Chapter 12.4, Problem 25AYU

To determine

To prove: The given statement $a - b$ is a factor of $a^n - b^n$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 25AYU

As the statement is true for the natural number $\left(k + 1\right)$ terms, it is true for all natural numbers by the theorem of mathematical induction.

Explanation of Solution

Given:

Statements says the series $a - b$ is a factor of $a^n - b^n$ is true for all natural number.

Formula used:

The Principle of Mathematical Induction

Suppose that the following two conditions are satisfied with regard to a statement about natural numbers:

CONDITION I: The statement is true for the natural number 1.

CONDITION II: If the statement is true for some natural number $k$, it is also true for the next natural number $k + 1$. Then the statement is true for all natural numbers.

Proof:

Consider the statement

$a - b$ is a factor of $a^n - b^n$   -----(1)

Step 1:

Show that statement (1) is true for the natural number $n = 1$.

That is $a^1 - b^1 = a - b$ which is obviously a factor of $a - b$. Hence the statement is true for the natural number $n = 1$.

Step 2:

Assume that the statement is true for some natural number $k$.

That is $a - b$ is a factor of $a^k - b^k$   -----(2)

Step 3:

Prove that the statement is true for the next natural number $k + 1$.

That is, to prove tha $a - b$ is a factor of $a^{k + 1} - b^{k + 1}$

Consider $a^{k + 1} - b^{k + 1}$

$a^{k + 1} - b^{k + 1} = a\left(a^k - b^k\right) + b^k\left(a - b\right)$

In here, the first term is a multiple of $\left(a^k - b^k\right)$, which has a factor $a - b$ and the second term is a multiple of $a - b$ , which obviously has the factor $a - b$. Therefore $a - b$ is a factor of $a^{k + 1} - b^{k + 1}$

As the statement is true for the natural number $\left(k + 1\right)$ terms, it is true for all natural numbers by the theorem of mathematical induction.