# 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:

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

First, suppose $0 < x$.

Then:

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

\(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \frac {x^n} {n!}\) | Real Number Ordering is Compatible with Multiplication | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\) | Ordering of Series of Ordered Sequences | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \exp x\) | Definition of $\exp$ |

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

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

Finally, suppose that $x < 0$.

Then:

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

\(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \map \exp {-x}\) | from above | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \frac 1 {\exp x}\) | Reciprocal of Real Exponential | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \exp x\) | Ordering of Reciprocals |

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:

- $\ds \exp x = \lim_{n \mathop \to \infty} \map {f_n} x$

where $\map {f_n} x = \paren {1 + \dfrac x n}^n$

First, fix $x \in \R$.

Let $N = \ceiling {\size x}$, where $\ceiling {\, \cdot \,}$ denotes the ceiling function.

Then:

\(\ds \exp x\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \map {f_n} x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \map {f_{n + N} } x\) | Tail of Convergent Sequence | |||||||||||

\(\ds \) | \(\ge\) | \(\ds \map {f_{n + N} } x\) | Exponential Sequence is Eventually Increasing and Limit of Bounded Convergent Sequence is Bounded | |||||||||||

\(\ds \) | \(>\) | \(\ds 0\) | Corollary to Exponential Sequence is Eventually Increasing |

This article needs to be linked to other articles.In particular: Corollary to Exponential Sequence is Eventually Increasing does not actually exist. The page it gets sent to does not give that result.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{MissingLinks}}` from the code. |

$\blacksquare$

## 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 \map \exp 0 = 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 $0 \in \openint {\exp \alpha} 1$.

From Intermediate Value Theorem:

- $\exists \zeta \in \openint \alpha 0: \map f \zeta = 0$

This contradicts the lemma.

$\blacksquare$