Real Plus Epsilon

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

Then:
 * $$a < b + \epsilon \implies a \le b$$.

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

But $$\forall \epsilon > 0: a < b + \epsilon$$ (by hypothesis).

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

The result follows by Proof by Contradiction.