Chapter 9.4, Problem 33E

To determine

To prove: The Property ${(x_1{x}_2{x}_3\mathrm{....}{x}_n)}^{-1}=x_1{}^{-1}{x}_2{}^{-1}{x}_3{}^{-1}\mathrm{.......}{x}_n{}^{-1},x_1\ne 0,x_2\ne \mathrm{0.....}{x}_n\ne 0$ for every integer $n$ . Using mathematical induction.

Explanation of Solution

Given: $\begin{array}{l}{x}_1\ne 0,x_2\ne \mathrm{0.....}{x}_n\ne 0\\ $ {(x_1{x}_2{x}_3\mathrm{....}{x}_n)}^{-1}=x_1{}^{-1}{x}_2{}^{-1}{x}_3{}^{-1}\mathrm{.......}{x}_n{}^{-1}$\end{array}$

The given formula, ${(x_1{x}_2{x}_3\mathrm{....}{x}_n)}^{-1}=x_1{}^{-1}{x}_2{}^{-1}{x}_3{}^{-1}\mathrm{.......}{x}_n{}^{-1}$

Let, $P\left(n\right)={(x_1{x}_2{x}_3\mathrm{....}{x}_n)}^{-1}=x_1{}^{-1}{x}_2{}^{-1}{x}_3{}^{-1}\mathrm{.......}{x}_n{}^{-1}$

When n =2

  ${(x_1{x}_2)}^{-1}=\frac{1}{x_1{x}_2}=\frac{1}{x_1}.\frac{1}{x_2}=x_1{}^{-1}{x}_2{}^{-1}$

So, $P\left(k\right)$ is true for $n=k$ where, $k\in Z^{+}$ .

  ${(x_1{x}_2{x}_3\mathrm{....}{x}_k)}^{-1}=x_1{}^{-1}{x}_2{}^{-1}{x}_3{}^{-1}\mathrm{.......}{x}_k{}^{-1}$

Now, prove for $P\left(k+1\right)$ .

Put $n=k+1$

  $\begin{array}{l}\text{Assume that}\\ $ {(x_1{x}_2{x}_3\mathrm{....}{x}_k)}^{-1}=x_1{}^{-1}{x}_2{}^{-1}{x}_3{}^{-1}\mathrm{.......}{x}_k{}^{-1}$then\\ $ {(x_1{x}_2{x}_3\mathrm{....}{x}_k{x}_{k+1})}^{-1}={((x_1{x}_2{x}_3\mathrm{....}{x}_k)x_{k+1})}^{-1}$\end{array}$

  $={(x_1{x}_2{x}_3\mathrm{....}{x}_k)}^{-1}{(x_{k+1})}^{-1}$=x1−1x2−1x3−1.......xk−1xk+1−1

Hence, $P\left(k+1\right)$ is true ,The formula is valid