We have encountered inequality signs many times before. In the discussion of the last section, we also applied mathematical operations to inequalities. In transforming (6.7') into (6.7"), for example, we subtracted 1 from each side of the inequality. What rules of operations apply to inequalities (as opposed to equations)?

To begin with, let us state an important property of inequalities: inequalities are transitive. This means that, if a > b and if b > c, then a > c. Since equalities (equations) are also transitive, the transitivity property should apply to "weak" inequalities (> or <) as well as to "strict" ones (> or <). Thus we have

This property is what makes possible the writing of a continued inequality, such as 3 < a < b < 8 or 7 < x < 24. (In writing a continued inequality, the inequality signs are as a rule arranged in the same direction, usually with the smallest number on the left.)

The most important rules of inequalities are those governing the addition (subtraction) of a number to (from) an inequality, the multiplication or division of an inequality by a number, and the squaring of an inequality. Specifically, these rules are as follows.

Rule I (addition and subtraction) a > b => a ± k > b ± k

An inequality will continue to hold if an equal quantity is added to or subtracted from each side. This rule may be generalized thus: If a > b > c, then a ± k > b±k>

Rule II (multiplication and division)

The multiplication of both sides by a positive number preserves the inequality, but a negative multiplier will cause the sense (or direction) of the inefluality to be

Example 1 Since 6 > 5, multiplication by 3 will yield 3(6) > 3(5), or 18 > 15; but multiplication by - 3 will result in (- 3)6 < (-3)5, or - 18 < - 15.

a > b a > b ka > kb (k > 0) ka < kb (k < 0)

reversed.

Division of an inequality by a number n is equivalent to multiplication by the number 1/«; therefore the rule on division is subsumed under the rule on

Rule III (squaring) ( a > b, (b > 0) => a2 > b2 )

If its two sides are both nonnegative, the inequality will continue to hold when both sides are squared.

Example 2 Since 4 > 3 and since both sides are positive, we have 42 > 32, or 16 > 9. Similarly, since 2 > 0, it follows that 22 > 02, or 4 > 0.

The above three rules have been stated in terms of strict inequalities, but their validity is unaffected if the > signs are replaced by > signs.

When the domain of a variable x is an open interval (a, b), the domain may be denoted by the set {x \ a < x < b) or, more simply, by the inequality a < x_< b. Similarly, if it is a closed interval [a,b], it may be expressed by the weak inequality a < x < b. In the special case of an interval of the form (- a, a)—say, (-10,10)—it may be represented either by the inequality -10 < x < 10 or, alternatively, by the inequality

where the symbol |x| denotes the absolute value (or numerical value) of x. For any real number n, the absolute value of n is defined as follows:*

Note that, if n = 15, then 1151 = 15; but if n = - 15, we find . __________

also. In effect, therefore, the absolute value of any real number is simply its numerical value after the sign is removed. For this reason, we always have_ \n\ = | - n\. The absolute value of n is also called the modulus of n.

Given the expression |x| = 10, we may conclude from (6.8) that x must be either 10 or -10. By the same token, the expression |x| < 10 means that (1) if x > 0, then x = |x| < 10, so that x must be less than 10; but also (2) if x < 0, then according to (6.8) we have —x = |x| < 10, or x > —10, so that x must be greater than - 10. Hence, by combining the two parts of this result, we see that x must lie within the open interval (- 10,10). In general, we can write " ----

(6.9) |x| < n <=> -n < x < n (n > 0)

* The absolute-value notation is similar to that of a first-order determinant, but these two concepts are entirely different. The definition of a first-order determinant is \a,j\ = air regardless of the sign of a,j. In the definition of the absolute value \n\, the sign of n will make a difference. The context of the discussion would normally make it clear whether an absolute value or a first-order determinant is under consideration.

which can also be extended to weak inequalities as follows: (6.10) (\x\ <n «> -n<x<ny (n > 0)

Because they are themselves numbers, the absolute values of two numbers m and n can be added, subtracted, multiplied, and divided. The following properties characterize absolute values:

The first of these, interestingly, involves an inequality rather than an equation. The reason for this is easily seen: whereas the left-hand expression \m\ + \n\ is definitely a sum of two numerical values (both taken as positive), the expression | m + «| is the numerical value of either a sum (if m and n are, say, both positive) or a difference (if m and n have opposite signs). Thus the left side may exceed the right side.

Example 3 If m = 5 and n = 3, then \m\ + |n| = \m + n\ =8. But if m = 5 and n = -3, then \m\ + |n| = 5 + 3 = 8, whereas

is a smaller number.

In the other two properties, on the other hand, it makes no difference whether m and n have identical or opposite signs, since, in taking the absolute value of the product or quotient on the right-hand side, the sign of the latter term will be removed in any case.

Example 4 If m = 1 and n = 8, then \m\ • |n| = \m ■ n\ = 7(8) = 56. But even if m — - 7 and n = 8 (opposite signs), we still get the same result from

M • W = I -7| • |8| = 7(8) = 56 and \m ■ n\ = | - 7(8)| = 7(8) = 56

Like an equation, an inequality containing a variable (say, x) may have a solution; the solution, if it exists, is a set of values of x which make the inequality a true statement. Such a solution will itself usually be in the form of an inequality.

Example 5 Find the solution of the inequality 3x - 3 > x + 1

As in solving an equation, the variable terms should first be collected on one side of the inequality. By adding (3 - x) to both sides, we obtain 3x-3 + 3- x>x+l + 3- x or 2x > 4

Multiplying both sides by \ (which does not reverse the sense of the inequality, because \ > 0) will then yield the solution x > 2

which is itself an inequality. This solution is not a single number, but a set of numbers. Therefore we may also express the solution as the set (x | x > 2} or as the open interval (2, oo).

Example 6 Solve the inequality 11 — x| < 3. First, let us get rid of the absolute-value notation by utilizing (6.10). The given inequality is equivalent to the statement that

or, after subtracting 1 from each side,

where the sense of inequality has been duly reversed. Writing the smaller number first, we may express the solution in the form of the inequality

or in the form of the set {x \ — 2 < x < 4} or the closed interval [ - 2,4].

Sometimes, a problem may call for the satisfaction of several inequalities in several variables simultaneously; then we must solve a system of simultaneous inequalities. This problem arises, for example, in mathematical programming, which will be discussed in the final part of the book.

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