User:Keith.U/Sandbox/Proof 5

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 definition of $\exp$ as the solution to an initial value problem.

That is, suppose $\exp$ satisfies:
 * $ (1): \quad D_x \exp x = \exp x$
 * $ (2): \quad \exp \left({ 0 }\right) = 1$

on $\R$.

Lemma
that, $\exists \alpha \in \R : \exp \alpha < 0$.

Then, from Intermediate Value Theorem: $\exists \zeta \in \left({ \alpha, \,.\,.\,. 0 }\right) : f \left({ \zeta }\right) = 0 \in \left({ \exp \alpha, \,.\,.\, 1 }\right)$

This contradicts the lemma.