# 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

$\blacksquare$

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

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

$\blacksquare$