# 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

\(\displaystyle \map \exp 0\) | \(=\) | \(\displaystyle 1\) | Exponential of Zero | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map \exp {x - x}\) | \(=\) | \(\displaystyle 1\) | as $\forall x \in \R : x - x = 0$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map \exp {-x} \, \map \exp x\) | \(=\) | \(\displaystyle 1\) | Exponential of Sum: Real Numbers | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map \exp {-x}\) | \(=\) | \(\displaystyle \dfrac 1 {\exp x}\) | dividing both sides by $\exp x$, which is non-zero by Exponential of Real Number is Strictly Positive |

$\blacksquare$