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

## Question

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

## 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.