Real Plus Epsilon

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

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

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

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

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

The result follows by Proof by Contradiction.