this definition is stated as follows: If
the base b raised to the x power equals N, then
xis the logarithm of the number N to the base b.
One of the many uses of logarithms may be
shown by an example in which the base is 2.
Table 8-2 shows the powers of 2 from through
20. Suppose that we wish to use logarithms to
multiply the numbers 512 and 256, as follows:
From table 8-2,
Then
and from the table again
512 = 2 9
256 = 2 8
512 x 256 = 2 9 x 2 (
= 2 17
2 17 = 131072
It is seen that the problem of multiplication
is reduced to the simple addition of the expo-
nents 9 and 8 and finding the corresponding
power in the table.
Table 8-2 (A) shows the base 2 in the expo-
nential form with its corresponding powers.
The actual computation in logarithmic work
does not require that we record the exponential
form. All that is required is that we add the
appropriate exponents and have available a
table in which we can look up the number cor-
responding to the new exponent after adding.
Therefore, table 8-2 (B) is adequate for our
purpose. Solving the foregoing example by this
table, we have the following:
Iog 2 512 = 9
Iog 2 256 = 8
log 2 of the product = 17
Therefore, the number we seek is the one in
the table whose logarithm is 17. This number
is 131,072. In this example, we found the expo-
nents directly, added them since this was a
Chapter 8 -LOGARITHMS AND THE SLIDE RULE
Table 8-2. Exponential and logarithmic
tables for the base 2.
(A) Powers of 2 from
through 20
(B) Logarithms for the
base 2 and corre-
sponding powers
Log
Number
2 = 1
1
2 1 = 2
1
2
2 2 = 4
2
4
2 3 = 8
3
8
2" = 16
4
16
2 s = 32
5
32
2 6 = 64
6
64
2 7 = 128
7
128
2 8 = 256
8
256
2 9 = 512
9
512
2 10 = 1024
10
1024
2 11 = 2048
11
2048
2 12 = 4096
12
4096
2 13 = 8192
13
8192
2 14 = 16384
14
16384
2 1S = 32768
15
32768
2 16 = 65536
16
65536
2 17 = 131072
17
131072
2 18 = 262144
18
262144
2 19 = 524288
19
524288
2 20 = 1048576
20
1048576
multiplication problem, and located the corre-
sponding power. This avoided the unnecessary
step of writing the base 2 each time.
Practice problems. Use the logarithms in
table 8-2 to perform the following multiplication:
1. 64 x 128
2. 1,024 x 256
Answers:
1. 8,192
3. 128 x 4,096
4. 512 x 2,048
3. 524,288
NATURAL AND COMMON LOGARITHMS
Many natural phenomena, such as rates of
growth and decay, are most easily described in
terms of logarithmic or exponential formulas.
Furthermore, the geometric patterns in which
certain seeds grow (for example, sunflower
seeds) is a logarithmic spiral. These facts ex-
plain the name "natural logarithms." Natural
logarithms use the base e, which is an irra-
tional number approximately equal to 2.71828.
This system is sometimes called the Napierian
system of logarithms, in honor of John Napier,
who is credited with the invention of logarithms.
To distinguish natural logarithms from other
logarithmic systems the abbreviation, In, is
sometimes used. When In appears, the base is
understood to be e and need not be shown. For
example, either log c 45 or In 45 signifies the
natural logarithm of 45.
COMMON LOGARITHMS
As has been shown in preceding paragraphs,
any number may be used as a base for a system
of logarithms. The selection of a base is a
matter of convenience. Briggs in 1617 found
that base 10 possessed many advantages not
obtainable in ordinary calculations with other
bases. The selection of 10 as a base proved so
satisfactory that today it is used almost exclu-
sively for ordinary calculations. Logarithms
with 10 as a base are therefore called COM-
MON LOGARITHMS.
When 10 is used as a base, it is not neces-
sary to indicate it in writing logarithms. For
example,
log 100 = 2
is understood to mean the same as
Iog 10 100 = 2
If the base is other than 10, it must be speci-
fied by the use of a subscript to the right and
below the abbreviation "log." As noted in the
foregoing discussion of natural logarithms, the
use of the distinctive abbreviation "In" elimi-
nates the need for a subscript when the base
is e.
It is relatively easy to convert common log-
arithms to natural logarithms or vice versa, if
necessary. It should be noted further that each
system has its peculiar advantages, but for
more often used. A simple relation connects
the two systems. If the common logarithm of a
number can be found, multiplying by 2.3026
gives the natural logarithm of the number. For
example,
log 1.60 = 0.2041
In 1.60 = 2.3026 x 0.2041
= 0.4700
Thus the natural logarithm of 1.60 is 0.4700,
correct to four significant digits.
Conversely, multiplying the natural loga-
rithm by 0.4343 gives the common logarithm of
a number. As might be expected, the conver-
sion factor 0.4343 is the reciprocal of 2.3026.
This is shown as follows:
1
2.3026
= 0.4343
Positive Integral Logarithms
The derivation of positive whole logarithms
is readily apparent. For example, we see in
table 8-3 (B) that the logarithm of 10 is 1. The
number 1 is simply the exponent of the base 10
which yields 10. This is shown in table 8-3 (A)
opposite the logarithmic equation. Similarly,
10 = 1 log 1 =
10 2 = 100 log 100 = 2
10 3 = 1,000 log 1,000 = 3
10 4 = 10,000 log 10,000 = 4
Table 8-3. Exponential and corresponding logarithmic notations using base 10.
A.
B.
10 ~ 4
1
~ 10 4
0.0001
log
0.0001 =
-4
10 " 3
1
0.001
log
0.001 =
-3
10 3
io- J
1
0.01
0.1
log
log
0.01
0.1
-2
-1
" 10 2
1
"10
10~ 1/2
1 NflO"
0.31623
log
0.31623 =
-0.5
N/TO" 10
=
0.5 -1
10
=
1
log
1
10 l/2
= \TIO
3.1623
log
3.1623 =
0.5
10 a
=
10
log
10
1
JQ 3/2
= 10 N/TO =
31.623
log
31.623 =
1.5
10 2
=
100
log
100
2
10 5/2
= 10 2 (N/T6) =
316.23
log
316.23
2.5
10 3
=
1,000
log
1,000
3
1Q7/2
= 103 (vio) =
3162.3
log
3162.3
3.5
10 4
=
10,000
log
10,000
4
82
Keiemng to taoie ts-a, notice mat tne ioga-
rithm of 1 is and the logarithm of 10 is 1.
Therefore, the logarithm of a number between
1 and 10 is between and 1. An easy way to
verify this is to consider some numbers be-
tween 1 and 10 which are powers of 10; the ex-
ponent in each case will then be the logarithm
we seek. Of course, the only powers of 10
which produce numbers between 1 and 10 are
fractional powers.
EXAMPLE: 10 1/2 = 3.1623 (approximately)
10
0.5 _
= 3.1623
Therefore, log 3.1623 = 0.5
Other examples are shown in the table for
10 3/2 , 10 5/2 , and 10 7/2 . Notice that the num-
ber that represents 10 3/2 , 31.623, logically
enough lies between the numbers representing
10 : and 10 2 -that is, between 10 and 100. No-
tice also that 10 s/2 appears between 10 2 and
10 3 , and 10 7/2 lies between 10 3 and 10 4 .
Negative Logarithms
Table 8-3 shows that negative powers of 10
may be fitted into the systpm of logarithms.
We recall that 10" 1 means -~, or the decimal
fraction, 0.1. What is the logarithm of 0.1?
SOLUTION: 10' 1 = 0.1; log 0.1 = -1
Likewise ID" 2 = 0.01; log 0.01 = -2
Negative Fractional Logarithms
Notice in table 8-3 that negative fractional
exponents present no new problem in loga-
rithmic notation. For example, 10~ 1/2 means
1
10'
/To
/ib"
10
= 0.31623
What is the logarithm of 0.31623?
SOLUTION:
10" 1/2 = 0.31623; log 0.31623 =
mere are only B integral logarithms m the en-
tire range. Excluding zero logarithms, the
logarithms for all other numbers in the range
are fractional or contain a fractional part. By
the year 1628, logarithms for all integers from
1 to 100,000 had been computed. Practically
all of these logarithms contain a fractional
part. It should be remembered that finding the
logarithm of a number is nothing more than ex-
pressing the number as a power of 10. Table
8-4 shows the numbers 1 through 10 expressed
as powers of 10. Most of the exponents which
comprise logarithms are found by methods be-
yond the scope of this text. However, it is not
necessary to know the process used to obtain
logarithms in order to make use of them.
Table 8-4. The numbers 1 through 10
expressed as powers of 10.
1 =
10
6 =
IQ 0.77815
o
JQ 0.30103
7 =
JQ 0.84510
3 =
10 0.47712
8 =
..00.90309
4 =
HQ 0.60206
9 =
IQ 0.95424
}Q 0.698 97
10 =
10 1
= -0.5
COMPONENTS OF LOGARITHMS
The fractional part of a logarithm is usually
written as a decimal. The whole number part
of a logarithm and the decimal part have been
given separate names because each plays a
special part in relation to the number which the
logarithm represents. The whole number part
of a logarithm is called the CHARACTERISTIC.
This part of the logarithm shows the position of
the decimal point in the associated number.
The decimal part of a logarithm is called the
MANTISSA.
For a particular sequence of digits making
up a number, the mantissa of a common loga-
rithm is always the same regardless of the
position of the decimal point in that number.
For example, log 5270 = 3.72181; the mantissa
is 0.72181 and the characteristic is 3.
CHARACTERISTIC
The characteristic of a common logarithm
shows the position of the decimal point in the
88
associated number. The characteristic for a
given number may be determined by inspection.
It will be remembered that a common logarithm
is simply an exponent of the base 10. It is the
power* of 10 when a number is written in
scientific notation.
When we write log 360 = 2.55630, we under-
stand this to mean 10 2 ' 55630 - 360. We know
that the number is 360 and not 36 or 3,600 be-
cause the characteristic is 2. We know 10 is
10 10 2 is 100, and 10 3 is 1,000. Therefore,
the number whose value is ID 2 - 55 " 30 must lie
between 100 and 1,000 and of course any num-
ber in that range has 3 digits.
Suppose the characteristic had been 1: where
would the decimal point in the number be
placed? Since 10 J is 10 and 10 2 is 100, any
number whose logarithm is between 1 and 2
must lie between 10 and 100 and will have 2
digits. Notice how the position of the decimal
point changes with the value of the character-
istic in the following examples:
log 36,000 = 4.55630
log 3,600 = 3.55630
log 360 = 2.55630
log 36 = 1.55630
log 3.6 = 0.55630
Note that it is only the characteristic
changes when the decimal point is moved,
advantage of using the base 10 is thus revea
If the characteristic is known, the decimal p
may easily be placed. If the number is km
the characteristic may be determined by
spection; that is, by observing the locatloi
the decimal point.
Although an understanding of the rela
of the characteristic to the powers of 1C
necessary for thorough comprehension of lc
rithms, the characteristic may be determi
mechanically by application of the follov
rules:
1. For a number greater than 1, the char
teristic is positive and is one less than
number of digits to the left of the decimal P'
in the number.
2. For a positive number less than 1,
characteristic is negative and has an abaci
value one more than the number of zeros
tween the decimal point and the first nonz
digit of the number.
Table 8-5 contains examples of each type
characteristic.
Practice problems. In problems 1 thro
4, write the characteristic of the logarithm
each number. In 5 through 8, place the decit
Table 8-5. Positive and negative characteristics.
Number
Power of 10
Digits in number
to the left of
decimal point
Characteristic
Between:
134
10 2 and 10 3
3
2
13.4
10 J and 10 2
2
1
1.34
0.134
10 and 10 l
10' 1 and 10
1
-1
Zeros between
decimal point
and first non-
zero digit
0.0134
ID' 2 and 10" 1
1
-2
0.00134
10 " 3 and 10~ 2
2
-3
84
point in each number as indicated by the char-
acteristic (c) given for each.
Table 8-6. -Effect of changes in the
location of the decimal point.
1. 4,321 2. 1.23
5. 123; c = 4
7. 8; c = -1
3. 0.05 4. 12
6. 8,210; c =
8. 321; c = -2
Answers:
1. 3
5. 12,300
2.
6. 8.210
3. -2
7. 0.8
4. 1
8. 0.0321
Negative Characteristics
When a characteristic is negative, such as
-2, we do not carry out the subtraction, since
this would involve a negative mantissa. There
are several ways of indicating a negative char-
acteristic. Mantissas as presented in the table
in the appendix are always positive and the sign
of the characteristic is indicated_ separately.
For example, where log 0.023 = 2.36173, the
bar over the 2 indicates that only the charac-
teristic is negative that is, the logarithm is
-2 + 0.36173.
Another way to show the negative character-
istic is to place it after the mantissa. In this
case we write : .36173-2.
A third method, which is used where possi-
ble throughout this chapter, is to add a certain
quantity to the characteristic and to subtract
the same quantity to the right of the mantissa.
In the case of the example, we may write:
2.36173
10
-10
8.36173-10
In this way the value of the logarithm remains
the same but we now have a positive character-
istic as well as a positive mantissa.
MANTISSA
The mantissa is the decimal part of a loga-
rithm. Tables of logarithms usually contain
only mantissas since the characteristic can be
readily determined as explained previously.
Table 8-6 shows the characteristic, mantissa,
and logarithm for several positions of the deci-
mal point using the sequence of digits 4, 5, 6.
It will be noted that the mantissa remains the
same for that particular sequence of digits, re-
gardless of the position of the decimal point.
Number
Charac-
teristic
Mantissa
Logarithm
45,600
4
0.6590
4.6590
4,560
3
0.6590
3.6590
456
2
0.6590
2.6590
45.6
1
0.6590
1.6590
4.56
0.6590
0.6590
0.456
-1
0.6590
0.6590-1
0.0456
-2
0.6590
0.6590-2
0.00456
-3
0.6590
0.6590-3
Appendix I of this training course is a table
which includes the logarithms of numbers from
1 to 100. For our present purpose in using this
table, we are concerned only with the first and
sixth columns.
The first column contains the number and
the sixth column contains its logarithm. For
example, if it is desired to find the logarithm
of 45, we would find the number 45 in the first
column, look horizontally across the page to
column 6 and read the logarithm, 1.65321. A
glance down the logarithm column will reveal
that the logarithms increase in value as the
numbers increase in value.
It must be noted in this particular table that
both the mantissa and the characteristic are
given for the number in the first column. This
is simply an additional aid, since the charac-
teristic can easily be determined by inspection.
Suppose that we wish to use the table of
Appendix I to find the logarithm of a number
not shown in the "number" column. By recall-
ing that the mantissa does not change when the
decimal point moves, we may be able to deter-
mine the desired logarithm. For example, the
number 450 does not appear in the number col-
umn of the table. However, the number 45 has
the same mantissa as 450; the only difference
between the two logs is in their characteristics.
Thus the logarithm of 450 is 2.65321.
Practice problems. Find the logarithms of
the following numbers:
85
1. 64
Answers:
1. 1.80618
3. 3.80618
2. 98 3. 64UU
2. 1.99123
4. 0.99123
THE SLIDE RULE
4. y.
In 1620, not long after the invention of loga-
rithms, Edmond Gunter showed how logarithmic
\ calculations could be carried out mechanically.
This is done by laying off lengths on a rule,
representing the logarithms of numbers, and by
combining these lengths in various ways. The
idea was developed and with the contributions
of Mannheim in 1851 the slide rule came into
being as we know it today.
The slide rule is a mechanical device by
which we can carry out any arithmetic calcula-
tion with the exception of addition and subtrac-
tion. The most common operations with the
slide rule are multiplication, division, finding
the square or cube of a number, and finding the
square root or cube root of a number. Also
trigonometric operations are frequently per-
formed. The advantage of the slide rule is that
it can be used with relative ease to solve com-
plicated problems. One limitation is that it
will give results with a maximum of only three
accurate significant digits. This is sufficient
in most calculations, however, since most phys-
ical constants are only correct to two or three
significant digits. When greater accuracy is
required, other methods must be used.
A simplified diagram of a slide rule is pic-
tured in figure 8-1. The sliding, central part
of the rule is called the SLIDE. The movable
glass or plastic runner with a hairline imprinted
on it is called the INDICATOR. There is a C
scale printed on the slide, and a D scale exactly
the same as the C scale printed on the BODY
or STOCK of the slide rule. The mark that is
associated with the primary number 1 on any
slide rule scale is called the INDEX. There is
HAIRLINE. SLIDE
an index at the extreme leit and at the extreme
right on both the C and D scales. There are
other scales, each having a particular use.
Some of these will be mentioned later.
SLIDE RULE THEORY
We have mentioned that the slide rule is
based on logarithms. Recall that, to multiply
two numbers, we simply add their logarithms.
Previously we found these logarithms in tables,
but if the logarithms are laid off on scales such
as the C and D scale of the slide rule, we can
add the lengths, which represent these loga-
rithms. To make such a scale we could mark
off mantissas ranging from to 1 on a rule as
in figure 8-2. We then find in the tables the
logarithms for numbers ranging from 1 to 10
and write the number opposite its correspond-
ing logarithm on the scale.
LOGARITHM
O.2 0.3 0.4 0.3 0.6 0.7 0.8 0.9
INOEV BODf' INDICATOR'
Figure 8-1. Simplified diagram of a slide rule.
Figure 8-2. Logarithms and corresponding
numbers on a scale.
Table 8-7 lists the numbers 1 through 10
and their corresponding logarithms to three
places. These numbers are written opposite
their logarithms on the scale shown in figure
8-2. If we have two such scales, exactly alike,
arranged so that one of them is free to slide
along the other, we can perform the operation
of multiplication, for example, by ADDING
LENGTHS; that is, by adding logarithms. For
example, if we wish to multiply 2x3, we find
the logarithm of 2 on the stationary scale and
move the sliding scale so that its index is over
that mark. We then add the logarithm of 3 by
finding that logarithm on the sliding scale and
by reading below it, on the stationary scale, the
logarithm that is the sum of the two.
Since we are not interested in the logarithms
themselves, but rather in the numbers they
represent, it is possible to remove the loga-
rithmic notation on the scale in figure 8-2, and
leave only the logarithmically spaced number
scale. The C and D scales of the ordinary slide
rule are made up in this manner. Figure 8-3
shows the multiplication of 2 x 3. Although the
logarithm scales have been removed, the num-
bers 2 and 3 in reality signify the logarithms of
86
Table 8-7. Numbers and their
corresponding logarithms.
Number
Logarithm
Number
Logarithm
1
0.000
6
0.778
2
0.301
7
0.845
3
0.477
8
0.903
4
0.602
9
0.954
5
0.699
10
1.000
1 5\6/7 a 9 1
Figure 8-3. Multiplication by use of
the slide rule.
2 and 3, namely, 0.301 and 0.477; the product 6
on the scale really signifies the logarithm of 6,
that is, 0.778. Thus, although logarithms are
the underlying principle, we are able to work
with the numbers directly.
It should be noted that the scale is made up
from mantissas only. The characteristic must
be determined separately as in the case where
tables are used. Since mantissas identify only
the digit sequence, the digit 3 on the slide rule
represents not only 3 but 30, 300, 0.003, 0.3,
and so forth. Thus, the divisions may repre-
sent the number multiplied or divided by any
power of 10. This is true also for numbers
that fall between the divisions. The digit se-
quence, 1001, could represent 100.1, 1.001,
0.01001, and so forth. The following example
shows the use of the same set of mantissas
which appear in the foregoing example, but with
a different characteristic and, therefore, a dif-
ferent answer:
EXAMPLE: Use logs (positions on the slide
rule) to multiply 20 times 30.
SOLUTION:
log 20 = 1.301 (2 on the slide rule)
log 30 = 1.477 (3 on the slide rule)
log of answer = 2.778 (6 on the slide rule)
Since the 2 in the log of the answer is
merely the indicator of the position of the deci-
mal point in the answer itself, we do not expect
to find it on the slide rule scale. As in the
foregoing example, we find the digit 6 opposite
the multiplier 3. This time, however, the 6
represents 600, because the characteristic of
the log represented by 6 in this problem is 2.
READING THE SCALES
Reading a slide rule is no more complicated
than reading a yard stick or ruler, if the dif-
ferences in its markings are understood.
Between the two indices of the C or D scales
(the large digit 1 at the extreme left and right
of the scales) are divisions numbered 2, 3, 4,
,5, &, 1, 8, and 9. Each length between two con-
secutive divisions is divided into 10 sections
and each section is divided into spaces. (See
fig. 8-4.)
DIVISION
SECTION
SPACED
r fi
II
C
1 i2 3
in mi
D
III III!
12 3
Figure 8-4. Division, section, and (space of
a slide rule scale.
Notice that the division between 1 and 2
occupies about one -third of the length of the
rule. This is sufficient space in which to write
a number for each of the section marks. The
sections in the remaining divisions are not
numbered, because the space is more limited.
Notice also that in the division between 1 and 2,
the sections are each divided into 10 spaces.
The sections of the divisions from 2 to 4 are
87
subdivided into only 5 spaces, and those from
4 to the right index are subdivided into only 2
spaces. These subdivisions are so arranged
because of tine limits of space.
Only the sequence of significant digits is
read on the slide rule. The position of the dec-
imal point is determined separately. For ex-
ample, if the hairline of the indicator is in the
left-hand position shown in figure 8-5, the sig-
nificant digits are read as follows:
1
'!" | 1 1 1 1 1 1 I 1 "
1 ^ 1456769
!
I 3
V
Figure 8-5. Readings in the first division
of a slide rule.
1. Any time the hairline falls in the first
division, the first significant digit is 1.
2 . Since the hairline lies between the index
and the first section mark, we know the number
lies between 1.0 and 1.1, or 10 and 11, or 100
and 110, etc. The second significant digit is 0.
3. We next find how far from the index the
hairline is located. It lies on the marking for
the third space.
4. The three significant digits are 103.
In the second example shown in figure 8-5,
the hairline is located in the first division, the
ninth section, and on the fourth space mark of
that section. Therefore, the significant digits
are 194.
Thus, we see that any number falling in the
first division of the slide rule will always have
1 as its first significant digit. It can have any
TEN SPACES IN EACH
SECTION
number from through 9 as its second digit, and
any number from through 9 as its third digit.
Sometimes a fourth digit can be roughly appr ox -
imated in this first division, but the number is
really accurate to only three significant digits .
In the second and third divisions, each sec-
tion is divided into only 5 spaces. (See fig. 8-6.)
Thus, each space is equal to 0.2 of the section.
Suppose, for example, that the hairline lies on
the third space mark after the large 2 indicat-
ing the second division. The first significant
digit is 2. Since the hairline lies between 2 and
the first section mark, the second digit is 0.
The hairline lies on the third space mark or
0.6 of the way between the division mark and
the first section mark, so the third digit is 6.
Thus, the significant digits are 206. Notice
that if the hairline lies on a space mark the
third digit can be written accurately; otherwise
it must be approximated.
From the fourth division to the right index,
each section is divided into only two spaces.
Thus, if the hairline is in the fourth division
and lies on the space mark between the sixth
and seventh sections, we would read 465. If the
hairline did not fall on a space mark, the third
digit would have to be approximated.
OPERATIONS WITH THE SLIDE RULE
There are two parts in solving problems
with a slide rule. In the first part the slide
rule is used to find the digit sequence of the
final result. The second part is concerned with
the placing of the decimal point in the result.
Let us consider first the digit sequence in mul-
tiplication and division.
Multiplication
Multiplication is performed on the C and D
scales of the slide rule. The following proce-
dure is used:
ONLY FIVE SPACES IN EACH
SECTION
1 2
HAIRLINE
Figure 8-6. Reading in the second division of a slide rule.
88
1. Locate one of the factors to be multiplied
on the D scale, disregarding the decimal point.
2. Place the index of the C scale opposite
that number.
3. Locate the other factor on the C scale
and move the hairline of the indicator to cover
this factor.
4. The product is on the D scale under the
hairline.
Sometimes in multiplying numbers, such as
25 x 6, the number on the C scale extends to
the right of the stock and the product cannot be
read. In such a case, we simply shift indices.
Instead of the left-hand index of the C scale,
the right-hand index is placed opposite the fac-
tor on the D scale. The rest of the problem
remains the same. By shifting indices, we are
simply multiplying or dividing by 10, but this
plays no part in reading the significant digits.
Shifting indices affects the characteristic only.
EXAMPLE:
252 x 3 = 756
1. Place the left index of the C scale over
252.
2. Locate 3 on the C scale and set the hair-
line of the indicator over it.
3. Under the hairline on the D scale read
the product, 756.
EXAMPLE:
4 x 64 = 256
1. Place the right index of the C scale
over 4.
2. Locate 64 on the C scale and set the
hairline of the indicator over it.
3. Under the hairline on the D scale read
the product, 256.
Practice problems. Determine the following
products by slide rule to three significant
digits:
1. 2.8 x 16
2. 7 x 1.3
Answers:
1. 44,8
2. 9.10
Division
3. 6 x 85
4. 2.56 x 3.5
3. 510
4. 8.96
Division being the inverse of multiplication,
the process of multiplication is reversed to
perform division on a slide rule. We subtract
the length representing the logarithm of the
divisor from the length representing the loga-
rithm of the dividend to get the logarithm of the
quotient.
The procedure is as follows:
1. Locate the dividend on the D scale and
place the hairline of the indicator over it.
2. Move the slide until the divisor (on the C
scale) lies under the hairline.
3. Read the quotient on the D scale opposite
the C scale index.
If the divisor is greater numerically than
the dividend, the slide will extend to the left. If
the divisor is less, the slide will extend to the
right. In either case, the quotient is the number
on the D scale that lies opposite the C scale in-
dex, falling within the limits of the D scale.
EXAMPLE:
6-3 =
1. Locate 6 on the D scale and place the
hairline of the indicator over it.
2. Move the slide until 3 on the C scale is
under the hairline.
3. Opposite the left C scale index, read the
quotient, 2, on the D scale.
EXAMPLE:
378 = 63 = 6
1. Locate 378 on the D scale and move the
hairline of the indicator over it.
2. Move the slide to the left until 63 on the
C scale is under the hairline.
3. Opposite the right-hand index of the C
scale, read the quotient, 6, on the D scale.
Practice problems. Determine the following
quotients by slide rule.
1. 126 * 3
2. 960 * 15
Answers:
1. 42
2. 64
3. 142 * 71
4. 459 * 17
3. 2
4. 27
PLACING THE DECIMAL POINT
Various methods have been advanced regard-
ing the placement of the decimal point in num-
bers derived from slide rule computations.
Probably the most universal and most easily
remembered method is that of approximation.
89
The method of approximation means simply
the rounding off of numbers and the mechanical
shifting of decimal points in the numbers of the
problem so that the approximate size of the
solution and the exact position of the decimal
point will be seen from inspection. The slide
rule may then be used to derive the correct se-
quence of significant digits. The method may
best be demonstrated by a few examples. Re-
member, shifting the decimal point in a number
one place to the left is the same as dividing by
10. Shifting it one place to the right is the
same as multiplying by 10. Every shift must
be compensated for in order for the solution to
be correct.
EXAMPLE:
0.573 x 1.45
SOLUTION: No shifting of decimals is neces-
sary here. We see that approximately 0.6 is to
be multiplied by approximately 1 1/2. Immedi-
ately, we see that the solution is in the neigh-
borhood of 0.9. By slide rule we find that the
significant digit sequence of the product is 832.
From our approximation we know that the deci-
mal point is to the immediate left of the first
significant digit, 8. Thus,
0.573x 1.45 = 0.832
EXAMPLE:
239 x 52.3
SOLUTION: For ease in multiplying, we shift
the decimal point in 52.3 one place to the left,
making it 5.23. To compensate, the decimal
point is shifted to the right one place in the
other factor. The new position of the decimal
point is indicated by the presence of the caret
symbol.
239. A x 5 A 2.3
Our problem is approximately the same as
2,400 x 5 = 12,000
By slide rule the digit sequence is 125. Thus,
239 x 52.3 = 12,500
EXAMPLE: 0.000134 x 0.092
SOLUTION:
Shifting decimal points, we have
A 00.000134 x 0.09 A 2
Approximation: 9 x 0.0000013 = 0.0000117.
By slide rule the digit sequence is 123. From
approximation the decimal point is located as
follows:
0.0000123
Thus,
0.000134 x 0.092 = 0.0000123
EXAMPLE: 53.1
42.4
SOLUTION: The decimal points are shifted so
that the divisor becomes a number between 1
and 10. The method employed is cancellation.
Shifting decimal points, we have
5A 3 ' 1
4 A 2.4
Approximation: 5_ _ 1
Digit sequence by slide rule:
1255
Placing the decimal point from the approxi-
mation:
1.255
Thus,
EXAMPLE:
53.1
42.4
= 1.255
0.00645
0.0935
SOLUTION:
Shifting decimal points
0.00 A 645
0.09 A 35
Approximation:
-j- = 0.07
Digit sequence by slide rule: 690
90
mation:
Thus,
^/\JJ.IIL xi uiu nit;
me 110.1
uvci oil uii LUC
0.0690
0.00645
0.0935
= 0.0690
Practice problems. Solve the following
problems with the slide rule and use the method
of approximation to determine the position of
the decimal point:
3. 0.0362 x 1.21
4. 67 - 316
3. 0.0438
4. 0.212
1. 0.00453 x 0.1645
2. 53.1 - 1.255
Answers:
1. 0.000745
2. 42.4
MULTIPLICATION AND
DIVISION COMBINED
In problems such as
0.644 x 330
161 x 12
it is generally best to determine the position of
the decimal point by means of the method of
approximation and to determine the significant
digit sequence from the slide rule. Such prob-
lems are usually solved by dividing and multi-
plying alternately throughout the problem. That
is, we divide 0.644 by 161, multiply the quotient
by 330, and divide that product by 12.
Shifting decimal points, we have
A 0.644 x 3 A 30
1 A 61 x 1 A 2
Since there is a combined shift of three places
to the left in the divisor, there must also be a
combined shift of three places to the left in the
dividend.
2
Approximation: 'v ** = 0.06 x 2 =0.12
The step-by-step process of determining the
significant digit sequence of this problem is as
follows:
2 . Draw the slide so that 16 1 of the C scale
lies under the hairline opposite 644.
3. Opposite the C scale index (on the D scale)
is the quotient of 644 +161. This is to be mul-
tiplied by 330, but 330 projects beyond the rule
so the C scale indices must be shifted.
4. After shifting the indices, find 330 on the
C scale and place the hairline over it. Opposite
330 under the hairline on the D scale is the
644
product of -
330.
5. Next, move the C scale until 12 is under
the hairline. Opposite the C scale index (on the
D scale) is the final quotient. The digit se-
quence is 110.
The decimal point is then placed according
to our approximation: 0.11. Thus,
0.644 x 330
161 x 12
= 0.11
Practice problems. Solve the following
problems, using a slide rule:
1. 22 x 78.5 x 157
17 x 18.3 x 85
2. 432 x 9,600
25,600 x 198
3. 2.77 x 0.064
0.17 x 1.97
Answers:
1. 10.2 2. 0.817 3. 0.529
SQUARES
Squares of numbers are found by reference
to the A scale. The number son the A scale are
the squares of those on the D Scale. The A
scale is really a double scale, each division
being one -half as large as the corresponding
division on the D scale. The use of a double
scale for squaring is based upon the fact that
the logarithm of the square of a number is twice
as large as the logarithm of the number itself.
In other words,
log N 2 = 2 log N
This is reasonable, since
log N 2 = log (N x N)
= log N + log N
91
MATHEMATICS, VOLUME 1
For a numerical example, suppose that we
seek to square 2 by means of logarithms.
log 2 = 0.301
log 2 2 = 2 log 2
= 2 x 0.301
= 0.602
Since each part of the A scale is half as
large as the corresponding part of the D scale,
the logarithm 0.602 on the A scale will be the
same length as the logarithm 0.301 on the D
scale. That is, these logarithms will be oppo-
site on the A and D scales. On the A scale as
on the D scale, the numbers are written rather
than their logarithms. Select several numbers
on the D scale, such as 2, 4, 8, 11, and read
their squares on the A scale, namely 4, 16,
64, 121.
Notice also that the same relation exists for
the B and C scales as for the A and D scales.
Of interest, also, is the fact that since the A
and B scales are made up as are the C and D
scales, they too could be used for multiplying
or dividing.
Placing the Decimal Point
Usually the decimal may be placed by the
method of approximation. However, close ob-
servation will reveal certain facts that elimi-
nate the need for approximations in squaring
numbers. Two rules suffice for squaring whole
or mixed numbers, as follows:
1. When the square of a number is read on
the left half of the A scale, that number will
contain twice the number of digits to the left of
the decimal point in the original number, less 1.
2. When the square of a number is read on
the right half of the A scale, that number will
contain twice the number of digits to the left of
the decimal point in the original number.
EXAMPLE: Square 2.5.
SOLUTION: Place the hairline over 25 on the
D scale. Read the digit sequence, 625, under
the hairline in the left half of the A scale.
By rule 1: (2xnumberof digits)-! = 2(1)-1=1.
There is one digit to the left of the decimal
point. Thus,
(2.5) 2 = 6.25
EXAMPLE: Square 6,340.
SOLUTION:
Digit sequence, right half A scale: 402.
By rule 2: 2 x number of digits = 2x4 = 8
(digits in answer). Thus,
(6,340) 2 = 40,200,000
Positive Numbers Less Than One
If positive numbers less than one are to be
squared, a slightly different version of the pre-
ceding rules must be employed. Count the
zeros between the decimal point and the first
nonzero digit. Consider this count negative.
Then the number of zeros between the decimal
point and the first significant digit of the
squared number may be found as follows:
1. Left half A scale: Multiply the zeros
counted by 2 and subtract 1.
2. Right half A scale: Multiply the zeros
counted by 2.
EXAMPLE: Square 0.0045
SOLUTION:
Digit sequence, right half A scale: 2025.
By rule 2: 2(-2) = -4. (Thus, 4 zeros be-
tween the decimal point and the first digit.)
(0.0045) 2 = 0.00002025
EXAMPLE: Square 0.0215
SOLUTION:
Digit sequence, left half A scale: 462.
By rule 1: 2(-l) -1 = -3
(0.02 15) 2 = 0.000462
SQUARE ROOTS
Taking the square root of a number with the
slide rule is the inverse process of squaring a
number. We find the number on the A scale,
set the hairline of the indicator over it, and
read the square root on the D scale under the
hairline.
Positioning Numbers on the A Scale
Since there are two parts of the A scale
exactly alike and the digit sequence could be
Chapter 8-LOGARITHMS AND THE SLIDE RULE
found on either part, a question arises as to
which section to use. Generally, we think of
the left half of the rule as being numbered from
1 to 10 and the right half as being numbered
from 10 to 100. The numbering continues- left
half 100 to 1,000, right half 1,000 to 10,000, and
so forth.
A simple process provides a check of the
location of the number from which the root is
to be taken. For whole or mixed numbers, be-
gin at the decimal point of the number and mark
off the digits to the left (including end zeros) in
groups of two. This is illustrated in the follow-
ing two examples:
1. ^40,300.21
V4'03'00.21
2. N/2, 034.1
N/20'34.1
Powers of 10
When the square root of 10, 1,000, 100,000,
and so forth, is desired, the center index is
used. That is, when the number of digits in a
power of 10 is even, use the center index.
The slide rule uses only the first three
significant digits of a number. Thus, if the
rule is used, \/23451.6 must be considered as
^23500.0. Likewise, 1.43567 would be consid-
ered 1.43000, and so forth. For greater accu-
racy, other methods must be used.
Practice problems. State which half of the
A scale should be used for each of the following:
4. \/0. 00045
Look at the left-hand group. If it is a 1-digit
number, use the left half of the A scale. If it
is a 2-digit number, use the right half of the A
scale. The number in example 1 is thus located
in the left half of the A scale and the number in
example 2 is located in the right half.
N" '.fibers Less Than One
For positive numbers less than one, begin at
the decimal point and mark off groups of two to
the right. This is illustrated as follows:
Answers:
1.
VO. 000245
VO.00'02'45
2. N/0. 00402
N/0.00'40'2
Looking from left to right, locate the first group
that contains a digit other than zero. If the
first figure in this group is zero, locate the
number in the left half of the A scale. If the
first figure is other than zero, locate the num-
ber in the right half of the A scale. Thus,
WO.00'02'45 is located left
and
1. Left
2. Left
3. Left
4. Left
5. Right
6. Right
7. Left
8. Left
Placing the Decimal Point
To place the decimal point in the square
root of a number, mark off the original number
in groups of two as explained previously.
For whole or mixed numbers, the number of
groups marked off is the number of digits in-
cluding end zeros to the left of the decimal
point in the root. The following problems il-
lustrate this:
N/2'34'15 Three digits to left of dec-
imal point in square root
2. \/421,562.4
^42'15'62.4 Three digits to left of dec-
imal point in square root
3. W231.321
A mo i -. i -._ i.j i
N/2'31.321 Two digits to left of deci-
*^s f\t vni vt ' 0m i n tf\ v*S\r\ (
VUijUMJi. i
For positive numbers less than one, there
will be one zero in the square root between the
decimal point and the first significant digit for
every pair of zeros counted between the deci-
mal point and the first significant digit of the
original number. This is illustrated as follows:
\/0.00'04 One zero before first digit
in square root
2. V 0.00008
VO.00'00'8 Two zeros before first digit
in square root
3. V0.08' No zeros before first digit
in square root
EXAMPLE:
(Two digits in left-hand group)
Place the hairline over 452 on the right half of
the A scale. Read the digit sequence of the
root, 672, on the D scale under the hairline.
Since there are two groups in the original num-
ber, there are two digits to the left of the deci-
mal point in the root. Thus,
EXAMPLE:
v/4~521 = 67.2
N/0.000741
VO.00'07'41
(First figure is zero in this group)
Place the hairline over 741 on the left half of
the A scale. Read the digit sequence of the
root, 272, under the hairline on the D scale.
Since there is one pair of zeros to the left of
the group containing the first digit, there is one
zero between the decimal point and the first
significant digit of the root. Thus,
^0.000741 = 0.0272
Practice problems. Evaluate each of
following by means of a slide rule:
1. (17.75) 2
2. (0.65) 2
the
Answers:
1. 315
2. 0.422
3. 3.07
4. 0.272
CUBES AND CUBE ROOTS
Cubes and cube roots are read on the K and
D scales of the slide rule. On the K scale are
compressed three complete logarithmic scales
in the same space as that of the D scale. Thus,
any logarithm on the K scale is three times the
logarithm opposite it on the D scale. To cube
a number by logarithms, we multiply its loga-
rithm by three. Therefore, the logarithms of
cubed numbers will lie on the K scale opposite
the numbers on the D scale.
As with the other slide rule scales men-
tioned, the numbers the logarithms represent,
rather than the logarithmic notations, are
printed on the rule. In the left-hand third of
the K scale, the numbers range from 1 to 10; in
the middle third they range from 10 to 100; and
in the right-hand third, they range from 100 to
1,000.
To cube a number, find the number on the D
scale, place the hairline over it, and read the
digit sequence of the cubed number on the K
scale under the hairline.
Placing the Decimal Point
The decimal point of a cubed whole or mixed
number may be easily placed by application of
the following rules:
1. If the cubed number is located in the left
third of the K scale, its number of digits to the
left of the decimal point is 3 times the number
of digits to the left of the decimal point in the
original number, less 2.
2. If the cubed number is located in the
middle third of the K scale, its number of digits
is 3 times the number of digits of the original
number, less 1.
3. If the cubed number is located in the
right third of the K scale, its number of digits
is 3 times the number of digits of the original
number.
EXAMPLE:
(1.6) 3
SOLUTION: Place the hairline over 16 on
scale. Read the digit sequence, 409, on the
scale under the hairline.
94
Number of digits to left of decimal point in the
number 1.6 is 1 and the cubed number is in the
left-hand third of the K scale.
3 x (No. of digits)- 2 = (3 x l)-2
= 1
Therefore,
(1.6) 3 - 4.09
(4. 1) 3
EXAMPLE:
Digit sequence = 689.
SOLUTION: Number of digits to left of decimal
point in the number 4.1 is 1, and the cubed
number is in the middle third of the K scale.
3 x (No. of digits)-! = (3 x 1)-1
= 2
Therefore,
EXAMPLE:
(4. 1) 3 = 68.9
(52) 3
SOLUTION: Digit sequence = 141.
Number of digits to left of decimal point in the
number 52 is 2, and the cubed number is in the
right-hand third of the K scale.
3 x No. of digits = 3x2
= 6
Therefore,
(52) 3 = 141,000
Positive Numbers Less Than One
If positive numbers less than one are to be
cubed, count the zeros between the decimal
point and the first nonzero digit. Consider the
count negative. Then the number of zeros be-
tween the decimal point and the first significant
digit of the cubed number may be found as
follows:
1. Left third of K scale: Multiply the zeros
counted by 3 and subtract 2.
2. Middle third of K scale: Multiply the
zeros counted by 3 and subtract 1.
3. Right third of K scale: Multiply the zeros
counted by 3.
EXAMPLE: Cube 0.034
SOLUTION: Digit sequence = 393
Zero count of 0.034 = -1, and 393 is in the mid-
dle third of the K scale.
3 x (No. of zeros)-! = (3 x -1)-1 = -4
Therefore,
(0.034) 3 = 0.0000393
Practice problems. Cube the following num-
bers using the slide rule.
1. 21
2. 0.7 3. 0.0128
4. 404
Answers:
1. 9260
2. 0.342
Cube Roots
3. 0.0000021
4. 66,000,000
Taking the cube root of a number on the
slide rule is the inverse process of cubing a
number. To take the cube root of a number,
find the number on the K scale, set the hairline
over it, and read the cube root on the D scale
under the hairline.
POSITIONING NUMBERS ON THE K SCALE .-
Since a given number can be located in three
positions on the K scale, the question arises as
to which third of the K scale to use when locat-
ing a number. Generally, the left index, the
left middle index, the right middle index, and
the right index are considered to be numbered
as shown in figure 8-7.
i
1,000
10
10,000
100
100,000
1,000
1,000,000
Figure -8- 7. Powers of 10 associated with
K-scale indices.
A system similar to that used with square
roots may be used to locate the position of a
number on the K scale. Groups of three are
used rather than groups of two. The grouping
for cube root is illustrated as follows:
95
MATHEMATICS, VOLUME 1
1. ^40,531.6
N/40'531.6
2. ^4,561.43
W561.43
3. ^0.000043
\'0.000'043
For whole or mixed numbers the following
rules apply:
1. If the left-hand group contains one digit,
locate the number in the left third of the K scale.
2. If the left group contains two digits, lo-
cate the number in the middle third of the K
scale.
3. If the left group contains three digits,
locate the number in the right third of the K
scale.
The following examples illustrate the fore-
going rules:
1. s/4'561.43
(One digit) -left third K scale.
2. K/40'531.6
(Two digits) -middle third K scale.
3. 3 '
(Three digits) right third of K scale.
For positive numbers less than one, look
from left to right and find the first group that
contains a digit other than zero.
1. If the first two figures in this group are
zeros, locate the number in the left third of the
K scale.
2. K only the first figure in this group is
zero, locate the number in the middle third of
the K scale.
3. If the first figure of the group is not zero,
locate the number in the right third of the K
scale.
The following examples illustrate these
rules:
1. 3 N/ 0.000' 004' 53
(Two zeros)-left third K scale.
2. N/0.000'050'43
(One zero) middle third K scale.
3. x/0.000'000'430
(No zero) right third K scale.
PLACING THE DECIMAL POINT.-To place
the decimal point in the cube root of a number,
we use the system of marking off in groups of
three as shown above.
For whole or mixed numbers, there is one
digit in the root to the left of the decimal point
for every group marked in the original number .
Thus,
(Two digits in root to left of decimal point.)
For positive numbers less than one, there
will be one zero in the root between the decimal
point and the first significant digit for every
three zeros counted between the decimal point
and the first significant digit of the original
number. Thus,
N/0.000'000'004
(Two zeros between decimal point and first sig-
nificant digit of root.)
EXAMPLE:
N/216000.4
3r
(Three digits in left group)
Place the hairline over 216 in the right third of
the K scale. Read the digit sequence, 6, under
the hairline on the D scale. Since there are
two groups in the original number, there are
two digits to the left of the decimal point in the
root. Thus,
V216000.4 = 60
EXAMPLE:
* 0.0000451
3r
VO.000'045'1
(Only first figure is zero in this group)
Place the hairline over 451 in the middle third
of the K scale. Read the digit sequence, 357,
under the hairline on the D scale. Since there
is one group of three zeros, there is one zero
between the decimal point and the first signifi-
cant digit of the root. Thus,
VO. 0000451 = 0.0357
96
POWERS OF 10. -To take the cube root of a 4. N/204,000 d. s. 589
power of 10, mark it off as explained in the _
preceding paragraphs. The number in the left 5. V 734,000, 000 d. s. 902
group will then be 1, 10, or 100. We know that
the cube root of 10 is a number between 2 and 6. V4,913 d. s. 17
3. Thus, for the cube root of any number whose
left group is 10, use the K scale index which Answers:
lies between 2 and 3 on the D scale. The cube
root of 100 lies between 4 and 5. Therefore, 1. 0.02844
for a number whose left group is 100, use the K
scale index between 4 and 5 on the D scale. 2. 0.371
Practice problems. Following are some
problems and the digit sequence (d. s.) of the 3. 5.026
roots. Locate the decimal point for each root.
3 4. 58.9
1. N/0.000023 d. s. 2844
2. N/TTU51 d. s. 371 5> 902
3. IfTZI d. s. 5026 6. 17
97
CHAPTER 9
FUNDAMENTALS OF ALGEBRA
The numbers and operating rules of arith-
metic form a part of a very important branch
of mathematics called ALGEBRA.
Algebra extends the concepts of arithmetic
so that it is possible to generalize the rules for
operating with numbers and use these rules in
manipulating symbols other than numbers. It
does not involve an abrupt change into a dis-
tinctly new field, but rather provides a smooth
transition into many branches of mathematics
with a continuation of knowledge already gained
in basic arithmetic.
The idea of expressing quantities in a gen-
eral way, rather than in the specific terms of
arithmetic, is fairly common. A typical exam-
ple is the formula for the perimeter of a rec-
tangle, P = 2L + 2W, in which the letter P rep-
resents perimeter, L represents length, and W
represents width. It should be understood that
2L = 2(L) and 2W = 2(W). If the L and the W
were numbers, parentheses or some other mul-
tiplication sign would be necessary, but the
meaning of a term such as 2L is clear without
additional signs or symbols.
All formulas are algebraic expressions, al-
though they are not always identified as such.
The letters used in algebraic expressions are
often referred to as LITERAL NUMBERS (lit-
eral implies "letteral").
Another typical use of literal numbers is in
the statement of mathematical laws of operation.
For example, the commutative, associative, and
distributive laws, introduced in chapter 3 with
respect to arithmetic, may be restated in gen-
eral terms by the use of algebraic symbols.
COMMUTATIVE LAWS
The word "commutative" is defined in chap-
ter 3. Remember that the commutative laws
refer to those situations in which the factors
and terms of an expression are rearranged in a
different order.
ADDITION
The algebraic form of the commutative law
for addition is as follows:
a + b = b + a
From this law, it follows that
a + (b + c) = a + (c + b) = (c + b) + a
In words, this law states that the sum of two or
more addends is the same regardless of the
order in which the addends are arranged.
The arithmetic example in chapter 3 shows
only one specific numerical combination in
which the law holds true. In the algebraic ex-
ample, a, b, and c represent any numbers we
choose, thus giving a broad inclusive example
of the rule. (Note that once a value is selected
for a literal number, that value remains the
same wherever the letter appears in that par-
ticular example or problem. Thus, if we give a
the value of 12, in the example just given, a's
value is 12 wherever it appears.)
MULTIPLICATION
The algebraic form of the commutative law
for multiplication is as follows:
ab = ba
In words, this law states that the product of
two or more factors is the same regardless of
the order in which the factors are arranged.
ASSOCIATIVE LAWS
The associative laws of addition and multi-
plication refer to the grouping (association) of
terms and factors in a mathematical expression.
ADDITION
The algebraic form of the associative law
for addition is as follows:
In words, this law states that the sum of three
or more addends is the same regardless of the
manner in which the addends are grouped.
MULTIPLICATION
The algebraic form of the associative law
for multiplication is as follows:
a b c = (a b) c = a (b c)
98
In words, this law states that the product of
three or more factors is the same regardless
of the manner in which the factors are grouped.
DISTRIBUTIVE LAW
The distributive law refers to the distribu-
tion of factors among the terms of an additive
expression. The algebraic form of this law is
as follows:
a(b + c) = ab + ac
From this law, it follows that: If the sum of two
or more quantities is multiplied by a third
quantity, the product is found by applying the
multiplier to each of the original quantities
separately and summing the resulting expres-
sions.
ALGEBRAIC SUMS
The word "sum" has been used several times
in this discussion, and it is important to realize
the full implication where algebra is concerned.
Since a literal number may represent either a
positive or a negative quantity, a sum of sev-
eral literal numbers is always understood to be
an ALGEBRAIC SUM. That is, it is the sum
that results when the algebraic signs of all the
addends are taken into consideration.
The following problems illustrate the proce-
dure for finding an algebraic sum;
Let a = 3, b = -2, and c = 4.
Then a + b + c = (3) + (-2) + (4)
= 5
Also, a-b-c = a + (-b) + (-c)
= 3 + (+2) + (-4)
= 1
The second problem shows that every expres-
sion containing two or more terms to be com-
bined by addition and subtraction may be re-
written as an algebraic sum, all negative signs
being considered as belonging to specific terms
and all operational signs being positive.
It should be noted, in relation to this subject,
that the laws of signs for algebra are the same
as those for arithmetic.
ALGEBRAIC EXPRESSIONS
An algebraic expression is made up of the
signs and symbols of algebra. These symbols
include the Arabic numerals, literal numbers,
the signs of operation, and &Q forth. Such an
expression represents one number or one quan-
tity. Thus, just as the sum of 4 and 2 is one
quantity, that is, 6, the sum of c and d is one
quantity, that is, c + d. Likewise p -vTB, ab,
a - b, and so forth, are algebraic expres-
sions each of which represents one quantity or
number.
Longer expressions may be formed by com-
binations of the various signs of operation and
the other algebraic symbols, but no matter how
complex such expressions are they still repre-
sent one number. Thus the algebraic expres-
-a + ^2a + b
sion.
6
-c is one number
The arithmetic value of any algebraic ex-
pression depends on the values assigned to the
literal numbers. For example, in the expres-
sion 2x 2 - 3ay, if x = -3, a = 5, and y = 1, then
we have the following:
2x 2 - Say = 2(-3) 2 -3(5)(1)
= 2(9) - 15 = 18 - 15 = 3
Notice that the exponent is an expression
such as 2x 2 applies only to the x. If it is de-
sired to indicate the square of 2x, rather than
2 times the square of x, then parentheses are
used and the expression becomes (2x) 2 .
Practice problems. Evaluate the following
algebraic expressions when a = 4, b = 2, c = 3,
x = 7, and y = 5. Remember, the order of op-
eration is multiplication, division, addition, and
subtraction.
ax
nr
1. 3x + 7y - c 3.
u
2. xy - 4a 2 4. c H
Answers:
1. 53 3. 19
2. -29 4. 53
TERMS AND COEFFICIENTS
The terms of an algebraic expression are
the parts of the expression that are connected
by plus and minus signs. In the expression
3abx + cy - k, for example, 3abx, cy, and k are
the terms of the expression.
99
An expression containing only one term, such
as Sab, is called a monomial (mono means one).
A binomial contains two terms; for example,
2r + by. A trinomial consists of three terms.
Any expression containing two or more terms
may also be called by the general name, poly-
nomial (poly means many). Usually special
names are not given to polynomials of more than
three times. The expression x 3 - 3x 2 + 7x + 1
is a polynomial of four terms. The trinomial
x 2 + 2x + 1 is an example of a polynomial which
has a special name.
Practice problems. Identify each of the fol-
lowing expressions as a monomial, binomial,
trinomial, or polynomial. (Some expressions
may have two names.)
1. x
3. abx
2. Sy + a-t-b 4. 4 + 2b + y
5. 3y 2 + 4
6.
Answers:
1. Monomial
3. Monomial
5. Binomial
(also polynomial)
2. Trinomial
(also polynomial)
4. Polynomial
6. Binomial
(also polynomial)
In general, a COEFFICIENT of a term is
any factor or group of factors of a term by
which the remainder of the term is to be multi-
plied. Thus in the term 2axy, 2ax is the coeffi-
cient of y, 2a is the coefficient of xy, and 2 is
the coefficient of axy. The word "coefficient"
is usually used in reference to that factor which
is expressed in Arabic numerals. This factor
is sometimes called the NUMERICAL COEF-
FICIENT. The numerical coefficient is cus-
tomarily written as the first factor of the term.
In 4x, 4 is the numerical coefficient, or simply
the coefficient, of x. Likewise, in 24xy 2 , 24 is
the coefficient of xy 2 and in 16(a + b), 16 is the
coefficient of (a + b). When no numerical coef-
ficient is written it is understood to be 1. Thus
in the term xy, the coefficient is 1.
COMBINING TERMS
When arithmetic numbers are connected by
plus and minus signs, they can always be com-
bined into one number. Thus,
Here three numbers are added algebraically
(with due regard for sign) to give one number.
The terms have been combined into one term.
Terms containing literal numbers can be
combined only if their literal parts are the
same. Terms containing literal factors in
which the same letters are raised to the same
power are called like terms. For example, 3y
and 2y are like terms since the literal parts
are the same. Like terms are added by adding
the coefficients of the like parts. Thus, 3y + 2y
= 5y just as 3 bolts + 2 bolts = 5 bolts. Also
3a 2 b and a 2 b are like; 3a 2 b + a 2 b = 4a 2 b and
3a 2 b - a 2 b = 2a 2 b. The numbers ay and by are
like terms with respect to y. Their sum could
be indicated in two ways: ay + by or (a + b)y.
The latter may be explained by comparing the
terms to denominate numbers. For instance,
a bolts + b bolts = (a + b) bolts.
Like terms are added or subtracted by add-
ing or subtracting the numerical coefficients
and placing the result in front of the literal
factor, as in the following examples:
7x 2 - 5x 2 = (7 - 5)x 2 = 2x 2
5b 2 x - Say 2 - 8b 2 x + lOay 2 = -3b 2 x + 7ay 2
Dissimilar or unlike terms in an algebraic
expression cannot be combined when numerical
values have not been assigned to the literal
factors. For example, -5x 2 + 3xy - 8y 2 con-
tains three dissimilar terms. This expression
cannot be further simplified by combining terms
through addition or subtraction. The expres-
sion may be rearranged as x(3y - 5x) - 8y 2 or
y(3x - 8y) - 5x 2 , but such a rearrangement is
not actually a simplification.
Practice problems. Combine like terms in
the following expression:
1. 2a + 4a
2. y + y 2 + 2y
3 4 ay - ay
3. 4 _
Answers:
1. 6a
2. y 2 + 3y
3. 3
ay
4. 2ay 2 - ay 2
5. bx 2 + 2bx 2
6. 2y + y 2
4. ay 2
5. 3bx 2
6. 2y + y 2
100
Chapter 9 - FUNDAMENTALS OF ALGEBRA
SYMBOLS OF GROUPING
Often it is desired to group two or more
terms to indicate that they are to be considered
and treated as though they were one term even
though there may be plus and minus signs be-
tween them. The symbols of grouping are pa-
rentheses ( ) (which we have already used),
brackets [ ] , braces { }, and the vinculum
The vinculum is sometimes called the "over-
score." The fact that -7 + 2-5 is to be sub-
tracted from 15, for example, could be indi-
cated in any one of the following ways:
15 - (-7 + 2 - 5)
15 - [-7 + 2 - 5]
15 - {-7 + 2 - 5}
15 - -7+2-5
Actually the vinculum is seldom used except
in connection with a radical sign, such as in
N' a + b, or in a Boolean algebra expression.
Boolean algebra is a specialized kind of sym-
bolic notation which is discussed in Mathe-
matics, Volume 3, NavPers 10073.
Parentheses are the most frequently used
symbols of grouping. When several symbols
are needed to avoid confusion in grouping, pa-
rentheses usually comprise the innermost sym-
bols, followed by brackets, and then by braces
as the outermost symbols. This arrangement
of grouping symbols is illustrated as follows:
2x - {3y + [- 8 - 5y - (x - 4)]}
REMOVING AND INSERTING
GROUPING SYMBOLS
Discussed in the following paragraphs are
various rules governing the removal and inser-
tion of parentheses, brackets, braces, and the
vinculum. Since the rules are the same for all
grouping symbols, the discussion in terms of
parentheses will serve as a basis for all.
Removing Parentheses
If parentheses are preceded by a minus sign,
the entire quantity enclosed must be regarded
as a subtrahend. This means that each term of
the quantity in parentheses is subtracted from
the expression preceding the minus sign. Ac-
cordingly, parentheses preceded by a minus
sign can be removed, if the signs of all terms
within the parentheses are changed.
This may be explained with an arithmetic
example. We recall that to subtract one num-
ber from another, we change the sign of the
subtrahend and proceed as in addition. To sub-
tract -7 from 16, we change the sign of -7 and
proceed as in addition, as follows:
16 - (-7) =16+7
= 23
It is sometimes easier to see the result of
changing signs in the subtrahend if the minus
sign preceding the parentheses is regarded as
a multiplier. Thus, the thought process in re-
moving parentheses from an expression such
as - (4 - 3 + 2) would be as follows: Minus
times plus is minus, so the first term of the
expression with parentheses removed is - 4.
(Remember that the 4 in the original expres-
sion is understood to be a +4, since it has no
sign showing.) Minus times minus is plus, so
the second term is +3. Minus times plus is
minus, so the third term is -2. The result is
-4+3-2, which reduces to -3.
This same result can be reached just as
easily, in an arithmetic expression, by combin-
ing the numbers within the parentheses before
applying the negative sign which precedes the
parentheses. However, in an algebraic expres-
sion with no like terms such combination is not
possible. The following example shows how the
rule for removal of parentheses is applied to
algebraic expressions:
2a - (-4x + 3by) = 2a + 4x - 3by
Parentheses preceded by a plus sign can be
removed without any other changes, as the fol-
lowing example shows:
2b + (a-b) = 2b+a-b = a + b
Many expressions contain more than one set
of parentheses, brackets, and other symbols of
grouping. In removing symbols of grouping, it
is possible to proceed from the outside inward
or from the inside outward. For the beginner,
it is simpler to start on the inside and work to-
ward the outside, collecting terms and simpli-
fying as one proceeds. In the following example
the inner grouping symbols are removed first:
= 2a - [7x - 12a]
= 2a - 7x + 12a
= 14a - 7x
Enclosing Terms in Parentheses
When it is desired to enclose a group of
terms in parentheses, the group of terms re-
mains unchanged if the sign preceding the pa-
rentheses is positive. This is illustrated as
follows:
3x - 2y + 7x - y = (3x - 2y) + (7x - y).
Note that this agrees with the rule for removing
parentheses preceded by a plus sign.
If terms are enclosed within parentheses
preceded by a minus sign, the signs of all the
terms enclosed must be changed as in the fol-
lowing example:
3x - 2y + 7x - y = 3x - (2y - 7x + y)
Practice problems. In problems 1 through 4,
remove the symbols of grouping and combine
like terms. In problems 5 through 8, enclose
the first two terms in parentheses preceded by
a plus sign (understood) and the last two in pa-
rentheses preceded by a minus sign.
1. 6a - (4a - 3)
2. 3x + [2x - 4y(6 - 4x)] + 2y - (3 - x + 3y)
3. -a + [-a - (2a + 3)] + 3
4. (7x - Say) - (4a - b) + 16
5. 4a - 3b - 2c + 4d
6. -2 -3x +4y - z
7. x + 4y + 3z + 7
8. -4 + 2a - 6c + 3d
Answers:
1. 2a + 3
2. 6x + 16xy - 25y - 3
3. -4a
4. 7x - Say - 4a + b + 16
7. (x + 4y) - (-3z - 7)
8. (-4 + 2a) - (6c - 3d)
EXPONENTS AND RADICALS
Exponents and radicals have the same mean-
ing in algebra as they do in arithmetic. Thus,
if n represents any number then n 2 = n . n,
n 3 = n n n, etc. By the same reasoning, n m
means that nis to be taken as a factor m times.
That is, n m is equal to n n . n . . . , with n
appearing m times. The series of dots, called
ellipsis (not to be confused with the geometric
figure having a similar name, ellipse), repre-
sents continuation of the same pattern or the
same symbol.
The rules of operation with exponents are
also the same in algebra as in arithmetic. For
example, n 2 n 3 = n 2+3 = n 5 . Some care is
necessary to avoid confusion over an expres-
sion such as 3 2 3 3 . In this example, n = 3 and
the product desired is 3 s , not 9 s . In general,
3 D 3 b = 3 a+b , and a similar result is reached
whether the factor which acts as a base for the
exponents is a number or a letter. Thus the
general form can be expressed as follows:
n b =
In words, the general rule for multiplication
involving exponents is as follows: When multi-
plying terms whose literal factors are like, the
exponents are added. This rule may be applied
to problems involving division, if all expres-
sions containing exponents in denominators are
rewritten as expressions with negative expo-
x 2 y
xy 2
rewritten as (x 2 y)(x" 1 y" 2 ), which is equal to
nents. For example, the fraction -
can be
(x 2 y)(x- 1 y- 2 ),
" 1
No-
(x 2 ~ 1 )(y 1 " 2 ). This reduces to xy" J , or.
tice that the result is the same as it would have
been if we had simply subtracted the exponents
of literal factors in the denominator from the
exponents of the same literal factors in the
numerator.
The algebraic rules for radicals also remain
the same as those of arithmetic. In arithmetic,
]/T = 4 1/2 = 2. Likewise, in algebra \Ta. = a 172
and
= a
1/n
102
MULTIPLYING MONOMIALS
If a monomial such as 3abc is to be multi-
plied by a numerical multiplier, for example 5,
the coefficient alone is multiplied, as in the
following example:
5 x 3abc = 15abc
When the numerical factor is not the initial
factor of the expression, as in x(2a), the result
of the multiplication is not written as x2a. In-
stead, the numerical factor is interchanged with
literal factors by use of the commutative law of
multiplication. The literal factors are usually
interchanged to place them in alphabetical or-
der, and the final result is as follows:
x(2a) = 2ax
The rule for multiplication of monomials
may be stated as follows: Multiply the numeri-
cal coefficients to form the coefficient of the
product. Multiply the literal factors, combining
exponents in like factors, to form the literal
part of the product. The complete process is
illustrated in the following example:
(2ab)(3a 2 )(2b 3 ) =
= 12a 3 b 4
Practice problems,
operations:
1. (2x 2 )(5x 5 )
2. (-5ab 2 )(2a 2 b)
3. (-4x 4 y)(-3xy 4 )
Answers:
1. 10x 7
2. -10a 3 b 3
3. 12x s y 5
Perform the indicated
4. (2")(2 b )
5. (-4a 3 ) 2
6. (3a 2 b) 2
4. 2 a+b
5. 16a 6
6. 9a 4 b 2
DIVIDING MONOMIALS
As may be expected, the process of dividing
is the inverse of multiplying. Because 3 x 2a
= 6a, 6a * 3 = 2a, or 6a + 2 = 3a. Thus, when
the divisor is numerical, divide the coefficient
of the dividend by the divisor.
When the divisor contains literal parts that
are also in the dividend, cancellation may be
performed as in arithmetic. For example,
6ab -r 3a may be written as follows:
(2)(3a)(b)
3a
Cancellation of the common literal factor, 3a,
from the numerator and denominator leaves 2b
as the answer for this division problem.
When the same literal factors appear in both
the divisor and the dividend, but with different
exponents, cancellation may still be used, as
follows:
14a 3 b 3 x
-21a 2 b s x
(7)(2)a 2 ab 3 x
(7)(-3)a 2 b 3 b 2 x
2a
2a
= - 3b 2
This same problem may be solved without
thinking in terms of cancellation, by rewriting
with negative exponents as follows:
14a 3 b 3 x 2a 3 " 2 b 3 ~ 5 x 1 " 1
-21a 2 b 5 x
-3
2ab~ 2 2a
= -
-3 ~ -3b 2
2a
3b 2
Practice problems,
operations:
Perform the indicated
x s
1'$?
6. Vx 4a y 2a
9 a 9 b 4
5a 4 b
2> a 6 b 3
' 10a 2 b 3
a 2 bc 2
10x 2 y 3 z 4
6 " abc
A a ' b
Bl -5xy 2 z 3
10. Va 6 b" 6 ""
4> ab 2
5. K16x 4 y 6
Answers:
1 V - 1
. A
6. x 2a y a
2. a 3 b
3. ac
7 a '
7< 2b 2
8. - 2xyz
4 '
9. 10a 4 b 2
5. 4x 2 y 3
10. a 3 b 3 "
103
OPERATIONS WITH POLYNOMIALS
Adding and subtracting polynomials is sim-
ply the adding and subtracting of their like
terms. There is a great similarity between the
operations with polynomials and denominate
numbers. Compare the following examples:
1. Add 5 qt and 1 pt to 3 qt and 2 pt.
3 qt + 2 pt
5 qt + 1 pt
8 qt + 3 pt
2. Add 5x + y to 3x + 2y.
3x
5x
2y
y
8x + 3y
One method of adding polynomials (shown in
the above examples) is to place like terms in
columns and to find the algebraic sum of the
like terms. For example, to add 3a + b - 3c,
3b + c - d, and 2a + 4d, we would arrange the
polynomials as follows:
3a + b - 3c
3b + c - d
2a + 4d
5a + 4b - 2c + 3d
Subtraction may be performed by using the
same arrangement that is, by placing terms of
the subtrahend under the like terms of the min-
uend and carrying out the subtraction with due
regard for sign. Remember, in subtraction the
signs of all the terms of the subtrahend must
first be mentally changed and then the process
completed as in addition. For example, sub-
tract lOa + b from 8a - 2b, as follows:
8a - 2b
lOa + b
-2a - 3b
Again, note the similarity between this type of
subtraction and the subtraction of denominate
numbers.
Addition and subtraction of polynomials also
can be indicated with the aid of symbols of
grouping. The rule regarding changes of sign
when removing parentheses preceded by a minus
sign automatically takes care of subtraction.
For example, to subtract lOa + b from 8a - 2b,
we can use the following arrangement:
(8a - 2b) - (lOa + b) = 8a - 2b - lOa - b
= -2a - 3b
Similarly, to add -3x + 2y to -4x - 5y, we can
write
(~3x + 2y) + (-4x - 5y) = -3x + 2y - 4x - 5y
= -7x - 3y
Practice problems. Add as indicated, in
each of the following problems:
1. 3a + b
2a_+ 5b
2. (6s 3 t + 3s 2 t + st + 5) + (s 3 t - 5)
3. 4a + b + c, a + c - d, and 3a + 2b + 2c
4. 4x + 2y
3x - y + z
_x - z
In problems 5 through 8, perform the indi-
cated operations and combine like terms.
5. (2a + b) - (3a + 5b)
6. (5x 3 y + 3x 2 y) - (x 3 y)
7. (x + 6) + (3x + 7)
8. {4a 2 - b) - (2a 2 + b)
Answers:
1. 5a + 6b
2. 7s 3 t + 3s 2 t + st
3. 8a + 3b + 4c - d
4. 8x + y
5. -(a + 4b)
6. 4x 3 y + 3x 2 y
7. 4x + 13
8. 2(a 2 - b)
MULTIPLICATION OF A POLY-
NOMIAL BY A MONOMIAL
We can explain the multiplication of a poly-
nomial by a monomial by using an arithmetic
example. Let it be required to multiply the
binomial expression, 7 - 2, by 4. We may write
this 4 x (7 - 2) or simply 4(7 - 2). Now 7 -2 = 5.
Therefore, 4(7 - 2) = 4(5) = 20. Now, let us
solve the problem a different way. Instead of
104
then subtract. Thus, 4(7 - 2) = (4 x 7) - (4 x 2)
= 20. Both methods give the same result. The
second method makes use of the distributive
law of multiplication.
When there are literal parts in the expres-
sion to be multiplied, the first method cannot
be used and the distributive method must be
employed. This is illustrated in the following
examples:
4(5 + a) = 20 + 4a
3(a + b) = 3a + 3b
ab(x + y - z) = abx + aby - abz
Thus, to multiply a polynomial by a monomial,
multiply each term of the polynomial by the
monomial.
Practice problems. Multiply as indicated:
1. 2a(a - b)
2. 4a 2 (a 2 + 5a + 2)
Answers:
1. 2a 2 - 2ab
2. 4a 4 + 20a 3
8a 2
3. -4x(-y - 3z)
4. 2a 3 (a 2 - ab)
3. 4xy + 12xz
4. 2a 5 - 2a 4 b
MULTIPLICATION OF A POLY-
NOMIAL BY A POLYNOMIAL
As with the monomial multiplier, we explain
the multiplication of a polynomial by a poly-
nomial by use of an arithmetic example. To
multiply (3 + 2)(6 - 4), we could do the opera-
tion within the parentheses first and then mul-
tiply, as follows:
(3 + 2)(6 - 4) = (5)(2) - 10
However, thinking of the quantity (3 + 2) as one
term, we can use the method described for a
monomial multiplier. That is, we can multiply
each term of the multiplicand by the multiplier,
(3 + 2), with the following result:
(3 + 2)(6 - 4) = [(3 + 2) x 6 - (3 + 2) x 4]
Now considering each of the two resulting
products separately, we note that each is a bi-
nomial multiplied by a monomial.
(3 + 2)6 = (3 x 6) + (2 x 6)
and the second is
-(3 + 2)4 = - [(3 x 4) + (2 x 4)]
= -(3 x 4) - (2 x 4)
Thus we have the following result:
(3 + 2)(6 - 4) = (3 x 6) + (2 x 6)
- (3 x 4) - (2x4)
=18+12-12-8
= 10
The complete product is formed by multiplying
each term of the multiplicand separately by
each term of the multiplier and combining the
results with due regard to signs.
Now let us apply this method in two exam-
ples involving literal numbers.
1. (a + b)(m + n) = am + an + bm + bn
2. (2b + c)(r + s + 3t - u) = 2br + 2bs
+ 6bt - 2bu + cr + cs + 3ct - cu
The rule governing these examples is stated as
follows: The product of any two polynomials is
found by multiplying each term of one by each
term of the other and adding the results alge-
braically.
It is often convenient, especially when either
of the expressions contains more than two
terms, to place the polynomial with the fewer
terms beneath the other polynomial and multi-
ply term by term beginning at the left. Like
terms of the partial products are placed one
beneath the other to facilitate addition.
Suppose we wish to find the product of
3x 2 - 7x - 9 and 2x - 3. The procedure is
27
27
Practice problems. In the following prob-
lems, multiply and combine like terms:
3x 2
2x
- 7x -
- 3
9
6x 3
- 14x 2 -
- 9x 2 +
18x
21x
6x 3
- 23x 2 +
3x
105
1. (2a - 3)(a + 2) 3. x 3 + 5x 2 - x + 2
2x + 3
2. (ax + b)(ax - b) 4. 2a 2 + Sab - b 2
a + b
Answers:
1. 2a 2 + a - 6 3. 2x 4 + 13x 3 + 13x 2 + x + 6
2. a 2 x 2 - b 2 4. 2a 3 + 7a 2 b + 4ab 2 - b 3
SPECIAL PRODUCTS
The products of certain binomials occur fre-
quently. It is convenient to remember the form
of these products so that they can be written
immediately without performing the complete
multiplication process. We present four such
special products as follows, and then show how
each is derived:
1. Product of the sum and difference of two
numbers .
EXAMPLE: (x - y)(x + y) = x 2 - y 2
2. Square the sum of two numbers.
EXAMPLE: (x + y) 2 = x 2 + 2xy + y 2
3. Square of the difference of two numbers.
EXAMPLE: (x - y) 2 = x 2 - 2xy + y 2
4. Product of two binomials having a com-
mon term.
EXAMPLE: (x + a)(x + b) = x 2 + (a + b)x + ab
Product of Sum and Difference
The product of the sum and difference of
two numbers is equal to the square of the first
number minus the square of the second number.
If, for example, x - y is multiplied by x + y, the
middle terms cancel one another. The result
is the square of x minus the square of y, as
shown in the following illustration:
x - y
x + y
x 2 - xy
+ xy - y 2
By keeping this rule in mind, the product of
the sum and difference of two numbers can be
written down immediately by writing the differ-
ence of the squares of the numbers. For ex-
ample, consider the following three problems:
(x + 3)(x - 3) = x 2 - 3 2 = x 2 - 9
(5a + 2b)(5a - 2b) = (5a) 2 - (2b) 2 = 25a 2 - 4b 2
(7x + 4y)(7x - 4y) = 49x 2 - 16y 2
RATIONALIZING DENOMINATORS.- The
product of the sum and difference of two num-
bers is useful in rationalizing a denominator
that is a binomial. For example, in a fraction
such as
\T2" - 6
the denominator can be altered so that no radi-
cal terms appear in it. (This process is called
rationalizing.) The denominator must be mul-
tiplied by N/T + 6, which is called the conjugate
of *J~2 - 6. Since the value of the original frac-
tion would be changed if we multiplied only the
denominator, our multiplier must be applied to
both the numerator and the denominator. Mul-
tiplying the original fraction by
is, in effect, the same as multiplying it by 1.
The result of rationalizing the denominator
of this fraction is as follows:
2 \T2 + 6 2(T2 + 6)
- 6
6 Ov/Tr - 6'
_ 2(V2 + 6)
2-36
2(\/2 + 6)
~ 2(1 - 18)
+ 6)
6
106
MENTAL MULTIPLICATION. -The product
of the sum and difference can be utilized to
mentally multiply two numbers that differ from
a multiple of 10 by the same amount, one
greater and the other less. For example, 67 is
3 less than 70 while 73 is 3 more than 70. The
product of 67 and 73 is then found as follows:
67(73) = (70 - 3)(70 + 3)
= 70 2 - 3 2 = 4,900 - 9 = 4,891
Square of Sum or Difference
The square of the SUM of two numbers is
equal to the square of the first number plus
twice the product of the numbers plus the square
of the second number. The square of the DIF-
FERENCE of the same two numbers has the
same form, except that the sign of the middle
term is negative.
These results are evident from multiplica-
tion. When x and y represent the two numbers,
we obtain
x + y
x + y
x 2 + xy
+ xy + y 2
x 2 + 2xy + y 2
- Y
x - xy
- xy + y
- 2xy +
Applying this rule to the squares of the bi-
nomials 3a + 2b and 3a - 2b, we have the fol-
lowing two cases:
1. (3a + 2b) 2 = (3a) 2 + 2(3a)(2b) + (2b) 2
= 9a 2 + 12ab + 4b 2
2. (3a - 2b) 2 = 9a 2 - 12ab + 4b 2
The square of the sum or difference of two
numbers is applicable to squaring a binomial
that contains one or two irrational terms, as in
the following examples:
1. KT+ 8) 2 = (*/3) 2 + 2(8)(^) + 64
= 3 + 16 -S/TH- 64 = 67 + 16 \T3
2.
3.
4.
- 8) 2 = (-/I) 2 - 2(8)(V3) + 64
= 3 - 16 \T3 + 64 = 67 - 16 -s/"3~
= 5 + 2 N/15 + 7 = 12 + 2 -735
- */T) 2 = 12 - 2 ^^55
The square of the sum or difference of two
numbers can be applied to the process of men-
tally squaring certain numbers. For example,
82 2 can be expressed as (80 + 2) 2 while 67 2
can be expressed as (70 - 3) 2 . We find that
(80 + 2) 2 = 80 2 + 2(80)(2) + 2 2
= 6,400 + 320 + 4 = 6,724
(70 - 3) 2 = 70 2 - 2(70)(3) + 3 2
= 4,900 - 420 + 9 = 4,489
Binomials Having a Common Term
The binomials x + 2 and x - 3 have a com-
mon term, x. They have two unlike terms,
+2 and -3. The product of these binomials is
x
x
+ 2
- 3
x 2 + 2x
- 3x - 6
x 2 - x - 6
Inspection of this product shows that it is
obtained by squaring the common term, adding
the sum of the unlike terms multiplied by the
common term, and finally adding the product of
the unlike terms.
Apply this rule to the product of 3y - 5 and
3y + 4. The common term is 3y; its square is
9y 2 . The sum of the unlike terms is -5 + 4 = -1;
the sum of the unlike terms multiplied by the
common term is -3y; and the product of the
unlike terms is -5(4) = -20. The product of the
two binomials is
(3y - 5)(3y + 4) = 9y 2 - 3y - 20
The product of two binomials having a com-
mon term is applicable to the multiplication of
numbers like /3 + 7 and *J~5 - 2 which contain
irrational terms. For example,
7)(V3" - 2) = (J5) 2 + 5 */3 - 14
= 3 + 5 *J~3 - 14
= -11 + 5 -/3
Practice problems. In problems 1 through 4,
multiply and combine terms. In 5 through 8,
simplify by using special products.
107
1. (x + 4)(x + 2)
2. (-J* - b) 2
3. (7a + 4b)(7a - 4b)
4. (ax + y) 2
Answers:
1. x 2 + 6x + 8
2. a - 2b\/~a + b 2
3. 49a 2 - 16b 2
4. a 2 x 2 + 2axy + y 2
5
v/lT - 2
6. 48(52)
7. (N/T+ 7) 2
8. (73) 2
5. -
+ 2)
6. (50 - 2)(50 + 2)
= 2496
7. 52 + \4\T5
8. (70 + 3)(70 + 3)
= 5329
DIVISION OF A POLY-
NOMIAL BY A MONOMIAL
Division, like multiplication, may be dis-
tributive. Consider, for example, the problem
(4 + 6 - 2) * 2, which may be solved by adding
the numbers within the parentheses and then
dividing the total by 2. Thus,
4 + 6-2 = 8 A
2 T
Now notice that the problem may also be solved
distributively.
4 + 6-2 _ 4_ 6_ 2^
2 2 2 " 2
= 2 + 3-1
= 4
CAUTION: Do not confuse problems of the
type just described with another type which is
similar in appearance but not in final result.
For example, in a problem such as 2 * (4 + 6 - 2)
the beginner is tempted to divide 2 successively
by 4, then 6, and then -2, as follows:
4 + 6-24 62
Notice that we have canceled the "equals" sign,
because 2 -r 8 is obviously not equal to 1/2 +
2/6. - 1. The distributive method applies only
in those cases in which several different nu-
merators are to be used with the same de-
nominator
When literal numbers are present in an ex-
pression, the distributive method must be used,
as in the following two problems:
1 2ax + aby + a _ 2 ax aby a
i. _ __ +___+_
2.
= 2x + by + 1
18ab 2 - 12bc 18ab 2 12bc
6b
6b " 6b
= 3ab - 2c
Quite often this division may be done men-
tally, and the intermediate steps need not be
written out.
DIVISION OF A POLY-
NOMIAL BY A POLYNOMIAL
Division of one polynomial by another pro-
ceeds as follows:
1. Arrange both the dividend and the divisor
in either descending or ascending powers of the
same letter.
2. Divide the first term of the dividend by
the first term of the divisor and write the re-
sult as the first term of the quotient.
3. Multiply the complete divisor by the quo-
tient just obtained, write the terms of the prod-
uct under the like terms of the dividend, and
subtract this expression from the dividend.
4. Consider the remainder as a new dividend
and repeat steps 1, 2, and 3.
EXAMPLE:
(10x 3 - 7x 2 y - 16xy 2 + 12y 3 ) + (5x - 6y)
SOLUTION:
2x 2 + xy - 2y 2
_
5x - 6y HOx 3 - 7x 2 y - 16xy 2 + 12y 3
10x 3 - 12x 2 y
5x 2 y - 16xy 2
5x 2 y - 6xy 2
- lOxy 2 + 12y 3
- IQxy 2 + 12y 3
In the example just shown, we began by di-
viding the first term, 10x 3 , of the dividend by
the first term, 5x, of the divisor. The result is
2x 2 . This is the first term of the quotient.
Next, we multiply the divisor by 2x 2 and
subtract this product from the dividend. Use
108
the remainder as a new dividend. Get the sec-
ond term, xy, in the quotient by dividing the
first term, 5x 2 y, of the new dividend by the
first term, 5x,of the divisor. Multiply the divi-
sor by xy and again subtract from the dividend.
Continue the process until the remainder is
zero or is of a degree lower than the divisor.
In the example being considered, the remainder
is zero (indicated by the double line at the bot-
tom). The quotient is 2x 2 + xy - 2y 2 .
The following long division problem is an
example in which a remainder is produced:
- x + 3
x + 3
1 X 3 -
X 3 -
h 2x 2
h 3x 2
+ 5
- x 2 - 3x
3x
3x
- 4
The remainder is -4.
Notice that the term -3x in the second step
of this problem is subtracted from zero, since
there is no term containing x in the dividend.
When writing down a dividend for long division,
leave spaces for missing terms which may en-
ter during the long division process.
In arithmetic, division problems are often
arranged as follows, in order to emphasize the
relationship between the remainder and the
divisor:
- 9 j.
2 - 2 + 2
This same type of arrangement is used in alge-
bra. For example, in the problem just shown,
the results could be written as follows:
2x 2
x + 3
= x 2 - x + 3 -
Remember, before dividing polynomials ar-
range the terms in the dividend and divisor
according to either descending or ascending
powers of one of the literal numbers. When
only one literal number occurs, the terms are
usually arranged in order of descending powers.
For example, in the polynomial 2x 2 + 4x 3 +
5 - 7x the highest power among the literal terms
is x 3 . If the terms are arranged according to de-
scending powers of x, the term in x 3 should ap-
pear first. The x 3 term should be followed by the
x 2 term, the x term, and finally the constant term.
The polynomial arranged according to descending
powers of x is 4x 3 + 2x 2 - 7x + 5.
Suppose that 4ab + b 2 + 15a 2 is to be divided
by 3a + 2b. Since 3a can be divided evenly into
15a 2 , arrange the terms according to descend-
ing powers of a. The dividend takes the form
15a 2 + 4ab + b 2
Synthetic Division
Synthetic division is a shorthand method of
dividing a polynomial by a binomial of the form
x - a. For example, if 3x 4 + 2x 3 + 2x 2 - x - 6
is to be divided by x - 1, the long form would
be as follows:
3x 3 + 5x 2 + 7x
x -
1 I 3x 4 + 2x 3 + 2x 2 - x - 6
3x -
+ 5x 3
+ 5x 3
+ 2x 2
- 5x 2
- x
- 7x
- 6
- 6
+ 7x 2
+ 7x 2
+ 6x
+ 6x
Notice that every alternate line of work in
this example contains a term which duplicates
the one above it. Furthermore, when the sub-
traction is completed in each step, these dupli-
cated terms cancel each other and thus have no
effect on the final result. Another unnecessary
duplication results when terms from the divi-
dend are brought down and rewritten prior to
subtraction. By omitting these duplications,
the work may be condensed as follows:
3x 3 +5x 2 +7x +6
x - 1 j 3x 4 +2x 3 +2x 2 -x ^
-3x 3 -5x 2 -7x -6
+5x 3 +7x 2 +6x
The coefficients of the dividend and the con-
stant term of the divisor determine the results
of each successive step of multiplication and
subtraction. Therefore, we may condense still
further by writing only the nonliteral factors,
as follows:
109
3 +5 +7 +6
- 1 |3 +2 +2 -1 -6
-3 -5 -7 -6
3 +5 +7 +6
Notice that if the coefficient of the first term
in the dividend is brought down to the last line,
then the numbers in the last line are the same
as the coefficients of the terms in the quotient.
Thus we do not really need to write a separate
line of coefficients to represent the quotient.
Instead, we bring down the first coefficient of
the dividend and make the subtraction "sub-
totals" serve as coefficients for the rest of the
quotient, as follows:
x - 1
3 2 2-1-6
-3 -5 -7 -6
The unnecessary writing of plus signs is also
eliminated here.
The use of synthetic division is limited to
divisors of the form x - a, in which the degree
of x is 1. Thus the degree of each term in the
quotient is 1 less than the degree of the corre-
sponding term in the dividend. The quotient in.
this example is as follows:
3x 3 + 5x 2 + 7x + 6
The sequence of operations in synthetic di-
vision may be summarized as follows, using as
an example the division of 3x - 4x 2 + x 4 - 3 by
x + 2:
1 0-43-3
2-406
1-2 03-9
First, rearrange the terms of the dividend
in descending powers of x. The dividend then
becomes x 4 - 4x 2 + 3x -3, with 1 understood
as the coefficient of the first term. No x 3 term
appears in the polynomial, but we supply a zero
as a place holder for the x 3 position.
Second, bring down the 1 and multiply it by
the +2 of the divisor. Place the result under
the zero, and subtract. Multiply the result (-2)
by the +2 of the divisor, place the product under
the -4 of the dividend, and subtract. Continue
this process, finally obtaining x 3 - 2x 2 + 3 as
the quotient. The remainder is -9.
Practice problems. In the following prob-
lems, perform the indicated operations. In 4,
5, and 6, first use synthetic division and then
check your work by long division:
1. (a 3 - 3a 2 + a) - a
2.
- 7x s
4x 4
3. (10x 3 - 7x 2 y - 16xy 2 + 12y 3 )
H- (2x 2 + xy - 2y 2 )
4. (x 2 + llx + 30) + (x + 6)
5. (12 + x 2 - 7x) + (x - 3)
6. (a 2 - lla + 30) - (a - 5)
Answers:
1. a 2 - 3a + 1 4. x + 5
2. x 4 - 7x 3 + 4x 2 5. x - 4
3. 5x - 6y
6. a - 6
110
CHAPTER 10
FACTORING POLYNOMIALS
A factor of a quantity N, as defined in chap-
ter 2 of this course, is any expression which
can be divided into N without producing a re-
mainder. Thus 2 and 3 are factors of 6, and
the factors of 5x are 5 and x. Conversely, when
all of the factors of N are multiplied together,
the product is N. This definition is extended to
include polynomials.
The factors of a polynomial are two or more
expressions which, when multiplied together,
give the polynomial as a product. For example,
3, x, and x 2 - 4 are factors of 3x 3 - 12x, as the
following equation shows:
(3)(x)(x 2 - 4) = 3x 3 - 12x
The factors 3 and x, which are common to both
terms of the polynomial 3x 3 - 12x, are called
COMMON FACTORS.
The distributive principle, mentioned in
chapters 3 and 9 of this course, is an important
part of the concept of factoring. It may be
stated as follows :
If the sum of two or more quantities is multi-
plied by a third quantity, the product is found
by applying the multiplier to each of the origi-
nal quantities separately and summing the re-
sulting expressions. It is this principle which
allows us to separate common factors from the
terms of a polynomial.
Just as with numbers , an algebraic expres-
sion is a prime factor if it has no other factors
except itself and 1. The factor x 2 - 4 is not
prime, since it can be separated into x - 2 and
x + 2. The factors x - 2 and x + 2 are both
prime factors, since they cannot be separated
into other factors.
The process of finding the factors of a poly-
nomial is called FACTORING. An expression
is said to be factored completely when it has
been separated into its prime factors. The
polynomial 3x 3 - 12x is factored completely as
follows:
3x 3 - 12x = 3x(x - 2)(x + 2)
COMMON FACTORS
Factoring any polynomial begins with the
removal of common factors. Notice that "re-
moval" of a factor does not mean discarding it.
To remove a factor is to insert parentheses and
move the factor outside the parentheses as a
common multiplier. The removal of common
factors proceeds as follows:
1. Inspect the polynomial and find the fac-
tors which are common to all terms. These
common factors, multiplied together, comprise
the "largest common factor."
2. Mentally divide each term of the poly-
nomial by the largest common factor and write
the quotients within a set of parentheses.
3. Write the largest common factor outside
the parentheses as a common multiplier.
For example, the expression x 2 y - xy 2 con-
tains xy as a factor of each term. Therefore,
it is factored as follows:
x 2 y - xy 2 = xy(x - y)
Other examples of factoring by the removal
of common factors are found in the following
expressions:
6m 4 n + 3m 3 n 2 - 3m 2 n 3 = 3m 2 n(2m 2 + mn - n 2 )
-5z 2 - 15z = -5z(z + 3)
7x - 7y + 7z = 7(x - y + z)
In selecting common factors, always remove
as many factors as possible from each term in
order to factor completely. For example, x is
a factor of 3ax 2 - Sax, so that 3ax 2 - Sax is
equal to x(3ax -3a). However, 3 and a are also
factors. Thus the largest common factor is3ax.
When factored completely, the expression is as
follows :
3ax 2 - 3ax = 3ax(x - 1)
Practice problems: Remove the common
factors:
111
i. y - y
4. 6mn 2 + 30m 2 n
22
4. 6mn(n + 5m)
2. a^ - a^b
3. 2b 3 - 8b 2 - 6b
Answers:
1. y(y - i)
2. a 2 b 2 (a - 1)
3. 2b(b 2 - 4b - 3)
LITERAL EXPONENTS
It is frequently necessary to remove com-
mon factors involving literal exponents; that is,
exponents composed of letters rather than num-
bers. A typical expression involving literal
2a
which x
exponents is x + x , in
factor. The factored form is x"(x n + 1).
is a common
An-
Re-
Thus
2a
a n
other example of this type is a m+n +
member that a m+n is equivalent to a m
the factored form is as follows:
a m+n + 2a m = a m a n + 2a m
= a m (a n + 2)
BINOMIAL FORM
The distinctions between monomial, bino-
mial, and trinomial factors are discussed in
detail in chapter 9 of this course. An expres-
sion such as a(x + y) + b(x + y) has a common
factor in binomial form. The factor (x + y) can
be removed from both terms, with the following
result:
a(x + y) + b(x + y) = (x + y)(a + b)
Sometimes it is easier to see this if a single
letter is substituted temporarily for the bino-
mial. Thus, let (x + y) = n, so that a(x + y) +
b(x + y) reduces to (an + bn). The factored
form is n(a + b),, which becomes (x + y)(a + b)
when n is replaced by its equal, (x + y).
Another form of this type is x(y - z) - w(z - y) .
Notice that this expression could be factored
easily if the binomial in the second term were
(y - z). We can show that -w(z - y) is equiva-
lent to +w(y - z), as follows:
-w(z - y) = -w [(-1) (-1) z + (-1) y]
= -w {(-1) [(-1) z + y]}
= (-w)(-l) [-z + y]
= +w(y - z)
Substituting +w(y - z) for -w(z - y) in the origi-
nal expression, we may now factor as follows:
x(y - z) -w(z - y) = x(y - z) + w(y - z)
= (y - z)(x + w)
In factoring an expression such as ax + bx +
ay + by, common monomial factors are re-
moved first, as follows:
ax
+ bx + ay + by = x(a + b) + y(a + b)
Having removed the common monomial factors,
we then remove the common binomial factor to
obtain (a + b)(x + y).
Notice that we could have rewritten the ex-
pression as ax + ay + bx + by, based on the
commutative law of addition, which states that
the sum of two or more terms is the same re-
gardless of the order in which they are ar-
ranged. The first step in factoring would then
produce a(x + y) + b(x + y) and the final form
would be (x + y)(a + b). This is equivalent to
(a + b)(x + y), by the commutative law of multi-
plication, which states that the product of two
or more factors is the same regardless of the
order in which they are arranged.
Practice problems. Factor each of the fol-
lowing:
3a
3x
2a
2. xy 2 + y + x 2 y + x
3. e x + 4e 4x
4. 7(x 2 + y 2 ) - 3z(x 2 + y 2 )
5. a 2 + ab - ac - cb
6. e
- ier 2
7. a x+2 + a 2
8. xy - 3x - 2y + 6
Answers:
1. x 2a (x a + 3)
2. (xy + l)(x + y)
3. e x (l + 4e 3x )
4. (x 2 + y 2 )(7 - 3z)
5. (a + b)(a - c)
6- |er(e - Jr)
7. aV + 1)
8. (y - 3)(x - 2)
112
Chapter 10- FACTORING POLYNOMIALS
BINOMIAL FACTORS
After any common factor has been removed
from a polynomial, the remaining polynomial
factor must be examined further for other fac-
tors. Skill in factoring is principally the ability
to recognize certain types of products such as
the square of a sum or difference. Therefore,
it is important to be familiar with the special
products discussed in chapter 9.
DIFFERENCE OF TWO SQUARES
In chapter 9 we learned that the product of
the sum and difference of two numbers is the
difference of their squares. Thus, (a + b)(a - b)
= a 2 - b 2 . Conversely, if a binomial is the dif-
ference of two squares, its factors are the sum
and difference of the square roots. For exam-
ple, in 9a 2 - 4b 2 both 9a 2 and 4b 2 are perfect
squares. The square roots are 3a and 2b, re-
spectively. Connect these square roots with a
plus sign to get one factor of 9a 2 - 4b 2 and with
a minus sign to get the other factor. The two
binomial factors are 3a - 2band 3a + 2b. There-
fore, factored completely, the binomial can be
written as follows:
9a 2 - 4b 2 = (3a - 2b)(3a + 2b)
We may check to see if these factors are
correct by multiplying them together to see if
their product is the original binomial.
The expression 20x 3 y - 5xy 3 reduces to the
difference of two squares after the common
factor 5xy is removed. Completely factored,
this expression produces the following:
20x 3 y - 5xy 3 = 5xy(4x 2 - y 2 )
= 5xy(2x - y)(2x + y)
Other examples that show the difference of
two squares in factored form are as follows:
49 - 16 = (7 + 4)(7 - 4)
16a 2 - 4x 2 = 4(4a 2 - x 2 )
= 4(2a + x)(2a - x)
4x 2 y - 9y = y(4x 2 - 9)
= y(2x + 3)(2x - 3)
Practice problems: Factor each of the fol-
lowing:
1. a 2 - b 2
2. b 2 - 9
3. a 2 b 2 - 1
4. a 2 - 144
Answers:
1. (a + b)(a - b)
2. (b + 3)(b - 3)
3. (ab + l)(ab - 1)
4. (a + 12)(a - 12)
5. x 2 - y 2
6. y 2 - 36
7. 1 - 4y 2
8. 9a 2 - 16
5. (x + y)(x - y)
6. (y + 6)(y - 6)
7. (1 + 2y)(l - 2y)
8. (3a + 4)(3a - 4)
SPECIAL BINOMIAL FORMS
Special cases involving binomial expressions
are frequently encountered. All such expres-
sions may be factored by reference to general
formulas, but these formulas are beyond the
scope of this course. For our purposes, anal-
ysis of some special cases will be sufficient.
Even Exponents
When the exponents on both terms of the bi-
nomial are even, the expression may be treated
as the sum or difference of two squares. For
example, x 6 - y 6 can be rewritten as (x 3 ) 2 -
(y 3 ) 2 which results in the following factored
form:
x 6 - y 6 = (x 3 - y 3 )(x 3 + y 3 )
In general, a binomial with even exponents
has the form x 2m y 2n , since all even numbers
have 2 as a factor. If the connecting sign is
positive, the expression may not be factorable;
for example, x 2 + y 2 , x 4 + y 4 , and x 8 + y 8 are
all nonfactorable binomials. If the connecting
sign is negative, a binomial with even exponents
is factorable as follows:
,2m
- y 2n = (x ra - y n )(x m + y n )
A special case which is particularly impor-
tant because it occurs so often is the binomial
which has the numeral 1 as one of its terms.
For example, x 4 - 1 is factorable as the differ-
ence of two squares, as follows:
x 4 - 1 = (x 2 - l){x 2 + 1)
= (x - l)(x + l)(x 2 + 1)
113
Odd Exponents
Two special cases involving odd exponents
are of particular importance. These are the
sum of two cubes and the difference of two
cubes. Examples of the sum and difference of
two cubes, showing their factored forms, are
as follows:
x 3 + y 3 = (x + y)(x 2 - xy + y 2 )
x 3 - y 3 = (x - y)(x 2 + xy + y 2 )
Notice that each of these factored forms in-
volves a first degree binomial factor ((x + y)
in the first case and (x - y) in the second). The
connecting sign in the first degree binomial
factor corresponds to the connecting sign in. the
original unfactored binomial.
We are now in a position to give the com-
pletely factored form of x 6 - y 6 , as follows:
x 6 - y 6 = (x 3 - y 3 )(x 3 + y 3 )
= (x - y)(x 2 + xy
(x + y)(x 2 - xy + y 2 )
y 2 )
In general, (x + y) is a factor of (x n + y n ) if
n is odd. If n is even, (x n + y") is not factor-
able unless it can be expressed as the sum of
two cubes. When the connecting sign is nega-
tive, the binomial is always factorable if n is
a whole number greater than 1. That is, (x - y)
is a factor of (x n - y n ) for both odd and even
values of n.
The special case in which one of the terms
of the binomial is the numeral 1 occurs fre-
quently. An example of this is x 3 + 1, which is
factorable as the sum of two cubes, as follows:
x 3 + 1 = (x + l)(x 2 - x + 1)
In a similar manner, 1 + x 6 can be treated
as the sum of two cubes and factored as follows:
1 + x 6 = 1 + (x 2 ) 3
= (1 + x 2 )(l - x 2 + x 4 )
Practice problems. In each of the following
problems, factor completely:
1. x 4 - y 4
2. m 3 + n 3
3. x 6 - y 6
4. x 3 - y 3
5. a 9 - b 9
6. x 2a - y 2b
7. 1 - x 4
8. x 6 + 1
9. 1 - x 3
Answers:
1. (x + y)(x - y)(x 2 + y 2 )
2. (m + n)(m 2 - mn + n 2 )
3. (x + y)(x - y)(x 2 + xy + y 2 )(x 2 - xy + y 2 )
4. (x - y)(x 2 + xy + y 2 )
5. (a - b)(a 2 + ab + b 2 )(a 6 + a 3 b 3 + b 6 )
6. (x a - y b )(x a + y b )
7. (1 + x 2 )(l - x)(l + x)
8. (x 2 + l)(x 4 - x 2 + 1)
9. (1 - x)(l + x + x 2 )
TRINOMIAL SQUARES
A trinomial that is the square of a binomial
is called a TRINOMIAL SQUARE. Trinomials
that are perfect squares factor into either the
square of a sum or the square of a difference.
Recalling that (x + y) 2 = x 2 + 2xy + y 2 and
(x - y) 2 = x 2 - 2xy + y 2 , the form of a trinomial
square is apparent. The first term and the last
term are perfect squares and their signs are
positive. The middle term is twice the product
of the square roots of these two numbers. The
sign of the middle term is plus if a sum has
been squared; it is minus if a difference has
been squared.
The polynomial 16x 2 - 8xy + y 2 is a trino-
mial in which the first term, 16x 2 , and the last
term, y 2 , are perfect squares with positive
signs. The square roots are 4x and y. Twice
the product of these square roots is 2(4x)(y) =
8xy. The middle term is preceded by a minus
sign indicating that a difference has been
squared. In factored form this trinomial is as
follows:
16x 2 - 8xy + y 2 = (4x - y) 2
To factor the trinomial, we simply take the
square roots of the end terms and join them
with a plus sign if the middle term is preceded
by a plus or with a minus if the middle term is
preceded by a minus.
The terms of a trinomial may appear in any
order. Thus, 8xy + y 2 + 16x 2 is a trinomial
square and may be factored as follows:
8xy + y 2 + 16x 2 = 16x 2 + 8xy + y 2 = (4x + y) 2
114
1U U1C CA."
2. 16y 2 + 30x + 9
3. 36 + 12x + x 2
4. a 2 + 2ab + b 2
Answers:
1. (y - 4) 2
2. Not a trinomial
square
3. (6 + x) 2
4. (a * b) 2
6. 4x 2 + y 2 + 4xy
7. 9 - 6cd
c 2 d 2
6.
x 4 + 4x 2 + 4
Not a trinomial
square
(2x + y) 2
1. (3 - cd) 2
8. (x 2 + 2) 2
SUPPLYING THE MISSING TERM
Skill in recognizing trinomial squares may
be improved by practicing the solution of prob-
lems which require supplying a missing term.
For example, the expression y 2 + (?) + 16 can
be made to form a perfect trinomial square by
supplying the correct term to fill the paren-
theses.
The middle term must be twice the product
of the square roots of the two perfect square
terms; that is, (2)(4)(y), or 8y. Check: y 2 + 8y
+ 16 = (y + 4) 2 . The missing term is 8y.
Suppose that we wish to supply the missing
term in 16x 2 + 24xy + (?) so that the three
terms will form a perfect trinomial square.
The square root of the first term is 4x. One-
half the middle term is 12xy. Divide 12xy by4x.
The result is 3y which is the square root of the
last term. Thus, our missing term is 9y 2 .
Checking, we find that (4x + 3y) 2 = 16x 2 +
24xy + 9y 2 .
Practice problems. In each of the following
problems, supply the missing term to form a
perfect trinomial square:
1.x 2 + (?) + y 2
2. t 2 + (?) + 25
3. 9a 2 - (?)
Answers:
1. 2xy
2. lot
3. 30ab
4. 4m 2 + 16m + (?)
5. x 2 + 4x + (?)
25b 2 6. c 2 - 6cd + (?)
4. 16
5. 4
6. 9d 2
pressions of which they are products:
1. (x + 3)(x + 4) = x 2 + 7x + 12
2. (x - 3)(x - 4) = x 2 - 7x + 12
3. (x - 3)(x + 4) = x 2 + x - 12
4. (x + 3)(x - 4) = x 2 - x - 12
It is apparent that trinomials like these may
be factored into binomials as shown. Notice
how the trinomial in each of the preceding ex-
amples is formed. The first term is the square
of the common term of the binomial factors.
The second term is the algebraic sum of their
unlike terms times their common term. The
third term is the product of their unlike terms.
Such trinomials may be factored as the prod-
uct of two binomials if there are two numbers
such that their algebraic sum is the coefficient
of the middle term and their product is the last
term.
For example, let us factor the expression
x 2 - 12x + 32. If the expression is factorable,
there will be a common term, x, in each of the
binomial factors. We begin factoring by placing
this term within each set of parentheses, as
follows:
(x
Next, we must find the other terms that are to
go in the parentheses. They will be two num-
bers such that their algebraic sum is -12 and
their product is +32. We see that -8 and -4
satisfy the conditions. Thus, the following ex-
pression results:
x 2 - 12x + 32 = (x - 8)(x - 4)
It is of value in factoring to note some use-
ful facts about trinomials. If both the second
and third terms of the trinomial are positive,
the signs of the terms to be found are positive
as in example 1 of this section. If the second
term is negative and the last is positive, both
terms to be found will be negative as in exam-
ple 2. If the third term of the trinomial is neg-
ative, one of the terms to be found is positive
and the other is negative as in examples 3 and 4.
Concerning this last case, if the second term is
115
MATHEMATICS, VOLUME 1
positive as in example 3, the positive term in
the factors has the greater numerical value. If
the second term is negative as in example 4,
the negative term in the factors has the greater
numerical value.
It should be remembered that not all trino-
mials are factorable. For example, x 2 + 4x + 2
cannot be factored since there are no two ra-
tional numbers whose product is 2 and whose
sum is 4.
Practice problems. Factor completely, in
the following problems:
1. y 2 + 15y + 50
2. y 2 - 2y - 24
3. x 2 + 8x - 48
4. x 2 - 4x - 60
Answers:
1. (y + 5)(y + 10)
2. (y - 6)(y + 4)
3. (x + 12)(x - 4)
4. (x - 10)(x + 6)
5. x 2 - 12x - 45
6. x 2 - 15x + 56
7. x 2 + 2x - 48
8. x 2 + 14x + 24
5. (x - 15)(x + 3)
6. (x - 7)(x - 8)
7. (x - 6)(x + 8)
8. (x + 12)(x + 2)
Thus far we have considered only those ex-
pressions in which the coefficient of the first
term is 1. When the coefficient of the first
term is other than 1, the expression can be fac-
tored as shown in the following example:
6x 2 - x - 2 = (2x + l)(3x - 2)
Although this result can be obtained by the trial
and error method, the following procedure
saves time and effort. First, find two numbers
whose sum is the coefficient of the second term
(-1 in this example) and whose product is equal
to the product of the third term and the coeffi-
cient of the first term (in this example, (6)(-2)
or -12). By inspection, the desired numbers
are found to be -4 and +3. Using these two
numbers as coefficients for x, we can rewrite
the original expression as 6x 2 - 4x + 3x - 2 and
factor as follows:
6x 2 - 4x + 3x - 2 = 2x(3x - 2) + l(3x - 2)
= (2x + l)(3x - 2)
1. 2x 2 + 13x + 21
2. 16x 2 + 26x + 3
3. 15x 2 - 16x - 7
4. 12x 2 - 8x - 15
1. (2x + 7)(x + 3)
2. (2x + 3)(8x + 1)
3. (3x + l)(5x
4. (6x + 5)(2x
7)
3)
REDUCING FRACTIONS TO
LOWEST TERMS
There are many useful applications of fac-
toring. One of the most important is that of
simplifying algebraic fractions. Fractions that
contain algebraic expressions in the numerator
or denominator, or both, can be reduced to
lower terms, if there are factors common to
numerator and denominator. If the terms of a
fraction are monomials, common factors are
immediately apparent, as in the following ex-
pression:
3x 2 y = 3xy(x) = x_
6xy 3xy(2) 2
If the terms of a fraction are polynomials,
the polynomials must be factored in order to
recognize the existence of common factors , as
in the following two examples:
1.
a - b
a - b
2.
a 2 - 2ab + b 2 (a - b)(a - b) (a - b)
4x 2 - 9 (2x + 3)(2x - 3) (2x + 3)
6x 2 - 9x
3x(2x - 3)
3x
Notice that without the valuable process of fac-
toring, we would be forced to use the fractions
in their more complicated form. When there
are factors common to both numerator and de-
nominator, it is obviously more practical to
cancel them (first using the factoring process)
before proceeding.
Practice problems. Reduce to lowest terms
in each of the following:
1.
12
6x + 12
4.
y 2 - 25
a 2 - b 2
2 ' a 2 - 2ab + b 2
Practice problems,
the following problems:
Factor completely, in 3 V 2 - 14y + 45
y - 8y - 9
6.
' 2 - 8y + 15
a 2 - 5a - 24
a 2 - 64
4x 2 y - 9y
4x 2 + 12x + J
116
Chapter 10- FACTORING POLYNOMIALS
Answers:
! 2 i y + 5
Altrcbra- a + b a a(a + b)
Aiecora. a _ b a . b - (a . b) a
Where possible, the work may be considerably
reduced by factoring, canceling, and then car-
rying out the multiplication, as in the following
example :
x 2 -2x + l x 2 + x-6
11 x + 2 y - 3
a + b R a + 3
a - b a + 8
o y - 5 _ y(2x - 3)
d< y + 1 b> 2x + 3
OPERATIONS INVOLVING FRACTIONS
Addition, subtraction, multiplication, and di-
vision operations involving algebraic fractions
are often simplified by means of factoring,
whereas they would be quite complicated with-
out the use of factoring.
MULTIPLYING FRACTIONS
Multiplication of fractions that contain poly-
nomials is similar to multiplication of fractions
that contain only arithmetic numbers. If this
fact is kept in mind, the student will have little
difficulty in mastering multiplication in algebra.
For instance, we recall that to multiply a frac-
tion by a whole number, we simply multiply the
numerator by the whole number. This is illu-
strated in the following example:
Arithmetic: 4 x 3 _ 12
17 IT
Aln.c.1 i'Q' tv A\ "
x 2 - 9 x 2 - 1
(x - l)j(x 17 j(x-^-3}(x - 2)
4x-K-3)(x - 3) (x + l)j(x -t)
(x - l)(x - 2) x 2 - 3x + 2
~ (x - 3)(x + 1) x 2 - 2x - 3
Although the factors may be multiplied to form
two trinomials as shown, it is usually sufficient
to leave the answer in factored form.
Practice problems. In the following prob-
lems, multiply as indicated:
1 5a 2 3b
1 - ^ * a + b
x + y x - y
"' x 2 x - 1
a 2 + 2ab + b 2 6a
J< a 2 - b 2 3a + 3b
4 a - 1 (a + I) 2
*' 2a 2 + 4a + 2 a - 1
Answers:
, 15a 2 b 2a
Aigebia. U-4; x 2 .. 5 ~ x 2 -5
Sometimes the work may be simplified by fac-
toring and canceling before carrying out the
multiplication. The following example illu-
strates this:
/, pN 3 2&--*}~ 3
*' a + b Jl (a - b)
v 2 - V 2 1
9 . ' . 4
*' X 3 - X 2 *' 2
DIVIDING FRACTIONS
^ a - b > a 2 - 8a + 16 - 1 " (a - 4)(ar- *r
2(3) 6
When the multiplier is a fraction, the rules of
arithmetic remain applicable that is, multiply
numerators together and denominators together.
This is illustrated as follows:
Arithmetic: l-x4= A-
o ID
The rules of arithmetic apply to the division
of algebraic fractions; as in arithmetic, simply
invert the divisor and multiply, as follows:
Arithmetic :
__ _ __
16 " 8 x
J[6
9
= x
117
AigeDra: X - Jy _._ x- - oxy + ay
x + 3y ' x 2 + 7xy
- x - 3y x 2 + 7xy + 12y 2
" x + 3y ' x^ - 6xy + 9y*
- ax. x - t
Practice problems. In the following prob-
lems, divide and reduce to lowest terms:
1.
x-2
1
x 2 + 4x +
4 x 2 -
4
2.
2a - 1 .
a + 1
a 2 + 3
a 3 + 3a
3.
a 3 - 4a 2 H
3a - (a
-3)
a + 2
4.
6t + 12
8t
- 12
9t 2 + 6t -
24 ' 15t
- 20
Answers:
1
(x - 2) 2
o a(a - 1)
4.
a + 2
5
4t - 6
ADDING AND SUBTRACTING
FRACTIONS
The rules of arithmetic for adding and sub-
tracting fractions are applicable to algebraic
fractions. Fractions that are to be combined
by addition or subtraction must have the same
denominator. The numerators are then com-
bined according to the operation indicated and
the result is placed over the denominator. For
example, in the expression
x - 4 2 - llx
x - 2
2 - x
the second denominator will be the same as the
first, if its sign is changed. The value of the
fraction will remain the same if the sign of the
numerator is also changed. Thus, we have the
following simplification:
X. - t __^
x - 2 "*" 2-x ~ x - 2 + -(2 - x)
_ x - 4 llx - 2
~ x - 2 x-2
- x - 4 + llx - 2
x - 2
12x - 6
x-2
- 6(2x -
~ x-2
When the denominators are not the same, we
must reduce all fractions to be added or sub-
tracted to a common denominator and then pro-
ceed.
Consider, for example,
x 2 - 4 + x 2 - 4x - 12
We first must find the least common denomina-
tor (LCD). Remember this is the least number
that is exactly divisible by each of the denomi-
nators. To find such a number, as in arithme-
tic, we first separate each of the denominators
into prime factors. The LCD will contain all of
the various prime factors, each one as many
times as it occurs in any of the denominators.
Factoring, we have
(x + 2){x - 2) T (x - 6)(x + 2)
and the LCD is (x + 2)(x - 2)(x - 6). Rewriting
the fractions with this denominator and adding
numerators, we have the following expression:
4(x - 6)
(x + 2)(x - 2)(x - 6) +
3(x - 2)
+ 2)(x - 2)(x - 6)
4(x - 6) + 3(x - 2)
EC15
4x - 24 + 3x - 6
LCD
7x - 30
= (x + 2)(x - 2)(x - 6)
As another example, consider
4 x + 2
118
Factoring the denominator of the second frac- x - 3 x + 2
tion, we find that the LCD is (x + 3)(x + 1). Re- 3x + 2x
writing the original fractions with the LCD as
denominator, we may now combine the fractions
as follows:
4(x + 1) _ (x + 2)
(x + 3)(x +1) (x + 3)(x + 1) J ' (a + 4) 2 " a(a + 4) + 6(a + 4)
4x + 4 - x - 2 Answers:
" (x + 3)(x + 1)
3x - x2
3x + 2 - (x + 2)(x -
(X + 3)(X + :) 6a + 9
Practice problems. Perform the indicated + " '
operations in each of the following problems: ,, 5
3 ' 6
3X-4X-2 o32
3 2
9 v 1 A 2 - a + a - a
4 ' (a* + l)(a + l)(a - 1)
3a 3 a 2 + IQa - 48
- 9 " 3 - a - 6a(a + 4) 2
119
CHAPTER 11
LINEAR EQUATIONS IN ONE VARIABLE
One of the principal reasons for an intensive
study of polynomials, grouping symbols, fac-
toring, and fractions is to prepare for solving
equations. The equation is perhaps the most
important tool in algebra, and the more skillful
the student becomes in working with equations,
the greater will be his ease in solving problems.
Before learning to solve equations, it is nec-
essary to become familiar with the words used
in the discussion of them. An EQUATION is a
statement that two expressions are equal in
value. Thus,
and
4 + 5 = 9
A = Iw
(Area of a rectangle = length x width)
are equations. The part to the left of the equal-
ity sign is called the LEFT MEMBER, or first
member, of the equation. The part to the right
is the RIGHT MEMBER, or second member, of
the equation.
The members of an equation are sometimes
thought of as corresponding to two weights that
balance a scale. (See fig. 11-1.) This com-
parison is often helpful to students who are
learning to solve equations. It is obvious, in
4 + 5
Figure 11-1. Equation compared to a
balance scale.
the case of the scale, that any change made in
one pan must be accompaniedby an equal change
in the other pan. Otherwise the scale will not
balance. Operations on equations are based on
the same principle. The members must be kept
balanced or the equality is lost.
CONSTANTS AND VARIABLES
Expressions in algebra consist of constants
and variables. A CONSTANT is a quantity
whose value remains the same throughout a
particular problem. A VARIABLE is a quan-
tity whose value is free to vary.
There are two kinds of constantsfixed and
arbitrary. Numbers such as 7, -3, 1/2, and IT
are examples of FIXED constants. Their values
never change. In 5x + 7 = 0, the numbers 0, 5,
and 7, are fixed constants.
ARBITRARY constants can be assigned dif-
ferent values for differentproblems. Arbitrary
constants are indicated by letters quite often
letters at the beginning of the alphabet such as
a, b, c, and d. In
ax + b = 0,
the letters a and b represent arbitrary con-
stants. The form ax + b = represent many
linear equations. If we give a and b particular
values, say a = 5 and b = 7, then these constants
become fixed, for this particular problem, and
the equation becomes
5x + 7 =
A variable may have one value or it may
have many values in a discussion. The letters
at the end of the alphabet, such as x, y, z, and w,
usually are used to represent variables. In
5x + 7, the letter x is the variable. If x = 1,
then
5x + 7 = 5 + 7 = 12
If x = 2, then
5x + 7 = 5(2) + 7 = 10 + 7 = 17
equality
5x + 7 = -23
holds true for just one value of x. The value is
-6, since
5(-6) + 7 = -23
In an algebraic expression, terms that con-
tain a variable are called VARIABLE TERMS.
Terms that do not contain a variable are CON-
STANT TERMS. The expression 5x + 7 con-
tains one variable term and one constant term.
The variable term is 5x, while 7 is the constant
term. In ax + b, ax is the variable term and b
is the constant term.
A variable term often is designated by nam-
ing the variable it contains. In 5x + 7, 5x is the
x-term. In ax + by, ax is the x-term, while by
is the y-term.
DEGREE OF AN EQUATION
The degree of an equation that has not more
than one variable in each term is the exponent
of the highest power to which that variable is
raised in the equation. The equation
3x - 17 =
is a FIRST-DEGREE equation, since x is raised
only to the first power.
An example of a SECOND-DEGREE equa-
tion is
5x - 2 x + 1 = 0.
The equation,
4x 3 -
= 0,
is of the THIRD DEGREE.
The equation,
3x - 2y = 5
is of the first degree in two variables, x and y.
When more than one variable appears in a term,
as in xy = 5, it is necessary to add the expo-
nents of the variables within a term to get the
Graphs are used in many different forms to
give visual pictures of certain related facts.
For example, they are used to show business
trends, production output, continued individual
attainment, and so forth. We find bar graphs,
line graphs, circle graphs, and many other
types, each of which is used for a particular
need. In algebra, graphs are also used to give
a visual picture containing a great deal of in-
formation about equations.
Sometimes many numerical values, when
substituted for the variables of an equation, will
satisfy the conditions of the equation. On a
particular type of graph (which will be explained
fully in chapter 12) several of these values are
plotted (located), and when enough are plotted,
a line is drawn through these points. For each
particular equation a certain type of curve re-
sults. For equations in the first degree in one
or two variables, the resulting shape of the
"curve" is a straight line. Thus, the name
LINEAR EQUATION is derived. Equations of
a higher degree form various other shapes.
The name "linear equation" now applies to
equations of the first degree, regardless of the
number of variables they contain. Chapter 12
shows how an equation may be pictured on a
graph. The purpose and value of graphing an
equation will also be developed.
IDENTITIES
If a statement of equality involves one or
more variables, it may be either an IDENTITY
(identical equation) or a CONDITIONAL EQUA-
TION. An identity is an equality that states a
fact, such as the following examples:
1. 9 + 5 = 14
2. 2n + 5n = 7n
3. 6(x - 3) = 6x - 18
Notice that equation 3 merely shows the fac-
tored form of 6x - 18 and holds true when any
value of x is substituted. For example, if x = 5,
it becomes
6(5-3) = 6(5) - 18
6(2) = 30-18
12 = 12
121
MATHEMATICS, VOLUME 1
If x assumes the negative value -10, this iden-
tity becomes
6(-10-3) = 6(-10)-18
6(-13) = -60-18
-78 = -78
An identity is established when both sides of
the equality have been reduced to the same
number or the same expression. When 5 is
substituted for x, the value of either side of
6(x-3) = 6x - 18 is 12. When -10 is substituted
for x, the value on either side is -78. The fact
that this equality is an identity can be shown
also by factoring the right side so that the
equality becomes
6(x-3) = 6(x-3)
The expressions on the two sides of the equality
are identical.
CONDITIONAL EQUATIONS
A statement such as 2x-l = is an equality
only when x has one particular value. Such a
statement is called a CONDITIONAL EQUA-
TION, since it is true only under the condition
that x = 1/2. Likewise, the equation y - 7 = 8
holds true only if y = 15.
The value of the variable for which an equa-
tion in one variable holds true is a ROOT, or
SOLUTION, of the equation. When we speak of
solving equations in algebra, we refer to condi-
tional equations. The solution of a conditional
equation can be verified by substituting for
the variable its value, as determined by the
solution.
The solution, is correct if the equality re-
duces to an identity. For example, if 1/2 is
substituted for x in 2x - 1 = 0, the result is
1-1 =
= j(an identity)
The identity is established for x =-i since the
t
value of each side of the equality reduces to
zero.
SOLVING LINEAR EQUATIONS
Solving a linear equation in one variable
means finding the value of the variable that
makes the equation true. For example, 11 is
the SOLUTION of x - 7 = 4, since 11-7 = 4.
The number 11 is said to SATISFY the equation.
Basically, the operation used in solving equa-
tions is to manipulate both members, by addi-
tion, subtraction, multiplication, or division
until the value of the variable becomes appar-
ent. This manipulation may be accomplished in
a straightforward manner by use of the axioms
outlined in chapter 3 of this course. These
axioms may be summed up in the following
rule: If both members of an equation are in-
creased, decreased, multiplied, or divided by
the same number, or by equal numbers, the re-
sults will be equal. (Division by zero is ex-
cluded.)
As mentioned earlier, an equation may be
compared to a balance. What is done to one
member must also be done to the other to main-
tain a balance. An equation must always be
kept in balance or the equality is lost. We use
the above rule to remove or adjust terms and
coefficients until the value of the variable is
discovered. Some examples of equations solved
by means of the four operations mentioned in
the rule are given in the following paragraphs.
ADDITION
Find the value of x in the equation
x - 3 = 12
As in any equation, we must isolate the variable
on either the right or left side. In this prob-
lem, we leave the variable on the left and per-
form the following steps:
1. Add 3 to both members of the equation,
as follows:
In effect, we are "undoing" the subtraction indi-
cated by the expression x - 3, for the purpose
of isolating x in the left member.
2. Combining terms, we have
x = 15
122
SUBTRACTION
Find the value of x in the equation
x + 14 = 24
1. Subtract 14 from each member. In effect,
this undoes the addition indicated in the expres-
sion x + 14.
x + 14 - 14 = 24 - 14
2. Combining terms, we have
x = 10
MULTIPLICATION
Find the value of y in the equation
Practice pr oble ms . Solve the following equa-
tions:
1. The only way to remove the 5 so that the
y can be isolated is to undo the indicated divi-
sion. Thus we use the inverse of division, which
is multiplication. Multiplying both members by
5, we have the following:
5(|) . 5(10)
2. Performing the indicated multiplications,
we have
y = 50
DIVISION
Find the value of x in the equation
3x = 15
1. The multiplier 3 may be removed from
the x by dividing the left member by 3. This
must be balanced by dividing the right member
by 3 also, as follows:
3x 15
T = T
2. Performing the indicated divisions, we
have
x = 5
1. m + 2 = 8
2. x - 5 = 11
3. 6x = -48
Answers:
1. m = 6
2. x = 16
3. x = -8
4 JL_ o
*' 14 " *
5. 2n = 5
6. ^y = 6
4. x = 28
5. n = 2^-
6. y = 36
SOLUTIONS REQUIRING MORE
THAN ONE OPERATION
Most equations involve more steps in their
solutions than the simple equations already de-
scribed, but the basic operations remain un-
changed. If the basic axioms are kept well in
mind, these more complicated equations will
not become too difficult. Equations may re-
quire one or all of the basic operations before
a solution can be obtained.
Subtraction and Division
Find the value of x in the following equation:
2x + 4 = 16
1. The term containing x is isolated on the
left by subtracting 4 from the left member.
This operation must be balanced by also sub-
tracting 4 from the right member, as follows:
2x +4-4= 16 -4
2. Performing the indicated operations, we
have
2x = 12
3. The multiplier 2 is removed from the x
by dividing both sides of the equation by 2, as
follows:
2x _ 12
2 " 2
x = 6
123
Addition, Multiplication, and Division
Find the value of y in the following equation:
1. Isolate the term containing y on the left
by adding 4 to both sides, as follows:
-4+4= 11 +4
3y
T
2. Since the 2 will not divide the 3 exactly,
multiply the left member by 2 in order to elim-
inate the fraction. This operation must be bal-
anced by multiplying the right member by 2, as
follows:
to eliminate the fraction. However, notice that
this multiplication cannot be performed on the
first term only; any multiplier which is intro-
duced for simplification purposes must be ap-
plied to the entire equation. Thus each term in
the equation is multiplied by 4, as follows:
+ 4(2x) = 4(12)
3x + 8x = 48
3. Add the terms containing x and then di-
vide both sides by 11 to isolate the x in the left
member, as follows:
llx = 48
48
x =
11
- 2 < 15
3y = 30
3. Divide both members by 3, in order to
isolate the y in the left member, as follows:
3y _ 30
~~ ~~
y = 10
Equations Having the Variable in
More Than One Term
Find the value of x in the following equation:
3x 10
-j- + x = 12 - x
1. Rewrite the equation with no terms con-
taining the variable in the right member. This
requires adding x to the right member to elim-
inate the -x term, and balance requires that we
also add x to the left member, as follows:
3x
~ + 2x = 12
2. Since the 4 will not divide the 3 exactly,
it is necessary to multiply the first term by 4
Practice problems. Solve each of the follow-
ing equations:
1.x- 1={
2.^ + y = 8
3. 4 + 3x = 7
Answers:
1. x = 3/2
2. y = 6
3. x = 28/13
4. 4 - 7x = 9 - 8x
5.-J + 6y = 13
6. -x - 2x = 25 + x
4. x = 5
5. y = 2
6. x = -10
EQUATIONS WITH LITERAL
COEFFICIENTS
As stated earlier, the first letters of the
alphabet usually represent known quantities
(constants), and the last letters represent un-
known quantities (variables). Thus, we usually
solve for x, y, or z.
An equation such as
ax - 8 = bx - 5
has letters as coefficients. Equations with lit-
eral coefficients are solved in the same way as
124
quations with numerical coefficients, except
lat when an operation cannot actually be per-
>rmed, it merely is indicated.
In solving for x in the equation
ax - 8 = bx - 5
ubtract bx from both members and add 8 to
oth members. The result is
ax - bx = 8 - 5
Since the subtraction on the left side cannot
ctually be performed, it is indicated. The
uantity, a - b, is the coefficient of x when
3rms are collected. The equation takes the
(a-b) x = 3
k>w divide both sides of the equation by a-b.
.gain the result can be only indicated. The
olution of the equation is
x =
a-b
In solving for y in the equation
ay + b = 4
ubtract b from both members as follows:
ay = 4 - b
Ividing both members by a, the solution is
4-b
Practice problems. Solve for x in each of
following:
. 3 + x = b
. 4x = 8 + t
Answers:
. x = b - 3
3. 3x + 6m = 7m
4. ax - 2(x + b) = 3a
4. x
3
3a
2b
a - 2
REMOVING SIGNS OF GROUPING
If signs of grouping appear in an equation
they should be removed in the manner indicated
in chapter 9 of this course. For example, solve
the equation
5 = 24 - [x-12(x-2) - 6(x-2)]
Notice that the same expression, x-2, occurs in
both parentheses. By combining the terms con-
taining (x-2), the equation becomes
5 = 24 - [x-18(x-2)]
Next, remove the parentheses and then the
bracket, obtaining
5 = 24 - [x-18x + 36]
= 24 - [36 - 17x]
= 24 - 36 + 17x
= -12 + 17x
Subtracting 17x from both members and then
subtracting 5 from both members, we have
-17x = - 12 - 5
-17x = - 17
Divide both members by -17. The solution is
x = 1
EQUATIONS CONTAINING FRACTIONS
To solve for x in an equation such as
12
first clear the equation of fractions. To do
this, find the least common denominator of the
fractions. Then multiply both sides of the equa-
tion by the LCD.
The least common denominator of 3, 12, 4,
and 2 is 12. Multiply both sides of the equation
by 12. The resulting equation is
8x + x - 12 = 3 + 6x
Subtract 6x from both members, add 12 to both
members, and collect like terms as follows:
9x - 6x = 12 + 3
3x = 15
125
The solution is
x = 5
To prove that x = 5 is the correct solution,
substitute 5 for x in the original equation and
show that both sides of the equation reduce to
the same value. The result of substitution is
2(5) + 5 l - i + 5
T" + 12 " 1 ~T + T
In establishing an identity, the two sides of
the equality are treated separately, and the op-
erations are performed as indicated. Some-
times, as here, fractions occur on both sides of
the equality, and it is desirable to find the least
common denominator for more than one set of
fractions. The same denominator could be used
on both sides of the equality, but this might
make some of the terms of the fractions larger
than necessary.
Proceeding in establishing the identity for
x = 5 in the foregoing equation we obtain
10 5 3 _ 1 10
Tl2~"3~T
7
5
11
3
12
4
28 .
5
11
. 4.
12
12 ~
4
33
li
12 ~
4
11 _
11
4
T
Each member of the equality has the value
11/4 when x = 5. The fact that the equation be-
comes an identity when x is replaced by 5
proves that x = 5 is the solution.
Practice problems. Solve each of the fol-
lowing equations:
A -T' ' -T
2. I- 1= I
2 v 3
Answers:
1. x = 24
2. v = 6
4. =
4x
3. y = 30
4. x = 1/8
GENERAL FORM OF A LINEAR
EQUATION
The expression GENERAL FORM, in mathe-
matics, implies a form to which all expressions
or equations of a certain type can be reduced.
The only possible terms in a linear equation in
one variable are the first-degree term and the
constant term. Therefore, the general form of
a linear equation in one variable is
ax + b =
By selecting various values for a and b, this
form can represent any linear equation in one
variable after such an equation has been simpli-
fied. For example, if a = 7 and b = 5, ax + b =
represents the numerical equation
7x + 5 =
If a = 2m - n and b = p - q, then ax + b = rep-
resents the literal equation
(2m-n)x + p - q =0
This equation is solved as follows:
(2m-n)x + (p-q) - (p-q) = - (p-q)
(2m-n)x = - (p-q)
x = -(p-q) ^ q-p
USING EQUATIONS TO
SOLVE PROBLEMS
To solve a problem, we first translate the
numerical sense of the problem into an equa-
tion. To see how this is accomplished, con-
sider the following examples and their solutions.
EXAMPLE 1: Together Smith and Jones have
$120. Jones has 5 times as much as Smith.
How much has Smith?
SOLUTION:
Step 1. Get the problem clearly in mind.
There are two parts to each problem what is
given (the facts) and what we want to know (the
question). In this problem we know that Jones
has 5 times as much as Smith and together they
have $120. We want to know how much Smith
has.
126
Chapter 11 -LINEAR EQUATIONS IN ONE VARIABLE
Step 2. Express the unknown as a letter.
Usually we express the unknown or number we
know the least about as a letter (conventionally
we use x). Here we know the least about Smith's
money. Let x represent the number of dollars
Smith has.
Step 3. Express the other facts in terms of
the unknown. If x is the number of dollars
Smith has and Jones has 5 times as much, then
5x is the number of dollars Jones has.
Step 4. Express the facts as an equation.
The problem will express or imply a relation
between the expressions in steps 2 and 3.
Smith's dollars plus Jones' dollars equal $120.
Translating this statement into algebraic sym-
bols, we have
x + 5x = 120
Solving the equation for x,
6x = 120
x = 20
Thus Smith has $20.
Step 5. Check: See if the solution satisfies
the original statement of the problem. Smith
and Jones have $120.
$20
(Smith's money)
$100
(Jones' money)
= $120
EXAMPLE 2: Brown can do a piece of work in
5 hr. If Olsen can do it in 4 hr how long will it
take them to do the work together ?
SOLUTION:
Step 1. Given: Brown could do the work in
5 hr. Olsen could do it in 4 hours.
Unknown: How long it takes them to do the
work together.
Step 2. Let x represent the time it takes
them to do the work together.
Step 3. Then -^ is the amount they do to-
gether in 1 hr. Also, in 1 hour Brown does -i of
p
the work and Olsen does -| of the work.
Step 4. The amount done in 1 hr is equal to
the part of the work done by Brown in 1 hr plus
that done by Olsen in 1 hr.
Solving the equation,
20x (-1) -
\x/
20 = 4x + 5x
20 = 9x
20x + 20x
^ = x, or x = 2 hours
y y
They complete the work together in 2^- hours.
y
2 1
Step 5. Check: 2-~x-= = amount Brown does
2 1
2^- x -T = amount Olsen does
20 i] [20 i\ 1 5.__9
9 x 5/\9 X 4/~99~9
Practice problems. Use a linear equation in
one variable to solve each of the following
problems:
1. Find three numbers such that the second is
twice the first and the third is three times as
large as the first. Their sum is 180.
2. A seaman drew $75.00 pay in dollar bills
and five-dollar bills. The number of dollar
bills was three more than the number of five-
dollar bills. How many of each kind did he
draw? (Hint: If x is the number of five-dollar
bills, then 5x is the number of dollars they
represent.)
3. Airman A can complete a maintenance task
in 4 hr. Airman B requires only 3 hr to do the
same work. If they work together, how long
should it take them to complete the job?
Answers:
1. First number is 30.
Second number is 60.
Third number is 90.
2. Number of five-dollar bills is 12.
Number of one-dollar bills is 15.
can be set up if they are related in some way,
even though the relationship may not be one of
equality.
The expression "number sentence" is often
used to describe a general relationship which
may be either an equality or an inequality. If
the number sentence states an equality, it is an
EQUATION; if it states an inequality, it is an
INEQUATION.
ORDER PROPERTIES
OF REAL NUMBERS
The idea of order, or relative rank accord-
ingto size, is based upon two intuitive concepts:
"greater than" and "less than." Mathematicians
use the symbol > to represent "greater than"
and the symbol < to represent "less than." For
example, the inequation stating that 7 is greater
than 5 is written in symbols as follows:
7 > 5
The inequation stating that x is less than 10 is
written as follows:
x < 10
A "solution" of an inequation involving a
variable is any number which may be substi-
tuted for the variable without changing the re-
lationship between the left member and the
right member. For example, the inequation
x < 10 has many solutions. All negative num-
bers zero, and all positive numbers less than 10,
may be substituted for x successfully. These
solutions comprise a set of numbers, called the
SOLUTION SET.
The SENSE of an inequality refers to the
direction in which the inequality symbol points.
For example, the following two inequalities
have opposite sense:
7 > 5
10 < 12
PROPERTIES OF INEQUALITIES
Inequations may be manipulated in accord-
ance with specific operational rules, in a man-
ner similar to that used with equations.
the same sense as the original inequation. The
following examples illustrate this:
1. 5 < 8
5 + 2 < 8 + 2
7 < 10
The addition of 2 to both members does not
change the sense of the inequation.
2. 5 < 8
5 + (-3) < 8 + (-3)
2 < 5
The addition of -3 to both members does not
change the sense of the inequation.
Addition of the same quantity to both mem-
bers is a useful method for solving inequations.
In the following example, 2 is added to both
members in order to isolate the x term on the
left:
x - 2 > 6
x-2 + 2>6 + 2
x> 8
Multiplication
The rule for multiplication is as follows: If
both members of an inequation are multiplied
by the same positive quantity, the sense of the
resulting inequation is the same as that of the
original inequation. This is illustrated as
follows:
1. -3 < -2
2(-3) < 2(-2)
-6 < -4
Multiplication of both members by 2 does not
change the sense of th inequation.
2. 10 < 12
5 < 6
128
Chapter 11- LINEAR EQUATIONS IN ONE VARIABLE
Multiplication of both members by 1/2 does not
;hange the sense of the inequation.
Notice that example 2 illustrates division of
ioth members by 2. Since any division can be
'ewritten as multiplication by a fraction, the
nultiplication rule is applicable to both multi-
lication and division.
Multiplication is used to simplify the solu-
ion of inequations such as the following:
~o~ ^ &
Multiply both members by 3:
3 () > 3(2)
x > 6
ense Reversal
If both sides of an inequation are multiplied
r divided by the same negative number, the
ense of the resulting inequation is reversed,
'his is illustrated as follows:
1. -4 < -2
(-3) (-4) > (-3) (-2)
12 > 6
2. 7 > 5
(-2) (7) < (-2) (5)
-14 < -10
Sense reversal is useful in the solution of an
lequation in which the variable is preceded by
negative sign, as follows:
2 - x < 4
Add -2 to both members to isolate the x term:
2 - x - 2 < 4 - 2
- x < 2
Multiply both members by -1:
x > -2
Practice problems. Solve each of the fol-
lowing inequations:
1. x + 2 > 3
2.
- 1 < 2
Answers:
1. x > 1
2. y < 9
3. 3 - x < 6
4. 4y > 8
3. x > -3
4. y > 2
GRAPHING INEQUALITIES
An inequation such as x > 2 can be graphed
on a number line, as shown in figure 11-2.
The heavy line in figure 11-2 contains all
values of x which comprise the solution set.
Notice that this line continues indefinitely in
the positive direction, as indicated by the arrow
head. Notice also that the point representing
x = 2 is designated by a circle. This signifies
that the solution set does not contain the num-
ber 2.
Figure 11-3 is a graph of the inequation
x 2 > 4. Since the square of any number greater
than 2 is greater than 4, the solution set con-
tains all values of x greater than 2. Further-
more, the solution set contains all values of x
less than -2. This is because the square of any
negative number smaller than -2 is a positive
number greater than 4.
-f
-4-3-2-1 I 234
Figure 11-2. Graph of the inequation x > 2.
-4 -3
-2-1 01 2
Figure 11-3. -Graph of x 2 > 4.
CHAPTER 12
LINEAR EQUATIONS IN TWO VARIABLES
Thus far in this course, discussions of equa-
tions have been limited to linear equations in
one variable. Linear equations which have two
variables are common, and their solution in-
volves extending some of the procedures which
have i Iready been introduced.
RECTANGULAR COORDINATES
An outstanding characteristic of equations in
two variables is their adaptability to graphical
analysis. The rectangular coordinate system,
which was introduced in chapters of this course,
is used in analyzing equations graphically. This
system of vertical and horizontal lines, meeting
each other at right angles and thus forming a
rectangular grid, is often called the Cartesian
coordinate system. It is named after the French
philosopher and mathematician, Rene Descartes,
who invented it.
COORDINATE AXES
The rectangular coordinate system is devel-
oped on a framework of reference similar to
figure 3-2 in chapter 3 of this course. On a
piece of graph paper, two lines are drawn in-
tersecting each other at right angles, as in
figure 12-1. The vertical line is usually labeled
with the capital letter Y and called the Y axis.
The horizontal line is usually labeled with the
capital letter X and called the X axis. The
point where the X and Yaxes intersect is called
the ORIGIN and is labeled with the letter o.
Above the origin, numbers measured along
or parallel to the Y axis are positive; below the
origin they are negative. To the right of the
origin, numbers measured along or parallel to
the X axis are positive; to the left they are
negative.
COORDINATES
A point anywhere on the graph may be lo-
cated by two numbers, one showing the distance
of the point from the Yaxis, and the other show-
ing the distance of the point from the X axis.
K-,+1
Y (+,+)
-8
-i
<O _
S(-8,5) ?-
-6
r 5
U p
2 !
1 1 1 1 III
- 1 ! x
1 I 1 1 1 1 1
1 1 1 1 III
-8 -7 -6 -5 -4 -3 -2 -1 -
1 1 1 1 1 I 1
L1 1 2 345678
R
-2
-3
-4 Q
--5 *
:6
-7
nri-.-i
=8
(+.-1
Figure 12-1. Rectangular coordinate system.
Point P (fig. 12-1) is 6 units to the right of the
Y axis and 3 units above the X axis. We call
the numbers that indicate the position of a point
COORDINATES. The number indicating the
distance of the point measured horizontally
from the origin is the X coordinate (6 in this
example), and the number indicating the dis-
tance of the point measured vertically from the
origin (3 in this example) is the Y coordinate.
In describing the location of a point by means
of rectangular coordinates, it is customary to
place the coordinates within parentheses and
separate them with a comma. The X coordinate
is always written first. The coordinates of
point P (fig. 12-1) are written (6, 3). The co-
ordinates for point Q are (4, -5); for point R,
they are (-5, -2); and for point S, they are
(-8, 5).
Usually when we indicate a point on a graph,
we write a letter and the coordinates of the
point. Thus, in figure 12-1, for point S, we
write S(-8, 5). The other points would ordinarily
130
Chapter 12-LINEAR EQUATIONS IN TWO VARIABLES
written, P(6, 3), Q(4, -5), and R(-5, -2). The
coordinate of a point is often called its ORDI-
VTE and the X coordinate is often called its
3SCISSA.
QUADRANTS
The X and Y axes divide the graph into four
rts called QUADRANTS. In figure 12-1, point
is in quadrant I, point S is in quadrant II, R
in quadrant HI, and Q is in quadrant IV. In
? first and fourth quadrants, the X coordinate
positive, because it is to the right of the
igin. In the second and third quadrant it is
gative, because it is to the left of the origin,
kewise, the Y coordinate is positive in the
'st and second quadrants, being above the
igin; it is negative in the third and fourth
adrants, being below the origin. Thus, we
ow in advance the signs of the coordinates of
point by knowing the quadrant in which the
int appears. The signs of the coordinates in
j four quadrants are shown in figure 12-1.
Locating points with respect to axes is called
jOTTING. As shown with point P (fig. 12-1),
)tting a point is equivalent to completing a
ctangle that has segments of the axes as two
its sides with lines dropped perpendicularly
the axes forming the other two sides. This
the reason for the name "rectangular co-
dinates."
PLOTTING A LINEAR EQUATION
A linear equation in two variables may have
my solutions. For example, in solving the
nation 2x - y = 5, we can find an unlimited
mber of values of x for which there will be a
rresponding value of y. When x is 4, y is 3,
ice (2 x 4) - 3 = 5. When x is 3, y is 1, and
en x is 6, y is 7. When we graph an equa-
n, these pairs of values are considered co-
iinates of points on the graph. The graph of
equation is nothing more than a line joining
* points located by the various pairs of num-
rs that satisfy the equation.
To picture an equation, we first find several
irs of values that satisfy the equation. For
imple, for the equation 2x - y = 5, we assign
reral values to x and solve for y. A conven-
it way to find values is to first solve the
aation for either variable, as follows:
2x - y = 5
-y = -2x + 5
y = 2x - 5
Once this is accomplished, the value of y is
readily apparent when values are substituted
for x. The information derived may be re-
corded in a table such as table 12-1. We then
lay off X and Y axes on graph paper, select
some convenient unit distance for measurement
along the axes, and then plot the pairs of values
found for x and y as coordinates of points on
the graph. Thus, we locate the pairs of values
shown in table 12-1 on a graph, as shown in
figure 12-2 (A).
Table 12-1. Values of x and y in the equation
2x - y = 5.
K v -
_p
i
3
5
R
7
g
Then y = ---
-9
-3
1
5
7
9
11
(8,11)
17,9)
16,7)
(5,5)
.0,1)
(V31
(A) IB)
Figure 12-2. Graph of 2x - y = 5.
Finally, we draw a line joining these points,
as in figure 12-2 (B). It is seen that this is a
straight line; hence the name "linear equation."
Once the graph is drawn, it is customary to
write the equation it represents along the line,
as shown in figure 12-2 (B).
It can be shown that the graph of an equation
is the geometric representation of all the points
whose coordinates satisfy the conditions of the
equation. The line represents an infinite num-
ber of pairs of coordinates for this equation.
For example, selecting at random the point on
the line where x is 2\ and y is and substitut-
^
ing these values in the equation, we find that
they satisfy it. Thus,
'(4) -
= 5
MATHEMATICS, VOLUME 1
If two points that lie on a straight line can
be located, the position of the line is known.
The mathematical language for this is "Two
points DETERMINE a straight line." We now
know that the graph of a linear equation in two
variables is a straight line. Since two points
are sufficient to determine a straight line, a
linear equation can be graphed by plotting two
points and drawing a straight line through these
points. Very often pairs of whole numbers
which satisfy the equation can be found by in-
spection. Such points are easily plotted.
After the line is drawn through two points, it
is well to plot a third point as a check. If this
third point whose coordinates satisfy the equa-
tion lies on the line the graph is accurately
drawn.
X AND Y INTERCEPTS
Any straight line which is not parallel to one
of the axes has an X intercept and a Y inter-
cept. These are the points at which the line
crosses the X and Y axes. At the X intercept,
the graph line is touching the X axis, and thus
the Y value at that point is 0. At the Y inter-
cept, the graph line is touching the Y axis; the
X value at that point is 0.
In order to find the X intercept, we simply
let y = and find the corresponding value of x.
The Y intercept is found by letting x = and
finding the corresponding value of y. For ex-
ample, the line
5x + 3y = 15
crosses the Y axis at (0,5). This may be veri-
fied by letting x = in the equation. The X in-
tercept is (3,0), since x is * when y is 0. Fig-
ure 12-3 shows the Lvne
5x + 3y = 15
graphedi by means of the X and Y intercepts.
EQUATIONS IN ONE VARIABLE
An equation containing only one variable is
easily graphed, since the line it represents lies
parallel to an axis. For example, in
the value of y is
2y = 9
f
H
-5
,5)
(3,o):
5 _
+t
\XJ> V
~Nr
10
X'
Figure 12-3. -Graph of 5x + 3y = 15.
The line 2y = 9 lies parallel to the X axis at a
distance of 4^ units above it. (See fig. 12-4.)
t
Notice that each small division on the graph
paper in figure 12-4 represents one-half unit.
The line 4x + 15 = lies parallel to the Y
axis. The value of x is - =|. Since this value is
4
negative, the line lies to the left of the Y axis
at a distance of 3j units. (See fig. 12-4.)
.-5
2y = 9
X"
Figure 12-4. Graphs of 2y = 9 and 4x + 15 = 0.
From the foregoing discussion, we arrive at
two important conclusions:
1. A pair of numbers that satisfy an equa-
tion are the coordinates of a point on the graph
of the equation.
2. The coordinates of any point on the graph
of an equation will satisfy that equation.
SOLVING EQUATIONS IN
TWO VARIABLES
A solution of a linear equation in two vari-
ables consists of a pair of numbers that satisfy
the equation. For example, x = 2 and y = 1
constitute a solution of
3x - 5y = 1
When 2 is substituted for x and 1 is substituted
for y, we have
3(2) - 5(1) = 1
The numbers x = -3 and y = -2 also form a
solution. This is true because substituting -3
for x and -2 for y reduces the equation to an
identity:
3(-3) -5(-2) = 1
-9 + 10 = 1
1 = 1
Each pair of numbers (x, y) such as (2, 1) or
(-3, -2) locates a point on the line 3x - 5y = 1.
Many more solutions could be found. Any two
numbers that constitute a solution of the equa-
tion are the coordinates of a point on the line
represented by the equation.
Suppose we were asked to solve a problem
such as: Find two numbers such that their sum
is 33 and their difference is 5. We could indi-
cate the problem algebraically by letting x rep-
resent one number and y the other. Thus, the
problem may be indicated by the two equations
x + y = 33
x - y = 5
Considered separately, each of these equations
represents a straight line on a graph. There
are many pairs of values for x and y which sat-
isfy the first equation, and many other pairs
which satisfy the second equation. Our problem
is to find ONE pair of values that will satisfy
BOTH equations. Such a pair of values is said
to satisfy both equations at the same time, or
simultaneously. Hence, two equations for which
we seek a common solution are called SIMUL-
TANEOUS EQUATIONS. ' The two equations,
taken together, comprise a SYSTEM of equa-
tions.
Graphical Solution
If there is a pair of numbers that can be substituted
for x and y in two different equations, the pair form
the coordinates of a point which lies on the graph of
each equation. The only way in which a point can lie
on two lines simultaneously is for the point to be at the
intersection of the lines. Therefore, the graphical
solution of two simultaneous equations involves
drawing their graphs and locating the point at which
the graph lines intersect.
For example, when we graph the equations
x + y = 33 and x - y = 5, as in figure 12-5, we
see that they intersect in a single point. There
is one pair of values comprising coordinates of
that point (19, 14), and that pair of values sat-
isfies both equations, as follows:
x + y = 33
19 + 14 = 33
x - y = 5
19 - 14 = 5
Y
'
k - -
1
s
\
~-f- u
2 "
Si
X
2
Vi t
2
\
*.
k
2
2
V
r
2
Z_
i
^
- ( 19 It*
*\
f
- "I
,
7 L -
. S- -
10--
V
S
\/
_ _ _ _s
*
*
ty
_ 2 - , . -
/
7
s
ul 5
"2*
Xr
J
f
'
f
i
o-
20 ---
-
i*-
--I 1
7
-?'
/
Figure 12-5. Graph of x + y = 33 and x - y = 5.
133
This pair of numbers satisfies each equation.
It is the only pair of numbers that satisfies the
two equations simultaneously.
The graphical method is a quick and simple
means of finding an approximate solution of
two simultaneous equations. Each equation is
graphed, and the point of intersection of the two
lines is read as accurately as possible. A high
degree of accuracy can be obtained but this, of
course, is dependent on the precision with which
the lines are graphed and the amount of accu-
racy possible in reading the graph. Sometimes
the graphical method is quite adequate for the
purpose of the problem.
figure 12-6 shows the graphs of x + y = 11
and x - y = -3. The intersection appears to be
the point (4, 7). Substituting x = 4 and y = 7
into the equations shows that this is the actual
point of intersection, since this pair of num-
bers satisfies both equations.
-57
.10
-5
?>
\X
$
10-
Figure 12-6. Graph of x + y = 11 and x - y = -3.
The equations 7x - 8y = 2 and 4x + 3y = 5
are graphed in figure 12-7. The lines intersect
where y is approximately 1/2 and x is approxi-
mately 5/6.
Practice problems. Solve the following si-
multaneous systems graphically:
1. x + y = 8
x - y = 2
2. 3x + 2y = 12
4x + 5y = 2
Answers:
1. x = 5
y = 3
Addition Method
2. x = 8
y = -6
The addition method of solving systems of
equations is illustrated in the following ex-
ample:
x - y = 2
x + y = 8
2x + = 10
x = 5
The result in the foregoing example is obtained
by adding the left member of the first equation
to the left member of the second, and adding the
right member of the first equation to the right
member of the second.
Having found the value of x, we substitute
this value in either of the original equations to
find the value of y, as follows:
x - y = 2
(5) - y = 2
-y = 2 - 5
-y = -3
y = s
Notice that the primary goal in the addition
method is the elimination (temporarily) of one
of the variables. If the coefficient of y is the
same in both equations, except for its sign,
adding the equations eliminates y as in the
foregoing example. On the other hand, suppose
that the coefficient of the variable which we de-
sire to eliminate is exactly the same in both
equations.
In the following example, the coefficient of x
is the same in both equations, including its sign:
x + 2y = 4
x - 3y = -1
Adding the equations would not eliminate either
x or y. However, if we multiply both members
of the second equation by -1, then addition will
eliminate x, as follows:
134
Chapter 12-LINEAR EQUATIONS IN TWO VARIABLES
r-.
|
1
f . -
C
J
s
^
\
2
(6,5):
2
^
*
\
J
jf
^ -
^
1 1 '
s
4
\
j
%
J
'
f
t-i
J
-1
j
"
1 1
V|
\-
t
^
\
/
|L
^
L
s
(
5
/
'X
?
<
X-
-4
-2
I
*
L. I
i
/
^
C2
J
i:
/
1
,
^
\
v
(-2, -2)
"*
\
^
"2 P "
h
/
s
i
\
\*
/
>
?
<
, s
/
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f r~~
-4-
L &
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v
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Figure 12-7. Graph of 7x - 8y = 2 and 4x + 3y = 5.
x + 2y = 4
-x + 3y = 1
5y ='5
y = i
'he value of x is found by substituting 1 for y
i either of the original equations, as follows:
x + 2(1) = 4
x = 2
As a second example of the addition method,
nd the solution of the simultaneous equations
3x + 2y = 12
4x + 5y = 2
ere both x and y have unlike coefficients. The
^efficients of one of the variables must be
lade the same, except for their signs.
The coefficients of x will be the same except
>r signs, if both members of the first equation
are multiplied by 4 and both members of the
second equation by -3. Then addition will elim-
inate x.
Following this procedure to get the value of
y, we multiply the first equation by 4 and the
second equation by -3, as follows:
12x + 8y = 48
-12x - 15y = -6
-7y = 42
y = -6
Substituting for y in the first equation to get the
value of x, we have
3x + 2(-6) = 12
x + 2(-2) = 4
x - 4 = 4
x = 8
llllo auiULiuu ID uucLn.cu itjgcui ctiLaiiy uy
substituting 8 for x and -6 for y in each of the
original equations, as follows:
1. 3x + 2y = 12
3(8) + 2(-6) = 12
24 - 12 = 12
2. 4x + 5y = 2
4(8) + 5(-6) = 2
32 - 30 = 2
Practice problems. Use the addition method
to solve the following problems:
1. x + y = 24
x - y = 12
2. 5t + 2v = 9
3t - 2v .= -5
Answers:
3. x - 2y = -1
2x + 3y = 12
4. 2x + 7y = 3
3x - 5y = 51
1.
x =
18
3.
X =
3
y =
6
y =
2
2.
t =
1/2
4.
X
12
V =
13
4
y =
-3
Substitution Method
In some cases it is more convenient to use
the substitution method of solving problems. In
this method we solve one equation for one of
the variables and substitute the value obtained
into the other equation. This eliminates one of
the variables, leaving an equation in one un-
known. For example, find the solution of the
following system:
4x + y = 11
x + 2y = 8
It is easy to solve for either y in the first equa-
tion or x in the second equation. Let us solve
for y in the first equation. The result is
y - 11 - 4x
we may substitute this value of y wherever y
appears in the second equation. Thus,
x + 2(11 - 4x) = 8
We now have one equation that is linear in x;
that is, the equation contains only the variable x.
Removing the parentheses and solving for x,
we find that
x + 22 - 8x = 8
-7x = 8 - 22
-7x = -14
x = 2
To get the corresponding value of y, we sub-
stitute x = 2 in y = 11 - 4x. The result is
y = 11 -4(2)
= 11-8
n
Thus, the solution for the two original equa-
tions is x = 2 and y = 3.
Practice problems. Solve the following sys-
tems by the substitution method:
1. 2x - 9y = 1
x - 4y = 1
2. 2x + y =
2x - y = 1
Answers:
3. 5r + 2s = 23
4r + s = 19
4. t - 4v = 1
2t - 9v = 3
1. x = 5
3. r = 5
y = i
s = -1
2. x = 1/4
4. t = -3
y = -1/2
v = -1
Literal Coefficients
Simultaneous equations with literal coeffi-
cients and literal constants may be solved for
the value of the variables just as the other
equations discussed in this chapter, with the
exception that the solution will contain literal
136
lumbers. For example, find the solution of the
system:
3x + 4y = a
4x + 3y = b
Ve proceed as with any other simultaneous
Lnear equation. Using the addition method, we
my proceed as follows: To eliminate the y
erm we multiply the first equation by 3 and the
econd equation by -4. The equations then
ecome
9x + 12y = 3a
-16x - 12y = -4b
-7x
x
= 3a - 4b
= 3a - 4b
-7
v _ 4b - 3a
x _
To eliminate x, we multiply the first equa-
on by 4 and the second equation by -3. The
juations then become
12x + 16y = 4a
-12x - 9y = -3b
7y = 4a - 3b
4a - 3b
y 7
We may check in the same manner as that
sed for other equations, by substituting these
ilues in the original equations.
INTERPRETING EQUATIONS
Recall that the general form for an equation
i the first degree in one variable is ax + b = 0.
he general form for first-degree equations in
ro variables is
ax + by + c = 0.
is interesting and often useful to note what
ippens graphically when equations differ, in
jrtain ways, from the general form. With this
formation, we know in advance certain facts
mcerning the equation in question.
LINES PARALLEL TO THE AXES
If in a linear equation the y term is miss-
ing, as in
2x - 15 =
the equation represents a line parallel to the Y
axis and 7^ units from it. Similarly, an equa-
*
tion such as
4y - 9 =
which has no x term, represents a line paral-
lel to the X axis and 2j units from it. (See
fig. 12-8.)
The fact that one of the two variables does
not appear in an equation means that there are
no limitations on the values the missing vari-
able can assume. When a variable does not ap-
pear, it can assume any value from zero to
plus or minus infinity. This can happen only if
the line represented by the equation lies paral-
lel to the axis of the missing variable.
Lines Passing Through the Origin
A linear equation, such as
4x + 3y =
that has no constant term, represents a line
passing through the origin. This fact is obvi-
ous since x = 0, y = satisfies any equation not
having a constant term. (See fig. 12-8.)
Lines Parallel to Each Other
An equation such as
3x - 2y = 6
has all possible terms present. It represents
a line that is not parallel to an axis and does
not pass through the origin.
Equations that are exactly alike, except for
the constant terms, represent parallel lines.
As shown in figure 12-8, the lines represented
by the equations
3x - 2y = -18 and 3x - 2y = 6
are parallel.
137
-9-0-
-10r
1-5
10
"S/
II
2
r-5
Figure 12-8. Interpreting equations.
Parallel lines have the same slope. Chang-
ing the constant term moves a line away from
or toward the origin while its various positions
remain parallel to one another. Notice in fig-
ure 12-8 that the line 3x - 2y = 6 lies closer to
the origin than 3x - 2y = -18. This is revealed
at sight for any pair of lines by comparing their
constant terms. That one which has the constant
term of greater absolute value will lie farther
from the origin. In this case 3x - 2y = -18 will
be farther from the origin since |-18l > |6|.
The fact that lines are parallel is indicated
by the result when we try to solve two equations
such as 3x - 2y = -18 and 3x - 2y = 6 simultane-
ously. Subtraction eliminates both x and y im-
mediately. If both variables disappear, we can-
find values for them such that both equations
are satisfied at the same time. This means that
there is no solution. No solution implies that
there is no point of intersection for the straight
lines represented by the equations. Lines that
do not intersect in the finite plane are parallel.
USING TWO VARIABLES IN
SOLVING WORD PROBLEMS
Many problems can be solved quickly and
easily using one equation with one variable.
Other problems that might be rather difficult to
solve in terms of one variable can easily be
solved using two equations and two variables.
The difference in the two methods is shown in
the following example, solved first by using one
variable and then using two.
138
EXAMPLE: Find the two numbers such that
half the first equals a third of the second and
twice their sum exceeds three times the second
by 4.
SOLUTION USING ONE VARIABLE:
1. Let x = the first number.
x 1
2. Then^ = ~ of the second number.
2 o
OY
3. Thus ^ = the second number.
Ct
From the statement of the problem, we then
have
3x\
2)
+ 4
2x
, 3x = f + 4
lOx = 9x + 8
x = 8
52 = 12
(first number)
(second number)
SOLUTION USING TWO VARIABLES:
If we let x and y be the first and second num-
bers, respectively, we can write two equations
almost directly from the statement of the prob-
lem. Thus,
2. 2(x + y) = 3y + 4
Solving for x in the first equation and sub
stituting this value in the second, we have
2 (f * y) - 3y * 4
4| + 2y = 3y + 4
4y + 6y = 9y + 12
y = 12
x 12
2 " 3
x = 8
(second number)
(first number)
Thus, we see that the solution using two vari-
ables is more direct and simple. Often it would
require a great deal of skill to manipulate a
problem so that it might be aolved using one
variable; whereas the solution using two vari-
ables might be very simple. The use of two
variables, of course, involves the fact that the
student must be able to form two equations
from the information given in the problem.
Practice problems. Solve the following prob-
lems using two variables:
1. A Navy tug averages 12 miles per hour down-
stream and 9 miles per hour upstream. How
fast is the stream flowing?
2. The sum of the ages of two boys is 18. If 4
times the younger boy's age is subtracted from
3 times the older boy's age, the difference is
12. What are the ages of the two boys?
Answers:
1. 4 mph.
i
2. 6 years and 12 years.
INEQUALITIES IN TWO VARIABLES
Inequalities in two variables are of the fol-
lowing form:
x + y > 2
Many solutions of such an inequation are ap-
parent immediately. For example, x could have
the value 2 and y could have the value 3, since
2 + 3 is greater than 2.
The existence of a large number of solutions
suggests that a graph of the inequation would
contain many points. The graph of an inequa-
tion in two unknowns is, in fact, an entire area
rather than just a line.
PLOTTING ON THE
COORDINATE SYSTEM
It would be extremely laborious to plot
enough points at random to define an entire
area of the coordinate system. Therefore our
method consists of plotting a boundary line and
shading the area, on one side of this line,
wherein the solution points lie.
The equation of the boundary line is formed
by changing the inequation to an equation. For
139
example, the equation of the boundary line for
the graph of
x + y > 2
is the equation
x + y
= 2
Figure 12-9 is a graph of x + y > 2. Notice
that the boundary line x + y = 2 is not solid.
This is intended to indicate that points on the
boundary line are not members of the solution
set. Every point lying above and to the right of
the boundary line is a member of the solution
set. Any solution point may be verified by sub-
stituting its X and Y coordinates for x and y in
the original inequation.
SIMULTANEOUS INEQUALITIES
The areas representing the solutions of two
different inequations may overlap. If such an
overlap occurs, the area of the overlap includes
all points whose coordinates satisfy both in-
equations simultaneously. An example of this
is shown in figure 12-10, in which the following
two inequations are graphed:
x + y > 2
x - y > 2
Figure 12-9. -Graph of x + y > 2.
Figure 12- 10. -Graph of x + y > 2 and x - y > 2.
The double crosshatchedarea in figure 12-10
contains all points which comprise the solution
set for the system.
140
CHAPTER 13
RATIO, PROPORTION, AND VARIATION
The solution of problems based on ratio,
roportion, and variation involves no new prin-
iples. However, familiarity with these topics
111 often lead to quick and simple solutions to
roblems that would otherwise be more com-
licated.
RATIO
The results of observation or measurement
ften must be compared with some standard
alue in order to have any meaning. For ex-
tnple, to say that a man can read 400 words
er minute has little meaning as it stands,
however, when his rate is compared to the 250
ords per minute of the average reader, one
an see that he reads considerably faster than
le average reader. How much faster? To
nd out, his rate is divided by the average
ite, as follows:
400
250
hus, for every 5 words read by the average
sader, this man reads 8. Another way of mak-
3
ig this comparison is to say that he reads lg
mes as fast as the average reader.
When the relationship between two numbers
i shown in this way, they are compared as a
ATIO. A ratio is a comparison of two like
lantities. It is the quotient obtained by divid-
ig the first number of a comparison by the
;cond.
Comparisons may be stated in more than
ie way. For example, if one gear has 40 teeth
id another has 10, one way of stating the com-
xrison would be 40 teeth to 10 teeth. This
smparison could be shown as a ratio in four
ays as follows:
1. 40:10
2. 40 * 10
3. 40
10
4. The ratio of 40 to 10.
When the emphasis is on "ratio," all of these
expressions would be read, "the ratio of 40 to
10." The form 40 - 10 may also be read "40
40
divided by 10." The form -rr- may also be read
"40 over 10."
Comparison by means of a ratio is limited
to quantities of the same kind. For example, in
order to express the ratio between 6 ft and 3 yd,
both quantities must be written in terms of the
same unit. Thus the proper form of this ratio
is 2 yd : 3 yd, not 6 f t : 3 yd. When the parts of
the ratio are expressed in terms of the same
unit, the units cancel each other and the ratio
consists simply of two numbers. In this exam-
ple, the final form of the ratio is 2 : 3.
Since a ratio is also a fraction, all the rules
that govern fractions may be used in working
with ratios. Thus, the terms may be reduced,
increased, simplified, and so forth, according
to the rules for fractions. To reduce the ratio
15:20 to lowest terms, write the ratio as a
fraction and then proceed as for fractions.
Thus, 15:20 becomes
15
20
4
Hence the ratio of 15 to 20 is the same as the
ratio of 3 to 4.
3
Notice the distinction in thought between ^
3
as a fraction and 7 as a ratio. As a fraction we
o
think of j as the single quantity "three-fourths."
3
As a ratio, we think of ^ as a comparison be-
tween the two numbers, 3 and 4. For example,
9
the lengths of two sides of a triangle are Ij-g ft
and 2 ft. To compare these lengths by means
of a ratio, divide one number by the other and
reduce to lowest terms, as follows:
i JL 25
116 _ 16 25
2 ~ 2 ~ 32
MATHEMATICS, VOLUME 1
The two sides of the triangle compare as 25
to 32.
INVERSE RATIO
It is often desirable to compare the numbers
of a ratio in the inverse order. To do this, we
simply interchange the numerator and the de-
nominator. Thus, the inverse of 15:20 is 20:15.
When the terms of a ratio are interchanged, the
INVERSE RATIO results.
Practice problems. In problems 1 through 6,
write the ratio as a fraction and reduce to low-
est terms. In problems 7 through 10, write the
inverse of the given ratio.
1. The ratio of 5 Ib to 15 Ib
2. $16 : $12
3. 18+4
4. One quart to one gallon
5. 5x to lOx
6. sj
4|
7. The ratio of 6 ft to 18 ft
members are ratios. In other words when two
ratios are set equal to each other, a proportion
is formed. The proportion may be written in
three different ways as in the following ex-
amples:
15:20 : : 3:4
15:20 = 3:4
15 _3
20 4
The last two forms are the most common. All
these forms are read, "15 is to 20 as 3 is to 4."
In other words, 15 has the same ratio to 20 as
3 has to 4.
One reason for the extreme importance of
proportions is that if any three of the terms
are given, the fourth may be found by solving a
simple equation. In science many chemical and
physical relations are expressed as propor-
tions. Consequently, a familiarity with propor-
tions will provide one method for solving many
applied problems. It is evident from the last
15 3
form shown, on = j> that a proportion is really
a fractional equation. Therefore, all the rules
for fraction equations apply.
9. 5 : 8
10. 15 to 21
Answers:
9 i
2 ' 3
i 4
3. --
' 27
7
' 1
*!
PROPORTION
Closely allied with the study of ratio is the
subject of proportion. A PROPORTION is
nothing more than an equation in which the
TERMS OF A PROPORTION
Certain names have been given to the terms
of the two ratios that make up a proportion. In
a proportion such as 3:8 = 9:24, the first and
the last terms (the outside terms) are called
the EXTREMES. In other words, the numerator
of the first ratio and the denominator of the
second are called the extremes. The second
and third terms (the inside terms) are called
the MEANS. The means are the denominator of
the first ratio and the numerator of the second.
In the example just given, the extremes are 3
and 24; the means are 8 and 9.
Four numbers, such as 5, 8, 15, and 24, form
a proportion if the ratio of the first two in the
order named equals the ratio of the second two.
When these numbers are set up as ratios with
the equality sign between them, the members
will reduce to an identity if a true proportion
exists. For example, consider the following
proportion:
5 _ 15
8 ~ 24
15 5
n this proportion, =7 must reduce to -g for the
iroportion to be true. Removing the same fac-
15
or from both members of we have
5 _ 3(5)
8 ~ 378]
The number 3 is the common factor that
aust be removed from both the numerator and
tie denominator of one fraction in order to show
tiat the expression
15
24
3 a true proportion. To say this another way,
: is the factor by which both terms of the ratio
must be multiplied in order to show that this
atio is the same as 57
Practice problems. For each of the follow-
ig proportions, write the means, the extremes,
nd the factor of proportionality.
JL - !>
' 16 " 80
o 25 1
d< 75 = 3
4. 12:3 :: 4:1
. 4:5 = 12:15
Answers:
. Means: 16 and 15
Extremes: 3 and 80
Factor of proportionality: 5
. M: 5 and 12
E: 4 and 15
FP: 3
. M: 75 and 1
E: 25 and 3
FP: 25
. M: 3 and 4
E: 12 and 1
FP: 3
PERATIONS OF PROPORTIONS
It is often advantageous to change the form
E a proportion. There are rules for changing
or combining the terms of a proportion without
altering the equality between the members.
These rules are simplifications of fundamental
rules for equations; they are not new, but are
simply adaptations of laws or equations pre-
sented earlier in this course.
Rule 1. In any proportion, the product of the
means equals the product of the extremes.
This is perhaps the most commonly used
rule of proportions. It provides a simple way
to rearrange a proportion so that no fractions
are present. In algebraic language the rule is
illustrated as follows:
a
b
c
cf
be = ad
To prove this rule, we note that the LCD of the
two ratios -| and ^ is bd. Multiplying both mem-
bers of the equation in its original form by this
LCD, we have
bd = bd
ad = be
d
The following numerical example illustrates
the simplicity of rule 1:
3 _9_
8 = 24
8(9) = 3(24)
If one of the terms of a proportion is a vari-
able to the first power as in
7:5 = x:6
the proportion is really a linear equation in one
variable. Such an equation can be solved for
the unknown.
Equating the products of the means and ex-
tremes produces the following:
5x = 42
'-I
Mean Proportional
When the two means of a proportion are the
same quantity, that quantity is called the MEAN
143
In the proportion
fL
x
JC
c
x is the mean proportional between a and c.
Rule 2. The mean proportional between two
quantities is the square root of their product.
This rule is stated algebraically as follows:
a _ x
x ~ c
X = '/llC
To prove rule 2, we restate the proportion
and apply rule 1, as follows:
=
x ~ c
x 2 = ac
x =
ac
Rule 2 is illustrated by the following nu
merical example:
2 _ JL
8 ~ 32
8 = \T2f32)
8 = \T64
d
c
_b
a
Note that the product of the means and the prod-
uct of the extremes still yield the same equal-
ity as in the original proportion.
The inversion relationship may be illustrated
by the following numerical example:
Therefore,
Alternation
10
16
16
10
The four selected numbers (a, b, c, and d)
are in proportion by ALTERNATION in the fol-
lowing form:
a
c
b
d
To prove the alternation relationship, first
multiply both sides of the original proportion
by , as follows:
C
OTHER FORMS FOR PROPORTIONS
If four numbers, for example, a, b, c, and d,
form a proportion, such as
a _c
b = d
they also form a proportion according to other
arrangements.
Inversion
The four selected numbers are in proportion
by INVERSION in the form
b _ d
a c
The inversion relationship is proved as fol-
lows, by first multiplying both members of the
original proportion by :
ciC
a. _
b ~ d
cb
b/c
c\d
3, D
c = d
The following numerical example illustrates
alternation:
10
16
Therefore,
10 ~ 16
SOLVING PROBLEMS BY
MEANS OF PROPORTION
One of the most common types of problems
based on proportions involves triangles with
144
roportional sides. Suppose that the corre-
ponding sides of two triangles are known to be
iroportional. (See fig. 13-1.) The lengths of
tie sides of one triangle are 8, 9, and 11. The
ength of the side of the second triangle corre-
ponding to side 8 in the first triangle is 10.
/e wish to find the lengths of the remaining
ides, b and c.
8 10
Figure 13-1. Triangles with corresponding
sides proportional.
Since the corresponding sides are propor-
onal, the pairs of corresponding sides may be
sed to form proportions as follows:
JL
10
b
11
c
__ _ 11
10 c
Si
b
To solve for b, we use the proportion
JL
10
i.
b
id obtain the following result:
8b = 90
4b = 45
The solution for c is similar to that for b,
sing the proportion
_8_
10
11
c
with the following result:
8c = 110
c = 13|
The sides of the second triangle are 10, 11 7,
3
and ISj. The result can also be obtained by
using the factor of proportionality. Since 8 and
10 are lengths of corresponding sides, we can
write
8k = 10
10 5
k =
8
The factor of proportionality is thus found to
bef.
Multiplying any side of the first triangle by
7 gives the corresponding side of the second
triangle, as follows:
b = 9 a = n
" & \ A I A *- J. .
\4/ 4 4
c -
C ~
55
4
Proportional sides of similar triangles may
be used to determine the height of an object by
measuring its shadow. (See fig. 13-2.)
20 FT.
16 FT.
12 FT.
Figure 13-2. Measuring height by
shadow length.
In figure 13-2, mast AC casts a shadow 20 ft.
long (AB). At the same time, DF (12 ft. long) casts a
shadow of 16 ft. long (DE). Assuming that both masts
are vertical and on level ground, triangle ABC is similar
to triangle DBF and their corresponding sides are
therefore proportional. Thus the height of AC may be
found as follows:
145
AC _
12 ~
AC =
20
16
= 15
Practice problems. In each of the following
problems, set up a proportion and then solve
for the unknown quantity:
1. Referring to figure 13-1, if the shortest side
of the larger triangle is 16 units long, rather
than 10, how long is side c ?
2. If a mast 8 ft high casts a shadow 10 ft long,
how high is a mast that casts a shadow 40 ft
long?
Answers:
1.
_8_
16 =
8c =
11
c
_
c
c = 22
Word Problems
JL A
10 40
(8)(40) _
10
h = 32
A knowledge of proportions often provides a
quick method of solving word problems. The
following problem is a typical example of the
types that lend themselves to solution by means
of proportion.
If an automobile runs 36 mi on 2 gal of gas,
how many miles will it run on 12 gal? Com-
paring miles to miles and gallons to gallons,
we have
36:x = 2:12
Rewriting this in fraction form, the solution is
as follows:
36 J_
x ~ 12
2x = 12(36)
x = 6(36)
= 216 mi
Practice problems. In each of the following
problems, first set up a proportion and then
solve for the unknown quantity:
1. The ratio of the speed of one aircraft to that
of another is 2 to 5. If the slower aircraft has
a speed of 300 knots, what is the speed of the
faster aircraft?
2. If 6 seamen can empty 2 cargo spaces in 1
day, how many spaces can 150 seamen empty in
1 day?
3. On a map having a scale of 1 in. to 50 mi,
how many inches represent 540 mi?
Answers:
1. 750 kt
2. 50
VARIATION
3. 10.8 in.
When two quantities are interdependent,
changes in the value of one may have a predict-
able effect on the value of the other. Variation
is the name given to the study of the effects of
changes among related quantities. The three
types of variation which occur frequently in the
study of scientific phenomena are DIRECT,
INVERSE, and JOINT.
DIRECT VARIATION
An example of direct variation is found in
the following statement: The perimeter (sum
of the lengths of the sides) of a square in-
creases if the length of a side increases. In
everyday language, this statement might be-
come: The longer the side, the bigger the
square. In mathematical symbols, using p for
perimeter and s for the length of the side, the
relationship is stated as follows:
p = 4s
Since the number 4 is constant, any varia-
tions which occur are the results of changes in
p and s. Any increase or decrease in the size
of s results in a corresponding increase or de-
crease in the size of p. Thus p varies in the
same way (increasing or decreasing) as s. This
explains the terminology which is frequently
used: p varies directly as s.
In general, if a quantity can be expressed in
terms of a second quantity multiplied by a con-
stant, it is said to VARY DIRECTLY AS the
second quantity. For example if x and y are
variables and k is a constant, x varies directly
as y, if x = ky. Thus, as y increases x increases,
146
and as y decreases, x decreases. There is a
direct effect on x caused by any change in y.
The fact that x varies as y is sometimes in-
dicated by x a, y ? or x ~ y. However, it is usu-
ally written in the form x = ky.
The relationship x = ky is equivalent to
^ = k. If one quantity varies directly as a sec-
ond quantity, the ratio of the first quantity to
the second quantity is a constant. Thus, what-
ever the value of x, where it is divided by y,
the result will always be the same value, k.
A quantity that varies directly as another
quantity is also said to be DIRECTLY PRO-
PORTIONAL to the second quantity. In x = ky,
the coefficient of x is 1. The relationship x = ky
can be written in proportion form as
- I
~
or
k_
x
J.
y
Notice that the variables, x and y, appear
either in the numerators or in the denominators
of the equal ratios. This implies that x and y
are directly proportional. The constant, k, is
the CONSTANT OF PROPORTIONALITY.
Practice problems. Write an equation show-
ing the stated relationship, in each of the fol-
lowing problems:
1. The cost, C of a dozen wrenches varies di-
rectly as the price, p, of one wrench.
2. X is directly proportional to Y (use k as the
constant of proportionality).
3. The circumference, C, of a circle varies
directly as its diameter, d (use n as the con-
stant of proportionality).
In the following problems, based on the formula
p = 4s, find the appropriate word or symbol to
fill the blank.
4. When s is doubled, p will be.
5. When s is halved, p will be_
Answers:
1. C = 12p
2. X = kY
3. C = Trd
4. doubled
5. halved
6. p
Variation as the Power of a Quantity
Another form of direct variation occurs
when a quantity varies as some power of an-
other. For example, consider the formula
A = ?rr 2
Table 13-1 shows the values of r and the cor-
responding values of A.
Table 13-1. Relation between values of
radius and area in a circle.
When r =
1
2
3
4
5
7
9
Then A =
7T
4?r
97T
167T
25n
497T
-817T
Notice how A changes as a result of a change
in r. When r changes from 1 to 2, A changes
from TT to 4 times ir or 2 2 times it. Likewise
when r changes from 3 to 4, A changes not as
r, but as the SQUARE of r. In general, one
quantity varies as the power of another if it is
equal to a constant times that quantity raised
to the power. Thus, in an equation such as
x = ky n , x varies directly as the n th power of y.
As y increases, x increases but more rapidly
than y, and as y decreases, x decreases, but
again more rapidly.
Practice problems.
1. In the formula V = e 3 , how does V vary?
2. In the formula A=s 2 , ifsis doubled how
much is A increased?
et 2
3. In the formula s = ^-, g is a constant. If t
is halved, what is the resulting change in s ?
Answers:
1. Directly as the cube of e.
2. It is multiplied by 4.
. ,. ,, ,. , , 3. It is multiplied by -r.
.is directly proportional to s. " J 4
147
INVERSE VARIATION
A quantity VARIES INVERSELY as another
quantity if the product of the two quantities is a
constant. For example, if x and y are variables
and k is a constant, the fact that x varies in-
versely as y is expressed by
xy = k
or
X
If values are substituted for x and y, we see
that as one increases, the other must decrease,
and vice versa. Otherwise, their product will
not equal the same constant each time.
If a quantity varies inversely as a second
quantity, it is INVERSELY PROPORTIONAL to
the second quantity. In xy = k, the coefficient
of k is 1. The equality xy = k can be written in
the form
x
k
1
y
or
Notice that when one of the variables, x or
y, occurs in the numerator of a ratio, the other
variable occurs in. the denominator of the sec-
ond ratio. This implies that x and y are in-
versely proportional.
Inverse variation may be illustrated by
means of the formula for area of a rectangle.
If A stands for area, L for length, and W for
width, the expression for the area of a rec-
tangle in terms of the length and width is
A = LW
Suppose that several rectangles, all having the
same area but varying lengths and widths, are
to be compared. Then LW = A has the same
form as xy = k, where A and k are constants.
Thus L is inversely proportional to W, and W
is inversely proportional to L.
If the constant area is 12 sq ft, this rela-
tionship becomes
LW = 12
If the length is 4 ft, the width is found as fol-
lows:
W = ^=^ = 3ft
If the length increases to 6 ft, the width de-
creases as follows:
W = ^ = 2 ft
If a constant area is 12, the width of a rec-
tangle decreases from 3 to 2 as the length in-
creases from 4 to 6. When two inversely pro-
portional quantities vary, one decreases as the
other increases.
Another example of inverse variation is
found in the study of electricity. The current
flowing in an electrical circuit at a constant
potential varies inversely as the resistance of
the circuit. Suppose that the current, I, is 10
amperes when the resistance, R, is 11 ohms
and it is desired to find the current when the
resistance is 5 ohms.
Since I and R vary inversely, the equation
for the relationship is IR = k, where k is the
constant voltage. Therefore, (10)(11) = k. Also,
when the resistance changes to 5 ohms, (5)(I) = k.
Quantities equal to the same quantity are equal
to each other, so we have the following equation:
51 = (10)(11)
. 22
D
The current is 22 amperes when the resistance
is 5 ohms. As the resistance decreases from
11 to 5 ohms, the current increases from 10 to
22 amperes.
One type of variation problem which tends to
be confusing to the beginner involves rates of
speed or rates of doing work. For example, if
1 men can complete a job in 20 days, how long
will 50 men require to complete the same job?
The strictly mechanical approach to this prob-
lem might result in the following false solution,
relating men to men and days to days:
7 men _ 20 days
50 men ~ T
However, a little thought brings out the fact
that we are dealing with an INVERSE relation-
ship rather than a direct one. In other words,
148
Chapter 13 -RATIO, PROPORTION, AMD VARIATION
the more men we have, the less time is re-
quired. Therefore, the correct solution re-
quires that we use an inverse proportion; that
is, we must invert one of the ratios as follows:
J7_
50
T =
_
20
= 2 | days
Practice problems. In problems 1 and 2,
express the given data as a proportion, using k
as the constant of proportionality.
1. The rate, r, at which a vessel travels in
going a certain distance varies inversely as the
time, t.
2. The volume, V, of a gas varies inversely as
the pressure, p.
3. A ship moving at a rate of 15 knots requires
10 hr to travel a certain distance. If the speed
is increased to 25 knots, how long will the ship
require to travel the same distance ?
Answers:
3. 6hr
JOINT VARIATION
A quantity VARIES JOINTLY as two or more
quantities, if it equals a constant times their
product. For example, if x, y, and z are vari-
ables and k is a constant, x varies jointly as
y and z, if x = kyz. Note that this is similar to
direct variation, except that there are two var-
iable factors and the constant with which to
contend in the one number; whereas in direct
variation, we had only one variable and the
constant. The equality, x = kyz, is equivalent to
[f a quantity varies jointly as two or more other
quantities, the ratio of the first quantity to the
product of the other quantities is a constant.
The formula for the area of a rectangle is
an example of joint variation. If A is allowed
to vary, rather than being constant as in the
example used earlier in this chapter, then A
varies jointly as L and W. When the formula is
written for general use, it is not commonly ex-
pressed as A = kLW, although this is a mathe-
matically correct form. Since the constant of
proportionality in this case is 1, there is no
practical need for expressing it.
Using the formula A = LW, we make the fol-
lowing observations: If L = 5 and W = 3, then
A = 3(5) = 15. If L = 5 and W = 4, then A =
4(5) = 20, and so on. Changes in the area of a
rectangle depend on changes in either the length
or the width or both. The area varies jointly
as the length and the width.
As a general example of joint variation,
consider the expression a oc be. Written as an
equation, this becomes a = kbc. If the value of
a is known for particular values of b and c, we
can find the new value of a corresponding to
changes in the values of b and c. For example,
suppose that a is 12 when b is 3 and c is 2.
What is the value of a when b is 4 and c is 5 ?
Rewriting the proportion,
Thus
Also,
12
= k
= k
Since quantities equal to the same quantity are
equal to each other, we can set up the following
proportion:
12
a =.40
Practice problems. Using k as the constant
of proportionality, write equations that express
the following statements:
1. Z varies jointly as x and y.
2. S varies jointly as b times the square of r.
3. The length, W, of a radio wave varies jointly
as the square root of the inductance, L, and the
capacitance, C.
Answers: v kw 2 L
~~
is an example of combined variation and is
read, "E varies jointly as L and the square of
W, and inversely as the square of p." Likewise,
COMBINED VARIATION V -
~ t
The different types of variation can be com-
bined. This is frequently the case in applied is read, "V varies jointly as r and s and in-
problems. The equation versely as t."
150
CHAPTER 14
DEPENDENCE, FUNCTIONS, AND FORMULAS
In chapter 13 of this course, use is made of
everal formulas, such as A = LW, E = ER, etc.
t is the purpose of this chapter to explain the
unction and dependency relationships which
aake formulas so useful.
DEPENDENCE AND FUNCTIONS
Dependence may be defined as any relation-
hip between two variables which allows the
redaction of change in one of them as a result
f change in the other. For example, the cost
f 200 bolts depends upon the price per hun-
red. If C represents cost and p represents
le price of 100 bolts, then the cost of 200 bolts
lay be expressed as follows:
C = 2p
In the example just given, C is called the
lEPENDENT VARIABLE because its value de-
ends upon the changing values of p. The EN-
IEPENDENT VARIABLE is p. It is standard
ractice to isolate the dependent variable on
ic left side of an equation, as in the example.
Consider the formula for the area of a rec-
rngle, A = LW. Here we have two independent
ariables, L and W.
Figure 14-1 (A) shows what happens if we
ouble the length. Figure 14-1 (B) ;snows the
esult of doubling the width. Figure 14-1 (C)
hows the effect of doubling both length and
idth. Notice that when the length or width
lone is doubled the area is doubled, but when
ath length and width are doubled the area is
>ur times as great.
In any equation showing a dependency rela-
onship, the dependent variable is said to be a
UNCTION of the independent variable. An-
ther use of the term "function" in describing
a equation such as C = 2p is to refer to the
hole expression as "the function C = 2p."
his terminology is especially useful when the
ight-hand expression has several terms. For
sample, consider the equation y = 2x 2 + 3x - 4.
[athematicians frequently use a shorthand no-
Figure 14-1. Changes in the area of a
rectangle resulting from changes in
length and width.
expression f(x) is understood to mean "a func-
tion of x" and reference to the function by call-
ing it f(x) saves the space and time that would
otherwise be required to write out all .three
terms.
Practice problems. Answer the following
questions concerning the function r = -r
1. When t increases and d remains the same,
doesr increase, decrease, or remain the same?
2. When d increases and t remains the same,
doesr increase, decrease, or remain the same?
3. When t decreases and d remains the same,
doesr increase, decrease, or remain the same?
4. When d decreases and t remains the same,
doesr increase, decrease, or remain the same?
5. When d is doubled "and t remains the same,
is r doubled or halved?
6. When t is doubled and d remains the same,
Answers:
1. Decreases.
2. Increases.
3. Increases.
4. Decreases.
5. Doubled.
6. Halved.
FORMULAS
P=2LH
h2W
One of the most common uses of algebra is
in the solution of formulas. Formulas have a
wide and varied use throughout the Navy. It is
important to know how formulas are derived,
how to translate them into words, how to make
them from word statements, and how to use
them to solve problems.
A formula is a general fact, rule, or princi-
ple expressed in algebraic symbols. It is a
shorthand expression of a rule in which letters
and signs of operation take the place of words.
The formula always indicates the mathematical
operations involved. For example, the formula
P = 2L + 2W indicates that the perimeter (sum
of the lengths of the sides) of a rectangle is
equal to twice its length plus twice its width.
(See fig. 14-2.)
W
Figure 14-2. Perimeter
of a rectangle.
A formula obtained by logical or mathemati-
cal reasoning is called a mathematical for-
mula. A formula whose reliability is based on
a limited number of observations, or on imme-
diate experience, and not necessarily on estab-
lished theories or laws is called an EMPIRI-
CAL formula. Empirical formulas are found
frequently in engineering and physical sciences.
They sometimes are valid for only a limited
number of values.
SUBJECT OF A FORMULA
Usually a formula is taken almost directly
from the verbal rule or law. For instance, the
perimeter of a rectangle is equal to twice the
length plus twice the width. Where possible,
letters are used as symbols for the words.
Thus, P = 2L + 2W. A simple formula such as
this is like a declarative sentence. The left
half is the SUBJECT and all the rest is the
predicate. The subject is P. It corresponds to
the part of the verbal rule that reads "the pe-
rimeter of a rectangle." This subject is usu-
ally a single letter followed by the equality sign.
All formulas are equations, but not all equa-
tions are formulas. Some distinctions between
a formula and an ordinary equation are worthy
of note. The equation may not have a subject,
while the formula typically does. In the for-
mula, the unknown quantity stands alone in the
left-hand member. No computation is per-
formed upon it, and it does not appear more
than once. In the equation, on the other hand,
the unknown quantity may appear once or more
in either or both members, and computation
may be performed with it or on it. We evaluate
a formula by substituting for the literal num-
bers in the right member. An equation is solved
by computation in either or both members until
all that remains is an unknown in one member
and a known quantity in the other. The solution
of an equation usually requires a knowledge of
algebraic principles, while the evaluation of a
formula may ordinarily be accomplished with
only a knowledge of arithmetic.
SYMBOLS
Letters that represent words have been
standardized in many cases so that certain for-
mulas may be written the same in various texts
and reference books. However, to avoid any
misunderstanding a short explanation often ac-
companies formulas as follows:
A = hw,
where
A = area in square units
h = height
w = width
Subscripts and Primes
In a formula in which two or more of the
same kind of letters are being compared, it is
desirable to make a distinction between them.
In electronics, for example, a distinction be-
tween resistances may be indicated by R a and
R b or Rj and R r These small numbers or
152
letters written to the right and below the R's
are called subscripts. Those shown here are
read: R sub a, R sub b, R sub one, and R sub
two. Primes are also used in the same manner
to distinguish between quantities of the same
kind. Primes are written to the right and above
the letters, as in S 1 , S", and S'". They are
read: S prime, S double prime, and S triple
prime.
CHANGING THE SUBJECT
OF A FORMULA
If values are given for all but one of its var-
iables, a formula can be solved to obtain the
value of that variable. The first step is usually
the rearranging of the formula so that the un-
known value is the subject that is, a new for-
mula is derived from the original. For exam-
ple, the formula for linear motion distance
equals rate times time is usually written
d = rt
Suppose that instead of the distance we wish
to know the rate, r, or the time, t. We simply
change the subject of the formula by the alge-
braic means developed in earlier chapters.
Thus, in solving the formula for r, we divide
both sides by t, with the following result:
d_
t
= r,
or r = -p
In words, this formula states that rate equals
distance divided by time . Likewise, in solving
for t, we have the following :
d_
r
rt
r
In words, this formula states that time equals
distance divided by rate.
We have in effect two new formulas, the
subject of one being rate and the subject of the
other being time. They are related to the orig-
inal formula because they were derived from
it, but they are different in that they have dif-
ferent subjects.
Practice problems. Derive new formulas
from the following expressions with subjects
as indicated:
1. A = -jbh, subject h
2. P = 2L + 2W, subject L
3. i = prt, subject r
4. p = br, subject b
5. E = IR, subject I
6. The modern formula for converting Fahren-
heit temperatures to Celsius (centigrade) is
/ 1;\
C = (F + 40)m - 40. Express the formula for
\ y /
converting Celsius (centigrade) temperatures
to Fahrenheit.
Answers:
2. L =
P - 2W
4. b =
r
6. F = (C + 40) I- - 40
EVALUATING FORMULAS
The first step in finding the value of the un-
known variable of a formula is usually the der-
ivation of a formula that has the unknown as its
subject. Once this is accomplished, the evalua-
tion of a formula consists of nothing more than
substituting numerical values for the letters
representing known quantities and performing
the indicated operations .
For example, suppose we wish to find the
time required to fly 1,250 nautical miles at the
rate of 250 knots. The formula is d = rt. We
can change the subject by dividing both sides of
the equation by r, as follows:
d_
r
t -
1 -
rt
"r"
125
250
- 5 hr
" & nr
Formulas can be solved for an unknown by
substituting directly in the original formula
153
even though that unknown is not the subject.
Generally, however, it is simpler to first make
the unknown the subject.
Formulas vary widely, from the simple type
such as we have been considering to some that
are very complex. All formulas have certain
characteristics in common. There is always a
subject, the quantity whose value is sought as a
final answer. This subject usually stands alone,
being placed equal to at least one and possibly
several literal numbers, which are combined
according to certain indicated operations. The
formula can always be evaluated for a specific
case when numerical values are known for all
these literal quantities.
Evaluating formulas may be facilitated by
developing a routine order of doing the work.
If someone else can read the work and clearly
understand wha j has been done, the work is in
good order. The original formula should be
written first, then the derived formula that is
going to be used in solving the problem, and
finally the actual substitutions. The indicated
operations may then be carried out. Care
should be taken to label answers with correct
units; that is, miles per hour, foot-pounds,
square feet, etc.
Practice problems.
1. E = IR. Solve for R in ohms if E is 110 volts
and I is 5 amperes.
2. d = rt. Solve for t in hours if d is 840 nauti-
cal miles and r is 25 knots.
3. F = (C + 40)(|) - 40. Solve for C if F is 32.
Answers:
1. 22 ohms
2. 33.6 hr
3. O c
DEVELOPING FORMULAS
Developing a formula from a verbal state-
ment is nothing more than reducing the state-
ment to a shorthand form and showing the math-
ematical relationships between the elements of
the statement.
For example, suppose that we wish to de-
velop a formula showing the distance, D, trav-
eled at the rate of 20 knots for t hours. If the
distance traveled in 1 hr is 20 nautical miles,
then the distance traveled in t hours is 20t.
Therefore, the formula is
D = 20t
Practice problems.
1. Write a formula for the cost, C of p pounds
of sugar at 15 cents per pound.
2. Write the formula for the cost, C, of one
article when the total cost, T, of n similar ar-
ticles is known.
3. Write a formula for the number of days, d,
in w weeks.
4. Write a formula for the number of ounces,
n, in p pounds.
Answers:
1. C = I5p
2. C = T/n
3. d = 7w
4. n = 16p
Developing Formulas from Tables
In technical work, instrument readings and
other data are often recorded in a tabular ar-
rangement. By careful observation of such
tables of data, it is frequently possible to find
values that are related in a definite pattern.
The table can thus be used in developing a for-
mula showing the relationship between the re-
lated quantities.
For example, table 14-1 shows the results
of time trials on a ship, with the data rounded
to the nearest whole hour and the nearest whole
mile.
Table 14-1. Time trials.
Nautical miles (d)
20
40
60
80
100
Hours (t)
1
2
3
4
5
By inspection of the table, it soon becomes
clear that the number of miles traveled is al-
ways 20 times the corresponding number of
hours. Therefore the formula developed from
this table is as follows :
d = 20t
154
A second example of the derivation of a for-
mula from a table is shown in figure 14-3.
Figure 14-3 (A) shows several polygons (many-
sided plane figures), each with one or more
diagonals. A diagonal is a straight line joining
one vertex (point where two sides meet) with
another.
DIAGONALS (d)
1
2
3
4
SIDES (n)
4
5
6
7
(B)
Figure 14-3. Diagonals of plane figures.
The table in figure 14-3 (B) compares the
number of sides of each polygon with the num-
ber of diagonals that can be drawn from any
one vertex. Using this table, we make a for-
mula for the number, d, of diagonals that can
be drawn from one vertex of a polygon of n
sides. In the table we note that the number of
diagonals is always 3 less than the number of
sides. Therefore the formula is d = n - 3.
Practice problems. Complete the following
tables and write formulas to show the relation-
ship between the numbers.
1.
2.
3.
4.
L
2
5
8
11
14
17
20
P
12
30
48
66
84
a
1
2
3
4
5
6
7
b
4
5
6
7
x
i
2
3
4
5
6
y
3
6
9
12
n
1
2
3
4
5
6
7
s
3
6
11
18
27
38
Answers:
1. P = 6L
2. b = a + 4
3. y = 3x
4. s = n 2 + 2
TRANSLATING FORMULAS
Thus far, we have been concerned primarily
with reducing verbal rules or statements to
formula form. It is also necessary to 'be able
to do the reverse, and translate a formula into
words. Technical publications frequently take
advantage of the fact that it is more convenient
to write formulas than longhand rules. Under-
standing is hampered if we are not able to
translate these formulas into words. As an ex-
ample of translation, we may translate the for-
mula V = Iwh into words, with the literal fac-
tors representing words as follows:
V = volume of a
rectangular solid
1 = length
w = width
h = height
This produces the following translation: The
volume of a rectangular solid equals the length
times the width times the height.
As a second example, we translate the alge-
braic expression 2 \Tx - 4 into words as fol-
lows: Twice the square root of a certain num-
ber, minus 4.
Practice problems. Translate each of the
following expressions into words.
1. PV = k, where P represents pressure of a
gas and V represents volume. (Assume con-
stant temperature.)
2. x = y + 4, where x and y are numbers.
3. A = LW, where A is the area of a rectangle,
L is its length, and W is its width.
4. d = rt, where d is distance, r is rate, and t
is time.
Answers:
1. The pressure of a gas multiplied by its vol-
ume is constant, if the temperature is constant.
155
2. A certain number, x, is equal to the sum of
another number, y, and 4.
3 . The area of a rectangle is equal to the prod-
uct of its length times its width.
4. Distance is equal to rate multiplied by time.
GRAPHING FORMULAS
We have seen that the formula is an equa-
tion. Since all formulas are equations they
may be graphed. Graphs of formulas have wide
use in the Navy in such fields as electronics
and engineering. In practical applications it is
often convenient to derive information from
graphs of formulas rather than from formulas
directly.
As an example, suppose that a fuel costs 30
cents per gallon. The formula for the cost in
dollars of n gallons is
C = 0.30n
We see that this is a linear equation, the re-
sulting curve of which passes through the origin
(no constant term). Since we are interested
only in positive values, we can eliminate three
quadrants of the graph and use only the first
quadrant. We already know one point on the
graph is (0,0). We need plot only one other
point to graph the formula. The result is shown
In figure 14-4.
We may read the cost directly from the
graph when the number of gallons is known, or
the number of gallons when the cost is known.
For instance, if 5-1/2 gal are sold, find 5-1/2
on the gallons scale and follow the vertical line
from that point to the point where it intersects
the graph of the formula. From this point, fol-
low the horizontal line to the cost scale. The
horizontal line intersects the cost scale at 1.65.
Therefore the cost of 5-1/2 gal is $1.65.
Likewise, to answer the question, "How many
gallons may be bought for $1.27," we would en-
large the graph enough to estimate to the exact
cent. Then we would follow a horizontal line
from 1.27 on the cost scale to the formula
graph and follow a vertical line from that point
to the gallons scale. Thus, 4-1/4 gal maybe
bought for $1.27.
Plotting two formulas on the same graph may
help to solve certain kinds of problems. For
example, suppose that two ships leave port at
the same time. One averages 10 knots and the
other averages 15 knots. How far has each
traveled at the end of 3 hr and at the end of
5 hr? A graph to relate the two ships' move-
ments at any time can be made as follows: Let
the vertical scale be in nautical miles and the
23456789 10
GALLONS
Figure 14-4. Graph for the formula C = O.SOn.
23456789
HOURS (t)
Figure 14-5. Graph of the formulas d = lOt
and d = 15t.
156
jrizontal scale be in hours. The formula for
ie first ship's distance related to time is
d = lOt
he formula for the second ship's distance re-
ited to time is
d = 15t
We see that these formulas are linear and
ieir curves pass through the origin. They are
raphed in figure 14-5.
With this graph we can now answer the ques-
ons originally posed, at a glance. Thus in 3
r the first ship traveled 30 mi and the second
traveled 45 mi. In 5 hr the first ship traveled
50 mi and the second traveled 75 mi.
We could also answer such questions as:
When the second ship has traveled 100 mi, how
far has the other traveled? We first find the
point on the graph of d = 15t where the ship has
traveled 100 mi. We then follow the vertical
line from that point to the point where it inter-
sects the graph of the other formula. From the
point of intersection we follow a horizontal line
to the distance axis and see that the first ship
has traveled about 67 mi when the second has
traveled 100 mi.
The foregoing examples serves to illustrate
the wide variety of applications in which graphs
of formulas are useful.
157
CHAPTER 15
COMPLEX NUMBERS
In certain calculations in mathematics and
related sciences, it is necessary to perform
operations with numbers unlike any mentioned
thus far in this course. These numbers, unfor-
tunately called "imaginary" numbers by early
mathematicians, are quite useful and have a
very real meaning in the physical sense. The
number system, which consists of ordinary
numbers and imaginary numbers, is called the
COMPLEX NUMBER system. Complex num-
bers are composed of a "real" part and an
"imaginary" part.
This chapter is designed to explain imagi-
nary numbers and to show how they can be com-
bined with the numbers we already know.
REAL NUMBERS
The concept of number, as has been noted in
previous chapters, has developed gradually. At
one time the idea of number was limited to
positive whole numbers.
The concept was broadened to include posi-
tive fractions; numbers that lie between the
whole numbers. At first, fractions included
only those numbers which could be expressed
with terms that were integers. Since any frac-
tion may be considered as a ratio, this gave
rise to the term RATIONAL NUMBER, which
is defined as any number which can be ex-
pressed as the ratio of two integers. (Remem-
ber that any whole number is an integer.)
It soon became apparent that these numbers
were not enough to complete the positive num-
ber range. The ratio, IT, of the circumference
of a circle to its diameter, did not fit the con-
cept of number thus far advanced, nor did such
numbers as *T2 and "JIT. Although decimal
values are often assigned to these numbers,
they are only approximations. That is, TT is not
exactly equal to 22/7 or to 3.142. Such num-
bers are called IRRATIONAL to distinguish
them from the other numbers of the system.
With rational and irrational numbers, the posi-
tive number system includes all the numbers
from zero to infinity in a positive direction.
Since the number system was not complete
with only positive numbers, the system was ex-
panded to include negative numbers. The idea
of negative rational and irrational numbers to
minus infinity was an easy extension of the
system.
Rational and irrational numbers, positive
and negative to infinity as they have been
presented in this course, comprise the REAL
NUMBER system. The real number system is
pictured in figure 15-1.
OPERATORS
As shown in a previous chapter, the plus
sign in an expression such as 5 + 3 can stand
for either of two separate things: It indicates
the positive number 3, or it indicates that +3
is to be added to 5; that is, it indicates the op-
eration to be performed on +3.
Likewise, in the problem 5 - 3, the minus
sign may indicate the negative number -3, in
which case the operation would be addition; that
is, 5 + (-3). On the other hand, it may indicate
the sign of operation, in which case +3 is to be
subtracted from 5; that is, 5 - (+3).
Thus, plus and minus signs may indicate
positive and negative numbers, or they may in-
dicate operations to be performed.
1
-4
r-1 1 i' -
r i - 2
~n -27
i 1 i
1-'
~/2
!_
1 1 1 1
PI
+-! ^Ja
2
ri
"1
+TT
+4
Figure 15-1. The real number system.
158
IMAGINARY NUMBERS
The number line pictured in figure 15-1 rep-
resents all positive and negative numbers from
plus infinity to minus infinity. However, there
is a type of number which does not fit into the
picture. Such a number occurs when we try to
solve the following equation:
x + 4 =
x 2 = -4
X =
Notice the distinction between this use of the
radical sign and the manner in which it was
used in chapter 7. Here, the symbol is in-
cluded with the radical sign to emphasize the
fact that two values of x exist. Although both
roots exist, only the positive one is usually
given. This is in accordance with usual mathe-
matical convention.
The equation
X =
raises an interesting question:
What number multiplied by itself yields -4 ?
The square of -2 is +4. Likewise, the square
of +2 is +4. There is no number in the system
of real numbers that is the square root of a
negative number. The square root of a nega-
tive number came to be called an IMAGINARY
NUMBER. When this name was assigned the
square roots of negative numbers, it was natu-
ral to refer to the other known numbers as the
REAL numbers.
IMAGINARY UNIT
To reduce the problem of imaginary num-
bers to its simplest terms, we proceed as far
as possible using ordinary numbers in the so-
lution. Thus, we may write \f -4 as a product
2
Likewise,
Also,
Thus, the problem of giving meaning to the
square root of any negative number reduces to
that of finding a meaning for '/^T.
The square root of minus 1 is designated i
by mathematicians. When it appears with a co-
efficient, the symbol i is written last unless
the coefficient is in radical form. This con-
vention is illustrated in the following examples:
2 xT^l = 2i
The
= i
T = 3i
symbol i stands for the imaginary unit
. An imaginary number is any real multi-
ple, positive or negative, of i. For example,
-7i, +7i, i NTiT, and bi are all imaginary num-
bers.
In electrical formulas the letter i denotes
current. To avoid confusion, electronic techni-
cians use the letter j to indicate '/"-T and call it
"operator j." The name "imaginary" should be
thought of as a technical mathematical term of
convenience. Such numbers have a very real
purpose in the physical sense. Also it can be
shown that ordinary mathematical operations
such as addition, multiplication, and so forth,
may be performed in exactly the same way as
for the so-called real numbers.
Practice problems. Express each of the
following as some real number times i:
1.
2. 2
Answers:
1. 4i
2. 2i
3.
5.
3. i
4. di
5. 5i
R 3 i
6. 1
Powers of the Imaginary Unit
The following examples illustrate the re-
sults of raising the imaginary unit to various
powers:
i =
1 3 = i 2 i = -li, or -i
1 4 =iV = -1
-1 = +1
- 3
159
MATHEMATICS, VOLUME 1
We see from these examples that an even
power of i is a real number equal to +1 or -1.
Every odd power of i is imaginary and equal to
i or -i. Thus, all powers of i reduce to one of
the following four quantities: N/^T, -1, --vTT,
or +1.
GRAPHICAL REPRESENTATION
Figure 15-1 shows the real numbers repre-
sented along a straight line, the positive num-
bers extending from zero to the right for an
infinite distance, and the negative numbers ex-
tending to the left of zero for an infinite dis-
tance. Every point on this line corresponds to
a real number, and there are no gaps between
them. It follows that there is no possibility of
representing imaginary numbers on this line.
Earlier, we noted that certain signs could be
used as operators. The plus sign could stand
for the operation of addition. The minus sign
could stand for the operation of subtraction.
Likewise, it is easy to explain the imaginary
number i graphically as an operator indicating
a certain operation is to be performed on the
number of which it is the coefficient.
H we graphically represent the length, n, on
the number line pictured in figure 15-2 (A), we
start at the point and measure to the right
(positive direction) a distance representing n
units. If we multiply n by -1, we may repre-
sent the result -n by measuring from in a
negative direction a distance equal to n units.
Graphically, multiplying a real number by
-1 is equivalent to rotating the line that repre-
sents the number about the point through 180
so that the new position of n is in the opposite
direction and a distance n units from 0. In this
case we may 'think of -1 as the operator that
rotates n through two right angles to its new
position (fig. 15-2 (B)).
As we have shown, i 2 = -l. Therefore, we
have really multiplied n by i 2 , or i x i. In other
words, multiplying by -1 is the same as multi-
plying by i twice in succession. Logically, if
we multiplied n by i just once, the line n would
be rotated only half as much as before that is,
through only one right angle, or 90. The new
segment ni would be measured in a direction
90 from the line n. Thus, i is an operator that
rotates a number through one right angle. (See
fig. 15-3.)
We have shown previously that a positive
number may have two real square roots, one
positive and one negative. For example, */ = 3 .
-n
(A)
\ ]
* n
-n
(B)
n
Figure 15-2. -Graphical multiplication by
- 1 and by operator i .
Figure 15-3. Graphical multi-
plication by operator i.
We also saw that an imaginary number may
have two roots. For example, \f^l is equal to
2i. When the operator -1 graphically rotates
a number, it may do so in a counterclockwise
or a clockwise direction. Likewise, the opera-
tor i may graphically rotate a number in either
direction. This fact gives meaning to numbers
such as 2i. It has been agreed that a number
multiplied by +i is to be rotated 90 in a coun-
terclockwise direction. A number multiplied
by -i is to be rotated 90 in a clockwise di-
rection.
In figure 15-4, +2i is represented by rotating
ae line that represents the positive real num-
er 2 through 90 in a counterclockwise direc-
ion. It follows that -2i is represented byrotat-
ig the line that represents the positive real
umber 2 through 90 in a clockwise direction.
Figure 15-4.-
-Graphical representation
of 2i.
In figure 15-5, notice that the idea of i as an
perator agrees with the concept advanced con-
erning the powers of i. Thus, i rotates a num-
er through 90; i 2 or -1 rotates the number
THE COMPLEX PLANE
All imaginary numbers may be represented
graphically along a line extending through zero
and perpendicular to the line representing the
real numbers. This line may be considered in-
finite in both the positive and negative direc-
tions, and all multiples of i may be represented
on it. This graph is similar to the rectangular
coordinate system studied earlier.
In this system, the vertical or y axis is
called the axis of imaginaries, and the horizon-
tal or x axis is called the axis of reals. In the
rectangular coordinate system, real numbers
are laid off on both the x and y axes and the
plane on which the axes lie is called the real
plane. When the y axis is the axis of imagi-
naries, the plane determined by the x and y axes
is called the COMPLEX PLANE (fig. 15-6).
In any system of numbers a unit is neces-
sary for counting. Along the real axis, the unit
is the number 1. As shown in figure 15-6,
along the imaginary axis the unit is i. Numbers
that lie along the imaginary axis are called
PURE IMAGINARIES. They will always be
some multiple of i, the imaginary unit. The
numbers 5i, 3i *f2, and */^7 are examples of
pure imaginaries.
irough 180, and the number is real and nega- NUMBERS IN THE COMPLEX PLANE
ive; i 3 rotates the number through 270, which
as the same effect as -i; and i 4 rotates the All numbers in the complex plane are complex
umber through 360, and the number is once numbers, including reals and pure imaginaries.
gain positive and real. However, since the reals and imaginaries have
= -bi
Figure 15-5. Operation with powers of i.
161
AXIS OF IMAGINARIES
Y
I"
-4 -3 -2 -1
AXIS OF REALS
rH I I -
1 234
Figure 15-6. The complex plane.
the special property of being located on the
axes, they are usually identified by their dis-
tinguishing names.
The term complex number has been defined
as the indicated sum or difference of a real
number and an imaginary number.
For example, 3 + 5 \l -1 or 3 + 5i, 2 - 6i,
and -2 + N/^5" are complex numbers. In the
complex number 7 - i ^/~2~ J 7 is the real part
and -i ^/Tis the imaginary part.
All complex numbers correspond to the gen-
eral form a + bi, where a and b are real num-
bers. When a has the value 0, the real term
disappears and the complex number becomes a
pure imaginary. When b has the value of 0, the
imaginary term disappears and the complex
number becomes a real number. Thus, 4 may
be thought of as 4 + Oi, and 3i may be consid-
ered + 3i. From this we may reason that the
real number and the pure imaginary number
are special cases of the complex number. Con-
sequently, the complex number may be thought
of as the most general form of a number and
can be construed to include all the numbers of
algebra as shown in the chart in figure 15-7.
Plotting Complex Numbers
Complex numbers may easily be plotted
in the complex plane. Pure imaginaries are
plotted along the vertical axis, the axis of imag-
inaries, and real numbers are plotted along the
horizontal axis, the axis of reals. It follows
that other points in the complex plane must
represent numbers that are part real and part
imaginary; in other words, complex numbers.
If we wish to plot the point 3 + 2i, we note that
the number is made up of the real number 3
and the imaginary number 2i. Thus, as in fig-
ure 15-8, we measure along the real axis in a
COMPLEX NUMBERS
(a + bi)
1
REAL NUMBERS
(FORM IS a WHERE
bIS o)
1
PURE IMAGINARIES
(FORM IS bi WHERE
a ISo)
(RATIONAL
IRRATIONAL
1
NEGATIVE
INTEGERS
POSITIVE
INTEGERS
s*
s*
V5
NEGATIVE
FRACTIONS
POSITIVE
FRACTIONS
IT
etc.
-TI
etc
Figure 15-7. The complex number system.
162
-3+2i
I"
_| 1 1 1 (_-
-5 -4 -3 -2 -I 9
-3-2i
3+2i
_ _____ |
H I 1 I H
12345
3-2i
Figure 15-8. Plotting complex numbers.
ositive direction. At point (3, 0) on the real
xis we turn through one right angle and meas-
re 2 units up and parallel to the imaginary
sis. Likewise, the number -3 + 2i is 3 units
) the left and up 2 units; the number 3 - 2i
5 3 units to the right and down 2 units; and the
umber -3 -2i is 3 units to the left and down
units.
omplex Numbers as Vectors
A vector is a directed line segment. A corn-
lex number represents a vector expressed in
le RECTANGULAR FORM. For example, the
omplex number 6 + 8i in figure 15-9 may be
onsidered as representing either the point P
r the line OP. The real parts of the complex
umber (6 and 8) are the rectangular compo-
ents of the vector. The real parts are the legs
E the right triangle (sides adjacent to the right
ngle),and the vector OP is its hypotenuse (side
pposite the right angle). If we merely wish to
idicate the vector OP, we may do so by writ-
igthe complex number that represents it along
le segment as in figure 15-9. This method not
nly fixes the position of point P, but also shows
hat part of the vector is imaginary (PA) and
hat part is real (OA).
If we wish to indicate a number that shows
le actual length of the vector OP, it is neces-
ary'to solve the right triangle OAP for its
ypotenuse. This may be accomplished by tak-
ig the square root of the sum of the squares of
I
53.1 AXIS OF REALS
Figure 15-9. A complex number
shown as a vector.
the legs of the triangle, which in this case are
the real numbers, 6 and 8. thus,
However, since a vector has direction as
well as magnitude, we must also show the di-
rection of the segment; otherwise the seg-
ment OP could radiate in any direction on the
complex plane from point 0. The expression
10/53.1 indicates that the vector OP has been
rotated counterclockwise from the initial posi-
tibn through an angle of 53.1. (The initial po-
sition in a line extending from the origin to the
right along OX.) This method of expressing the
vector quantity is called the POLAR FORM.
The number represents the magnitude of the
quantity, and the angle represents the position
of the vector with respect to the horizontal ref-
erence, OX. Positive angles, represent coun-
terclockwise rotation of the vector, and nega-
tive angles represent clockwise rotation. The
polar form is generally simpler for multiplica-
tion and division, but its use requires a knowl-
edge of trigonometry.
163
ADDITION AND SUBTRACTION OF
COMPLEX NUMBERS
Pure imaginaries are added and subtracted
in the same way as any other algebraic quanti-
ties. The coefficients of similar terms are
added or subtracted algebraically, as follows:
4i + 3i = 7i
4i - 3i = i
4i - (-3i) = 71
Likewise, complex numbers in the rectangular
form are combined like any other algebraic
polynomials. Add or subtract the coefficients
of similar terms algebraically. If parentheses
enclose the numbers, first remove the paren-
theses. Next, place the real parts together and
the imaginary parts together. Collect terms.
As examples, consider the following:
1. (2 - 3i) + (5 -f 4i) = 2 - 3i + 5 + 4i
= 2 + 5 - 3i + 4i
= 7 + i
2. (2 - J3) - (5 + j4) = 2 - J3 - 5 - J4
= 2 - 5 - j3 - j4
= - 3 - J7
In example 2, notice that the convention for
writing operator j (the electronics form of the
imaginary unit) with numerical coefficients is
to place j first.
If the complex numbers are placed one un-
der the other, the results of addition and sub-
traction appear as follows:
ADDITION
3 + 4
2-7
5 - 3
SUBTRACTION
I:
a + jb
-c + jd
(a - c) + j(b - d)
Practice problems. Add or subtract as in-
dicated, in the following problems:
1. (3a + 4i) + (0 - 2i)
2. (3 + 2i) + (-3 + 3i)
3. (a + bi) + (c + di)
4. (1 + 2 -s/^I) + (-2 - 2
5. (-5 + 3i) - (4 - 2i)
6. (a + bi) - (-c + di)
Answers:
1. 3a + 2i
2. 5i
3. a + c + (b + d)i
4. -1
5. -9 + 51
6. a + c + (b - d)i
MULTIPLICATION OF
COMPLEX NUMBERS
Generally, the rules for the multiplication of
complex numbers and pure imaginaries are the
same as for other algebraic quantities. How-
ever, there is one exception that should be
noted: The rule for multiplying numbers under
radical signs does not apply to TWO NEGA-
TIVE numbers. When at least one of two radi-
cands is positive, the radicands can be multi-
plied immediately, as in the following examples:
When both radicands are negative, however,
as in N/~^2~ N/ -3, an inconsistent result is ob-
tained if we multiply both numbers under the
radical signs immediately. To get the correct
result, express the imaginary numbers first in
terms of i, as follows:
= (-1)
Multiplying complex numbers is equivalent
to multiplying binomials in the manner ex-
plained previously. After the multiplication is
performed, simplify the powers of i as in the
following examples:
1.
12 + i - i 2 = 12 + i - (-1)
= 13 + i
164
. (-6 + 5 \/^7) (8-2
= (-6 + 5i N/77 (8 - 2i
= -48 + 40i N/T + 12i
= -48 + 52i N/T+ 70
= 22 + 52i VY
- 10(7)i
Practice problems. Perform the indicated
Derations:
(2 + 5i) (3 - 2i)
(a + \/-b) (a -
(-2 + /T45 (-1
(8 - ^7] ( 6 +
Answers:
-12 5. 16 + Hi
6i 6. a 2 + b
-6 7. -2 - 61
-ab \fa~ 8. 55 + 2i
DNJUGATES AND
'ECIAL PRODUCTS
Two complex numbers that are alike except
r the sign of their imaginary parts are called
)NJUGATE COMPLEX NUMBERS. For ex-
(iple, 3 + 5i and 3 - 5i are conjugates. Either
mber is the conjugate of the other.
If one complex number is known, the conju-
te can be obtained immediately by changing
3 sign of the imaginary part. The conjugate
-8 + -V -10 is -8 - N/-10. The conjugate of
The sum of two conjugate complex numbers
a real number, as illustrated by the following:
(3 + j5) + (3 - J5) = 2(3) = 6
^
(4* )*(-!--)
Product of Two Conjugates
The product of two conjugate complex num-
bers is a real number. Multiplying two conju-
gates is equivalent to finding the product of the
sum and difference of two numbers.
Consider the following examples:
1. (3 + J5) ( 3 - J5) = 3 2 - (J5) 2
= 9 - 25(-l)
= 9 + 25
= 34
1 -v/3
i = s- -
4
= 1
Squaring a Complex Number
Squaring a complex number is equivalent to
raising a binomial to the second power. For
example :
(-6 - ^25) 2 = (-6 - J5) 2
= [(-1) (6 + J5)] 2
= (-1) 2 (6 2 + J60 + J 2 25)
= 36 + J60 - 25
= 11 + J60
DIVISION OF COMPLEX NUMBERS
When dividing by a pure imaginary, the de-
nominator may be rationalized and the problem
thus simplified by multiplying both numerator
and denominator by the denominator. Thus,
12
= -6i
165
Division of complex numbers can be accom-
plished by multiplying the numerator and de-
nominator by the number that is the conjugate
of the denominator. This process is similar to
the process of rationalizing a denominator in
the case of real numbers that are irrational.
As an example, consider
5 - 2i
3 + i
The denominator is 3 + i. Its conjugate is
3 - i. Multiplying numerator and denominator
by 3 - i gives
5-21 3 - i _ 15 - Hi + 2i 2
9 - 12 '
3 + i 3 - i
= 15 - 111 - 2
9 + 1
_ 13 - 111
To"
13.
10
ill
10 l
Practice problems. Rationalize the denomi-
nators and simplify:
1.-
2.
3.
4 + 2
-2 + 4i
-1 + 41
3
1.
2.
3.
3 -
Answers:
21 + 1
5
18 + 41
17
7 + 61 */
11
4.
5.
5.
6.
3 - i
8 - 41
3
166
CHAPTER 16
QUADRATIC EQUATIONS IN ONE VARIABLE
The degree of an equation in one variable is
the exponent of the highest power to which the
variable is raised in that equation. A second-
degree equation in one variable is one in which
the variable is raised to the second power. A
second-degree equation is often called a QUAD-
RATIC EQUATION. The word quadratic is de-
rived from the Latin word quadratus, which
means "squared." In a quadratic equation the
terra of highest degree is the squared term.
For example, the following are quadratic equa-
tions:
x 2 + 3x + 4 =
3m + 4m 2 = 6
The terms of degree lower than the second
may or may not be present. The possible terms
of lower degree than the squared term in a
quadratic equation are the first- degree term
and the constant term. In the equation
3x 2 - 8x - 5 =
-5 is the coefficient of x. If we wished to
emphasize the powers of x in this equation, we
could write the equation in the form
3x 2 - Sx 1 - 5x =
Examples of quadratic equations in which either
the first-degree term or the constant term is
missing are:
1. 4x 2 = 16
2. y 2 + 16y =
3. e 2 + 12 =
GENERAL FORM OF A
QUADRATIC EQUATION
Any quadratic equation can be arranged in
the general form:
If it has more than three terms, some of them
will be alike and can be combined, after which
the final form will have at most three terms.
For example,
2x 2 + 3 + 5x - 1 + x 2 = 4 - x 2 - 2x - 3
reduces to the simpler form
4x 2 + 7x + 1 =
In this form, it is easy to see that a, the coef-
ficient of x 2 , is 4; b, the coefficient of x, is 7;
and c, the constant term, is 1.
Sometimes the coefficients of the terms of
a quadratic appear as negative numbers, as
follows:
2x 2 - 3x - 5 =
This equation can be rewritten in such a way
that the connecting signs are all positive, as in
the general form. This is illustrated as follows:
2x 2 + (-3)x + (-5) =
In this form, the value of a is seen to be 2,
b is -3, and c is -5.
An equation of the form
x 2 +
2 =
has no x term. This can be considered as a
case in which a is 1 (coefficient of x 2 under-
stood to be 1), b is 0, and c is 2. For the pur-
pose of emphasizing the values of a, b, and c
with reference to the general form, this equa-
tion can be written
x
2 + Ox + 2 =
The coefficient of x 2 can never be 0; if it
were 0, the equation would not be a quadratic.
If the coefficients of x and x are , then those
terms do not normally appear. To say that the
coefficient of x is is the same as saying that
the constant term is or is missing.
167
nignesi power, in umer woraa, me numoer ui
roots is the same as the degree of the equation.
A fourth-degree equation has four roots, a
cubic (third-degree) equation has three roots, a
quadratic equation has two roots, and a linear
equation has one root.
As an example, 6 and -1 are roots of the
quadratic equation
x 2 - 5x - 6 =
This can be verified by substituting these val-
ues into the equation and noting that an identity
results- in each case.
Substituting x = 6 gives
6 2 - 5(6) -6 =
36 - 36 =
=
Substituting x = - 1 gives
(-1) 2 - 5(-l) -6 =
1+5-6 =
6-6 =
=
Several methods of finding the roots of quad-
ratic equations (SOLVING) are possible. The
most common methods are solution by FAC-
TORING and solution by the QUADRATIC FOR-
MULA. Less commonly used methods of solu-
tion are accomplished by completing the square
and by graphing.
SOLUTION BY FACTORING
The equation x 2 - 36 = is a pure quadratic
equation. There are two numbers which, when
substituted for x, will satisfy the equation as
follows:
also
(+6) 2
36
(-6) 2
36
36 =
36 =
36 =
36 =
rauc ;one in wmcn no x term appears ana me
constant term is a perfect square) involves re-
writing with the constant term in the right
member, as follows:
x 2 = 36
Taking square roots on both sides, we have
x = 6
The reason for expressing the solution as both
plus and minus 6 is found in the fact that both
+6 and -6, when squared, produce 36.
The equation
x 2 - 36 =
can also be solved by factoring, as follows:
x 2 - 36 =
(x + 6)(x - 6) =
We now have the product of two factors equal
to zero. According to the zero factor law, if a
product is zero, then one or more of its factors
is zero. Therefore, at least one of the factors
must be zero, and it makes no difference which
one. We are free to set first one factor and
then the other factor equal to zero. In so doing
we derive two solutions or roots of the equation.
If x + 6 is the factor whose value is 0, then
we have
x + 6 =
x = -6
If x - 6 is the zero factor, we have
x - 6 =
x = 6
When a three-term quadratic is put into
simplest form, it is customary to place all the
terms on the left side of the equality sign with
the squared term first, the first-degree term
next, and the constant term last, as in
9x 2 - 2x + 7 =
168
If the trinomial in the left member is readily
factorable, the equation can be solved quickly
by separating the trinominal into factors. Con-
sider the equation
3x 2 - x - 2 =
By factoring the trinominal, the equation be-
comes
(3x + 2)(x - 1) =
Once again we have two factors, the product of
which is 0. This means that one or the other of
them (or both) must have the value 0. If the
zero factor is 3x + 2, we have
3x + 2 =
3x = -2
If the zero factor is x - 1, we have
x - 1 =
x = 1
Substituting first x = 1 and then x = - in
o
the original equation, we see that both roots
satisfy it. Thus,
3(1) 2 - (1) - 2 =
3-1-2 =
[-!]'-[-!]-*-
=
In summation, when a quadratic may be
readily factored, the process for finding its
roots is as follows:
1. Arrange the equation in the order of the
descending powers of the variable so that all
the terms appear in the left member and zero
appears in the right.
2. Factor the left member of the equation.
3. Set each factor containing the variable
equal to zero and solve the resulting equations.
4. Check by substituting each of the derived
roots in the original equation.
EXAMPLE: Solve the equation x 2 - 4x = 12
forx.
1. x 2 - 4x - 12 =
2. (x - 6)(x + 2) =
3. x - 6 =
x = 6
x + 2 =
x = -2
4. (6) 2 - 4(6) = 12 (x = 6)
36 - 24 = 12
12 = 12
(-2) 2 - 4 (-2) = 12 (x = -2)
4 + 8 = 12
12 = 12
Practice problems. Solve the following equa-
tions by factoring:
1. x2 + lOx - 24 = 4. 7y 2 - 19y - 6 =
5. m 2 - 4m = 96
4. y = 3
2
y = -7
5. m = -8
m = 12
2. a - a - 56 =
3. y 2 - 2y = 63
Answers:
1. x = -12
x = 2
2. a = 8
a = -7
3. y = -7
y = 9
SOLUTION BY
COMPLETING THE SQUARE
When a quadratic cannot be solved by fac-
toring, or the factors are not readily seen, an-
other method of finding the roots is needed. A
method that may always be used for quadratics
in one variable involves perfect square trino-
mials. These, we recall, are trinomials whose
factors are identical. For example,
x 2 - lOx + 25 = (x - 5)(x - 5) = (x - 5) 2
Recall that in squaring a binomial, the third
term of the resulting perfect square trinomial
169
MATHEMATICS, VOLUME 1
is always the square of the second term of the
binomial. The coefficient of the middle term of
the trinomial is always twice the second term
of the binomial. For example, when (x + 4) is
squared, we have
x + 4
x + 4
4. Take the square root of both members.
x 2 + 4x
+ 4x + 16
x 2 + 8x + 16
Hence if both the second- and first-degree
terms of a perfect square trinomial are known,
the third may be written by squaring one-half
the coefficient of the first- degree term.
Essentially, in completing the square, cer-
tain quantities are added to one member and
subtracted from the other, and the equation is
so arranged that the left member is a perfect
square trinomial. The square roots of both
members may then be taken, and the subsequent
equalities may be solved for the variable.
For example,
+ 5x -
11
=
cannot be readily factored. To solve for x by
completing the square, we proceed as follows:
1. Leave only the second- and first-degree
terms in the left member.
X 2 4- 5X = ^
(If the coefficient of x 2 is not 1, divide through
by the coefficient of x 2 .)
2. Complete the square by adding to both
members the square of half the coefficient of
the x term. In this example, one -half of the
g
coefficient of the x term is -, and the square
-
V- Thus >
25 11
T = T +
25
4
3. Factor the left member and simplify the
right member,
2
x + | = 3
Remember that, in taking square roots on both
sides of an equation, we must allow for the fact
that two roots exist in every second-degree
equation. Thus we designate both the plus and
the minus root of 9 in this example.
5. Solve the resulting equations.
1=3
x ~ 2 " 2
x= 2
x -
X ~ " "
x = -
n
2
6. Check the results.
(i)
'*!-
11
4
5
2 "
10
4
11
2
=
(B(-M-
121
4
55
2
11
=
110 55 _ n
~4~ " T ~
=
The process of completing the square may
always be used to solve a quadratic equation.
However, since this process may become com-
plicated in more complex equations, a formula
based on completing the square has been devel-
oped in which known quantities may be substi-
tuted in order to derive the roots of the quad-
ratic equation. This formula is explained in
the following paragraphs.
SOLUTION BY THE
QUADRATIC FORMULA
The quadratic formula is derived by apply-
ing the process of completing the square to
solve for x in the general form of the quadratic
equation, ax 2 + bx + c = 0. Remember that the
general form represents every possible quad-
ratic equation. Thus, if we can solve this equa-
tion for x, the solution will be in terms of a, b,
and c. To solve this equation for x by complet-
ing the square, we proceed as follows:
1. Subtract the constant term, c, from both
members.
ax + bx = -c
2. Divide all terms by a so that the coeffi-
cient of the x 2 term becomes unity.
o b c
x 2 + -x = - -
a a
3. Add the square of one-half the coefficient
of the x term, , to both members,
a
Square 7:
4a 2
2
Add: x 2 + -x + 77 = -r-
a 4a^ 4a- a.
4. Factor the left member and simplify the
right member.
x +
b - 4ac
2a/ " 4a 2
5. Take the square root of both members,
b
X + 7T"~ i
- 4ac
2a
6. Solve for x.
- 4ac
~2a
-b
2a
- 4ac
2a
Thus, we have solved the equation repre-
senting every quadratic for its unknown in terms
of its constants a, b, and c. Hence, in a given
quadratic we need only substitute in the ex-
pression
-b \/b 2 - 4ac
2a
the values of a, b, and c, as they appear in the
particular equation, to derive the roots of that
equation. This expression is called the QUAD-
RATIC FORMULA. The general quadratic
equation, ax 2 + bx + c = 0, and the quadratic
formula should be memorized. Then, when a
quadratic cannot be solved quickly by factoring,
it may be solved at once by the formula.
EXAMPLE: Use the quadratic formula to solve
the equation
x 2 + 30 - llx = 0.
SOLUTION:
1. Set up the equation in standard form.
x 2 - llx + 30 =
Then a (coefficient of x 2 ) =1
b (coefficient of x) = -11
c (the constant term) = 30
2. Substituting,
x =
-b vb 2 - 4ac
2a
- (-11) >/(-! I) 2 - 4(1)(30)
= _
11 N/121 - 120
11 1
3. Checking:
When
= 6 or 5
When
x = 6, x = 5,
(6) 2 - 11(6) + 30 = (5) 2 - 11(5) + 30 =
36 - 66 + 30 = 25 - 55 + 30 =
0=0 0=0
EXAMPLE: Find the roots of
2x 2 - 3x - 1 =
Here, a = 2, b = -3, and c = -1.
171
MATHEMATICS, VOLUME 1
Substituting into the quadratic formula gives
When
3 - N/T7
- (-3) V(-3)^ - 4(2)(-l)
^2 - 4
then
/3 - -/T7\ 2 /3 - -JTJ] j _
3 "^9 + 8
4
3 N/T7
"I 4 / ~*( 4 / ' * U
9 -6 's/Tf +17 9-3 N/T7
4
The two roots are
and x = -
These roots are irrational numbers, since the
radicals cannot be removed.
If the decimal values of the roots are de-
sired, the value of the square root of 17 can be
taken from appendix I of this course. Substi-
tuting ^/T7"= 4.1231 and simplifying gives
x l -
3 + 4.1231
and
x 2 =
3 - 4.1231
_ 7.1231
4
= 1.781
x, =
-1.1231
4
= -0.281
In decimal form, the roots of 2x 2 - 3x - 1 =
to the nearest tenth are 1.8 and -0.3.
Notice that the subscripts, 1 and 2, are used
to distinguish between the two roots of the equa-
tion. The three roots of a cubic equation in x
might be designated x 1( x 2 , and x 3 . Sometimes
the letter r is used for root. Using r, the roots
of a cubic equation could be labeled r l} r 2 ,
and r 3 .
Checking:
When Xj =
2x 2 - 3x - 1 =
then
4 / -3^^^-; -1 =
(3 + N/17) 2 9 + 3 N/17 1
8 4 1 =
9 + 6 'JW + 17-18-6 ^/l7 - 8
8
=
8
Multiplying both members of the equation by 8,
the LCD, we have
- 8(1) = o
9-6 -/IT + 17 - 2 (9 - 3 \/T7) -8 =
9 - 6 -S/T7 + 17-18 + 6 N/T? -8 =
=
Practice problems. Use the quadratic for-
mula to find the roots of the following equations:
1. 3x 2 - 20 - 7x =
2. 4x 2 - 3x - 5 =
Answers:
1. Xi = 4
3. 15x 2 - 22x - 5 =
4.
3.
+ 7x = 8
x 9 = --
1
5
2.
4.
Xo =
3 -
GRAPHICAL SOLUTION
A fourth method of solving a quadratic equa-
tion is by means of graphing. In graphing lin-
ear equations using both axes as reference, we
recall that an independent variable, x, and a
dependent variable, y, were needed. The co-
ordinates of points on the graph of the equation
were designated (x, y).
Since the quadratics we are considering con-
tain only one variable, as in the equation
=
- 8x + 12 =
we cannot plot values for the equations in the
present form using both x and y axes. A de-
pendent variable, y, is necessary.
K we think of the expression
x 2 - 8x + 12
is a function, then this function can be consid-
sred to have many possible numerical values,
iepending on what value we assign to x. The
particular value or values of x which cause the
/alue of the function to be are solutions for
:he equation
x 2 - 8x + 12 =
For convenience, we may choose to let y
'epresent the function
x 2 - 8x + 12
f numerical values are now assigned to x, the
:orresponding values of y may be calculated.
Vhen these pairs of corresponding values of x
md y are tabulated, the resulting table pro-
'ides the information necessary for plotting a
;raph of the function.
5XAMPLE: Graph the equation
x 2 + 2x - 8 =
(-5,7).
J-4.01
-5
10
(3,7)
(2,0)
and from the graph write the roots of the equa-
tion.
SOLUTION:
1. Let y = x 2 + 2x - 8.
2. Make a table of the y values corresponding
to the value assigned x, as shown in table 16-1.
Table 16-1. Tabulation of x and y values
for the function y = x 2 +2x-8.
ifx=
-5
-4
-3
-2
-1
1
2
3
then y
7
-5
-8
-9
-8
-5
7
3. Plot the pairs of x and y values that ap-
pear in the table as coordinates of points on a
rectangular coordinate system as in figure
16-1 (A).
4. Draw a smooth curve through these points,
as shown in figure 16-1 (B).
Notice that this curve crosses the X axis in
two places. We also recall that, for any point
on the X axis, the y coordinate is zero. Thus,
in the figure we see that when y is zero, x is
-4 or +2. When y is zero, furthermore, we
have the original equation,
(V5)
(A)
Figure 16-1. Graph of the equation y = x 2 + 2x - 8. (A) Points plotted;
(B) curve drawn through plotted points.
173
Thus, the values of x at these points where
the graph of the equation crosses the X axis
(x = -4 or +2) are solutions to the original equa-
tion. We may check these results by solving
the equation algebraically. Thus,
x 2 + 2x - 8 =
(x + 4)(x - 2) =
Xj + 4 =
Xl = -4
Check:
(-4) 2 + 2(-4) -8 =
16 - 8 - 8 =
=
me A coordinate, or aoscissa, 01 me maximum
or minimum value is
x 2 - 2 =
x, = 2
(2) 2 + 2(2) -8=0
4 + 4-8 =
=
The curve in figure 16-1 (B) is called a
PARABOLA. Every quadratic of the form
ax 2 + bx + c = y will have a graph of this gen-
eral shape. The curve will open downward if a
is negative, and upward if a is positive.
Graphing provides a fourth method of finding
the roots of a quadratic in one variable. When
the equation is graphed, the roots will be the X
intercepts (those values of x where the curve
crosses the X axis). The X intercepts are the
points at which y is 0.
Practice problems. Graph the following
quadratic equations and read the roots of each
equation from its graph
1. x 2 - 4x - 8 =
2. 6x - 5 - x 2 =
Answers:
1. See figure 16-2. x = 5.5; x = -1.5
2. See figure 16-3. x = 1; x = 5
MAXIMUM AND MINIMUM POINTS
It will be seen from the graphs of quadratics
in one variable that a parabola has a maximum
or minimum value, depending on whether the
curve opens upward or downward. Thus, when
a is negative the curve passes through a maxi-
mum value; and when a is positive, the curve
passes through a minimum value. Often these
maximum or minimum values comprise the only
information needed for a particular problem.
x =
_
2a
In other words, if we divide minus the coeffi-
cient of the x term by twice the coefficient of
the x 2 term, we have the X coordinate of the
maximum or minimum point. If we substitute
this value for x in the original equation, the
result is the Y value or ordinate, which corre-
sponds to the X value.
For example, we know that the graph of the
equation
x 2 +
2x - 8 = y
passes through a minimum value because a is
positive. To find the coordinates of the point
where the parabola has its minimum value, we
note that a = 1, b = 2, c = -8. From the rule
given above, the X value of the minimum point is
2a
X =
_
" 2(1)
x = -1
Substituting this value for x in the original
equation, we have the value of the Y coordinate
of the minimum point. Thus,
(.1)2 + 2 (-l) - 8 = y
1 - 2 - 8 = y
-9 = y
The minimum point is (-1, -9). From the graph
in figure 16-1 (A), we see that these coordi-
nates are correct. Thus, we can quickly and
easily find the coordinates of the minimum or
maximum point for any quadratic of the form
ax 2 + bx + c = 0.
Practice problems. Without graphing, find
the coordinates of the maximum or minimum
points for the following equations and state
whether they are maximum or minimum.
1. 2x 2 - 5x + 2 =
2. 68 - 3x - x 2 =
3. 3 + 7x - 6x 2 =
4. 24x 2 - 14x = 3
174
Figure 16-2. Graph of x 2 - 4x - 8 = 0.
Answers:
5 . .
x = -j Minimum
7
= yjj Maximum
Maximum
y = 281
y 4
121
14
7
4. x = gi Minimum
121
175
1 1 1 i ,';:! ;; : i : ; ; ['.4 ' '..! , 1} ; :'.i ; ;.. f^Li/*j-!-' : ''lijli-i
m &M$m$$
Figure 16-3. Graph of 6x - 5 - x 2 = 0.
THE DISCRIMINANT
The roots of a quadratic equation may be
classified in accordance with the following
criteria:
1. Real or imaginary.
2. Rational or irrational.
3. Equal or unequal.
The task of discriminating among these possi-
ble characteristics to find the nature of the
roots is best accomplished with the aid of the
quadratic formula. The part of the quadratic
formula which is used is called the DISCRIMI-
NANT.
If the roots of a quadratic are denoted by the
symbols r : and r 2 , then the following relations
may be stated:
-b -
- 4ac
-b +
- 4ac
2a
We can show that the character of the roots
is dependent upon the form taken by the expres-
sion
b 2 - 4ac
which is the quantity under the radical in the
formula. This expression is the DISCRIMI-
NANT of a quadratic equation.
IMAGINARY ROOTS
Since there is a radical in each root, there
is a possibility that the roots could be imagi-
nary. They are imaginary when the number
under the radical in the quadratic formula is
negative (less than 0). In other words, when
the value of the discriminant is less than 0, the
roots are imaginary.
176
"~r*
EXAMPLE:
x 2 + x + 1 =
a = 1, b = 1, c = 1
b 2 - 4ac = (I) 2 - 4(1)(1)
= 1-4
= -3
rhus, without further work, we know that the
roots are imaginary.
CHECK: The roots are
r, =
-1
-i -
fi = -
1 i -V5
2 + ~2~
Ve recognize both of these numbers as being
m aginary.
We may also conclude that when one root is
maginary the other will also be imaginary,
rhis is because the pairs of imaginary roots
ire always conjugate complex numbers. If one
oot is of the form a + ib, then a - ib is also
i root. Knowing that imaginary roots always
>ccur in pairs, we can conclude that a quad-
atic equation always has either two imaginary
oots or two real roots.
Practice problems. Using the discriminant,
itate whether the roots of the following equa-
ions are real or imaginary:
.. x 2 - 6x - 16 =
!. x 2 - 6x = -12
i. 3x 2 - lOx + 50 =
;. 6x 2 + x = 1
Answers:
. Real
. Imaginary
. Imaginary
. Real
1QUAL OR DOUBLE ROOTS
If the discriminant b 2 - 4ac equals zero, the
adical in the quadratic formula becomes zero.
In this case the roots are equal; such roots are
sometimes called double roots.
Consider the equation
9x 2 + 12x + 4 =
Comparing with the general quadratic, we no-
tice that
a = 9, b = 12, and c = 4
The discriminant is
b 2 - 4ac = 12 2 - 4(9) (4)
= 144 - 144
=
Therefore, the roots are equal.
CHECK: From the formula
-12 -b -12-0
I
2(9)
2(9)
'.--I
The equality of the roots is thus verified.
The roots can be equal only if the trinomial
is a perfect square. Its factors are equal.
Factoring the trinomial in
we see that
9x 2 + 12x + 4 =
(3x + 2) =
Since the factor 3x + 2 is squared, we actu-
ally have
3x + 2 =
twice , and we have
x ~ "3
twice.
The fact that the same root must be counted
twice explains the use of the term "double
root." A double root of a quadratic equation is
always rational because a double root can oc-
cur only when the radical vanishes.
177
wnen me discriminant is positive, me roots
must be real. Also they must be unequal since
equal roots occur only when the discriminant
is zero.
Rational Roots
If the discriminant is a perfect square, the
roots are rational. For example, consider the
equati on
3x 2 - x - 2 =
in which
a = 3, b = -1, and c = -2
The discriminant is
b 2 - 4ac = (-1) 2 - 4(3) (-2)
= 1 + 24
= 25
We see that the discriminant, 25, is a per-
fect square. The perfect square indicates that
the radical in the quadratic formula can be re-
moved, that the roots of the equation are ra-
tional, and that the trinomial can be factored.
In other words, when we evaluate the discrimi-
nant and find it to be a perfect square, we know
that the trinomial can be factored.
Thus,
3x 2 - x - 2 =
(3x + 2)(x - 1) =
from which
3x + 2 =
x - 1 =
x = 1
We see that the information derived from the
discriminant is correct. The roots are real,
unequal, and rational.
Irrational Roots
If the discriminant is not a perfect square,
the radical cannot be removed and the roots
are irrational.
2x z - 4x + 1 =
in which
a = 2, b = -4, and c = 1.
The discriminant is
b 2 - 4ac = (-4) 2 - 4(2) (1)
= 16-8
This discriminant is positive and not a perfect
square. Thus the roots are real, unequal, and
irrational.
To check the correctness of this information,
we derive the roots by means of the formula.
Thus,
x =
-b
- 4ac
2a
4
2
1
x = 1 +
or x = 1 -
This verifies the conclusions reached in
evaluating the discriminant. When the dis-
criminant is a positive number, not a perfect
square, it is useless to attempt to factor the
trinomial. The formula is needed to find the
roots. They will be real, unequal, and irrational.
SUMMARY
The foregoing information concerning the
discriminant may be summed up in the follow-
ing four rules:
1. If b 2 - 4ac is a perfect square or zero,
the roots are rational; otherwise they are
irrational.
2. If b 2 - 4ac is negative (less than zero),
the roots are imaginary.
3. If b 2 - 4ac is zero, the roots are real,
equal, and rational.
4. If b 2 - 4ac is greater than zero, the roots
are real and unequal.
178
1. x" - 7x + 12 =
2. 9x 2 - 6x + 1 =
3. 2x 2 - x + 1 =
4. 2x - 2x 2 + 6 =
Answers:
1. Real, unequal, rational
2. Real, equal, rational
3. Imaginary
4. Real, unequal, irrational
GRAPHICAL INTERPRETATION
OF ROOTS
When a quadratic is set equal to y and the
resulting equation is graphed, the graph will
reveal the character of the roots, but it may
not reveal whether the roots are rational or
irrational.
Consider the folio-wing equations:
1. x 2 + 6x - 3 = y
2.x 2 +6x+9=y
3. x 2 + 6x + 13 = y
The graphs representing these equations are
shown in figure 16-4.
We recall that the roots of the equation are
the values of x at those points where y is zero.
Y is zero on the graph anywhere along the X
axis. Thus, the roots of the equation are the
positions where the graph crosses the X axis.
In parabola No. 1 (fig. 16-4) we see immedi-
ately that there are two roots to the equation
and that they are unequal. These roots appear
to be -6.5 and 0.5. Algebraically, we find them
to be the irrational numbers
-3 + 2-^/3 and -3 -2 /5".
For equation No. 2 (fig. 16-4), the parabola
just touches the X axis atx = -3. This means
that both roots of the equation are the same
that is, the root is a double root. At the point
where the parabola touches the X axis, the two
roots of the quadratic equation have moved
Figure 16-4. Graphical interpretation of roots.
together and the two points of intersection of the
parabola and the X axis are coincident. The
quantity -3 as a double root agrees with the
algebraic solution.
When the equation No. 3 (fig. 16-4) is solved
algebraically, we see that the roots are -3 + 2i
and -3 - 2i. Thus they are imaginary. Para-
bola No. 3 does not cross the X axis. When this
situation occurs, imaginary roots are implied.
Only equations having real roots will have
graphs that cross or touch the X axis. Thus we
may determine from the graph of an equation
whether the roots are real or imaginary.
VERBAL PROBLEMS
INVOLVING QUADRATIC EQUATIONS
Many practical problems give rise to quad-
ratic equations. In such problems it often hap-
pens that one of the roots will have no meaning.
We must select the root that satisfies the con-
ditions of the problem .
Consider the following example: The length
of a plot of ground exceeds its width by 7 ft and
the area of the plot is 120 sq ft. What are the
dimensions?
179
SOLUTION:
then
and
Let x = length
y = width
x - y = 7
xy = 120
Solving (1) for y, y = x - 7
Substituting (x - 7) for y in (2)
x(x - 7) = 120
Therefore
x 2 - 7x - 120 =
(x - 15){x + 8) =
x = 15, x = -8
Thus, length = +15 or -8.
(1)
(2)
But the length obviously cannot be a negative
value. Therefore, we reject -8 as a value for
x and use only the positive value, +15. Then
from equation (1),
15 - y = 7
y = 8
Length =15, Width = 8
Practice problems. Solve the following
problems by forming quadratic equations:
1. A rectangular plot is 8 yd by 24 yd. If the
length and width are increased by the same
amount, the area is increased by 144 sq yd.
How much is each dimension increased?
2. Two cars travel at uniform rates of speed
over the same route a distance of 180 mi. One
goes 5 mph slower than the other and takes
1/2 hr longer to make the run. How fast does
each car travel?
Answers:
1. Length and width are each increased by 4yd.
2. Faster car: 45 mph.
Slower car: 40 mph.
180
CHAPTER 17
PLANE FIGURES
The discussion of lines and planes in chap-
ter 1 of this course was limited to their con-
sideration as examples of sets. The present
chapter is concerned with lines, angles, and
areas as found in various plane (flat) geometric
figures.
LINES
In the strictly mathematical sense, the term
"line segment" should be used whenever we re-
fer to the straight line joining some point A to
some other point B. However, since the straight
lines comprising geometric figures have clearly
designated end points, we may simplify our
terminology. Throughout the remaining chap-
ters of this course, the general term "line" is
used to designate straight line segments, unless
stated otherwise.
TYPES OF LINES
The two basic types of lines in geometry are
straight lines and curved lines. A curved line
joining points A and B is designated as "curve
AB." (See fig. 17-1.) If curve AB is an arc of
a circle, it may be designated as "arc AB."
BROKEN LINE DASHED LINE
Figure 17-2. Broken and dashed lines.
ORIENTATION
Straight lines may be classified in terms of
their orientation to the observer's horizon or
in terms of their orientation to each other. For
example, lines in the same plane which run be-
side each other without meeting at any point,
no matter how far they are extended, are PAR-
ALLEL. (See fig. 17-3 (A).) Lines in the same
plane which are not parallel are OBLIQUE.
Oblique lines meet to form angles (discussed in
the following section). If two oblique lines
cross or meet in such a way as to form four
equal angles, as in figure 17-3 (B), the lines
are PERPENDICULAR. This definition includes
the case in which only one angle is formed,
such as angle AEC in figure 17-3 (C). By ex-
tending line AE to form line AD, and extending
CE to form CB, four equal angles (AEC, CED,
DEB, and BEA) are formed.
A B A
LINE AB CURVE AB
Figure 17-1. Straight and curved lines.
The term "broken line" in mathematics
means a series of two or more straight seg-
ments connected end-to-end but not running In
the same direction. In mathematics, a series
of short, straight segments with breaks be-
tween them, which would form a single straight
line if joined end-to-end, is a DASHED LINE.
(See fig. 17-2.)
(A)
(B)
A
)--- I ..(
I
A
(0
Figure 17-3. -(A) Parallel lines; (B) and (C)
perpendicular lines.
Lines parallel to the horizon are HORIZON-
TAL. Lines perpendicular to the horizon are
VERTICAL.
181
MATHEMATICS, VOLUME 1
ANGLES
Lines which meet or cross each other are
said to INTERSECT. Angles are formed when
two straight lines intersect. The two lines
which form an angle are its SIDES, and the point
where the sides intersect is the VERTEX. In
figure 17-4, the sides of the angles are AV and
BV, and the vertex is V in each case. Figure
17-4 (A) is an ACUTE angle; (B) is an OBTUSE
angle.
(A)
(B)
Figure 17-4. (A) Acute angle; (B) obtuse angle.
CLASSIFICATION BY SIZE
When the sides of an angle are perpendicular
to each other, the angle is a RIGHT angle. This
term is related to the Latin word "rectus,"
which may be translated "erect 11 or "upright. 11
Thus, if one side of a right angle is horizontal,
the other side is erect or upright.
The size of an angle refers to the amount of
separation between its sides, and the unit of
angular size is the angular DEGREE. A right
angle contains 90 degrees, abbreviated 90. An
angle smaller than a right angle is acute; an
angle larger than a right angle is obtuse. There-
fore, acute angles are angles of less than 90,
and obtuse angles are angles between 90 and
180.
If side AV in figure 17-5 (A) is moved down-
V
(A)
Figure 17 -5. -(A) LJ
(B) straigh
figure 17-6 are VERTIC
because they share a con
and 4 are opposite eac
vertical angles. Lines wt
17-6, always form two pi
and the vertical angles th
pairs; that is, angle 1 eq
2 equals angle 4.
Figure 17-6. -V<
Angles 1 and 2 infigui
angles. Other pairs of i
ure 17-6 are 2 and 3, 3
the sense used here, ac
side, not merely close
For example, angles 1 i
angles even though they t
COMPLEMENTS AND S 1
Two angles whose su
Chapter 17 -PLANE FIGURES
: two equal angles are complementary, each
ains how many degrees?
'ind the size of an angle which is twice as
e as its own complement.
t: If x is the angle, then 90 - x is its
plement.)
definition assumes that the standard po
a triangle drawn for general discussi<
shown in figure 17-7, in which the tri
lying on one of its sides. The vertex
the base is the highest point of a tri;
standard position, and is thus called th<
nswers:
.cute
ight
5
GEOMETRIC FIGURES
'he discussion of geometric figures in this
iter is limited to polygons and circles. A
lYGON is a plane closed figure, the sides of
;h are all straight lines. Among the poly-
i discussed are triangles, parallelograms,
trapezoids.
ANGLES
. triangle is a polygon which has three sides
three angles. In general, any polygon has
lany angles as it has sides, and conversely.
:s of a Triangle
ach of the three angles of a triangle is a
TEX; therefore, every triangle has three
ices. The three straight lines joining the
ices are the SIDES (sometimes called legs),
the side upon which the triangle rests is its
E, often designated by the letter b. This
APEX
BASE
Figure 17-7. Triangle in standard pc
A straight line perpendicular to the
a triangle, joining the base to the apex
ALTITUDE, often designated by the ]
The altitude is sometimes referred tc
height, and is then designated by the 1
Figure 17-8 (B) shows that the apex ma
situated directly above the base. In th
the base must be extended, as shown
dashed line, in order to drop a perpe
from the apex to the base. Mathem
often use the term "drop a perpend
The meaning is the same as "draw a i
perpendicular line."
In general, the geometrical term "
from a point to a line" means the ler
perpendicular dropped from the poin
line. Many straight lines could be dra
a line to a point not on the line, but the
of these is the one we use in measu:
distance from the point to the line. The short-
est one is perpendicular to the line.
Perimeter and Area
The PERIMETER of a triangle is the sum of
the lengths of its sides. In less precise terms,
this is sometimes stated as "the distance
around the triangle." If the three sides are
labeled a, b, and c, the perimeter P can be
found by the following formula:
P = a + b + c
/
The area of a triangle is the space bounded
(enclosed) by its sides. The formula for the
area can be found by using a triangle which is
part of a rectangle. In figure 17-9, triangle
ABC is one -half of the rectangle. Since the
area of the rectangle is a times b (that is, ab),
the area of the triangle is given by the follow-
ing formula:
Area = ^ ab
Written in terms of h, representing height,
th- formula is:
A =
bh
This formula is valid for every triangle, in-
cluding those with no two sides perpendicular.
Figure 17-10. Perime
of triangles
Answers:
1. P = 12 units 3.
A = 6 square units
2. P = 16 units 4.
A = 12 square units
CAUTION: The concept c
lj-io-i if 4-U,, ..-.:t-~ ^e xi i
Chapter 17-PLANE FIGURES
.1 Triangles
3 classification of triangles depends upon
special characteristics, if any. For ex-
, a triangle may have all three of its sides
in length; it may have two equal sides and
I side which is longer or shorter than the
two; it may contain a right angle or an
i angle. If it has none of these special
iteristics, it is a SCALENE triangle. A
ie triangle has no two of its sides equal
1 two of its angles equal.
}HT TRIANGLE. -If one of the angles of a
le is a right angle, the figure is a right
le. The sides which form the right angle
e LEGS of the triangle, and the third side
;ite the right angle) is the HYPOTENUSE.
2 area of a right triangle is always easy
3rmine. If the base of the triangle is one
legs, as in figure 17-10 (4), the other leg
altitude. If the hypotenuse is acting as
ise, as in figure 17-10 (3), the triangle
s turned until one of its legs is the base,
figure 17-10 (1). If the triangle is not
to be a right triangle, then the altitude
36 given, as in figure 17-10 (2), in order
culate the area.
y triangle whose sides are in the ratio of
is a right triangle. Thus, triangles with
as follows are right triangles:
>ide 1
3
6
12
3x
Side 2
4
8
16
4x
Side 3
5
10
20
5x
Figure 17 -11. (A) Isoceles triangh
(B) equilateral triangle.
Figure 17-11 (B) illustrates an EQTJ
ERAL triangle, which is a special case
isosceles triangle. An equilateral trian^
all three of its sides equal in length. Sii
lengths of the sides are directly related
size of the angles opposite them, an equi
triangle is also equiangular; that is, all
of its angles 'are equal.
OBLIQUE TRIANGLES. -Any triangh
taining no right angle is an OBLIQUE tr
Figure 17-12 illustrates two possible c<
rations, both of which are oblique trij
An oblique triangle which contains an
angle is often called an OBTUSE triangle
(A) ACUTE
(B) OBTUSE
(x is any positive number)
Figure 17-12. Oblique triangles.
(A) Acute; (B) obtuse.
MATHEMATICS, VOLUME 1
QUADRILATERALS
A QUADRILATERAL is a polygon with four
sides. The parts of a quadrilateral are its
sides, its four angles, and its two DIAGONALS.
A diagonal is a straight line joining two alter-
nate vertices of a polygon. Figure 17-13 illus-
trates the parts of a quadrilateral, in which
AC and DB are the diagonals.
Figure 17-13. Parts of a quadrilateral.
Perimeter and Area
The perimeter of a quadrilateral is the sum
of the lengths of its sides. For example, the
perimeter of the quadrilateral in figure 17-13
is 30 units.
Figure 17-14. A p
Since lines AB and CI
DE and CF (both perpend
figure 17-14) are equal,
in figure 17-14 are equal
line cutting two parallel 1
BC, forms equal angles w
Thus, triangles AED and
line AD equals line BC.
proved that the opposite
gram are equal. If all fou
same length, the parallelo
In addition to the equ
sides, the opposite angle
are also equal. For exam]
angle BCD in figure 17-
equals angle ABC.
RECTANGLES AND SC
the angles of a parallelogi
it is a RECTANGLE. A r
of its sides the samelengt
a square is a rhombus hav
square is a rectangle, anc
parallelogram. Notice th;
statement is not true.
The area of a rectang!
plying its length times its
each side of a square has
thp smiarp* is s2_
Chapter 17 -PLANE FIGURES
gle AED and figure EBCD. Since triangle
is equal to triangle BFC, the sum of AED
EBCD is equal to the sum of BFC and
D. Thus the area of parallelogram ABCD
,e same as the area of rectangle EFCD.
j the area of EFCD is DC multiplied by
and DC has the same length as AB, we
lude that the area of a parallelogram is the
act of its base times its altitude. Written
formula, this is
A = ba
A = bh, where h is height
iczoids
TRAPEZOID is a quadrilateral in which
sides are parallel and the other two sides
not parallel. By orienting a trapezoid so
its parallel sides are horizontal, we may
the parallel sides bases. Observe that the
s of a trapezoid are not equal in length,
fig. 17-15.)
(A) (B)
Figure 17-15. Typical trapezoids.
he area of a trapezoid may be found by
rating it into two triangles and a rectangle,
i figure 17-16. The total area A of the
^zoid is the sum of A x plus A 2 plus A 3 , and
Jculated as follows:
bi
Figure 17-16. Area of a trapezoi
Practice problems. Find the area of
the following figures:
1. Rhombus; base 4 in., altitude 3 in.
2. Rectangle; base 6 ft, altitude 4 ft
3. Parallelogram; base 10 yd, altitude 1
4. Trapezoid; bases 6 ft and 4 ft, altitu
Answers:
1. 12 sq in.
2. 24 sq ft
CIRCLES
3. 40 sq yd
4. 30 sq ft
The mathematical definition of a circ
that it is a plane figure bounded by a
line, every point of which is equally
from the center of the figure. The pa
circle are its circumference, its radi
its diameter.
MATHEMATICS VOLUME 1
Figure 17-17. Parts of a circle.
An ARC is a portion of the circumference of
a circle. A CHORD is a straight line joining
the end points of any arc. The portion of the
area of a circle cut off by a chord is a SEG-
MENT of the circle, and the portion of the
circle's area cut off by two radii (radius lines)
is a SECTOR. (See fig. 17-18.)
Formulas for Circumference and Area
The formula for the circumference of a
circle is based on the relationship between the
circumference and the diameter. This com-
parison can be made experimentally by mark-
ing the edge of a circular object, such as a
coin, and rolling it (without slippage) along a
Figure 17-18. Arc, chori
C=3.I4
INITIAL POSITION
Figure 17-19. Measuri
of a cii
This formula states that
flat
firr 17-1Q
Chapter 17 -PLANE FIGURES
:tice problems. Calculate the circum-
of each of the following circles, using
the value of n:
us = 21 in. 3. Radius = 14 ft
neter = 7.28 in. 4. Diameter = 2.8 yd
vers:
in.
8 in.
3. 88 ft
4. 8.8 yd
)A. The area of a circle is found by
ying the square of its radius by TT. The
i is written as follows:
A = Trr 2
?LE: Find the area of a circle whose
er is 4 ft, using 3.14 as the value of n.
ION: The radius is one-half the diam-
rherefore,
r = |(4 ft)
= 2 f t
A = ?rr 2 = ?r(2 ft) 2
A = 3.14 (4 sq ft)
= 12.56 sq ft
Then
Figure 17 -20. Concentric circles.
Let R = radius of large circle
r = radius of small circle
A R = area of large circle
A r = area of small circle
A = area of ring
A = A R - A
= TfR 2 - TTr 2
= Tr(R 2 - r 2 )
CHAPTER 18
GEOMETRIC CONSTRUCTIONS AND SOLID F
Many ratings in the Navy involve work which
requires the construction or subdivision of
geometric figures. For example, materials
must be cut into desired shapes, perpendicular
lines must be drawn, etc. In addition to these
skills, some Navy ratings require the ability to
recognize various solid figures and calculate
their volumes and surface areas.
CONSTRUCTIONS
From the standpoint of geometry, a CON-
STRUCTION may involve either the process of
building up a figure or that of breaking down a
figure into smaller parts. Some typical con-
structions are listed as follows:
1. Dividing a line into equal segments.
2. Erecting the perpendicular bisector of a
line.
3. Erecting a perpendicular at any point on
a line.
4. Bisecting an angle.
5. Constructing an angle.
6. Finding the center of a circle.
7. Constructing an ellipse.
EQUAL DIVISIONS ON A LINE
A line may be divided into any desired num-
ber of equal segments by the method shown in
figure 18-1.
off seven spaces of some co
1/2 inch, on it. Extend A
order to get seven intervals
on it. This produces the p<
and g, as shown in figure
from g to B, and then draw
starting at each of the points
The segments of AB cut ofi
equal in length.
It is frequently necess
determined number of lines
material. This may be don
on the foregoing discussion,
pose that the sheet of typi
18-2 is to be divided into 24
The 12-inch ruler is la
at an angle, in such a way
Chapter IB-GEOMETRIC CONSTRUCTIONS AND SOLID FIGURES
coincide with the top and bottom edges of
Der. There are 24 spaces, each 1/2 inch
on a 12-inch ruler. Therefore, we mark
Der beside each 1/2-inch division marker
ruler. After removing the ruler, we
i line through each of the marks on the
parallel to the top and bottom edges of
ENDICULAR BISECTOR
LINE
bisect a line or an angle means to divide
two equal parts. A line may be bisected
tctorily by measurement, or by a geo-
: method. If the measuring instrument
iot reach the full length of the line, pro-
s follows:
Starting at one end, measure about half
igth of the line and make a mark.
Starting at the other end, measure exactly
cne distance as before and make a second
rhe bisector of the line lies halfway be-
these two marks.
j geometric method of bisecting a line is
"pendent on measurement. It is based
le fact that all points equally distant from
ds of a straight line lie on the perpen-
r bisector of the line.
ecting a line geometrically requires the
a mathematical compass, which is an in-
snt for drawing circles and comparing
:es. If a line AB is to be bisected as in
18-3, the compass is opened until the
:e between its points is more than half as
s AB. Then a short arc is drawn above
proximate center of the line and another
using A as the center of the arcs' circle.
g. 18-3.)
"> mnrp shnrt nrr.s arp rirawn nnp above
A
V
Figure 18-3. Bisecting a line geometric
Figure 18-4.-
-Erecting a perpendicul
at a point.
1. Using any convenient point above t
(such as O) as a center, draw a circle wi
dius OC. This circle cuts AB at C and
2. Draw line DO and extend it to int
the circle at E.
3. Draw line EC. This line is perpenc
MATHEMATICS, VOLUME 1
Figure 18-5. Bisecting an angle.
SPECIAL ANGLES
Several special angles may be constructed
by geometric methods, so that an instrument
for measuring angles is not necessary in these
special cases.
Figure 18-4 illustrates a method of con-
structing a right angle, DCE, by inscribing a
right triangle in a semicircle. But an alternate
method is needed for those situations in which
drawing circles is inconvenient. The method
described herein makes use of a right triangle
having its sides in the ratio of 3 to 4 to 5. It is
often used in laying out the foundations of build-
ings. The procedure is as follows:
1. A string is stretched as shown in figure
18-6, forming line AC. The length of AC is
3 feet.
2. A second string is stretched, crossing
line AC at A, directly above the point intended
as the corner of the foundation. Point D on this
line is 4 feet from A.
3. Attach a third string, 5 feet long, at C
and D. When AC and AD are spread so that line
CD is taut, angle DAC is a right angle.
A 60 angle is constructed as shown in fig-
ure 18-7. With AB as a radius and A and B as
centers, draw arcs intersec
and B are connected to C by
three angles of triangle AI
The special angles alre
used in constructing 45 and
angle is bisected to form twi
60 angle is bisected to fo]
FINDING THE CENTER
OF A CIRCLE
It is sometimes necessar
of a circle of which only an
given. (See fig. 18-8.)
From any point on the ar
two chords intersecting th
points, such as B and C. W
and C as centers, use any
and draw short intersecting
perpendicular bisectors of c
Join the intersecting arcs o
obtaining line MP, and join
side of AB, obtaining line N<^
of MP and NQ is point O,
circle.
ELLIPSES
An ellipse of specified 1
constructed as follows:
1. Draw the major axis,
axis, CD, as shown in figure
2. On a straightedge or r
(labeled a in the figure) an
measure one -half the length
and make a second mark (
From point a, measure one
Chapter IB-GEOMETRIC CONSTRUCTIONS AND SOLID FIGURES
A B
ure 18-7. Constructing 60 angles.
M
Figure 18-8. Finding the center
of a circle.
SOLID FIGURES
The plane figures discussed in chapter
this course are combined to form solid fig
For example, three rectangles and two tria
may be combined as shown in figure 1
The flat surfaces of the solid figure ai
FACES; the top and bottom faces are theB^
and the faces forming the sides are the
ERAL FACES.
EDGE
UPPER
BASE
LATERAL
FACE '
LATERAL
EDGE
LOWER
BASE
Figure 18-10. -Parts of a solid figure
Some solid figures do not have any flat :
and some have a combination of curved su]
and flat surfaces. Examples of solids
curved surfaces include cylinders, cones
spheres.
PRISMS
The solid shown in figure 18-10 is a P
A prism is a solid with three or more 1
faces which intersect in parallel lines.
Types of Prisms
MATHEMATICS, VOLUME 1
prism, and if its bases are rectangles, it is a
rectangular solid. A CUBE is a rectangular
solid in which all of the six rectangular faces
are squares.
Parts of a Prism
The parts of a prism are shown in figure
18-10. The line formed by the joining of two
faces of a prism is an EDGE. If the two faces
forming an edge are lateral faces, the edge
thus formed is a LATERAL EDGE.
Surface Area and Volume
The SURFACE AREA of a prism is the sum
of the areas of all of its faces, including the
bases. The VOLUME of a prism may be con-
sidered as the sum of the volumes of many thin
wafers, each having a thickness of one unit and
a shape that duplicates the shape of the base.
(See fig. 18-11.)
Figure 18-12. -He
which is not a :
CIRCULAR CYLINDERS
Any surface may be cc
of moving a straight line
angles to its length. For
the stick of charcoal in J
from position CD to posi
across the paper. The bi
charcoal represents a pis
face is said to be "gene:
AB.
Figure 18-11. Volume of a prism.
The wafers which comprise the prism in
figure 18-11 all have the same area, which is
the area of the base. Therefore, the volume of
the prism is found by multiplying the area of
CHARCOAL
STICK
Fieure 18- 13. -Surf
Chapter 18-GEOMETRIC CONSTRUCTIONS AND SOLID FIGURES
Figure 18-14. (A) Line generating a cylinder;
(B) elliptical cylinder;
(C) circular cylinder.
are circles, the cylinder is a CIRCULAR
FDER. Figure 18-14 (C) illustrates a
circular cylinder. Line O-O T , joining the
s of the bases of a right circular cylin-
5 the AXIS of the cylinder.
e Area and Volume
5 lateral area of a cylinder is the area of
rved surface, excluding the area of its
Figure 18-15 illustrates an experimen-
thod of determining the lateral area of a
circular cylinder.
The card of length L and width W in
18-15 is rolled into a cylinder. The he
the cylinder is W and the circumference
The lateral area is the same as the o:
area of the card, LW. Therefore, the ]
area of the cylinder is found by multiply
height by the circumference of its base,
ten as a formula, this is
A = Ch
EXAMPLE: Find the lateral area of a
circular cylinder whose base has a rad
4 inches and whose height is 6 inches.
W
MATHEMATICS, VOLUME 1
SOLUTION: The circumference of the base is
C = ?rd
C = 3.14 x 8 in.
= 25.12 in.
Therefore,
A = 25.12 in. x 6 in.
= 151 sq in. (approximately)
The formula for the volume of a cylinder is
obtained by the same reasoning process that
was used for prisms. The cylinder is consid-
ered to be composed of many circular wafers,
or disks, each one unit thick. The area of each
disk, multiplied by the number of disks, is the
volume of the cylinder. With V representing
volume, A ..representing the area of each disk,
and n representing the number of disks, the
formula is as follows:
Since the number of disks is the same as the
height of the cylinder, the formula for the vol-
ume of a cylinder is normally written
V = Bh
In this formula, B is the area of the base and h
is the height of the cylinder.
EXAMPLE: Determine the volume of a circular
cylinder with a base of radius 5 inches and a
height of 14 inches.
SOLUTION:
2. Determine the volumi
problem 1.
Answers:
1. 88 sq in.
2.
REGULAR PYRAMIDS AI
RIGHT CIRCULAR CONE
A PYRAMID is a sol
faces of which are triang!
A REGULAR PYRAMID
faces equal.
Figure 18-16. -(A) Ir:
(B) regular j
A regular pyramid with
of lateral faces would ha\
many sides. If the numt
ciently large, the base pol
able from a circle and tl
the many lateral faces
curved surface. The soli
is a RIGHT CIRCULAR C(
A
Chapter 18-GEOMETRIC CONSTRUCTIONS AND SOLID FIGURES
igth of line AV in figure 18-18 (A) is the
leight. The slant height of a right circu-
le is the length of any straight line join-
5 vertex to the circumference line of the
Such a line is perpendicular to a line
t to the base at the point where the slant
intersects the base. (See fig. 18-18 (B).)
AV, BV, and CV in figure 18-18 (B) are
nt heights.
i 18-18. (A) Slant height of a regular
tmid; (B) slant height of a right circular
.1 Area
i lateral area of a pyramid is the sum of
5as of its lateral faces. If the pyramid is
r, its lateral faces have equal bases;
rmore, the slant height is the altitude of
ice. Therefore, the area of each lateral
3 one-half the slant height multiplied by
gth of one side of the base polygon. Since
m of these sides is the perimeter of the
the total lateral area of the pyramid is
Volume
The volume of a pyramid is determii
its base and its altitude, as is the cas
other solid figures. Experiments show tl
volume of any pyramid is-one-thirdof the
uct of its base and its altitude. This n
stated as a formula with V representin
ume, B representing the area of the bas
h representing height (altitude), as follow
V-i.Bh
The formula for the volume of a pj
does not depend in any way upon the nun-
faces. Therefore, we use the same fornn
the volume of a right circular cone. Sir
base is a circle, we replace B with Trr 2
r is the radius of the base). The formi
the volume of a right circular cone is the
Practice problems:
1. Find the lateral area of a regular pj
with a 5 -sided base measuring 3 inches c
side, if the slant height is 12 inches.
2. Find the lateral area of a right circula
whose base has a diameter of 6 cm and
slant height is 14 cm.
3. Find the volume of a regular pyramid
square base measuring 4 cm on each s
the vertex is 9 cm above the base.
MATHEMATICS, VOLUME 1
Figure 18-19. Parts of a sphere.
In figure 18-19, the center of the sphere is
point O.
A RADIUS of a sphere is a straight line seg-
ment joining the center of the sphere to a point
on the surface. Lines OA, OB, OC, OD, OE,
and OF in figure 18-19 are radii. A DIAMETER
of a sphere is a straight line segment joining
two points on the surface and passing through
the center of the sphere. Lines AB, CD, and
EF in figure 18-19 are diameters. A HEMI-
SPHERE is half of a sphere.
Circles of various sizes may be drawn on
the surface of a sphere. The largest circle
that may be so drawn is one with a radius equal
to the radius of the sphere. Such a circle is a
GREAT CIRCLE. In figure 18-19, circles
AEBF, ACBD, and CEDF are great circles.
On the surface of a sphere, the shortest dis-
tance between two points is an arc of a great
circle drawn so that it passes through the two
points. This explains the importance of great
circles in the science of navigation, since the
earth is approximately a sphere.
The formula for the sur
may be rewritten as follov
A = (2-nr]
When the formula is facto
easy to see that the surfac
simply its circumference
Volume
The volume of a sphe]
is given by the formula
EXAMPLE: Find the volu
diameter is 42 inches.
SOLUTION:
= I x 3. 14 x (21 i
= | x 3.14x21 x
= 4.187x21 x21 >
= 38,776 cu in. (aj
Practice problems. C
area and the volume of t
the following problems:
1. Radius = 7 inches
Answers:
CHAPTER 19
NUMERICAL TRIGONOMETRY
The word "trigonometry" means "measure-
ment by triangles." As it is presented in many
textbooks, trigonometry includes topics other
than triangles and measurement. However, this
chapter is intended only as an introduction to
the numerical aspects of trigonometry as they
relate to measurement of lengths and angles.
SPECIAL PROPERTIES OF
RIGHT TRIANGLES
A RIGHT TRIANGLE has been defined as
any triangle containing a right angle. The side
opposite the right angle in a right triangle is a
HYPOTENUSE. (See fig. 19-1.) In figure 19-1,
s ; e AC is the hypotenuse.
x
{A)
Figure 19-2. The Pythagorc
(A) General triangle; (B) t]
sides of specific len
labeled as in figure 19-2 (A),
Theorem is stated in symbols a
x 2 + y 2 = r 2
An example of the use of
Theorem in a problem follows:
EXAMPLE: Find the length o
in the triangle shown in figure
SOLUTION:
EXAMPLE: An observer on 5
figure 19-3, knows that his dis
C is 1,200 yards and that the
MATHEMATICS, VOLUME 1
. A
Figure 19-3. Using the Pythagorean Theorem.
SIMILAR RIGHT TRIANGLES
Two right triangles are SIMILAR if one of
the acute angles of the first is equal to one of
the acute angles of the second. This conclusion
is supported by the following reasons:
1. The right angle in the first triangle is
equal to the right angle in the second, since all
right angles are equal.
2. The sum of the angles of any triangle is
180. Therefore, the sum of the two acute
angles in a right triangle is 90.
3. Let the equal acute angles in the two tri-
angles be represented by A and A' respectively.
(See fig. 19-4.) Then the other acute angles,
B and B', are as follows:
B = 90 - A
B' = 90 - A'
A
B'
angle of the second have
spending angles equal. Th
are similar.
Practical situations freqi
similar right triangles are
lems. For example, the he
determined by comparing
shadow with that of a nearbj
in figure 19-5.
TREE
SHADOW
Figure 19-5.-Calculat
comparison of s
Assume that the rays of
and that the tree and flag
angles with the ground. r .
and A T B T C f are right trian
equal to angle B' . Thereto:
similar and their correspoi
portional, with the following
B'C'
Suppose that the flagpol
feet high, the shadow of the
fho aViaslmir rrf
-fla
Chapter 19 -NUMERICAL TRIGONOMETRY
^igure 19-6 represents an L-shaped build-
with dimensions as shown. On the line of
it from A to D, a stake is driven at C, a
it 8 feet from the building and 10 feet from
If ABC is a right angle, find the length of
and the length of AD. Notice that AE is 18
and ED is 24 feet.
-24 FT-
Figure 19-6. Using similar triangles.
Answers:
> feet 2. AB = 6 feet
AD = 30 feet
TRIGONOMETRIC RATIOS
rhe relationships between the angles and the
js of a right triangle are expressed in terms
^RIGONOMETRIC RATIOS. For example, in
re 19-7, the sides of the triangle are named
.ccordance with their relationship to angle 9 .
;rigonometry, angles are usually named by
ins of Greek letters. The Greek name of
fivmhnl f) is theta.
Ul P
SIDE ADJACENT
TO ANGLE
(A)
X
(B!
Figure 19 -7. Relationship of sides an<
in a right triangle. (A) Names of th
(B) symbols used to designate the sid<
Table 19-1. Trigonometric ratio;
Name of ratio
Abbreviat
sine of 9
sin 6
cosine of Q
cos 9
tangent of Q
tan B
cotangent of B
cot B
secant of 9
sec 8
cosecant of 9
esc 9
hypotenuse
side adjacent to 9
hypotenuse
side opposite to 9
The other acute angle in figure 19
labeled a fGreek alDha}. The side or
MATHEMATICS, VOLUME 1
be expressed as a common fraction or as a
decimal. For example,
sin 6 = I- = 0.800
o
sin a = ~ = 0.600
o
Decimal values have been computed for
ratios of angles between and 90, and values
for angles above 90 can be expressed in terms
of these same values by means of conversion
formulas. Appendix II of this training course
gives the sine, cosine, and tangent of angles
from to 90. The secant, cosecant, and
cotangent are calculated, when needed, by using
their relationships to the three principal ratios.
These relationships are as follows:
1
secant =
cosecant 6 =
cotangent d =
cosine 9
1
sine 6
1
tangent 6
TABLES
Tables of decimal values for the trigono-
metric ratios may be constructed in a variety
of ways. Some give the angles in degrees, min-
utes, and seconds; others in degrees and tenths
of a degree. The latter method is more com-
pact and is the method used for appendix II.
The "headings" at the bottom of each page in
appendix II provide a convenient reference
showing the minute equivalents for the decimal
fractions of a degree. For example, 12' (12
minutes) is the equivalent of 0.2.
the angle plus 0.0; in 01
0.0, or simply 35.0. ^
sine of 35.0 is 0.5736. B;
the sine of 42.7 is 0.67!
32.3 is 0.6322.
A typical problem in t:
the value of an unknown s
when only one side and
known. EXAMPLE: In tri;
find the length of AC if A
angle CAB is 34.7.
Figure 19-8. Using t
ratios to evaluat
SOLUTION:
AC
= cos 34
13
AC = 13 cos
= 13 x
= 10.69 i
Thp antrlpsnf a triano-l*
Chapter 19 NUMERICAL TRIGONOMETRY
9
13,
12
Figure 19-9. Using trigonometric
ratios to evaluate angles.
t triangle. The only information given,
'ning angle 9, is the ratio of sides in the
le. The size of is calculated as follows:
9 = ll = - 4167
9 = the angle whose tangent is 0.4167
mining that the sides and angles in figure
ire in approximately the correct proper -
we estimate that angle 9 is about 20.
.ble entries for the tangent in the vicinity
are slightly too small, since we need a
r near 0.4167. However, the tangent of
is 0.4163 and the tangent of 2242 T is
The following arrangement of numbe
recommended for interpolation:
ANGLE TANGENT
2236 1
2242'
.0004
0.4183
The spread between 22 36' and 2242'
and we use the comparison of the tangent
to determine how much of this 6' spread
eluded in 6 , the angle whose value is s
Notice that the tangent of 9 is different
tan 22 36' by only 0.0004, and the total s
in the tangent values is 0.0020. Therefoi
tangent of 9 is ' Q20 of the way betwei
tangents of the two angles given in the
This is 1/5 of the total spread, since
0.0004 _4_
0.0020 " 20
Another way of arriving at this res
to observe that the total spread is 2(
thousandths, and that the partial spreac
responding to angle 9 is 4 ten-thousa
Since 4 out of 20 is the same as 1 out of 5 3
e is 1/5 of the way between 22 36' and 2:
Taking 1/5 of the 6' spread betwee
angles, we have the following calculation:
A- V fi' - JL v Ei'fift"
^f A D ~fr X DU
= 1'12" (1 minute and 12 secor
The 12" obtained in this calculation caus
answer to appear to have greater accurac
MATHEMATICS, VOLUME 1
0.1 -<
ANGLE TANGENT
'22.60 0.41631
f 0.0004
B 0.4167 J
22.70 0.4183
2. Find the angle which cc
the following decimal value
a. sin B - 0.2790
> 0.0020 b. cos 9 = 0.9018
Answers:
In this example, we are concerned with an
angular spread of 0.10 and 9 is located 1/5 of
the way through this spread. Thus we have
9 = 22.60 -f i x 0.10'
\o
= 22.60 + 0.02
9 = 22.62
' Interpolation must be approached with com-
mon sense, in order to avoid applying correc-
tions in the wrong direction. For example, the
cosine of an angle decreases in value as the
angle increases from to 90. If we need the
value of the cosine of an angle such as 2239',
the calculation is as follows:
ANGLE
2242'
COSINE
0.9232
0.9225
0.0007
In this example, it is easy to see that 22 39'
is halfway between 2236' and2242'. There-
fore the cosine of 2239' is halfway between the
cosine of 2236' and that of 2242'. Taking
one-half of the spread between these cosines,
we then SUBTRACT from 0.9232 to find the
cosine of 2239', as follows:
1. a. 1
b. 0.8660
c. 0.7420
2. a. B = 16.2
b. 9 = 2536'
RIGHT TRIANG
SPECIAL ANGLES A1S
Three types of right tri;
significant because of th
rence. These are the 30-
45 -90 triangle, and the 3
THE 30-60-90 TRIANC
The 30-60-90 trian
cause these are the sizes
The sides of this trianglt
1 to N/T to 2 , as shown in f
cos 2239 T = 0.9232 - Mr x 0.0007
Chapter 19 -NUMERICAL TRIGONOMETRY
^.B is 2 units long and, by the rule of
oras, AC is found as follows:
AC = N/(AB)2 - (BC) 2
= \/4 - i = \rr
jardless of the size of the unit, a 30-
} triangle has a hypotenuse which is 2
as long as the shortest side. The short-
ie is opposite the 30 angle. The side op-
the 60 angle is \T~3 times as long as the
ist side. For example, suppose that the
muse of a 30 -60 -90 triangle is 30 units
then the shortest side is 15 units long,
e length of the side opposite the 60 angle
*>/3~ units,
ictice problems. Without reference to
or to the rule of Pythagoras, find the
ing lengths and angles in figure 19-11:
igth of AC .
e of angle A.
e of angle B.
4. Length of RT.
5. Length of RS.
6. Size of angle T.
A
90 <
45 C
Figure 19-12. -A 45-90 triangle.
measures 90. Since angles A and B are
the sides opposite them are also equal. 1
fore, AC equals CB. Suppose that CB is
long; then AC is also 1 unit long, and the
of AB is calculated as follows:
(AB) 2 = I 2 + I 2 = 2
AB =
Regardless of the size of the triangl
has two 45 angles and one 90 angle, its
are in the ratio 1 to 1 to \T2~. For exam
sides AC and CB are 3 units long, AB is
units long.
Practice problems. Without referei
tables or to the rule of Pythagoras, fi
following lengths and angles in figure
1. AB
2. BC
3. A
Answers:
MATHEMATICS, VOLUME 1
Figure 19-13.- Finding unknown parts
in a 45 -90 triangle.
It is interesting to note U
figure 19-15 (B) is N/T. Thi
coincidence, in which one s
angle is the square root of t
two sides.
Related to the basic 3-4
triangles whose sides are i
to 5 but are longer (proport
basic lengths. For example
tured in figure 19-6 is a 3-4
10.
Figure 19-14. A 3-4-5 triangle.
8
Figure 19-16. Triangle i
are multiples of 3,
The 3-4-5 triangle is ver
tions of distance. If the dat
fit a 3-4-5 configuration, nc
tion of square root (Pythago
needed.
EXAMPLE: An observer at
vertical tower knows that th
is 30 feet from a target or
does he calculate his slant r
sight) from the target?
SOLUTION: Figure 19-17
sired length, AB, is the hy
Chapter 19 -NUMERICAL TRIGONOMETRY
gure 19-17. Solving problems with a
3-4-5 triangle.
;uy wire 15 feet long is stretched from
D of a pole to a point on the ground 9 feet
he base of the pole. Calculate the height
pole.
wers:
feet
2. 12 feet
OBLIQUE TRIANGLES
ique triangles were defined in chapter 17
; training course as triangles which con-
D right angles. A natural approach to the
>n of problems involving oblique triangles
construct perpendicular lines and form
riangles which subdivide the original tri-
Then the problem is solved by the usual
is for right triangles.
ON INTO RIGHT TRIANGLES
35
D
Figure 19-18. Finding the unknown ps
of an oblique triangle.
CAUTION: A careless appraisal of this
lem may lead the unwary trainee to rep
the ratio AC/AB as the cosine of 40.
error is avoided only by the realization t
trigonometric ratios are based on RIG!
angles.
2. In order to find the length of DC
calculate BD.
BD
= sin 40 C
35
BD = 35 sin 40
= 35 (0.6428)
= 22.4 (approximately)
3. Find the length of DC.
22.4 , _.o
-^ = tan 75
- 22.4 _ 22.4
" ~~ 1 ~
MATHEMATICS, VOLUME 1
Figure 19-19. Calculation of unknown
quantities by means of oblique triangles.
Suppose that point B is the top of a hill, and
point D is inaccessible. Then the only meas-
urements possible on the ground are those
shown in figure 19-19. If we let h represent
BD and x represent CD, we can set up the fol-
lowing system of simultaneous equations:
= tan 70
x
50 + x
= tan 30 C
Solving these two equations for h in terms of
x, we have
h = x tan 70
and
h = (50 + x) tan 30'
Since the two quantities which are both equal
to h must be equal to each other, we have
x tan 70 = (5
x (2.748) = 5C
x (2.748) - x (0.5774) = 28
x (2.171) = 2
_ 28.8 ..g
~ 2.171
Knowing the value of x,
compute h as follows:
h = x tan 70
= 13.3 (2.748)
= 36.5 feet (ap
Practice problems:
1. Find the length of sideE
2. Find the height of poii
figure 19-20 (B).
Answers:
1. 21.3 feet
LAW OF SINES
2.
The law of sines provi
to the solution of oblique t
necessity of subdividing
Let the triangle in figure
any oblique triangle with 2
The labels used in fig
ardized. The small lette
side opposite angle A; sm;
B; small c is opposite ang
Chapter 19 -NUMERICAL TRIGONOMETRY
Figure 19-21.- (A) Acute oblique triangle with standard labels;
(B) obtuse triangle with standard labels.
The law of sines states that in any triangle,
whether it is acute as in figure 19-21 (A) or
obtuse as in figure 19-21 (B), the following is
true:
EXAMPLE: In figure 19-21 (A), let angle A be
15 and let angle C be 85. If BC is 20 units,
find the length of AB.
SOLUTION: By the law of sines,
c =
20 sin 85 C
sin 15
_ 20 (0.9962) = r
c 0.2588
APPENDIX I
SQUARES, CUBES, SQUARE ROOTS, CUBE
LOGARITHMS, AND RECIPROCALS OF NU
No.
Squire
Cub*
Squirt
Root
Cuba
Boot
Log.
1000
x Reoip.
No. -Dim.
Circum.
Area
1
1
1
1.0000
1 . 0000
0.00000
1000.000
3.142
0.7854
2
4
8
1.4142
1 2599
0.30103
500.000
6.283
3.1416
3
9
27
1.7321
1.4422
0.47712
333.333
9.425
7.0*86
4
16
64
2.0000
1.5874
. 60206
250.000
12.566
12.5664
S
25
125
2.2361
1.7100
0.69897
200.000
15.708
19.6350
6
36
216
2.4495
1.8171
0.77815
166.667
18.850
28.2743
7
49
343
2.6458
1.9129
0.84510
142.857
21.991
38.4845
8
64
512
2.8284
2.0000
0.90308
125.000
25.133
50.265=
9
81
729
3.0000
2.0801
0.95424
111.111
28.274
63.6173
10
100
1000
3.1623
2.1544
1.00000
100.000
31.416
78.5398
11
121
1331
3.3166
2.2240
1.04139
90.9091
34.558
95.0332
12
144
1728
3.4641
2.2894
1.07918
83.3333
37.699
113.097
13
169
2197
3.6056
2.3513
1.11394
76.9231
40.841
132.732
14
196
2744
3.7417
2.4101
1.14613
71.4286
43.982
153 938
15
22S
3375
3.8730
2.4662
1.17609
66 6667
47.124
176.715
16
256
4096
4.0000
2.5198
1.20412
62.5000
50.265
201.062
17
289
4913
4.1231
2.5713
1.23045
58.8235
53.407
226.980
18
324
5832
4.2426
2.6207
1.25527
55.5556
56.549
254.469
19
361
6859
4.3S89
2.6684
1.27875
52.6316
59 . 690
283.529
20
400
8000
4.4721
2.7144
1.30103
50.0000
62.832
314.159
21
441
9261
4.5826
2.7589
1.32222
47.6190
65.973
346.361
22
484
10648
4.6904
2.8020
1.34242
45.4545
69.115
380.133
23
529
12167
4.7958
2.8439
1.36173
43.4783
72.257
415.476
24
576
13824
4.8990
2.8845
1.38021
41.6667
75.398
452.389
25
62S
15625
5.0000
2.9240
1 .39794
40.0000
78.540
490.874
26
676
17576
5.0990
2.9625
1.41497
38.4615
81.681
530.929
27
729
19683
5.1962
3.0000
1.43136
37.0370
84.823
572.555
28
784
21952
5.2915
3.0366
1.44716.
35.7143
87.965
615.752
29
841
24389
5.3852
3.0723
1 .46240
34.4828
91.106
660.520
30
900
27000
5.4772
3.1072
1.47712
33.3333
94.248
706.858
31
961
29791
5.5678
3.1414
1.49136
32.2SR1
97.389
754.768
32
1024
32768
5.6569
3.1748
1.50515
31.2500
100.531
804.248
33
1089
35937
5.7446
3.2075
1.51851
30.3030
103.673
855.299
Appendix I- POWERS, ROOTS, LOGARITHMS, ETC.
No.
iqiura
Cub*
Square
Root
Cube
Root
Ix*.
1000
x Kacip.
No.-DU.
Cuxum.
/klW
45
2025
9U2S
6.7082
3.5569
1.65321
22.2222
141.37
1590.43
46
2116
97336
6.7823
3.5830
1.66276
21.7391
144.51
1661.90
47
2209
103823
6.8557
3.6088
1 67210
21.2766
147.65
1734.94
48
2304
110592
6.9282
3.6342
1.68124
20 8333
150.80
1809.56
49
2401
117649
7.0000
3.6593
1.69020
20.4082
153.94
1885.74
50
2500
125000
7.0711
3.6840
1.69897
20.0000
157.08
1963.50
51
2601
132651
7.1414
3.7084
1 70757
19.6078
160 22
2042.82
52
2704
140(508
7.2111
3.7325
1.71600
19.2308
163.36
2123.72
53
2809
148877
7.2801
3.7563
1.72428
18.8679
166 50
2206.18
54
2916
157464
7.3485
3.7798
1.73239
18 5185
169.65
2290.22
55
3025
166375
7.4162
3.8030
1.74036
18.1818
172.79
2375.83
56
3136
175616
7.4833
3.8259
1.74819
17.8571
175 93
2463.01
57
3249
185193
7.5498
3.8485
1.75587
17.5439
179.07
2551.76
58
3364
195112
7.6158
3.8709
1.76343
17.2414
182.21
2642.08
59
3481
205379
7.6811
3.8930
1.77085
16.9492
185.35
2733.97
60
3600
216000
7.7460
3.9140
1.77815
16.6667
188.50
2827.43
61
3721
226981
7.8102
3.9365
1.78533
16.3934
191.64
2922 47
62
3844
238328
7.8740
3.9579
1.79239
16.1290
194.78
3019.07
63
3969
250047
7.9373
3.9791
1.79934
15.8730
197.92
3117.25
64
4096
262144
8.0000
4.0000
1.80618
15.6250
201.06
3216.99
65
4225
274625
8.0623
4.0207
1.81291
15.3846
204.20
3318.31
66
4356
287496
8.1240
4.0412
1.81954
15.1515
207 . 35
3421.19
67
4489
300763
8.18S4
4.0615
1.82607
14.9254
210.49
3525.65
68
4624
314432
8.2462
4.0817
1.83251
14.7059
213.63
3631.68
69
4761
328509
8.3066
4.1016
1.83885
14.4928
216.77
3739.28
70
4900
343000
8.3666
4 1213
1.84510
14.2857
219.91
3848.45
71
5041
357911
8.4261
4.1408
1.85126
14.0845
223 05
3959.19
7>
5184
373248
8.4853
4.1602
1.85733
13.8889
226.19
4071.50
73
5329
389017
8.5440
4.1793
1.86332
13.6986
229.34
4185 39
74
5476
405224
8.6023
4.1983
1.86923
13.5135
232.48
4300.84
75
5625
421875
8.6603
4.2172
1.87506
13.3333
235.62
4417.86
76
5776
438976
8.7178
4.2358
1.88081
13.1579
238.76
4536.4(5
77
5929
456533
8.7750
4.2543
1.88649
12.9870
241.90
4656.63
78
6084
474552
8.8318
4.2727
1.89209
12.8205
245.04
4778. 3C
79
6241
493039
8.8882
4.2908
1.89763
12.6582
248.19
4901.61
80
6400
512000
8.9443
4.3089
1.90309
12.5000
251.33
5026.5.
8
6561
531441
9.0000
4.3267
1.90849
12.3457
254.47
5153. 0(
82
6724
551368
9.0554
4.3445
1.91381
12.1951
257.61
5281.0
83
6889
571787
9.1104
4.3621
1.91908
12.0482
260.75
5410.6
84
7056
592704
9.1652
4.3795
1.92428
11.9048
263 . 89
5541.7
MATHEMATICS, VOLUME 1
No.
Bqusr*
Cube
Square
Root
Cub*
Root
Loi.
1000
z Recip.
No.-DU.
Circuin.
Arts
90
8100
729000
9.4868
4.4814
.95424
11.1111
282.74
6361.73
91
8281
753571
9.5394
4.4979
.95904
10.9890
285.88
6503.8!
92
8464
778688
9.5917
4.5144
.96379
10.8696
289.03
6647.61
93
8649
804357
9.6437
4.5307
.96848
10.7527
292.17
6792.9)
94
8836
830584
9.6954
4.5463
.97313
10.6383
295.31
6939. 71
95
9025
857375
9.7468
4.5629
.97772
10.5263
298.45
7088.21
96
9216
884736
9.7980
4.5789
.98227
10.4167
301.59
7238.2.
97
9409
912673
9.8489
4.5947
.98677
10.3093
304.73
7389.8
98
9604
941192
9.8995
4.6104
.99123
10.2041
307.88
7542.9'
99
9801
970299
9.9499
4.6261
.99564
10.1010
311.02
7697.6
100
10000
1000000
10.0000
4.6416
2.00000
10.00000
314.16
7853.9
101
10201
1030301
10.0499
4.6570
2.00432
9.90099
317.30
8011.8
102
10404
1061208
10.0995
4.6723
2.00860
9.80392
320.44
8171.2
103
10609
1092727
10.1489
4.6875
2.01284
9.70874
323.58
8332.2
104
10816
1124864
10.1930
4.7027
2.01703
9.61538
326.73
8494. 8
105
11025
1157625
10.2470
4.7177
2.02119
9.52381
329.87
8659. C
106
11236
1191016
10.2956
4.7326
2.02531
9.43396
333.01
8824.;
107
11449
1225043
10.3441
4.7475
2.02938
9.34579
336.15
8992. (
108
11664
1259712
10.3923
4.7622
2.03342
9.25926
339.29
9160.1
109
11881
1295029
10.4403
4.7769
2.03743
9.17431
342.43
9331.;
110
12100
1331000
10.4881
4.7914
2.04139
9.09091
345.58
9503.;
111
12321
1367631
10.5357
4.8059
2.04532
9.00901
348.72
9676.
112
12544
1404928
10.5830
4.8203
2.04922
8.92857
351.86
9852. C
113
12769
1442897
10.6301
4.8346
2.05308
8.84956
355.00
10028.7
114
12996
1481544
10.6771
4.8488
2.05690
8.77193
358.14
10207. C
115
13225
1520875
10.7238
4.8629
2.06070
8.69565
361.28
10386.9
116
13456
1560896
10.7703
4.8770
2.06446
8.62069
364.42
10568.3
117
13689
1601613
10.8167
4.8910
2.06819
8.54701
367.57
10751.3
118
13924
1643032
10.8628
4.9049
2.07188
8.47458
370.71
10935.9
119
14161
1685159
10.9087
4.9187
2.07555
8.40336
373.85
11122.0
120
14400
1728000
10.9545
4.9324
2.07918
8.33333
376.99
11309.7
121
14641
1771561
11.0000
4.9461
2.08279
8.26446
380. 13
11499.0
122
14884
1815848
11.0454
4.9597
2.08636
8.19672
383.27
11689.9
123
15129
1860867
11.0905
4.9732
2.08991
8.13008
386.42
11882.3
124
15376
1906624
11.1355
4.9866
2.09342
8.06452
389.56
12076.3
125
15625
1953125
11.1803
5.0000
2.09691
8.00000
392.70
12271.8
126
15876
2000376
11.2250
5.0133
2 . 10037
7.93651
395.84
12469.0
127
16129
2048383
11.2694
5.0265
2 . 10380
7.87402
398.98
12667.7
128
16384
2097152
11.3137
5.0397
2.10721
7.81250
402.12
12868.0
129
16641
2146689
11.3578
5.0528
2.11050
7 75104
405 . 27
13060 a
APPENDIX II
NATURAL SINES, COSINES, AND TANGENTS
OF ANGLES FROM to 90
0-14.9
DB.
Function
0.0
0.1
0.2
0.3
0.4
0.6
0.6
0.7
0.8
0.9
Bin
COB
tan
0.0000
1.0000
0.0000
0.0017
1 0000
0.0017
0.0035
1.0000
0.0035
0.0052
1.0000
0.0052
0.0070
1.0000
0.0070
0.0087
1.0000
0.0087
0.0105
0.9999
0.0105
0.0122
0.9999
0.0122
0.0140
0.9999
0.0140
0.0157
0.9999
0.0157
1
Bin
COB
tan
0.0175
0.9998
0.0175
0.0192
0.9998
0.0192
0.0209
0.9998
0.0209
0.0227
0.9997
0.0227
0.0244
0.9997
0.0244
0.0262
0.9997
0.0262
0.0279
0.9990
0.0279
0.0297
0.9996
0.0297
0.0314
0.9995
0.0314
0.0332
0.9995
0.0332
2
sin
COB
tan
0.0349
0.9994
0.0349
0.0366
0.9993
0.0367
0.0384
0.9993
0.0384
0.0401
0.9992
0.0402
0.0419
0.9991
0.0419
0.0436
O.990
0.0437
0.0454
0.9990
0.0454
0.0471
0.9989
0.0472
0.0488
0.9988
0.0489
0.050f
0.9987
0.0507
3
sin
COB
tan
0.0523
0.9986
0.0524
0.0541
0.9985
0.0542
0.0558
0.9984
0.0559
0.0576
0.9983
0.0577
0.0593
0.9982
0.0594
0.0610
0.9981
0.0612
0.0628
0.9980
0.0629
0.0645
0.9979
0.0647
0.0663
0.9978
0.0664
0.0680
0.9977
0.0682
4
sin
COB
tan
0.0698
0.9976
0.0699
0.0715
0.9974
0.0717
0.0732
0.9973
0.0734
0.0750
0.9972
0.0752
0.0767
0.9071
0.0769
0.0785
0.9969
0.0787
0.0802
0.9968
0.0805
0.0819
0.9966
0.0822
0.0837
0.9965
0.0840
0.0854
0.9963
0.0857
6
Bin
COB
taa
0.0872
0.9962
0.0875
0.0889
0.9960
0.0892
0.0906
0.9959
0.0910
0.0924
0.9957
0.0928
0.0941
0.9956
0.0945
0.0958
0.9954
0.0963
0.0976
0.9952
0.0981
0.0993
0.9951
0.0998
0.1011
0.9949
0.1016
0.1028
0.9947
0.1033
6
ain
COB
tan
0.1045
0.9945
0.1051
0.1063
0.9943
0.1069
0.1080
0.9942
0.1086
0.1097
0.9940
0.1104
0.1115
0.9938
0.1122
0.1132
0.9936
0.1139
0.1149
0.9934
0.1157
0.1167
0.9932
0.1175
0.1184
0.9930
0.1192
0.1201
0.9928
0,1210
7
Bin
COB
tan
0.1219
0.9925
0.1228
0.1236
0.9923
0.1246
0.1253
0.9921
0.1263
0.1271
0.9919
0.1281
0.1288
0.9917
0. 1299
0.1305
0.9914
0-.1317
0.1323
0.9912
0.1334
0.1340
0.9910
0.1352
0.1357
0.9907
0.1370
0.1374
0.9905
0.1388
8
Bin
COB
tan
0.1392
0.9903
0.1405
0.1409
0.9900
1423
0.1426
0.9898
1441
0.1444
0.9895
0.1459
0.1461
0.9893
0.1477
0.1478
0.9890
0.1495
0. 1495
0.9888
0.1512
0.1513
0.9885
0.1530
0.1530
0.9882
0.1548
0.1547
0.9880
0.1566
9
ain
COB
tan
0.1564
0.9877
0.1584
0.1582
0.9874
0.1602
0.1599
0.9871
0.1620
0.1616
0.9869
0.1638
0.1633
0.9866
0.1055
0.1650
0.9863
0.1673
0.1668
0.9860
0.1691
0.1685
0.9857
0.1709
0.1702
0.9854
0.1727
0.1719
0.9851
0.1745
10
Bin
COB
tan
0.1736
0.9848
0.1763
0.1754
0.9845
0.1781
0.1771
0.9842
0.1799
0.1788
0.9839
0.1817
0.1805
0.9836
0.1835
0. 1822
0.9833
0.1853
0.1840
0.9829
0.1871
0.1857
0.9826
0.1890
0.1874
0.9823
0.1908
0.1891
0.9820
0.1926
11
sin
COB
tan
0.1908
0.9816
0.1944
0.1925
0.9813
0.1962
0.1042
0.9810
0.1980
0.1959
0.9806
0.1998
0.1977
0.9803
0.2016
0. 1894
0.9799
0.2035
0.2011
0.9796
0.2053
0.2028
0.9792
0.2071
0.2045
0.9789
0.2089
0.2062
0.9785
0.2107
12
sin
COB
tan
0.2079
0.9781
0.2126
0.2096
0.9778
0.2144
2113
0.9774
0.2162
0.2130
0.9770
0.2180
0.2147
0.9767
0.2199
0.2164
0.9763
0.2217
0.2181
0.9759
0.2235
0.2198
0.9755
0.2254
0.2215
0.9751
0.2272
0.2232
0.9748
0.2290
MATHEMATICS, VOLUME 1
16-29.9
Degs.
Function
0.0
0.1
0.2
0.8
0.4,
0.5
o.
0.7
0.9
o.
16
sin
cos
tan
0.2588
0.9659
0.2679
0.2605
. 9655
0.2698
0.2622
0.9650
0.2717
0.2639
0.9646
0.2736
0.2866
0.9641
0.2754
0.2672
0.9636
0.2773
0.2689
0.9632
0.2792
0.2706
0.9627
0.2811
0.2723
0.9622
0.2830
0.2740
0.0617
0.2840
16
in
COB
tan
0.2756
0.9613
0.2867
0.2773
0.9608
0.2886
0.2790
0.9603
0.2905
0.2807
0.9598
0.2924
0.2823
0.9593
0.2943
0.2840
9.9588
0.2962
0.2857
0.9583
0.2981
0.2874
0.9578
0.3000
0.2890
0.9673
0.3019
0.2907
0.9568
0.3038
17
sin
COB
tan
0.2924
0.9563
0.3057
0.2940
0.9558
0.3076
0.2957
0.9553
0.3096
0.2974
0.9548
0.3116
0.2990
0.9542
0.3134
0.3007
0.9537
0.3153
0.3024
0.9532
0.3172
0.3040
0.9527
0.3191
0.3057
0.0621
0.3211
0.3074
0.9516
0.3230
18
Bin
cos
tan
0.3090
0.9511
0.3249
0.3107
0.9505
0.3269
0.3123
0.9500
0.3288
0.3140
0.9494
0.3307
0.3156
0.9489
0.3327
0.3173
0.9483
0.3346
0.3190
0.9478
0.3365
0.3206
0.9472
0.3386
0.3223
0.0466
0.3404
0.3230
0.9461
0.3424
19
in
COB
tan
0.3256
0.9455
0.3443
0.3272
0.9449
0.3463
0.3289
0.9444
0.3482
0.3305
0.9438
0.3502
0.3322
0.9432
0.3522
0.3338
0.9426
0.3541
0.3366
0.9421
0.3561
0.3371
0.9416
0.3681
0.3387
0.0400
0.3600
0.3404
0.9403
0.3620
20
sin
COB
tan
0.3420
0.9397
0.3640
0.3437
0.9391
0.3659
0.3453
0.9385
0.3679
0.3469
0.9379
0.3699
0.3486
0.9373
0.3719
0.3502
0.9367
0.3739
0.3518
0.0361
0.3759
0.3636
0.6364
0.3779
0.3661
0.0348
0.8790
0.3567
0.9342
0.3810
21
Bin
CO*
tan
0.3584
0.9336
0.3839
0.3600
0.9330
0.3859
0.3616
0.9323
0.3879
0.3633
0.9317
0.3899
0.3649
0.9311
0.3919
0.3665
0.9304
0.3939
0.3661
0.9298
0.3969
0.3697
0.9291
0.3979
0.3714
0.0285
0.4000
0.3730
0.0278
0.4020
22
aln
COB
tan
0.3746
0.9272
0.4040
0.3762
0.9265
0.4061
0.3778
0.9259
0.4081
0.3795
0.9252
0.4101
0.3811
0.9245
0.4122
0.3827
0.9239
0.4142
0.3843
0.9232
0.4163
0.3859
0.9226
0.4183
0.3875
0.9210
0.4204
0.3801
0.9212
0.4224
23
in
COB
tan
0.3907
0.9205
0.4245
0.3923
0.9198
0.4265
0.3939
0.9191
0.4286
0.3955
0.9184
0.4307
0.3971
0.9178
0.4327
0.3987
0.9171
0.4348
0.4003
0.9164
0.4369
0.4019
0.9167
0.4360
0.4035
0.0150
0.4411
0.4061
0.0143
0.4431
24
in
COB
tan
0.4067
0.9135
0.4452
0.4083
0.9128
0.4473
0.4099
0.9121
0.4494
0.4115
0.9114
0.4515
0.4131
0.9107
0.4536
0.4147
0.9100
0.4657
0.4163
0.9092
0.4578
0.4179
0.9086
0.4609
0.4195
0.0078
0.4621
0.4210
0.0070
0.4642
26
in
COB
tan
0.4226
0.9063
0.4663
0.4242
0.9056
0.4684
0.4258
0.9048
0.4706
0.4274
0.9041
0.4727
0.4289
0.9033
0.4748
0.4306
0.9026
0.4770
0.4321
0.9018
0.4791
0.4337
0.9011
0.4813
0.4352
0.0003
0.4834
0.4368
0.8006
0.4866
26
in
COB
tan
0.4384
0.8988
0.4877
0.4399
0.8980
0.4899
0.4415
0.8973
0.4921
0.4431
0.8965
0.4942
0.4446
0.8957
0.4904
0.4462
0.8949
0.4986
0.4478
0.8942
0.6008
0.4403
0.8934
0.6020
0.4600
Q.8926
0.6061
0.4624
0.8018
0.6073
27
sin
COB
tan
0.4540
0.8910
0.5095
0.4555
0.8902
0.5117
0.4571
0.8894
0.5139
0.4586
0.8886
0.5161
0.4603
0.8878
0.5184
0.4617
0.8870
0.6206
0.4633
0.8862
0.6228
0.4648
0.8864
0.6260
0.4664
0.8846
0.6272
0.4670
0.8838
0.5296
no
in
0.4695
0.4710
0.4726
0.4741
0.4766
0.4772
0.4787
0.4802
0.4818
0.4833
Appendix II- NATURAL SINES, COSINES, AND TANGENTS
30-44.9
Degs.
Function
0.0 C
0.1
0.2
0.3
0.4
0.5
0.6
0.70
0.8
0.9
30
sin
cos
tan
0.5000
0.8000
0.5774
0.5015
0.8052
0.5797
0.5030
0.8043
0.5S20
0.5045
8034
0.5844
0.5060
O.K025
0.5867
0.5075
O.K<51
0.5890
0.5090
0.8007
0.5014
0.5105
0.8509
0.5038
0.5120
0.8590
0.5961
0.5135
0.8581
0.5080
31
sin
cos
tan
0.6150
0.8572
0.0009
0.5165
0.8563
0.0032
0.5180
0.8554
0.0056
0.5195
0>545
0.0080
0.5210
0.8530
0.6104
0.5225
0.8526
0.0128
0.5240
0.8517
0.0152
0.5255
0.8508
0.0176
5270
8499
0.0200
0.5284
0.8400
0224
32
sin
cos
tan
0.5209
O.S4SO
0.0249
0.5314
O.X471
0.0273
0.5329
O.S4t>2
0.6297
0.5344
O.K453
0.6322
0.535S
0.8443
0.0346
0.5373
0.8434
0.6371
0.5388
0.8425
0.6395
0.5402
0.8415
0.0420
0.5417
0.8406
0.6445
0.5432
0.8390
0460
33
sin
cos
tan
5440
0.8387
0.0494
0.5401
0.8377
0.0519
5470
8308
0.0544
0.5490
O.K3S8
0.0569"
0.5505
O.N348
0.6594
0.5519
0.8330
0019
5534
0.832!i
0.0044
0.5548
0.8320
0.60! 9
0.5503
0.8310
0.6694
0.5577
8300
0.0720
34
sin
cos
tan
0.555)2
8290
0.0745
0.5600
O.S2KI
0.6771
0.5621
0.8271
0.071)0
0.56351 0.5050
().K261| O.S251
O.HN22] O.ON47
0.5004
S241
0.6N73
0.5678 0.5GU3
O.S231; 0.8221
0.08991 O.SD24
0.5707
0.8211
O.M50
0.5721
O.S202
0.6976
35
sin
cos
tan
0.5730
0.8192
0.7002
5750
0.8181
0.7028
0.5704
0.8171
0.7054
0.577'.) 0.5703
O.Sllil 1 O.S151
0.70SOJ 0.7107
0.5807
0.8141
0.7133
0.5821
0.8131
0.715!)
5835
0.8121
0.7180
0.5850
0.8111
0.7212
0.5864
O.S100
0.7239
36
sin
cos
tan
0.5878
SOUO
0. 72155
0.5802
O.NO.XO
0.7202
0.5006
0.8070
0.7319
0.5920
0.8050
0.7340
0.5034
0.804'.!
0.7373
0.5948
0.8039
0.7400
0.5062
0.802K
0.7427
3.5076
0.8018
0.7454
5990
0.8007
0.7481
6004
7097
0.7508
37
sin
cos
tan
(I 001.S
0.79*0
0.753C
M.12
7!) 70
0.7503
O.fi04fl
0.71)05
0.7590
0.0000
0.7955
0.7018
0.0074
0.7944
0.704b
O.OOSS
0.7934
0.7073
0.0101
0.7023
0.7701
0. 1)115
0.7912
0.7729
0.6129
0.7902
0.7757
0.0143
0.7801
0.7785
38
sin
cos
tan
0.0157
0.78W
0.7813
0.0170
0.7,Mi!
0.7S41
0.01S4
0.7,Vi!>
0.7809
0.0198
0.7,x4fc
0.789S>
0.6211
0.7.S37
0.7020
0.0225
0.7S20
0.7054
0.0239
0.7815
0.7SIS3
0.6252
0.7804
8012
0.6200
0.7793
0.8040
0.6280
0.7782
0.8069
39
sin
cos
tan
0. 02fl3
0.7771
0.8098
0.0307
0.7700
0.8127
0.0320
774!)
0.8150
0334
0.773.S
O.S1S5
0.0347
0.7727
0.8214
0.0301
0.7716
0.8243
0.0374J 0.6388
0.7705! 0.7604
0.8273 0.8302
0.6401
0.7683
0.8332
0.6414
0.7672
0.8361
40
sin
cos
tan
0.642.si 0441
7UOO' (1 71)4!)
0.8391 O.S421
0.0455
o.7i ax
1 S451
0.6468
0.7627
0.84S1
0.048
0.7615
0.8511
0.0494
0.71KI4
0.8541
0.6508
0.7593
0.8571
0.6521
0.75S1
0.8601
0.6534
0.7570
0.8032
0.6547
0.7559
0.8662
41
sin
cos
tan
O.flftiil
0.7547
O.S693
6574| 0.fi5S7l 0.0000
0.7."i3ti| 7M4- 0.7513
S7241 O.S754I 0.8785
O.tiHi:
0.750
O.SSll)
002f
0.7491
0.8847
0.6039
0.747.X
0.8878
0.6652
0.7466
0.8010
0.0065
0.7455
0.8041
6678
7443
0.8972
42
Aft
sin
cos
tan
sin
Midi! i>7(H' O.H717
7431 0.74201 0.7408
0. H0041 O.tKWUi O.fiOO?
O.H.S20 1 0X33 0.084.1
0.6730
0.73!Hi
o.oooa
0.iiS5>
0.0743
0.7385
0.913
0.6H7
O.fiTSh
0.7373
0.9103
0884
0.0700
1 0.7301
0.0105
O.f.S'.H
0.0782
0.7349
0.9228
0.6009
n 79'jn
0.0794
0.7337
0.9200
6921
n 791N
0807
0.7325
0.9293
6934
n 7n
MATHEMATICS, VOLUME 1
46-69.9
Dess.
Function
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
45
sin
cos
tan
0.7071
0.7071
1.0000
0.7083
0.7059
1.0035
0.7096
0.7046
1.0070
0.7108
0.7034
1.0105
0.7120
0.7022
1.0141
0.7133
0.7009
1.0176
0.7145
0.6997
1.0212
0.7157
0.6984
1.0247
0.7169
0.6972
1.0283
0.7181
0.6959
1.0319
46
sin
cos
tan
0.7183
0.6947
1.0365
0.7206
0.6934
1.0392
0.7218
0.6921
1.0428
0.7230
0.6909
1.0464
0.7242
0.6896
1.0501
0.7254
0.6884
1.0538
0.7266
0.6871
1.0575
0.7278
0.6858
1.0612
0.7290
0.6845
1.0649
0.7302
0.6833
1.0686
47
sin
cos
tan
0.7314
0.0820
1.0724
0.7325
0.6807
1.0761
0.7337
0.6794
1.0799
0.7349
0.6782
1.0837
0.7361
0.6769
1.0875
0.7373
0.6756
1.0913
0.7385
0.6743
1.0951
0.7396
0.6730
1.0990
0.7408
0.6717
1 . 1028
0.7420
0.6704
1.1067
48
sin
cos
tan
0.7431
0.6G91
1.1106
0.7443
0.6678
1.1145
0.7455
0.6665
1.1184
0.7466
0.6652
1.1224
0.7478
0.6639
1 . 1263
0.7490
0.6626
1.1303
0.7501
0.6613
1.1343
0.7513
0.6600
1.1383
0.7524
0.6587
1.1423
0.7536
0.6574
1.1463
49
sin
cos
tan
0.7547
0.6561
1.1504
0.7559
0.6547
1.1544
0.7670
0.6534
1.1585
0.7581
0.6521
1.1626
0.7593
0.6508
1.1667
0.7604
0.6494
1.1708
0.7615
0.6481
1.1750
0.7627
0.6468
1.1792
0.7638
0.6455
1.1833
0.7649
0.6441
1.1875
50
sin
cos
tan
0.7660
0.6428
1.1918
0.7672
0.6414
1.1960
0.7683
0.6401
1.2002
0.7694
0.6388
1.2045
0.7705
0.6374
1.2088
0.7716
0.6361
1.2131
0.7727
0.6347
1.2174
0.7738
0.6334
1.2218
0.7749
0.6320
1.2261
0.7760
0.6307
1.2305
51
sin
cos
tan
0.7771
0.6293
1.2349
0.7782
0.6280
1.2393
0.7793
0.6266
1.2437
0.7804
0.6252
1.2482
0.7815
0.6239
1.2527
0.7826
0.6225
1.2572
0.7837
0.6211
1.2617
0.7848
0.6198
1.2662
0.7859
0.6184
1.2708
0.7869
0-.6170
1.2753
52
sin
cos
tan
0.7880
0.6157
1.2799
0.7891
0.6143
1.2846
0.7902
0.6129
1.2892
0.7912
0.6115
1.2938
0.7923
0.6101
1.2985
0.7934
0.6088
1.3032
*0.7944
0.6074
1.3079
0.7955
0.6060
1.3127
0.7965
0.6046
1.3176
0.7976
0.6032
1.3222
53
sin
cos
tan
0.7986
0.6018
1.3270
0.7997
0.6004
1.3319
0.8007
0.5990
1.3367
0.8018
0.5976
1.3416
0.8028
0.5962
1.3465
0.8039
0.5948
1.3514
0.8049
0.5934
1.3564
0.8059
0.5920
1.3613
0.8070
0.5906,
1.3663
0.8080
0.5892
1.3713
64
sin
cos
tan
O.S090
0.5878
1.3764
0.8100
0.5864
1.3814
0.8111
0.5850
1.3865
0.8121
0.5835
1.3916
0.8131
0.5821
1.3968
0.8141
0.5807
1.4019
0.8151
0.5793
1.4071
0.8181
0.5779
1.4124
0.8171
0.5764
1.4176
0.8181
0.5750
1.4229
55
sin
cos
tan
0.8192
0.5736
1.4281
0.8202
0.5721
1.4335
0.8211
0.5707
1.4388
0.8221
0.5093
1.4442
0.8231
0.5678
1.4496
0.8241
0.5664
1.4550
0.8251
0.5650
1.4605
0.8261
0.5635
1.4659
0.8271
0.5621
1.4715
0.8281
0.5606
1.4770
56
sin
cos.
tan
0.8290
0.5592
1.4826
0.8300
0.5577
1.4882
0.8310
0.5563
1.4938
0.8320
0.5548
1.4994
0.8329
0.5534
1.5051
0.8339
0.5519
1.5108
0.8348
0.5505
1.5166
0.8358
0.5490
1.5224
0.8368
0.5476
1.5282
0.8377
0.5461
1.5340
57
sin
cos
tan
0.8387
0.5446
1.5399
0.8396
0.5432
1.5458
0.8406
0.5417
1.5517
0.8415
0.5402
1.5577
0.8425
0.5388
1.5637
0.8434
0.5373
1.5697
0.8443
0.5358
1.5757
0.8453
0.5344
1.5818
0.8462
0.5329
1.5880
0.8471
0.5314
1.5941
K.Q
sin
0.8480
0.8490
0.8499
0.8508
0.8517
0.8526
0.8536
0.8545
0.8554
0.8563
Appendix H-NATURAL SINES, COSINES, AND TANGENTS
60-74.9
Begs.
function
0.0
0.1
o.a
0.3
0.4
0.8
0.6
0.7
0.8
0.9
60
tin
cos
tan
0.8660
0.6000
1.7321
0.8669
0.4985
1.7391
0.8678
0.4970
1.7461
0.8686
0.4955
1.7532
0.8695
0.4939
1.7603
0.8704
0.4924
1.7675
0.8712
0.4909
1.7747
0.8721
0.4894
1.7820
0.8729
0.4879
1.7893
0.8738
0.4863
1.7966
61
sin
COB
tan
0.8746
0.4848
1.8040
0.8755
0.4833
1.8115
0.8763
0.4818
1.8190
0.8771
0.4802
1.8265
0.8780
0.4787
1.8341
0.8788
0.4772
1.8418
0.8796
0.4756
1.8495
0.8805
0.4741
1.8572
0.8813
0.4726
1.8650
0.8821
0.4710
1.8728
62
sin
COB
tan
0.8829
0.4695
1.8807
0.8838
0.4679
1.8887
0.8846
0.4664
1.8967
0.8854
0.4G48
1.9047
0.8862
0.4633
1.9128
0.8870
0.4617
1.9210
0.8878
0.4602
1.9292
0.8886
0.4586
1.9375
0.8894
0.4571
1.9458
0.8902
0.4555
1.9542
63
sin
COB
tan
0.8910
0.4540
1.9626
0.8918
0.4524
1.9711
0.8926
0.4509
1.9797
0.8934
0.4493
1.9883
0.8942
0.4478
1.9970
0.8949
0.4462
2.0057
0.8957
0.4446
2.0145
018965
0.4431
2.0233
0.8973
0.4415
2.0323
0.8980
0.4399
2.0413
64
ain
COB
tan
0.8988
0.4384
2.0503
0.8996
0.4368
2.0594
0.9003
0.4352
2.0686
0.9011
0.4337
2.0778
0.9018
0.4321
2.0872
0.9026
0.4305
2.0965
0.9033
0.4289
2.1060
0.9041
0.4274
2. 1155
0.9048
0.4258
2.1251
0.9056
0.4242
2.1348
65
sin
cos
tan
0.9063
0.4226
2.1445
0.9070
0.4210
2.1543
0.9078
0.4195
2.1642
0.9085
0.4179
2.1742
0.9092
0.4163
2.1842
0.9100
0:4147
2.1943
0.9107
0.4131
2.2045
0.9114
0.4115
2.2148
0.9121
0.4099
2.2251
0.9128
0.4083
2.2355
66
sin
COB
tan
0.9135
0.4067
2.2460
0.9143
0.4051
2.2566
0.9150
0.4035
2.2673
0.9157
0.4019
2.2781
0.9164
0.4003
2.2889
0.9171
0.3987
2.2998
0.9178
0.3971
2.3109
0.9184
0.3955
2.3220
0.9191
0.3939
2.3332
0.9198
0.3923
2.3445
67
Bin
COB
tan
0.9205
0.390
2.355
0.921
0.389
2.3673
0.921
0.387
2.378
0.9225
0.385
2.390
0.9232
0.3843
2.4023
0.9239
0.3827
2:4142
0.9245
0.381
2.4262
0.9252
0.3795
2.4383
0.9259
0.3778
2.4504
0.9265
0.3762
2.4627
68
sin
COB
tan
0.927
0.374
'2.475
0.927
0.373
2.487
0,928
0.371
2.500
0.929
0.369
2.512
0.929
0.368
2.525
0.930
0.366
2.538
0.931
0.364
2.551
0.931
0.3633
2.564
0.9323
0.361
2.578
0.9330
0.3600
2.5916
69
Bin
COB
tan
0.933
0.358
2.605
0.934
0.356
2.618
0.934
0.355
2.632
0.935
0.353
2.6464
0.936
0.351
2.660
0.936
0,350
2.674
0.937
0.348
2.688
0.937
0.346
2.703
0.9385
0.3453
2.717
0.9391
0.3437
2.7326
70
sin
cos
tan
0.939
0.342
2.747
0.9403
0.3404
2.7625
0.9409
0.338
2.777
0.941
0.33?
2.792
0.942
0.335
2.8083
0.942
0.333
2.823
0.943
0.332
2.839
0.943
0.330
2.855
0.944
0.328
2.871
0.9449
0.3272
2.8878
71
sin
COB
tan
0.945
0.325
2.904
0.946
0.323
2.920
0.946
0.322
2.937
0.947
0.320
2.954
0.947
0.3190
2.971
0.9483
0.317
2.988
0.948
0.3156
3.006
0.949
0.314
3.023
0.9500
0.3123
3.041
0.9505
0.3107
3.0695
72
sin
COB
tan
0.951
0.309
3.077
0.951
0.307
3.096
0.952
0.305
3.114
0.952
0.304
3.133
0.953
0.302
3.152
0.953
0.300
3.171
0.954
0.299(
3.191
0.954
0.297
3.210
0.955
0.295
3.230J
0.9658
0.2940
3.2506
73
Bin
COB
0.9563
0.292
0.956
0.290
0.957"
0.289(
0.957
0.287
0.9583
0.285
0.958*
0.284
4). 959
0.282.
0.969
0.280
0.960,
0.279(
0.9608
0.27'
MATHEMATICS, VOLUME 1
75-89.9
Degs.
Function
0.0
0.1
0.2
O.S
0.4
0.8
0.6
0.7
o.
O.ft
75
Bin
cos
tan
0.9659
0.2588
3.7321
0.9664
0.2571
3.7583
0.9668
0.2554
3.7848
0.9673
0.2538
3.8118
0.9677
0.2521
3.8391
0.9681
0.2504
3.8667
0.9686
0.2487
3.8947
0.9690
0.2470
3.9232
0.9694
0.2453
3.9520
0.9099
0.2436
3.9812
76
sin
COB
tan
0.9703
0.2419
4.0108
0.9707
0.2402
4.0408
0.9711
0.2385
4.0713
0.9715
0.2368
4.1022
0.9720
0.2351
4.1335
0.9724
0.2334
4.1653
0.9728
0.2317
4.1976
0.9732
0.2300
4.2303
0.9736
0.2284
4.2635
0.9740
0.2267
4.2972
77
sin
COB
tan
0.9744
0.2250
4.3315
0.9748
0.2232
4.3662
0.9751
0.2215
4.4015
0.9755
0.2198
4.4374
0.9759
0.2181
4.4737
0.9763
0.2164
4.5107
0.9767
0.2147
4.5483
0.9770
0.2130
4.5864
0.9774
0.2113
4.6252
0.9778
0.2096
4.6646
78
sin
COB
tan
0.9781
0.2070
4.7040
0.9785
0.2062
4.7453
0.9789
0.2045
4.7867
0.9792
0.2028
4.8288
0.9796
0.2011
4.8716
0.9799
0.1994
4.9152
0.9803
0.1977
4.9594
0.9806
0.1959
5.0045
0.9810
0.1942
5.0504
0.9813
0.1925
5.0970
79
sin
COB
tan
0.9816
0.1908
5.1446
0.9820
0.1891
5.1929
0.9823
0.1874
5.2422
0.9826
0.1857
5.2924
0.9829
0.1840
5.3435
0.9833
0.1822
5.3955
0.9836
0.1805
5.4486
0.9839
0.1788
5.5026
0:9842
0.1771
5.5578
0.9845
0.1754
5.6140
80
Bin
COB
tan
0.9848
0.1735
5.6713
0.9851
0.1719
5.7297
0.9854
0.1702
5.7894
0.9857
0.1685
5.8502
0.9860
0. 1668
5.9124
0.9863
0.1650
5.9758
0.9866
0.1633
6.0405
0.9869
0.1616
6. 1066
0.9871
0.1599
6. 1742
0.9874
0.1582
6.2432
81
sin
COB
tan
0.9877
0.1564
6.3138
0.9880
0.1547
6.3859
0.9882
0.1530
6.4596
0.9885
0.1513
6.5350
0.9888
0.1495
6.6122
0.9890
0.1478
6.6912
0.9893
0.1461
6. 7720
0.9895
0.1444
6.8548
0.9898
0.1426
6.9395
0.9900
0.1409
7.0264
82
sin
COB
tan
0.9903
0.1392
7.1154
0.9905
0.1374
7.2066
0.9907
0.1357
7.3002
0.9910
0.1340
7.3962
0.9912
0.1323
7.4947
0.9914
0.1305
7.5958
0.9917
0.1288
7. 6986
0.9919
0.1271
7.8062
0.9921
0.1253
7.9158
0.9923
0.1236
8.0285
83
sin
COB
tan
0.9925
0.1219
8.1443
0.9928
0.1201
8.2636
0.9930
0.1184
8.38G3
0.9932
0.1167
8.5126
0.9934
0.1149
8.6427
0.9936
0.1132
8.7769
0.9938
0.1115
8.9152
0.9940
0.1097
9.0579
0.9942
0.1080
9.2052
0.9943
0.1063
9.3572
84
sin
COB
tan
0.9945
0.1045
9.5144
0.9947
0.1028
9.6768
0.9949
0.1011
9.8448
0.9951
0.0993
10.02
0.9952
0.0976
10.20
0.9954
0.0958
10.39
0.9956
0.0941
10.58
0.9957
0.0924
10.78
0.9959
0.0906
10.99
0.9960
0.0889
11.20
85
sin
COB
tan
0.9962
0.0872
11.43
0.9963
0.0854
11.66
0.9965
0.0837
11.91
0.9966
0.0819
12.16
0.9968
0.0802
12.43
0.9069
0.0785
12.71
0.9971
0.0767
13.00
0.9972
0.0750
13.30
0.9973
0.0732
13.62
0.9974
0.0715
13.95
86
Bin
COB
tan
0.9976
0.0698
14.30
0.9977
0.0680
14.67
0.9978
0.0663
15.06
0.9979
0.0645
15.46
0.9980
0.0628
15.89
0.9981
0.0610
16.35
0.9982
0.0593
16.83
0.9983
0.0576
17.34
0.9984
0.0558
17.89
0.9985
0.0541
18.46
87
Bin
COB
tan
0.9986
0.0523
19.08
0.9987
0.0506
19.74
0.9988
0.0488
20.45
0.9989
0.0471
21.20
0.9990
0.0454
22.02
0.9990
0.0436
22.90
0.9991
0.0419
23.86
0.9992
0.0401
24.90
0.9993
0.0384
26.03
0.9993
0.0366
27.27
oo
Bin
0.9994
0.9995
0.9995
0.9996
0.9996
0.9997
0.9997
0.9997
n no97
0.9998
n none
0.9998
n moo
APPENDIX III
MATHEMATICAL SYMBOLS
SYMBOL NAME OR MEANING
+ Addition or positive value
Subtraction or negative value
Positive or negative value
Multiplication dot (Centered; not to
be mistaken for decimal point.)
x Multiplication symbol
( ) Parentheses
[ ] Brackets
{ } Braces
Vinculum (overscore)
% Percent
-r Division symbol
: Ratio symbol
: : Proportion symbol
= Equality symbol
"Not equal" symbol
Grouping
symbols
SYMBOL NAME OR ME1
N/~ Square root symbol
\/ Square root symbol w:
Vinculum is made L
cover all factors o:
whose square root i
NT Radical symbol. Lett
sents a number ind
root is to be taken.
i or j Imaginary unit; operat
tronics; represents
oc Infinity symbol
Ellipsis. Used in se
bers in which suo
bers are predicts
conformance to a p;
ing is approximatec
log N Logarithm of N to the
log N Logarithm of N to the
(understood)
In N Natural or Napierian k
Base of the natura
lop-arithm svstem.
APPENDIX IV
WEIGHTS AND MEASURES
Dry Measure
ips = 1 pint (pt)
.nts = 1 quart (qt)
larts = 1 gallon (gal)
larts = 1 peck (pk)
jcks = 1 bushel (bu)
Liquid Measure
:aspoons (tsp) = 1 tablespoon (tbsp)
;able spoons = 1 cup
ips = 1 pint
:luid ounces (oz) = 1 pint
.nts = 1 quart
larts = 1 gallon
> gallons = 1 barrel (bbl)
cubic inches = 1 gallon
3 gallons = 1 cubic foot (cu ft)
Weight
Dunces = 1 pound (Ib)
)0 pounds = 1 short ton (T)
Area
144 square inches = 1 square foot (sq
9 square feet = 1 square yd (sq yd)
30-1/4 square yards = 1 square rod
160 square rods = 1 acre (A)
640 acres = 1 square mile (sq mi)
Volume
1,728 cubic inches = 1 cubic foot
27 cubic feet = 1 cubic yard (cu yd)
Counting Units
12 units = 1 dozen (doz)
12 dozens = 1 gross
144 units = 1 gross
24 sheets = 1 quire
480 sheets = 1 ream
Equivalents
1 cubic foot of water weighs 62
(approx) = 1,000 ounces
1 gallon of water weighs 8-1/3 pound
APPENDIX V
FORMULAS
- a 2
A = s
A-fh
A = Trr 2
A = Iw
A = Ch
Areas
The area of a square is equal to
the square of a side.
The area of a triangle is equal to
one half the base times the
height.
The area of a circle is equal to
the radius squared times pi.
The area of a rectangle is equal
to the length times the width.
The lateral area of a cylinder is
equal to the circumference of
the base times the height.
A = 47rr 2
V = e 3
V = Bh
'!
Areas
The square area c
equal to 4 times
radius squared.
Volumes
The volume of a c 1
cube of an edge.
The volume of a rec
or cylinder equj
the base times tl
The volume of a S]
pi times the rad:
INDEX
Absolute value, 21
Accuracy, 15, 59
Addend, 7
Adding:
complex numbers, 164
decimals, 51
fractions, 118
signed numbers, 21
unlike fractions, 35
Addition:
and subtraction, 7
method for solving simultaneous equations,
135
Adjacent angles, 182
Algebraic:
expressions, 99
fractions, 117
sum, 99
Alternation in a proportion, 144
Altitude of a triangle, 183
Angles, 182
Apex of a triangle, 183
Approximate numbers, 61
Arabic numerals, 1
Arbitrary constant, 120
Areas:
circle, 189
quadrilateral, 186
triangle, 184
Associative laws, 26, 98
Axioms of equality, 25
Base of:
exponent, 65
Centigrade thermometer, 19
Changing:
common fractions to decimal
fractions to percent, 55
integers to percent, 55
percent to a decimal, 56
Characteristic, logarithms, 83
Checking accuracy, 14
Chord of a circle, 188
Circle, 187
Circular cylinder, 194-195
Circumference of a circle, 187
Coefficients, literal, 125, 136
Combined variation, 150
Combining:
radicals, 74
terms, 100
Common:
denominator, 34
factors, 111
fractions, 28, 49
logarithms, 81
Commutative laws, 26, 98
Complement of an angle, 182
Completing the square, 169
Complex:
decimal, 47
fraction, 43-44
numbers, 158-163
plane, 161
Components of logarithms, 83
Composite number, 17
Concentric circles, 189
Conditional equation, 121-122
INDEX
Decimal Continued :
complex, 47
divisors, 53
equivalent, 47
fractions, 45
mixed, 47
multiplying, 51-52
nonterminating, 50
number system, 2, 45
points, 13, 15
power of, 66
reducing, 47
system, 2
Degree:
angular, 182
of an equation, 121
Denominate numbers, 9, 15
Denominator, definition, 28
Dependence, 151
Dependent variable , 151
Developing formulas, 154
Diameter:
circle, 187
sphere, 198
Difference:
answer in subtraction, 7
of two squares, 113
Digit positions:
binary, 3
decimal, 2
Digits, significant, 60
Direction of measurement, 19
Directly proportional, 147
Direct variation, 146
Discriminant, 176
Distributive law, 27', 99
Dividend, 11
Dividing:
a line into equal segments, 190
approximate numbers, 61
VVCT r\r\m*-r*a n-f fo R4
Edge of a prism, 194
Element:
cylinder, 194
set, 4
Ellipses, 192
Ellipsis, definition, 5
End zeros in multiplication, 13
Equality axioms, 25
Equal or double roots, 177
Equations, plotting, 131
Equilateral triangle, 185
Equivalent:
decimal, 47
fraction, 29
Error:
percent of, 59
relative, 60
Estimation, 14, 58
Evaluating:
formulas, 153
radicals, 78
Exponential form, 80
Exponents:
and radicals, 102
definition, 65
fractional, 70
laws of, 67
literal, 112
Extremes of a proportion, 142
Faces of a solid, 193
Factor, 11, 17
Factoring:
definition, 111
method of solving quadratic e
radicals, 75
trinomials, 115
Fixed constant, 120
Formulas:
developing, 154
evaluating. 153
MATHEMATICS, VOLUME 1
Fractions Continued:
fundamental rule, 30
improper, 28, 31
in equations, 125
measurement, 29
negative, 32
partitive, 29
power of, 66
proper, 28
reducing, 31, 116
Function:
general, 151
trigonometric, 202, 213
Fundamental rule of fractions, 30
General form of a linear equation, 126
Geometric:
classification of angles, 182
figures, 183-190
Graphical:
interpretation of roots, 179
representation of complex numbers, 160
solution of quadratic equations, 172
Graphing:
formulas, 156
general, 20
inequalities, 129, 139
Great circle, 198
Greater than (symbol), 20, 128
Greatest common divisor, 34
Grouping:
for multiplication, 11
symbols, 101
Hemisphere, 198
Highest common factor, 34
Horizontal lines, 181
Hypotenuse, 199
Identity, 121
Imaginary:
number, 66. 159
Intersecting lines, 182
Inversely proportional, 148
Inverse ratio, 142
Inverse variation; 148
Inversion in a proportion, 144
Irrational:
number, 77, 158
root, 178
Irregular pyramid, 196
Isosceles triangle, 185
Joint variation, 149
Lateral:
area, pyramid, 197
edge, prism, 194
Laws:
associative, 26
commutative, 26
distributive, 27
exponents, 67
sines, 208
Least common multiple, 34
Less than (symbol), 20, 128
Like:
fractions, 33
signs, adding, 21
Line:
general, 161
parallel, 137
segment, 5
Linear equation, 121, 126
Literal:
coefficient, 124, 136
exponent, 112
Logarithm:
definition, 80
natural, 81
Lowest common denominator,
Mantissa, 83, 85
Mathematical svmbols. 219
INDEX
Minuend, 7
Mixed:
decimal, 47
number, 28, 32
Monomial multiplication, 103
Multiples, 17
Multiplicand, 11
Multiplication:
fractions, 37
general, 10
grouping, 11
Multiplier, 11
Multiplying:
approximate numbers, 61
complex numbers, 164
decimals, 51-52
denominate numbers, 15
signed numbers, 23
Natural logarithms, 81
Negative:
exponents, 69
fractions, 32
logarithms, 83
numbers, 19
Nonterm mating decimals, 50
Number:
set, 4
systems, 2, 3
Number line:
fractions, 28
general, 5, 20
Numerals, 1
Numerator, definition, 28
Numerical coefficient, definition, 100
Oblique:
line, 181
triangle, 185, 207
Obtuse :
Parallel lines, 181
Parallelogram, 186
Parentheses, removing, 101
Partial products, 12
Partitive fractions, 29
Percent:
changing numbers to, 55
changing to decimal, 56
definition, 55
fractional, 57
of error, 59
Percentage cases, 56
Perimeter:
quadrilateral, 186
triangle, 184
Perpendicular:
at any point on a line, 191
bisector of a line, 191
lines, 181
Pi (TT) , 188
Place value, 1, 2, 46
Placing decimal points, 13, 15
Plotting:
complex numbers, 162
coordinates, 131
equations, 131
inequalities, 139
Points and lines, 5
Polar form, 163
Polynomials, 104-106
Positional notation, 2
Positive:
and negative numbers, 20
integers, 4
Powers and roots, 65
Powers of:
fractions, 66
negative integers, 65
ten, 52, 54, 71-73
Precision, 58
Prime:
MATHEMATICS, VOLUME 1
Quadrant, definition, 131
Quadratic:
definition, 167
equations, 172, 179
formula, 170-172
Quadrilateral, 186
Quotient, 11
Radical, 73, 102
Radicand, definition, 74
Radius:
circle, 187
sphere, 198
Ratio:
definition, 141
trigonometric, 201
Rational:
number, 28, 77, 158
roots, 178
Rationalizing denominators, 77, 106
Ray, geometric, 5
Reading:
decimals, 47
micrometers, 62
slide rule scales, 87
Real numbers, 66, 158
Reciprocals, 73
Rectangle, 186
Rectangular;
coordinates, 19, 130
prism, 193
Reducing:
decimals, 47
fractions, 31, 116
Regrouping, 7
Regular pyramid, 196
Relative error, 60
Remainder, 14
Removing parentheses, 101
Rhombus, 186
Rip-ht-
Segment of a circle, 188
Sense reversal, inequalitic
Sets:
comprising points and 1
elements of, 4
infinite, 6
Sides of a triangle, 183
Signed numbers, 19, 23
Significant digits, 60, 73
Similar triangles, 200
Simplifying radicals, 75
Simultaneous:
equations, 133
inequalities, 140
Sines, law of, 208
Slide rule:
description, 86
operation, 88-97
Solid figures, 193
Solving:
linear equations, 122-1!
oblique triangles, 208
Special:
exponents, 69
products, 106
triangles, 204-250
Spheres, 197-198
Square:
geometric, 186
of a sum or difference,
root, 78, 92
Squaring:
by slide rule, 91
complex numbers, 165
Straight and curved lines,
Subject of a formula, 152
Subscripts, 152
Subsets, 4
Substitution method for sol
equations, 136
INDEX
Surface area Continued:
sphere, 198
Symbols:
grouping, 101
in formulas, 152
mathematical, 219
Synthetic division, 110
System of equations, 133
Tangent to a circle, 187
Terms:
and coefficients, 99
of a proportion, 142
Test for divisibility, 18
Thermometer, 19
Three percentage cases, 56
Translating formulas, 155
Trapezoid, 187
Trial quotients, 14
Triangles:
general, 183-186
similar, 200
special, 204-205
Triangular prism, 193
Trigonometric:
ratios, 201
tables, 202, 213
Trinomial:
factoring, 115
squares, 114
Uneven division, 14
Unit, imaginary, 159
Unlike:
fractions, 33
signs, adding, 21
Variable, 120, 151
Variation:
combined, 150
general, 146
joint, 149
Vector representation of comp
Verbal problems, 138-139, 179
Vernier:
caliper, 64
general, 61-64
measurements, 63
micrometer, 64
principle, 63
Vertex:
angle, 182
triangle, 183
Vertical:
angle, 182
line, 181
Volume:
prism, 194
pyramid, 197
sphere, 198
Weights and measures, 220
Whole numbers, 1
Zero as an exponent, 69
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