Question

Prove or disprove that if m and n are integers such that\(mn = 1\), then either\(m = 1\)and\(n = 1\), or else\(m = - 1\)and\(n = - 1\).

Short answer

If mn = 1 then either m = 1 and n = 1, or else m = -1 and n = -1

Step-by-step solution

Introduction

Consider the product of any two integers \(m\)and\(n \) is 1.

That is\(mn = 1\).

The purpose is to prove or disprove that when\(mn = 1\)then either \(m = 1\)and(n = 1), or else \(m = - 1\)and\(n = - 1\).

Proof of the above statement

The statement “if \(mn = 1\)then either \(m = 1\)and\(n = 1\), or else \(m = - 1\)and\(n = - 1\)” is true.

Suppose that \(m\)is neither 1 nor -1.

Then \(mn \)has a factor\(m\)longer than 1.

On the other hand, \(mn = 1\)and 1 has no such factor.

Hence, \(m = 1\)or \(m = - 1\)

Taking \(m  = 1\)in \(mn  = 1\).

Case 1:

Put \(m = 1\)in \(mn = 1\).

\(\left( 1 \right)n = 1\)

\( \Rightarrow n = 1\)

Taking (m  =  - 1) in mn = 1

Case 2: Put \(m = - 1\) in \(mn = 1\)

\(\begin{array}{l}\left( { - 1} \right)n = 1\\ \Rightarrow - n = 1\\ \Rightarrow n = - 1\end{array}\)

Hence, in the first case\(n = 1\)and in the second case \(n = - 1\)because\(n = \frac{1}{m}\).

Therefore, if\(mn = 1\)then either \(m = 1\)and\(n = 1\), or else \(m = - 1\)and\(n = - 1\)