Exponential of Real Number is Strictly Positive

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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:

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

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:

\(\displaystyle \exp x\) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \map {f_n} x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \map {f_{n + N} } x\) $\quad$ Tail of Convergent Sequence $\quad$
\(\displaystyle \) \(\ge\) \(\displaystyle \map {f_{n + N} } x\) $\quad$ 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.

$\blacksquare$


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