User:Keith.U/Sandbox/Lemma

Theorem
Let $x$ be a real number.

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

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

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 $\R$.

that $\exists \alpha \in \R : \exp \alpha = 0$.

Suppose that $\alpha > 0$.

Let $J = \left[{ 0 \,.\,.\, \alpha }\right]$.

From Exponential Function is Continuous, $\exp$ is  continuous on $J$.

From Max and Min of Function on Closed Real Interval:
 * $\exists K \in \R : \forall x \in J : \left\vert{ \exp x }\right\vert < K$

Then, $\forall n \in \N : \exists c_n \in J$ such that:

So $\forall n \in \N : 1 \leq K \dfrac{ \alpha^{n} }{ n! }$

That is, dividing both sides by $K$:
 * $\forall n \in \N : \dfrac{1}{K} \leq \dfrac{ \alpha^{n} }{ n! }$

But from Power over Factorial, $\dfrac{ \alpha^{n} }{ n! } \to 0$.

This contradicts our assumption.

The same argument, mutatis mutandis proves the result for $\alpha < 0$.

By hypothesis $(2)$:

Hence the result.