Sum Of The Powers

Theorem: $\;\;a^{x+y} = a^x.a^y \quad\forall\; x, y \in \mathbb{R} \text{ and } a > 0$

Prerequisites:
Exponentiation (definition)
Continuity of exponential function

Proof:
Case I: when $x, y \in\mathbb{Z^+}$

$\qquad\quad \begin{align} a^{x+y} &= a.a.a.a......a\quad \text{($x+y$ times)}\\
&= \underbrace{(a.a.a.....a)}_\text{$x \text{ times}$}.\underbrace{(a.a.a.....a)}_\text{$y\text{ times}$}\\
&= a^x.a^y\end{align}$

Case II: when $x, y \in\mathbb{Z^-}$

Let $x_1 = -x, \;\; y_1 = -y$ and $1/a = b$

$\qquad\quad \begin{align} a^{x+y} &= a^{-(x_1+y_1)}\\\\
&= \left(\dfrac{1}{a}\right)^{(x_1+y_1)}\\\\
&= b^{(x_1+y_1)}\end{align}$

Hence the case is reduced to the previous case and can be proved in a similar manner.

Case III: when $x \in\mathbb{Z^+}$ and $y \in\mathbb{Z^-}$
Let $y_1 = -y$

$\qquad\quad \begin{align} a^{x+y} &= a^{x-y_1}\\
&= a.a.a....a\quad\text{($x-y_1$ times)}\\\
&= \dfrac{a.a.a....a}{a.a.a....a}\begin{matrix}\text{($x$ times)}\\
\text{($y_1$ times)}\end{matrix}\\\
&= \dfrac{a^x}{a^{y_1}}\\\
&= a^x.a^{-y_1}\\
&= a^x.a^y\end{align}$

Case IV: When atleast one of $x, y$ equal to $0$
Let $y = 0$

$\qquad\quad \begin{align}a^0 &= a^{n-n} \qquad\quad\;\;\;\text{ where $n\in\mathbb{Z^+}$}\\
&= a^n.a^{-n}\qquad\quad\text{(from case III)}\\
&= \dfrac{a^n}{a^n} = 1\end{align}$

$\begin{align}\text{Now, }\quad  a^{x+y} &= a^{x+0} = a^x\\
&= a^x.1 = a^x.a^0\\
&= a^x.a^y\end{align}$

Case V: When $x, y \in\mathbb{Q}$, i.e., $x, y$ are rational numbers

As a corollary of case I, for $c \in\mathbb{Z^+}$

$\qquad\quad\begin{align}\big(a^{b}\big)^c &= a^b.a^b.a^b....a^b\qquad\text{($c$ times)}\\
&= a^{(b+b+b...b)}\\
&= a^{\displaystyle (bc)}\end{align}$

Let $x = p_1/q_1$ and $y = p_2/q_2$ where $p_1, p_2, q_1, q_2 \in\mathbb{Z}\;\text{ and }\; q_1, q_2\neq 0$
Let $q' = q_1q_2$,

$ \begin{align} \text{Now, }\quad  a^{x+y} &=  a^{\tfrac{p_1}{q_1}+ \tfrac{p_2}{q_2}}\\\\
&=  a^{\tfrac{p_1q_2 + p_2q_1}{q'}}\\\\
&= \Big( a^{(1/q')}\Big) ^{(p_1q_2 + p_2q_1)}\qquad\qquad\qquad\text{(from the above corollary)}\\\\
&= \Big( a^{(1/q')}\Big) ^{(p_1q_2)}\Big( a^{(1/q')}\Big) ^{(p_2q_1)}\qquad\text{(from case I)}\\\\
&= a^{\left( \tfrac{p_1q_2}{q'}\right)}a^{\left( \tfrac{p_2q_1}{q'}\right)}\\\\
&= a^{\tfrac{p_1}{q_1}}.a^{\tfrac{p_2}{q_2}}\\\\
&= a^x.a^y\end{align}$

Case VI: When $x, y\in\mathbb{R}$

If $x$ and $y$ are irrational, then in any neighbourhood of $x$ and $y$, there exists infinitely many rational numbers. Since at these rational numbers, the given equality holds, hence by the continuity of the exponential functions, the equality must also hold at $x$ and $y$.

Therefore the result holds at all real values of $x$ and $y$.


Recommended:
Product of the powers
Power of the product
Log of product
Log of powers