Reciprocal of Real Exponential

Theorem

Let $x \in \R$.

Let $\exp$ denotes the real exponential function.

Then:

$\dfrac 1 {\map \exp x} = \map \exp {-x}$

Proof

\(\ds \map \exp 0\)

\(=\)

\(\ds 1\)

Exponential of Zero

\(\ds \leadsto \ \ \)

\(\ds \map \exp {x - x}\)

\(=\)

\(\ds 1\)

as $\forall x \in \R : x - x = 0$

\(\ds \leadsto \ \ \)

\(\ds \map \exp {-x} \, \map \exp x\)

\(=\)

\(\ds 1\)

Exponential of Sum of Real Numbers

\(\ds \leadsto \ \ \)

\(\ds \map \exp {-x}\)

\(=\)

\(\ds \dfrac 1 {\exp x}\)

dividing both sides by $\exp x$, which is non-zero by Exponential of Real Number is Strictly Positive

$\blacksquare$