# Another proof

• Dec 3rd 2009, 04:58 PM
Pinkk
Another proof
Let $a,b \in \mathbb{R}$ Show that if $a \le b + \frac{1}{n}$ for all $n\in \mathbb{N}$, then $a \le b$.

• Dec 3rd 2009, 05:01 PM
Drexel28
Quote:

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 $a \le b$.

Suppose that $a>b\implies a-b>0$. By the Archimedean principle there exists some $n'\in\mathbb{N}$ such that $\frac{1}{n'}. Therefore $a\le b+\frac{1}{n'}\implies a-b\le\frac{1}{n'} contradiction.
• Dec 3rd 2009, 05:02 PM
tonio
Quote:

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 $a \le b$.

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. $\frac{1}{n}<\epsilon$.