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4130sols2

# 4130sols2 - 4130 HOMEWORK 2 Due Thursday February 11(1 In...

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4130 HOMEWORK 2 Due Thursday February 11 (1) In this exercise, we will show that the ordered field Q is not complete. (a) Suppose q Q and q 2 < 2. Let n N and suppose that n > max { 2 | q | +1 2 - q 2 , 1 } . Show that ( q + 1 n ) 2 < 2. If n > max { 2 | q | +1 2 - q 2 , 1 } then n > 2 | q | +1 2 - q 2 , so 2 - q 2 > 2 | q | +1 n and thus 2 > q 2 + 2 | q | +1 n . We therefore have: ( q + 1 n ) 2 = q 2 + 2 q n + 1 n 2 < q 2 + 2 q n + 1 n because n > 1 . q 2 + 2 | q | + 1 n because q ≤ | q | . < 2 . (b) Suppose r Q and r 2 > 2. Let n N and suppose n > 2 r r 2 - 2 . Show that ( r - 1 n ) 2 > 2. Since n > 2 r r 2 - 2 , we have r 2 - 2 > 2 r n and so r 2 - 2 r n > 2. We therefore have: ( r - 1 n ) 2 = r 2 - 2 r n + 1 n 2 > r 2 - 2 r n > 2 . (c) Using the results of (a) and (b), together with the fact that there is no s Q with s 2 = 2 (do not prove this), show that Q is not complete. (Hint: show that S = { x Q : 0 < x 2 < 2 } is bounded above but has no least upper bound.) It is clear that S is bounded above, for example by b = 2. We must show that there is no least upper bound.

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