Exponential of Real Number is Strictly Positive/Proof 1
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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
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.
$\blacksquare$