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

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## Question

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

Video Solution
Real Numbers
Ex 1.3 | Question 1

## 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.

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