Reciprocal of Real Exponential

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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

$\blacksquare$