# Exponential of Real Number is Strictly Positive

## Theorem

Let $x$ be a real number.

Let $\exp$ denote the (real) exponential function.

Then:

- $\forall x \in \R : \exp x > 0$

## Proof 1

This proof assumes the series definition of $\exp$.

That is, let:

- $\displaystyle \exp x = \sum_{n \mathop = 0}^\infty \dfrac {x^n} {n!}$

First, suppose $0 < x$.

Then:

\(\displaystyle 0\) | \(<\) | \(\displaystyle x^n\) | $\quad$ Power Function is Strictly Increasing over Positive Reals: Natural Exponent | $\quad$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle 0\) | \(<\) | \(\displaystyle \frac {x^n} {n!}\) | $\quad$ Real Number Ordering is Compatible with Multiplication | $\quad$ | ||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle 0\) | \(<\) | \(\displaystyle \sum_{n \mathop = 0}^\infty \frac{x^n}{n!}\) | $\quad$ Ordering of Series of Ordered Sequences | $\quad$ | ||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle 0\) | \(<\) | \(\displaystyle \exp x\) | $\quad$ Definition of $\exp$ | $\quad$ |

So $\exp$ is strictly positive on $\R_{>0}$.

From Exponential of Zero, $\exp 0 = 1$.

Finally, suppose that $x < 0$.

Then:

\(\displaystyle 0\) | \(<\) | \(\displaystyle -x\) | $\quad$ Order of Real Numbers is Dual of Order of their Negatives | $\quad$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle 0\) | \(<\) | \(\displaystyle \exp \left({-x}\right)\) | $\quad$ from above | $\quad$ | ||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle 0\) | \(<\) | \(\displaystyle \frac 1 {\exp x}\) | $\quad$ Reciprocal of Real Exponential | $\quad$ | ||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle 0\) | \(<\) | \(\displaystyle \exp x\) | $\quad$ Ordering of Reciprocals | $\quad$ |

So $\exp$ is strictly positive on $\R_{<0}$.

Hence the result.

$\blacksquare$

## Proof 2

This proof assumes the limit definition of $\exp$.

That is, let:

- $\displaystyle \exp x = \lim_{n \to \infty} f_n \left({x}\right)$

where $f_n \left({x}\right) = \left({1 + \dfrac x n}\right)^n$

First, fix $x \in \R$.

Let $N = \left\lceil{\left\vert{x}\right\vert}\right\rceil$, where $\left\lceil{\cdot}\right\rceil$ denotes the ceiling function.

Then:

\(\displaystyle \exp x\) | \(=\) | \(\displaystyle \lim_{n \to \infty} f_n \left({x}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lim_{n \to \infty} f_{n + N} \left({x}\right)\) | $\quad$ Tail of Convergent Sequence | $\quad$ | |||||||||

\(\displaystyle \) | \(\ge\) | \(\displaystyle f_{n + N} \left({x}\right)\) | $\quad$ Exponential Sequence is Eventually Increasing and Limit of Bounded Convergent Sequence is Bounded | $\quad$ | |||||||||

\(\displaystyle \) | \(>\) | \(\displaystyle 0\) | $\quad$ Corollary to Exponential Sequence is Eventually Increasing | $\quad$ |

$\blacksquare$

## Proof 3

This proof assumes the definition of $\exp x$ as the unique continuous extension of $e^x$.

Since $e > 0$, the result follows immediately from Power of Positive Real Number is Positive over Rationals.

$\blacksquare$

## Proof 4

This proof assumes the definition of $\exp$ as the inverse mapping of extension of $\ln$, where $\ln$ denotes the natural logarithm.

Recall that the domain of $\ln$ is $\R_{>0}$.

From the definition of inverse mapping, the image of $\exp$ is the domain of $\ln$.

That is, the image of $\exp$ is $\R_{>0}$.

Hence the result.

$\blacksquare$

## Proof 5

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

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

$\Box$

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

Then $0 \in \left({\exp \alpha \,.\,.\, 1}\right)$.

From Intermediate Value Theorem:

- $\exists \zeta \in \left({\alpha \,.\,.\, 0}\right): f \left({\zeta}\right) = 0$

This contradicts the lemma.

$\blacksquare$