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 feld 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 and y. 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 n is an even integer if and only if n2 is an even integer. (Hint: prove the “if” part by contraposition, that is, prove that if n is 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) if0 < x < 1, then x2 < x. (b) if x > 1, then x < 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. 2 (x 2 (a) xy ≤ 1 + y2). √ (b) If x ≥ 0 and y ≥ 0, then xy ≤ 1 2 (x + y).
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