Question

Prove the rule of exponents $(ab{)}^n=a^n{b}^n$ by using principle of mathematical induction for every natural number.

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i. e.,
$P\left(n\right):(ab{)}^n=a^n{b}^n$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):(ab{)}^1=a^1{b}^1$
$\Rightarrow ab=ab$
Thus $P\left(n\right)$ is true for $n=1$.

Step $\left(3\right):$ $P\left(n\right)$ for $n=1$
Put $n=K$ in $P\left(n\right)$ and assume this is true for
some natural number $K$ i.e.,
$P\left(K\right):(ab{)}^K=a^K{b}^K\cdots (1)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now we shall prove that $P(K+1)$ is true
whenever $P\left(K\right)$ is true. Now, we have
$(ab{)}^{(K+1)}$
$=\left(ab{)}^K\right(ab)$
$=\left(a^K{b}^K\right)\left(ab\right)$ (Using $\left(1\right)$)
$=(a^K\cdot a^1)(b^K\cdot b^1)$
$=a^{K+1}\cdot b^{K+1}$
Therefore, $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Final Answer:
Hence, by principle of mathematical induction,
$P\left(n\right)$ is true for all $n\in N$.