Precalculus (6th Edition) Blitzer

Let ${{S}_{n}}$ 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 its first term on left and substituting n with $1$ on the right. \begin{align} & {{S}_{1}}: \\ & {{\left( \frac{a}{b} \right)}^{1}}=\frac{{{a}^{1}}}{{{b}^{1}}} \\ & =\frac{a}{b} \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. ${{S}_{k}}:{{\left( \frac{a}{b} \right)}^{k}}=\frac{{{a}^{k}}}{{{b}^{k}}}$ Now, one has to prove ${{S}_{k+1}}$ is true ${{S}_{k+1}}:{{\left( \frac{a}{b} \right)}^{k+1}}=\frac{{{a}^{k+1}}}{{{b}^{k+1}}}$ Since, one assume that ${{S}_{k}}$ is true, ${{\left( \frac{a}{b} \right)}^{k}}=\frac{{{a}^{k}}}{{{b}^{k}}}$ And multiply $\left( \frac{a}{b} \right)$ on the both sides, \begin{align} & {{\left( \frac{a}{b} \right)}^{k}}\left( \frac{a}{b} \right)=\frac{{{a}^{k}}}{{{b}^{k}}}\left( \frac{a}{b} \right) \\ & {{\left( \frac{a}{b} \right)}^{k+1}}=\frac{{{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 ${{\left( \frac{a}{b} \right)}^{n}}=\frac{{{a}^{n}}}{{{b}^{n}}}$ is true for every positive integer $n$.