# Exponential of Real Number is Strictly Positive/Proof 5/Lemma

## Theorem

Let $x$ be a real number.

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

Then:

$\forall x \in \R: \exp x \ne 0$

## Proof

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

That is, suppose $\exp$ satisfies:

$(1): \quad D_x \exp x = \exp x$
$(2): \quad \map \exp 0 = 1$

on $\R$.

Aiming for a contradiction, suppose that $\exists \alpha \in \R: \exp \alpha = 0$.

Suppose that $\alpha > 0$.

Let $J = \closedint 0 \alpha$.

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

$\exists K \in \R: \forall x \in J: \size {\exp x} < K$

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

 $\displaystyle 1$ $=$ $\displaystyle \exp 0$ by $(2)$ $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 0}^{n - 1} \frac {\map {\exp^j} \alpha} {j!} \paren {-\alpha}^j + \frac {\exp c_n} {n!} \paren {-\alpha}^n$ Taylor's Theorem for Univariate Functions $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 0}^{n - 1} \frac {\map \exp \alpha} {j!} \paren {-\alpha}^j + \frac {\exp c_n} {n!} \paren {-\alpha}^n$ by $(1)$ $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 0}^{n - 1} \frac 0 {j!} \paren {-\alpha}^j + \frac {\exp c_n} {n!} \paren {-\alpha}^n$ from our assumption aiming at a contradiction $\displaystyle$ $=$ $\displaystyle \frac {\exp c_n} {n!} \paren {-\alpha}^n$

So:

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

That is, dividing both sides by $K$:

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

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

The same argument, mutatis mutandis proves the result for $\alpha < 0$.
By hypothesis $(2)$:
 $\displaystyle \alpha = 0 \ \$ $\displaystyle \implies \ \$ $\displaystyle \exp \alpha$ $=$ $\displaystyle 1$ $\displaystyle$ $\ne$ $\displaystyle 0$
$\blacksquare$