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.
$\blacksquare$