Question

$(ab{)}^n=a^n{b}^n$ for all n $nϵN$.

Answer 1

Let p(n) : $(ba{)}^n=a^n{b}^n$

For n = 1

$(ab{)}^1=a^1{b}^1$

$ab=ab$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k

$(ab{)}^k=a^k{b}^k$ ..........(1)

We have to show that,

$(ab{)}^{k+1}=a^{k+1}.b^{k+1}$

Now,

$(ab{)}^{k+1}$

$=\left(ab{)}^k\right(ab)$

$=\left(a^k{b}^k\right)\left(ab\right)$ [Using equation (1)]

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

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all n $ϵ$ N by PMI.