Question

We can define the nonnegative powers of a number \(a\) by the rules \(a^{0}=1\) and \(a^{n+1}=a^{n} \cdot a\). Explain why this defines \(a^{n}\) for all nonnegative integers \(n\). From this definition, prove the rule of exponents \(a^{m+n}=a^{m} a^{n}\) for nonnegative integers \(m\) and \(n\).

Short answer

The nonnegative powers of \(a\) are defined recursively, and induction verifies the rule \(a^{m+n} = a^m \cdot a^n\) holds.

Step-by-step solution

Understanding the Power Definition

The exercise defines nonnegative powers of a number \(a\). The base case is \(a^0 = 1\). For any nonnegative integer \(n\), if we know \(a^n\), then \(a^{n+1} = a^n \cdot a\) allows us to find \(a^{n+1}\). This recursive definition means we can find \(a^n\) for any nonnegative \(n\) by starting at 0 and working upwards.

Using Mathematical Induction for Exponent Addition Rule

To prove \(a^{m+n} = a^m \cdot a^n\) for all nonnegative integers \(m\) and \(n\), we'll use induction on \(n\).

Base Case

First, consider \(n = 0\). Then \(m + 0 = m\) and we have \(a^{m+0} = a^m\), which equals \(a^m \cdot a^0 = a^m \cdot 1\). Thus, the base case holds: \(a^{m+0} = a^m \cdot a^0\).

Inductive Step

Assume the property holds for \(n = k\), so \(a^{m+k} = a^m \cdot a^k\). Now, consider \(n = k + 1\). Then \(m + (k + 1) = m + k + 1\), which by the definition is \(a^{m+k+1} = a^{m+k} \cdot a\). By the inductive hypothesis, \(a^{m+k} = a^m \cdot a^k\), so \(a^{m+k+1} = (a^m \cdot a^k) \cdot a = a^m \cdot (a^k \cdot a) = a^m \cdot a^{k+1}\). This completes the induction.

Concluding the Proof

Since the base case holds and the induction step is valid, we have proven that for all nonnegative integers \(m\) and \(n\), the property \(a^{m+n} = a^m \cdot a^n\) holds.

Key concepts

Exponentiation

Exponentiation is a mathematical operation involving two numbers. It consists of a base and an exponent. The base is a number that is being multiplied, and the exponent is how many times that multiplication occurs. For example, in the expression \(a^n\), \(a\) is the base, and \(n\) is the exponent, indicating that \(a\) is multiplied by itself \(n\) times. This process is essential in numerous fields of study, including algebra, physics, and computer science.Understanding this concept helps to simplify expressions and solve problems involving large powers. With repeated multiplication, exponentiation provides a quick way to write down products of the same numbers.

Recursive Definition

A recursive definition is a way of defining a concept where the concept itself appears in its definition or calculation. It's a method used for items that are defined by a repeated process. In the case of exponentiation, we can define nonnegative powers of a number \(a\) recursively:

• The base case is \(a^0 = 1\).
• If we know \(a^n\), then \(a^{n+1} = a^n \cdot a\).

This way, by starting from 0 and applying the rule as many times as necessary, we can find \(a^n\) for any nonnegative integer \(n\). Recursive definitions are handy for breaking down complex problems into simpler, repeated steps.

Base Case

The base case is a foundational step in a proof involving recursion or mathematical induction. It verifies that a statement or concept is true for an initial value, usually the smallest or simplest case. In proving properties by induction, the base case often concerns the smallest number in the set of numbers we are considering.For exponentiation, the base case given is \(a^0 = 1\). Starting with \(n = 0\), we assert that any non-zero number raised to the power of 0 is 1. Confirming the base case ensures the validity of a recursive or inductive proof, establishing a strong foundation before moving to more complex cases.

Inductive Step

The inductive step is a critical component in mathematical induction. It involves proving that if a statement holds for a certain case, it must also hold for the next case. This step bridges the base case to all subsequent cases.To accomplish this, assume the statement is true for an integer \(n = k\), and show it holds for \(n = k+1\).In the example of proving \(a^{m+n} = a^m \cdot a^n\), we suppose it is true for some integer \(n = k\), that is, \(a^{m+k} = a^m \cdot a^k\). Then, prove that under this assumption it holds for \(n = k+1\), i.e., \(a^{m+k+1} = a^m \cdot a^{k+1}\). Once this step is complete, combined with the base case, the statement is shown to be true for all nonnegative integers.