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Possible Duplicate:
Proof with inequalities

I've just started reading a book on real analysis and a lot of my proofs reduce to proving this fact over and over again:

For all $\epsilon > 0$, if $a < b + \epsilon $ then $a \leq b$. How do I do this. Are there different ways of doing it?


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marked as duplicate by Clayton, rschwieb, Ittay Weiss, Marvis, Henry T. Horton Jan 27 '13 at 3:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Duplicate hmmmm – Rustyn Jan 27 '13 at 1:23
up vote 1 down vote accepted

The statement should probably read something like "If for all $\epsilon > 0$, $a < b + \epsilon$, then $a \le b$."

There are numerous ways to convince yourself of this. One is to examine the contrapositive: if $a > b$, take $\epsilon > 0$ so that $a > b + \epsilon$ (i.e., $\epsilon < a-b$). Then it's not true that $a < b + \epsilon$ for all $\epsilon > 0$. An easy modification of this will give you a proof by contradiction.

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