Chapter 9.4, Problem 37E

To determine

To prove that if $x_1\ne 0,x_2\ne 0,\dots ,x_n\ne 0,$ then ${\left(x_1{x}_2{x}_3\dots x_n\right)}^{-1}=x_1{}^{-1}{x}_2{}^{-1}{x}_3{}^{-1}\dots x_n{}^{-1}$ for all integers $n\ge 1$ by using mathematical induction.

Answer to Problem 37E

The property that if $x_1\ne 0,x_2\ne 0,\dots ,x_n\ne 0,$ then ${\left(x_1{x}_2{x}_3\dots x_n\right)}^{-1}=x_1{}^{-1}{x}_2{}^{-1}{x}_3{}^{-1}\dots x_n{}^{-1}$ for all integers $n\ge 1$ is valid.

Explanation of Solution

Given:

Let $X_n={\left(x_1{x}_2{x}_3\dots x_n\right)}^{-1}$ .

Concept used:

The Principle of Mathematical Induction states that for a statement $P_n$ involving the positive integer n . If

1. $P_1$ is true, and

2. for every positive integer k , the truth of $P_k$ implies the truth of $P_{k+1}$ then the statement $P_n$ must be true for all positive integers n .

Calculation:

First to show that the property is valid for $n=1$ . When $n=1$ , the property is valid because:

  $\begin{array}{c}{X}_1={\left(x_1\right)}^{-1}\\ =x_1{}^{-1}\end{array}$

Again, show that the property is valid for $n=2$ . When $n=2$ , the property is valid because:

  $\begin{array}{c}{X}_2={\left(x_1{x}_2\right)}^{-1}\\ =\frac{1}{x_1{x}_2}\\ =\frac{1}{x_1}\cdot \frac{1}{x_2}\\ =x_1{}^{-1}{x}_2{}^{-1}\dots (1)\end{array}$

Now, let us assume that for $n=k$ ,

  $X_k={\left(x_1{x}_2{x}_3\dots x_k\right)}^{-1}=x_1{}^{-1}{x}_2{}^{-1}{x}_3{}^{-1}\dots x_k{}^{-1}$ ,

we need to show that for $n=k+1$ ,

  $X_{k+1}={\left(x_1{x}_2{x}_3\dots x_k{x}_{k+1}\right)}^{-1}=x_1{}^{-1}{x}_2{}^{-1}{x}_3{}^{-1}\dots x_k{}^{-1}{x}_{k+1}{}^{-1}$ .

For $n=k+1$ , the expression will be:

  $X_{k+1}={\left(x_1{x}_2{x}_3\dots x_k{x}_{k+1}\right)}^{-1}$

Here let two terms be $\left(x_1{x}_2{x}_3\dots x_k\right)$ and $x_{k+1}$ , and apply the result (1), and the assumption taken:

  $\begin{array}{r}{X}_{k+1}={\left(x_1{x}_2{x}_3\dots x_k\right)}^{-1}{\left(x_{k+1}\right)}^{-1}\\ =x_1{}^{-1}{x}_2{}^{-1}{x}_3{}^{-1}\dots x_k{}^{-1}{x}_{k+1}{}^{-1}\end{array}$

Therefore, by combining all the results, we can conclude by the mathematical induction that the property is valid for all $n\ge 1$ .