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.

$\blacksquare$