CHAPTER IV

RELATIVE MAGNITUDE OF POSITIVE AND NEGATIVE QUANTITIES— INEQUALITIES BETWEEN TWO ALGEBRAIC EXPRESSIONS CONTAINING UNKNOWN QUANTITIES

324. The Conventions Concerning the Relative Magnitudes of Positive Numbers.-Let a and b be two positive numbers; then, in case the difference a-b is positive, a is said to be greater than b, and is written a >b (24); and if the difference a b is negative, a is less than b, and is written a<b (?4).

This definition is extended to the case when a and b are any positive or negative numbers, and it is agreed to consider a as greater than b when the difference a-b is positive, and as less than b when the difference is negative.

In accordance with this convention,

Every positive number is greater than 0, and than every negative number; and reciprocally,

Every negative number is less than 0, and than every positive number.

Of two negative numbers, the greater is that number which has the less absolute value.

For example,

—4, -3, -2, -1, 0, 1, 2, 3, 4, form a series of increasing numbers, and

It

-4-3-2<-1<0<1<2<3<4

be remarked, moreover, that if the difference a

may -b is positive, the difference b-a is negative; and it follows, therefore, that each of the inequalities,

a-b> 0, a>b, b-a<0, b <a,

expresses the same fact.

The preceding can be illustrated as follows: Let O be a fixed, and M a variable point on the indefinite straight line X'OX.

Let x be the distance of M from O reckoned positively to the right (21, 23), and negatively to the left. If the point M is at first to the left of O, at negative infinity, and moves continually from left to right, to O and through it, the M moving on to the right to positive infinity, the number x, which is the measure, in magnitude and sign, of the distance OM, increases from - to 0, then from 0 to + ∞, and it is said that the distance OM increases from - to zero, then from zero to∞.

325. General Definitions of Equality and greater or lesser Inequality of Negatives.

The same reasoning, used in proving VII, 37, proves that

as a+c> or < b + c

(2) { a> or <b.

Hence, it follows from (1), on account of (2) and Law VII, 37, that

Similarly, (3)

<-b.

a< 0 <b.

Q. E. D.

INEQUALITIES BETWEEN TWO ALGEBRAIC EXPRESSIONS WHICH CONTAIN ONE, TWO, OR THREE UNKNOWN QUANTITIES

326. Suppose A and B are two algebraic expressions involving one unknown quantity ; by the solution of the inequality

A> B

is meant the values of x for which the numerical values A, and B, of A and B, found by substituting in A and B these values of x, will satisfy the inequality

Two inequalities are said to be equivalent when they have the same solutions.

327. The theorems which were demonstrated for the solutions of equations (solutions of equations and the transformations resulting from them) are applicable, with certain slight modifications, to inequalities. These rules are the following.

THEOREM I.—If the same finite quantity is added to or subtracted from both members of an inequality, a new inequality is formed, equiv alent to the first; and reciprocally.

Hence, (1) and (2) are equivalent and (2) can be derived from (1) by adding m to both members of (1), which is the proof for the first part of the rule.

Hence, inequality (5) is equivalent to (1), and (5) can be derived from (1) by subtracting m from both members of (1).

328. General Proof of Theorem I.-Consider the inequality

in which A and B are algebraic expressions in x; let C be any algebraic expression in x which is finite for all finite values of x. adding to both members of (1),

(2) A+ C> B+ C.

By

It is now necessary to show that (1) and (2) are equivalent inequalities, and the converse.

Let A and B, be the values which A and B

1

A, B, >0.

or, what is the same thing written differently,

(3)

1

[324]

For this same value of x, C takes a finite value C1, and it is necessary

to show that (4) A1+C1> B1+ C1.

That the relation in (4) should hold, it is necessary and sufficient that the difference A1 + C1− (B1+ C) is positive (2324); that is, (5) A,+C-(B+C)=A,B,>0.

that

1

The result (5) is true on account of the truth of relation (3).

Reciprocally, let x, be a solution of inequality (2); then by hypo

Therefore, any solution of inequality (2) is a solution of inequality (1). Hence, inequalities (1) and (2) are equivalent.

Similarly, by adding C to both members of inequality (1), inequality (6) is formed:

Application. This theorem justifies the removal of a term from one member of the inequality to the other member on changing the sign of this term.

329. THEOREM II.—By multiplying or dividing the two members of an inequality by the same quantity whose value is always finite and positive, a new inequality is formed equivalent to the first.

For example, suppose a + 1 > 5.

If both members of the inequality be multiplied by 4, then it is true that

in which A and B are certain algebraic expressions involving one or

more unknown quantities r

Let be a positive number

that, for every finite value assigned to the unknown, the expression C takes a finite and positive value. If both members of inequality (1) are multiplied by C a new inequality (2) is formed:

and inequality (2) is equivalent to the inequality

C(A-B) > 0;

ad, since the factor C is by hypothesis always positive, in order that the product CA

factor AB be positive. alent to the first.

B) may be positive, it is necessary that the

Therefore the second inequality is equiv

330. In case the factor is negative and both members of the inequality

A> B

C, the sign of the inequality will be reversed,

For example, suppose

(3) x+3>7,

and multiply both members of this inequality by 5, then will we

The sign of inequality will be reversed, because, by definition

which is true since, by definition, from inequality (3) we have

x+3 70, or x 4 > 0;

-

and in general, the inequality (2) is equivalent to

Since C is positive, and the product C(A-B) is positive, therefore AB is positive and A> B. Therefore the inequality (2) is equivalent to (1).