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16 CHAPTER 2. THE REAL AND COMPLEX NUMBERS

By the above theorem on order, (a+b)−1 ∈ R+ and so using the associative law,

(a+b)−1 ((b+a)(b−a)) = (b−a) ∈ R+

Now

|x+ y|2 = (x+ y)2 = x2 +2xy+ y2

≤ |x|2 + |y|2 +2 |x| |y|= (|x|+ |y|)2

and so the first of the inequalities follows. Note I used xy ≤ |xy| = |x| |y| which followsfrom the definition.

To verify the other form of the triangle inequality, x = x−y+y so |x| ≤ |x− y|+ |y| andso |x| − |y| ≤ |x− y| = |y− x| . Now repeat the argument replacing the roles of x and y toconclude |y|− |x| ≤ |y− x| .Therefore, ||y|− |x|| ≤ |y− x| .

Example 2.4.9 Solve the inequality 2x+4≤ x−8

Subtract 2x from both sides to yield 4≤−x−8. Next add 8 to both sides to get 12≤−x.Then multiply both sides by (−1) to obtain x ≤ −12. Alternatively, subtract x from bothsides to get x+4≤−8. Then subtract 4 from both sides to obtain x≤−12.

Example 2.4.10 Solve the inequality (x+1)(2x−3)≥ 0.

If this is to hold, either both of the factors, x+ 1 and 2x− 3 are nonnegative or theyare both non-positive. The first case yields x+1 ≥ 0 and 2x−3 ≥ 0 so x ≥−1 and x ≥ 3

2yielding x≥ 3

2 . The second case yields x+1≤ 0 and 2x−3≤ 0 which implies x≤−1 andx≤ 3

2 . Therefore, the solution to this inequality is x≤−1 or x≥ 32 .

Example 2.4.11 Solve the inequality (x)(x+2)≥−4

Here the problem is to find x such that x2 + 2x + 4 ≥ 0. However, x2 + 2x + 4 =(x+1)2 +3≥ 0 for all x. Therefore, the solution to this problem is all x ∈ R.

Example 2.4.12 Solve the inequality 2x+4≤ x−8

This is written as (−∞,−12].

Example 2.4.13 Solve the inequality (x+1)(2x−3)≥ 0.

This was worked earlier and x ≤ −1 or x ≥ 32 was the answer. In terms of set notation

this is denoted by (−∞,−1]∪ [ 32 ,∞).

Example 2.4.14 Solve the equation |x−1|= 2

This will be true when x−1 = 2 or when x−1 =−2. Therefore, there are two solutionsto this problem, x = 3 or x =−1.

Example 2.4.15 Solve the inequality |2x−1|< 2

From the number line, it is necessary to have 2x− 1 between −2 and 2 because theinequality says that the distance from 2x−1 to 0 is less than 2. Therefore, −2 < 2x−1 < 2and so −1/2 < x < 3/2. In other words, −1/2 < x and x < 3/2.