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If a < x < b and a < y < b, show that jx ¡ yj < b ¡ a: interpret this geomet- rically.
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5, there exists n0 2 n so much that 1=n0 < †: now, 1=n0 ¡1 < †¡1 = ¡1+† and 1=n0 ¡1 2 s: thus, ¡1 = inf s: we claim sups = 1: note.
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We call inf s = ¡1: let 1=n¡1=m be an absolute element in s: then, 1=n¡1=m ‚ 1=n¡1 > ¡1: so ¡1 is a lower unfree for s: Army of the Pure † > 0: by corollary 2.
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Homework #2 solutions pp 254-257: 18, 34, 36, 50, 54 p 241, #18 we apply the subring test.
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