Exploring the sum of exponents law

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. It’s easy to reason intuitively as to why this property holds for integral exponents:

However, I wanted to see how difficult it’d be to formalize this argument with induction. Let’s see!

The sum of exponents law

The statement we’ll 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. A recursive definition is convenient for induction:

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

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

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 is:

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:

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:

Otherwise we have $|x| \geq |y|$, in which case taking the reciprocal again admits Lemma 1.2:

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. $\square$