22 1.4 Ordered Field Axioms For the converse, suppose−M<x<M. Again, we consider two cases. If x≥0, then|x| =x<M as desired. Next suppose x <0. Now, −M<x implies M>−x. Then |x| =−x <M. The result follows. □ Note that as a consequence of part (iv) above, since |x| ≤ |x| we get −|x| ≤x ≤ |x|. The next theorem will play an important role in the study of limits. Theorem 1.4.3 — Triangle Inequality. Given x,y ∈R, |x+y| ≤ |x| +|y|. Proof: From the observation above, we have −|x| ≤x ≤ |x| and − |y| ≤y ≤ |y|. Adding up the inequalities gives −|x| − |y| ≤x+y ≤ |x| +|y|. Since −|x| − |y| =−(|x| +|y|), the conclusion follows from Proposition 1.4.2 (iv). □ Corollary 1.4.4 For any x,y ∈R, ||x| − |y|| ≤ |x−y|. Remark 1.4.1 The absolute value has a geometric interpretation when considering the numbers in an ordered field as points on a line. The number |x| denotes the distance from the number x to 0. More generally, the number d(x,y) =|x−y| is the distance between the points x andy. It follows easily from Proposition 1.4.2 that d(x,y) ≥0, and d(x,y) =0 if and only if x =y. Moreover, the triangle inequality implies that d(x,y) ≤d(x,z)+d(z,y) for all real numbers x,y,z. Exercises 1.4.1 Prove that nis an even integer if and only if n2 is an even integer. (Hint: prove the “if” part by contraposition, that is, prove that if nis odd, then n2 is odd.) 1.4.2 Prove parts (iii) and (iv) of Proposition 1.4.1 1.4.3 Let x ∈R. Prove that (a) if 0<x <1, thenx2 <x. (b) if x >1, thenx <x2. 1.4.4 Let x,y,z,w∈R. Suppose 0<x <y and 0<z <w. Prove that xz <yw. 1.4.5 Let x,y ∈R. Prove the following. (a) xy ≤1 2(x 2 +y2). (b) If x ≥0 and y ≥0, then√xy ≤1 2(x+y).
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