Another proof

**Pinkk** · December 3rd 2009, 03:58 PM

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

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

**Drexel28** · December 3rd 2009, 04:01 PM

Originally Posted by Pinkk

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

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

Suppose that $a>b\implies a-b>0$ . By the Archimedean principle there exists some $n'\in\mathbb{N}$ such that $\frac{1}{n'}<a-b$ . Therefore $a\le b+\frac{1}{n'}\implies a-b\le\frac{1}{n'}<a-b\implies a-b<a-b$ contradiction.

**tonio** · December 3rd 2009, 04:02 PM

Originally Posted by Pinkk

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

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

Suppose $a>b$ and put $\epsilon=a-b>0$ . As $\frac{1}{n}\xrightarrow [n\to\infty] {}0$ , there exists $n\in \mathbb{N}$ s.t. $\frac{1}{n}<\epsilon$ .
Get now your contradiction...

Tonio

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