# Solve the following

Question:

(ab)n = anbn for all n ∈ N.

Solution:

Let P(n) be the given statement.

Now,

$P(n):(a b)^{n}=a^{n} b^{n}$ for all $n \in N$.

Step 1:

$P(1):(a b)^{1}=a^{1} b^{1}=a b$

Thus, $P(1)$ is true.

Step 2 :

Let $P(m)$ be true.

Then,

$(a b)^{m}=a^{m} b^{m}$

We need to show that $P(m+1)$ is true whenever $P(m)$ is true.

Now,

$P(m+1):(a b)^{m+1}=(a b)^{m} \cdot a b$

$=a^{m} b^{m} \cdot a b$

$=a^{m} a \cdot b^{m} b$

$=a^{m+1} b^{m+1}$

Hence, $P(m+1)$ is true.

By the $p$ rinciple of mathematical $i$ nduction, $P(n)$ is true for all $n \in N$.