Real Plus Epsilon

[edit] Theorem

Let $a, b, \epsilon \in \mathbb{R}$ , where $ε > 0$ .

Then $a < b + \epsilon \Longrightarrow a \le b$ .

Suppose $a > b$ . Then $a - b > 0$ .

But $\forall \epsilon > 0, a < b + \epsilon$ .

Let $ε = a - b$ . Then $a < b + \left({a - b}\right) \Longrightarrow a < a$ .

The result follows by Proof by Contradiction.