# 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 \exp \left({0}\right) = 1$

on $\R$.

Aiming for a contradiction, suppose 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:

\(\displaystyle 1\) | \(=\) | \(\displaystyle \exp 0\) | $\quad$ By hypothesis $(2)$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{j \mathop = 0}^{n-1} \frac{\exp^{j} \left({\alpha}\right)}{j!} \left({-\alpha}\right)^{j} + \frac{\exp c_n}{n!} \left({-\alpha}\right)^{n}\) | $\quad$ Taylor's Theorem for Univariate Functions | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{j \mathop = 0}^{n-1} \frac{\exp \left({\alpha}\right)}{j!} \left({-\alpha}\right)^{j} + \frac{\exp c_n}{n!} \left({-\alpha}\right)^{n}\) | $\quad$ By hypothesis $(1)$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{j \mathop = 0}^{n-1} \frac 0 {j!} \left({-\alpha}\right)^{j} + \frac{\exp c_n}{n!} \left({-\alpha}\right)^{n}\) | $\quad$ From our assumption aiming at a contradiction | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac{\exp c_n}{n!} \left({-\alpha}\right)^{n}\) | $\quad$ | $\quad$ |

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, $\dfrac{\alpha^{n}}{n!} \to 0$.

This contradicts our assumption.

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

By hypothesis $(2)$:

\(\displaystyle \alpha = 0 \ \ \) | \(\displaystyle \implies \ \ \) | \(\displaystyle \exp \alpha\) | \(=\) | \(\displaystyle 1\) | $\quad$ | $\quad$ | |||||||

\(\displaystyle \) | \(\ne\) | \(\displaystyle 0\) | $\quad$ | $\quad$ |

Hence the result.

$\blacksquare$