# Another proof

• Dec 3rd 2009, 03:58 PM
Pinkk
Another 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 $\displaystyle a \le b$.

• Dec 3rd 2009, 04:01 PM
Drexel28
Quote:

Originally Posted by Pinkk
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 $\displaystyle a \le b$.

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 PM
tonio
Quote:

Originally Posted by Pinkk
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 $\displaystyle a \le b$.

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