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

Suppose that $\alpha > 0$.

Let $J = \closedint 0 \alpha$.

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

From Max and Min of Function on Closed Real Interval:

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

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

\(\ds 1\)

\(=\)

\(\ds \exp 0\)

by $(2)$

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 0}^{n - 1} \frac {\map {\exp^j} \alpha} {j!} \paren {-\alpha}^j + \frac {\exp c_n} {n!} \paren {-\alpha}^n\)

Taylor's Theorem for Univariate Functions

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 0}^{n - 1} \frac {\map \exp \alpha} {j!} \paren {-\alpha}^j + \frac {\exp c_n} {n!} \paren {-\alpha}^n\)

by $(1)$

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 0}^{n - 1} \frac 0 {j!} \paren {-\alpha}^j + \frac {\exp c_n} {n!} \paren {-\alpha}^n\)

from our assumption aiming for a contradiction

\(\ds \)

\(=\)

\(\ds \frac {\exp c_n} {n!} \paren {-\alpha}^n\)

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

This contradicts our assumption.

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

By hypothesis $(2)$:

\(\ds \alpha\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \exp \alpha\)

\(=\)

\(\ds 1\)

\(\ds \)

\(\ne\)

\(\ds 0\)

Hence the result.

$\blacksquare$