Question

Show that if \(r\)is an irrational number, there is a unique integer \(n\)such that the distance between \(r\)and \(n\) is less than\(\frac{1}{2}\).

Short answer

There is to be a unique integer n such that the distance between r and n is less than\(\frac{1}{2}\).

Step-by-step solution

Introduction

As r is irrational implies r is not an integer and lies between two integers.

So there is to be an integer n such that\(n < r < n + 1\).

As r is irrational,\(r \ne \frac{{n + 1}}{2}\).

There exist two cases,

Case 1: \(r < n + \frac{1}{2}\)

Here\(\frac{1}{2} > r - n > 0\), so distance between r and n is less than\(\frac{1}{2}\).

Case 2: \(r > \frac{{n + 1}}{2}\)

Here\(\frac{1}{2} > \left( {n + 1} \right) - r > 0\), so distance between r and \(n + 1\) is less than\(\frac{1}{2}\).

Therefore, the distance between an irrational number r and an integer n is less than\(\frac{1}{2}\).

Uniqueness

Suppose there were two distinct integer n and m such that

\(\left| {r - n} \right| < \frac{1}{2}\)and.\(\left| {r - m} \right| < \frac{1}{2}\)

Then,

\(\begin{aligned}{}\left| {n - m} \right| \le \left| {n - r + r - m} \right|\\ \le \left| {r - n} \right| + \left| {r - m} \right|\\ < \frac{1}{2} + \frac{1}{2}\\ = 1\end{aligned}\)

This is contradiction to the fact that n and m is two distinct integers.

Therefore, there is to be a unique integer n such that the distance between r and n is less than\(\frac{1}{2}\).