104_Fall09_hw_1_sol-1 - C C . For > 0, choose...

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Homework 1 for MATH 104 Solutions to selected problems Pugh, Chapter 1, Exercise 12 (a) Assume no such s S exists. Then b is an upper bound for S and no element of S is greater than or equal b - ± . Hence b - ± is an upper for S , contradicting the fact that b is the l.u.b. of S . (b) No, not necessarily. Consider for instance S = { 0 * } Then sup ( S ) = 0 * , and for no ± > 0 there exists s S with - ± < s < 0 * . (c) x = A | B is an upper bound of A : If r A , then r * = C | D 6 x since C A . x is also a l.u.b.( A ), since y = E | F < x implies that E A and A \ E is infinite and thus for any r A \ E , we have r * > y . Pugh, Chapter 1, Exercise 13 Prove 2 R by showing that x · x = 2 where x = A | B is the cut in Q with A = { r Q : r 6 0 or r 2 < 2 } . Proof: Since 0 A , we have x > 0. By the definition of multiplication of cuts on p. 14, we let x · x = C | D such that C = { r Q : r 6 0 or a , c A , a > 0, c > 0, r = ac } . Let D = Q \ C . Let C 0 = { r : r < 2 } be the left cut of 2. We want to show that C = C 0 . First, we see that if r C , either r 6 0, in which case r < 2, so r C 0 , or r = ac , where a , c A are positive. Then a 2 < 2, c 2 < 2, so r 2 = a 2 c 2 < 4 by the multiplicative property for inequalities. Then r = ac < 2 by taking (rational) square roots, which also preserves inequalities. So we see that
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Unformatted text preview: C C . For &gt; 0, choose &gt; 0 such that if | x-y | &lt; , then | x 2-y 2 | &lt; . For example, we may take = min { / 3 x , x } . Then-x &lt; y-x &lt; x so x &lt; x + y &lt; 3 x , so | x + y | &lt; 3 x . Then | x 2-y 2 | = | x-y || x + y | &lt; 3 x &lt; . Now, for b B , we have b &gt; 0 and b 2 &gt; 2. By Exercise 12, there is a such that x- &lt; a &lt; x , and there is b such that x &lt; b &lt; x + . Then | b 2-a 2 | 6 | b 2-x 2 | + | x 2-a 2 | &lt; 2 . But a 2 &lt; 2 &lt; b 2 , so a 2 &gt; b 2-2 &gt; 2-2 . Since was arbitrary, we conclude that for any , there is a A such that a &gt; and 2 &gt; a 2 &gt; 2-2 . Suppose that r C , so r &lt; 2. By the previous paragraph, letting 2 = 2-r , we may choose a A such that a 2 &gt; 2-2 = r . Letting c = r/a , c 2 = r 2 /a 2 &lt; r 2 /r &lt; 2, so c A . Thus, we see that r C , so C C ....
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