User:Keith.U/Sandbox/Proof 1

Theorem
Let $x$ be a real number.

Let $\exp$ denote the (real) Exponential Function.

Then:
 * $\forall x \in \R : \exp x > 0$

Proof
This proof assumes the series definition of $\exp$.

That is, let:
 * $\displaystyle \exp x = \sum_{n \mathop = 0}^{\infty} \dfrac{x^n}{n!}$

First, suppose $0 \leq x < y$.

Then, $\forall n \in \N$:

So $\exp$ is strictly increasing on $\R_{\geq 0}$.

Suppose instead that $y < x < 0$.

Then, $\forall n \in \N$:

So $\exp$ is strictly increasing on $\R_{< 0}$.

Hence the result.

Hence the result.