Question

(ab)ⁿ = aⁿbⁿ for all n ∈ N. 

 

Answer 1

Let P(n) be the given statement.
Now, 

\[P(n): (ab )^n = a^n b^n \text{ for all } n \in N . \]
\[\text{ Step } 1: \]
\[P(1): (ab )^1 = a^1 b^1 = ab\]
\[\text{ Thus, P(1) is true }  . \]
\[\text{ Step 2 } : \]
\[\text{ Let P(m) be true } . \]
\[\text{ Then } , \]
\[(ab )^m = a^m b^m \]
\[ \text{ We need to show that P(m + 1) is true whenever P(m) is true }  . \]
\[\text{ Now, }  \]
\[ P(m + 1): (ab )^{m + 1} = (ab )^m . ab\]
\[ = a^m b^m . ab\]
\[ = a^m a . b^m b\]
\[ = a^{m + 1} b^{m + 1} \]
\[\text{ Hence, P(m + 1) is true .}  \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n }  \in N .\]