# Ex.1.3 Q2 Real Numbers Solution - NCERT Maths Class 10

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## Question

Prove that \begin{align} 3{{ }} + {{ }}2\sqrt 5 \end{align} is irrational.

Video Solution
Real Numbers
Ex 1.3 | Question 2

## Text Solution

What is unknown?

\begin{align} \,\,3\,\, + \,\,2\sqrt 5 \end{align} is irrational

Reasoning:

In this question you have to prove that \begin{align} 3{{ }} + {{ }}2\sqrt 5 \end{align} is irrational. Solve this question with the help of contradiction method, suppose that that \begin{align} 3{{ }} + {{ }}2\sqrt 5 \end{align}is rational. If \begin{align} 3{{ }} + {{ }}2\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 $$q$$ 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 $$1.$$ First find out the value of \begin{align} \sqrt 5 \end{align} i.e. \begin{align} \sqrt 5 = \frac{{a - 3b}}{{2b}},\end{align} where \begin{align} \frac{{a - 3b}}{{2b}}\end{align} a rational number and \begin{align} b{{ }} \ne 0.\end{align} If \begin{align} \frac{{a - 3b}}{{2b}}\end{align} is a rational number that means \begin{align} \sqrt 5 \end{align} is also a rational number. But, we know that \begin{align} \sqrt 5 \end{align} is irrational this contradicts the fact that \begin{align} \sqrt 5 \end{align} is irrational.

Therefore, our assumption was wrong that \begin{align} 3{{ }} + {{ }}2\sqrt 5 \,\, + \end{align} is rational.

So, \begin{align} 3{{ }} + {{ }}2\sqrt 5 \,\, + \end{align} is irrational.

Steps:

Let us assume, to the contrary that \begin{align} 3{{ }} + {{ }}2\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} {3 + 2\sqrt 5} &={ \frac{a}{b}}\\{b(3 + 2\sqrt 5 )} &=a\\{\,\,3b + 2\sqrt {5b}}&= a\\{2\sqrt {5b}}& = a - 3b\\{\sqrt 5} &={ \frac{{a - 3b}}{{2b}}}\end{align}

(where $$a$$ and $$b$$ are co - primes and have no common factor other than $$1$$)

Since, \begin{align} \frac{{a - 3b}}{{2b}}\end{align} is a rational number then \begin{align} \sqrt 5 \end{align} is also a rational number.

But we know that \begin{align} \sqrt 5 \end{align} is irrational this contradicts the fact that \begin{align} \sqrt 5 \end{align} is rational.

Therefore, our assumption was wrong that \begin{align} 3{{ }} + {{ }}2\sqrt 5 \end{align} is rational. So, \begin{align} 3{{ }} + {{ }}2\sqrt 5 \end{align} is irrational.

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