# Exponential of Sum/Real Numbers/Proof 4

## 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:

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

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 $(2)$.

$f = \exp$

That is:

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

$\blacksquare$