Question

Prove that between every two rational numbers there is an irrational number.

Short answer

There is an irrational number \(a + \frac{{\sqrt 2 }}{n}\)between the rational numbers\(a\,and\,b\).

Step-by-step solution

Introduction

A rational value is a value that can be represented in the form of\(\frac{p}{q}\), of two integral values such that\(q \ne 0\).

Proof using generality

Let\(a\,and\,b\)be any two rational numbers.

Without loss of generality, let us assume that

\(\begin{aligned}{l}a < b\\ \Rightarrow b - a > 0\end{aligned}\)

Clearly, for some large enough \(n\)belonging to positive real numbers,

\(\begin{aligned}{l}n\left( {b - a} \right) > \sqrt 2 \\\, \Rightarrow b - a > \frac{{\sqrt 2 }}{n}\\ \Rightarrow a + \frac{{\sqrt 2 }}{n} < b\end{aligned}\)

It is also clear that \(\frac{{\sqrt 2 }}{n}\)is irrational, since \(\frac{{\sqrt 2 }}{n} = \sqrt 2 \left( {\frac{1}{n}} \right)\)which is a multiplication of irrational\(\sqrt 2 \)and rational\(\left( {\frac{1}{n}} \right)\).

Also, \(a + \frac{{\sqrt 2 }}{n}\)is irrational, since it is addition of rational and an irrational.

Clearly, \(a < a + \frac{{\sqrt 2 }}{n} < b\)

So, there is an irrational number \(a + \frac{{\sqrt 2 }}{n}\)between the rational numbers\(a\,and\,b\).