Exponential of Sum/Real Numbers/Proof 4

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Theorem

Let $x, y \in \R$ be real numbers.

Let $\exp x$ be the exponential of $x$.


Then:

$\map \exp {x + y} = \paren {\exp x} \paren {\exp y}$


Proof

This proof assumes the definition of $\exp$ as defined by an initial value problem.

That is, suppose $\exp$ satisfies:

$(1): \quad D_x \exp x = \exp x$
$(2): \quad \exp 0 = 1$

on $\R$.


Consider the real function $f: \R \to \R$ defined by:

$\map f x := \dfrac {\map \exp {x + y} } {\map \exp y}$

From Exponential of Real Number is Strictly Positive, $f$ is well-defined.


So:

\(\ds D_x \, \map f x\) \(=\) \(\ds D_x \frac {\map \exp {x + y} } {\map \exp y}\)
\(\ds \) \(=\) \(\ds \frac 1 {\map \exp y} D_x \, \map \exp {x + y}\) Derivative of Constant Multiple
\(\ds \) \(=\) \(\ds \frac {\map \exp {x + y} } {\map \exp y}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \map f x\)


Thus $f$ satisfies $(1)$.


Further:

\(\ds \map f 0\) \(=\) \(\ds \frac {\map \exp {0 + y} } {\map \exp y}\)
\(\ds \) \(=\) \(\ds \frac {\map \exp y} {\map \exp y}\)
\(\ds \) \(=\) \(\ds 1\)


So $f$ satisfies $(2)$.


From Exponential Function is Well-Defined:

$f = \exp$

That is:

\(\ds \frac {\map \exp {x + y} } {\map \exp y}\) \(=\) \(\ds \map \exp x\)
\(\ds \leadsto \ \ \) \(\ds \map \exp {x + y}\) \(=\) \(\ds \map \exp x \, \map \exp y\)

$\blacksquare$


Sources