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
\(\ds \map \exp 0\) | \(=\) | \(\ds 1\) | 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 |
$\blacksquare$