# Ex.1.3 Q1 Real Numbers Solution - NCERT Maths Class 10

## Question

Prove that \( \begin{align} \sqrt 5 \end{align} \) is irrational.

## Text Solution

**What is unknown?**

\( \begin{align} \text{ }\sqrt{5}\end{align} \) is irrational

**Reasoning:**

In this question you have to prove that \( \begin{align} \sqrt 5 \end{align} \) is irrational. This question can be solved with the help of contradiction method. Suppose that \( \begin{align} \sqrt 5 \end{align} \) is rational. If \( \begin{align} \sqrt 5 \end{align} \) is rational that means it can be written in the form of \( \begin{align} \frac{p}{q},\end{align} \) where *\(p\)* and *\(q\)* are integers and \( \begin{align} q \ne 0.\end{align} \)

Now, *\(p\)* and *\(q\)* have common factors, when you cancel them you will get \( \begin{align} \frac{a}{b}\end{align} \) where *\(a\)* and *\(b\)* are co-primes and have no common factor other than \(1.\)

Now square both the sides, if \( \begin{align} {a^2}\end{align} \) is divisible by \(5\) that means *\(a\)* is also divisible by \(5\) (Let *\(p\) *be a prime number. If *\(p\)* divides \( \begin{align} {a^2},\end{align} \) then *\(p\)* divides \(a,\) where *\(a\)* is a positive integer). So, you can write \(a = 5c.\) Again on squaring you will get the value of \( \begin{align} {a^2}\end{align} \) substitute the value of \( \begin{align} {a^2}\end{align} \) in the above equation, you will get \( \begin{align} \frac{{{b^2}}}{5} = {c^2},\end{align} \) this means \( \begin{align} {b^2}\end{align} \) is divisible by \(5\) and so *\(b\)* is also divisible by \(5.\) Therefore, *\(a\)* and *\(b\)* have at least \(5 \) as a common factor.

But this contradicts the fact that *\(a\)* and *\(b\)* are coprime. This contradiction has arisen because of our incorrect assumption that \( \begin{align} \sqrt 5 \end{align} \) is a rational number. So, we conclude that \( \begin{align} \sqrt 5 \end{align} \) is an irrational number.

**Steps:**

Let us assume, to the contrary that \( \begin{align} \sqrt 5 \end{align} \) is a rational number. Let *\(p\)* and *\(q\)* have common factors, so by cancelling them we will get \( \begin{align} \frac{a}{b},\end{align} \) where *\(a\)* and *\(b\)* are co-primes.

\( \begin{align} \frac{{\sqrt 5 }}{1} = \frac{a}{b}\end{align} \) (where \(a\) and \(b\) are co-primes and have no common factor other than \(1\))

\[\begin{align} \sqrt {5b} = a\end{align} \]

Squaring both sides,

\[ \begin{align} 5{b^2} &= {a^2}\\\,\,{b^2} &= \frac{{{a^2}}}{5} \qquad {\text{(1)}}\end{align} \]

\(5\) divides \(a^2\),

That means it also divide *\(a,\)*

\[ \begin{align}{\frac{a}{5}}& = c\\a &= 5c\end{align} \]

on squaring,

\[ \begin{align} {a^2} = {{ }}25{c^2}\end{align} \]

put the value of \(a^2\)^{ }in equation (\(1\))

\[ \begin{align} {5{b^2}} &= 25{c^2}\\{{b^2}} &= 5{c^2}\\{\frac{{{b^2}}}{5}}& = {c^2}\end{align} \]

This means \(b^2\)^{ }is divisible by \(5\) and so *\(b\)* is also divisible by \(5.\) Therefore, \(a \) and *\(b\)* have at least \(5\) as a common factor. But this contradicts the fact that *\(a\)* and *\(b\)* are coprime. This contradiction has arisen because of our incorrect assumption that \( \begin{align} \sqrt 5 \end{align} \) is a rational number. So, we conclude that \( \begin{align} \sqrt 5 \end{align} \) is irrational.