## Precalculus (6th Edition) Blitzer

Let us assume ${{S}_{n}}$ be a statement involving the positive integer n. If 1. ${{S}_{1}}$ is true, and 2. Let us assume that the ${{S}_{k}}$ be true for some integer $k$. Then, one has to use ${{S}_{k}}$ to prove ${{S}_{k+1}}$ is also true. If one is able to prove ${{S}_{k+1}}$ then by the principle of mathematical induction the statement ${{S}_{n}}$ is true for all positive integers $n$. Firstly, show that ${{S}_{1}}$ is true. For ${{S}_{1}}$ one has And write ${{S}_{1}}$ by taking the first term on left and substituting n with $1$ on the right. \begin{align} & {{S}_{1}}: \\ & {{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}} \\ & {{\left( ab \right)}^{1}}={{a}^{1}}{{b}^{1}} \\ & =ab \end{align} Therefore, the statement ${{S}_{1}}$ is true. Then, let us assume that ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ should also be true by the method of mathematical induction. ${{S}_{k}}:{{\left( ab \right)}^{k}}={{a}^{k}}{{b}^{k}}$ Now, one has to prove ${{S}_{k+1}}$ is true ${{S}_{k+1}}:{{\left( ab \right)}^{k+1}}={{a}^{k+1}}{{b}^{k+1}}$ Since, one assume that ${{S}_{k}}$ is true, ${{\left( ab \right)}^{k}}={{a}^{k}}{{b}^{k}}$ And multiplying $\left( ab \right)$ on the both sides, \begin{align} & {{\left( ab \right)}^{k}}\left( ab \right)={{a}^{k}}{{b}^{k}}\left( ab \right) \\ & {{\left( ab \right)}^{k+1}}={{a}^{k+1}}{{b}^{k+1}} \end{align} So, the final statement is ${{S}_{k+1}}$. Thus, by the method of principal of mathematical induction, the statement ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$ is true for every positive integer $n$.