# Exponent Combination Laws/Product of Powers/Proof 2

## Theorem

Let $a \in \R_{> 0}$ be a positive real number.

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

Let $a^x$ be defined as $a$ to the power of $x$.

Then:

$a^x a^y = a^{x + y}$

## Proof

Let $x, y \in \R$.

From Rational Sequence Decreasing to Real Number, there exist rational sequences $\sequence {x_n}$ and $\sequence {y_n}$ converging to $x$ and $y$, respectively.

 $\displaystyle a^{x + y}$ $=$ $\displaystyle a^{\displaystyle \paren {\lim_{n \mathop \to \infty} x_n + \lim_{n \mathop \to \infty} y_n} }$ $\displaystyle$ $=$ $\displaystyle a^{\displaystyle \paren {\lim_{n \mathop \to \infty} \paren {x_n + y_n} } }$ Sum Rule for Real Sequences $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} a^{x_n + y_n}$ Sequential Continuity is Equivalent to Continuity in the Reals $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \paren {a^{x_n} a^{y_n} }$ Sum of Indices of Real Number: Rational Numbers $\displaystyle$ $=$ $\displaystyle \paren {\lim_{n \mathop \to \infty} a^{x_n} } \paren {\lim_{n \mathop \to \infty} a^{y_n} }$ Product Rule for Real Sequences $\displaystyle$ $=$ $\displaystyle a^x a^y$ Sequential Continuity is Equivalent to Continuity in the Reals

$\blacksquare$