Let $\displaystyle a,b \in \mathbb{R}$ Show that if $\displaystyle a \le b + \frac{1}{n}$ for all $\displaystyle n\in \mathbb{N}$, then $\displaystyle a \le b$.

I have no idea how to go about this. Any help would be appreciated.

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- Dec 3rd 2009, 03:58 PMPinkkAnother proof
Let $\displaystyle a,b \in \mathbb{R}$ Show that if $\displaystyle a \le b + \frac{1}{n}$ for all $\displaystyle n\in \mathbb{N}$, then $\displaystyle a \le b$.

I have no idea how to go about this. Any help would be appreciated. - Dec 3rd 2009, 04:01 PMDrexel28
Suppose that $\displaystyle a>b\implies a-b>0$. By the Archimedean principle there exists some $\displaystyle n'\in\mathbb{N}$ such that $\displaystyle \frac{1}{n'}<a-b$. Therefore $\displaystyle a\le b+\frac{1}{n'}\implies a-b\le\frac{1}{n'}<a-b\implies a-b<a-b$ contradiction.

- Dec 3rd 2009, 04:02 PMtonio