The sum of exponents law, $b^x \cdot b^y = b^{x+y}$, is well-known—indeed, it was one of the first properties that I learnt after being introduced to the concept of exponentiation. Though it’s easy to reason intuitively as to why this property holds for integral exponents,
$$ 2^{3 + 2} = 2^5 = \underbrace{2 \cdot 2 \cdot 2}_{2^3} \cdot \underbrace{2 \cdot 2}_{2^2}, $$
I wanted to see how difficult it’d be to formalize this argument with induction. Let’s take a look!
The sum of exponents law
The statement we aim to prove is the following:
Sum of Exponents Law for Integer Exponents. For positive real $b$, the equation $$ b^x \cdot b^y = b^{x + y} $$ holds for all integer $x$, $y$.
This law holds for $x$, $y$ real, but we concern ourselves only with the easiest integer case in this blog. Perhaps someday there will be a follow-up post…
Defining exponentiation
We first need a definition of exponentiation for integer powers to work with. A recursive definition is convenient for induction:
$$ b^x = \begin{cases} \dfrac{1}{b^{-x}}, & \text{if $x < 0$} \\ 1, & \text{if $x = 0$} \\ b \cdot b^{x-1}, & \text{if $x > 0$} \end{cases} $$
Adding positive exponents
We can immediately address the trivial case where both exponents are positive integers.
Lemma 1.1. For all $x, y \in \mathbb{Z^+}$, we have $b^{x + y} = b^x \cdot b^y$.
Proof. We induct on $x$.
Base case. If $x = 1$, then $b^{x + y} = b^{y + 1} = b \cdot b^y$ by definition, which is equal to ${b^x \cdot b^y = b^1 \cdot b^y = b \cdot b^y}$.
Induction step. Let $x \in \mathbb{Z^+}$ be arbitrary and suppose $b^{x + y} = b^x \cdot b^y$ for all $y \in \mathbb{Z^+}$. Then
$$ \begin{align*} b^{(x + 1) + y} &= b^{x + y + 1} \\ &= b \cdot b^{x + y} \\ &= b \cdot b^{x} \cdot b^y && \text{(by induction hypothesis)} \\ &= b^{x + 1} \cdot b^y \end{align*} $$
as desired. $\square$
Subtracting positive exponents
Lemma 1.1 covers the case of two positive exponents (and, indirectly, the case of two negative exponents); we still have no way to deal with exponents of different signs. The following lemma addresses this gap:
Lemma 1.2. For all $x, y \in \mathbb{Z^+}$ where $x \geq y$, we have $b^{x - y} = b^x \cdot b^{-y}$.
Equivalently, this lemma claims that $b^{x + y} = b^x \cdot b^y$ where $y$ is negative and $|x| \geq |y|$; however, the original presentation in terms of subtracting positive exponents is more amenable to induction.
Proof. We induct on $x$.
Base case. If $x = 1$, then the condition $x \geq y$ forces $y = 1$. Substituting, we have ${b^{1 - 1} = b^0 = 1}$ which is equal to $b^1 \cdot b^{-1} = 1$.
Induction step. Let $x \in \mathbb{Z^+}$ be arbitrary and suppose $b^{x - y} = b^x \cdot b^{-y}$ for all $y \in \mathbb{Z^+}$ such that $x \geq y$. Then
$$ \begin{align*} b^{(x + 1) - y} &= b^{x - y + 1} \\ &= b \cdot b^{x - y} \\ &= b \cdot b^x \cdot b^{-y} && \text{(by induction hypothesis)} \\ &= b^{x + 1} \cdot b^{-y} \end{align*} $$
as desired. $\square$
The full proof
We are now equipped to tackle the full proof where $x$ and $y$ are unrestricted aside from being integers. Broadly speaking, Lemma 1.1 covers the case where $x$ and $y$ are both positive or both negative, and Lemma 1.2 covers the case where one is negative and the other positive; all that remains is to carefully invoke them and handle the straightforward case where one exponent is zero.
To recall, the statement we aim to prove was:
Sum of Exponents Law for Integer Exponents. For positive real $b$, the equation $$ b^x \cdot b^y = b^{x + y} $$ holds for all integer $x$, $y$.
Proof. We consider the following cases.
Case 1. Both $x$ and $y$ are positive. Lemma 1.1 discusses precisely this case.
Case 2. Both $x$ and $y$ are negative. Then $-x$ and $-y$ are both positive, so taking the reciprocal allows us to apply Lemma 1.1 as follows:
$$ \begin{align*} b^x \cdot b^y &= \frac{1}{b^{-x}} \cdot \frac{1}{b^{-y}} \\ &= \frac{1}{b^{-x} \cdot b^{-y}} \\ &= \frac{1}{b^{-x - y}} && \text{(by Lemma 1.1)} \\ &= \frac{1}{b^{-(x + y)}} \\ &= b^{x + y}. \end{align*} $$
Case 3. Exactly one of $x, y$ is negative.
Without loss of generality, suppose $x < 0$ and $y > 0$. We consider two sub-cases: ${|x| < |y|}$ and ${|x| \geq |y|}$.
First consider the case where $|x| < |y|$. We manipulate the expression into a form suitable for Lemma 1.2:
$$ \begin{align*} b^x \cdot b^y &= b^{-(-x)} \cdot b^y \\ &= b^y \cdot b^{-(-x)} \\ &= b^{y - (-x)} && \text{(by Lemma 1.2)} \\ &= b^{x + y}. \end{align*} $$
Otherwise we have $|x| \geq |y|$, in which case taking the reciprocal again admits Lemma 1.2:
$$ \begin{align*} \frac{1}{b^x \cdot b^y} &= \frac{1}{b^x} \cdot \frac{1}{b^y} \\ &= b^{-x} \cdot b^{-y} \\ &= b^{-x - y} && \text{(by Lemma 1.2)} \\ &= b^{-(x + y)} \\ &= \frac{1}{b^{x + y}}. \end{align*} $$
Case 4. At least one of $x$, $y$ is 0.
Without loss of generality, suppose $x = 0$. Then $b^x \cdot b^y = 1 \cdot b^y = b^y$ which is equal to $b^{x + y} = b^{0 + y} = b^y$ as desired.
In all four cases, the law holds and so we are done. $\square$