Let . If for every , then . I'm fairly certain I have to apply some of the ordered field properties but I am not sure how. Thanks!
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Suppose ; let . Let . Then so by assumption . So , which is absurd.
So the contradiction you arrived at is , if I'm not mistaken. Thanks!
That is correct! There may be a more abstract approach but this is pretty straightforward.
Originally Posted by Pinkk Let . If for every , then . I'm fairly certain I have to apply some of the ordered field properties but I am not sure how. Thanks! http://www.mathhelpforum.com/math-he...her-proof.html. Look familiar? Clearly, if it is true for every then it is true for the above post.
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