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Unformatted text preview: C C . For > 0, choose > 0 such that if  xy  < , then  x 2y 2  < . For example, we may take = min { / 3 x , x } . Thenx < yx < x so x < x + y < 3 x , so  x + y  < 3 x . Then  x 2y 2  =  xy  x + y  < 3 x < . Now, for b B , we have b > 0 and b 2 > 2. By Exercise 12, there is a such that x < a < x , and there is b such that x < b < x + . Then  b 2a 2  6  b 2x 2  +  x 2a 2  < 2 . But a 2 < 2 < b 2 , so a 2 > b 22 > 22 . Since was arbitrary, we conclude that for any , there is a A such that a > and 2 > a 2 > 22 . Suppose that r C , so r < 2. By the previous paragraph, letting 2 = 2r , we may choose a A such that a 2 > 22 = r . Letting c = r/a , c 2 = r 2 /a 2 < r 2 /r < 2, so c A . Thus, we see that r C , so C C ....
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 Fall '08
 RIEMAN
 Math

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