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Let x and y be real numbers. Prove that if x <= y + b for every positive number b, then x <= y.

- stringzoffury
**Posts:**2**Joined:**Mon Feb 17, 2014 10:04 pm

stringzoffury wrote:Let x and y be real numbers. Prove that if x <= y + b for every positive number b, then x <= y.

Proof by contradiction might be a useful way to proceed. Assume that x is less than or equal to y for every positive value b, but also that x is greater than y. Where does this lead?

If you prefer using a different technique, kindly please reply showing your progress thus far. Thank you.

- nona.m.nona
**Posts:**256**Joined:**Sun Dec 14, 2008 11:07 pm

Thanks a bunch. I found that if x > y then there would exist a b such that x > y + b, which would contradict the original given statement that x <= y + b for all positive real b. Although I'm not quite sure how to prove the existence of such a b or if I even need to.

- stringzoffury
**Posts:**2**Joined:**Mon Feb 17, 2014 10:04 pm

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