Real Plus Epsilon

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[edit] Theorem

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

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


[edit] Proof

Suppose a > b. Then ab > 0.

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

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

The result follows by Proof by Contradiction.

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