# NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.3

Go back to  'Real Numbers'

## Question 1

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

### 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.

## Question 2

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

### 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.

## Question 3

Prove that the following are irrationals:

(i) \begin{align}\frac{1}{{\sqrt 2 }}\end{align}

(ii) \begin{align}7\sqrt 5 \end{align}

(iii)  \begin{align}6 +\sqrt 2 \end{align}

### Solution

What is unknown?

(i) \begin{align}\frac{1}{{\sqrt 2 }}\end{align}

(ii) \begin{align}7\sqrt 5 \end{align}

(iii)  \begin{align}6 +\sqrt 2 \end{align} are Irrationals

Steps:

(i) \begin{align} \frac{1}{{\sqrt 2 }}\end{align}

Let us assume, to the contrary \begin{align} \frac{1}{{\sqrt 2 }}\end{align} that is a rational number.

\begin{align} \frac{1}{{\sqrt 2 }} = \frac{p}{q}\end{align}

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{1}{{\sqrt 2 }}& = \frac{a}{b} \\\sqrt 2 {a} &= {b}\\\sqrt 2 &= \frac{b}{a}\end{align}
(where $$a$$ and $$b$$ are co - primes and have no common factor otherthan $$1$$)

Since, $$b$$ and $$a$$ are integers, \begin{align} \frac{b}{a}\end{align} is rational number and so, \begin{align} \sqrt 2 \end{align} is rational.

But we know that \begin{align} \sqrt 2 \end{align} is irrational this contradicts the fact that \begin{align} \sqrt 2 \end{align} is rational.

So, our assumption was wrong. Therefore, \begin{align} \frac{1}{{\sqrt 2 }}\end{align} is a rational number.

(ii) \begin{align} 7\sqrt 5 \end{align}

Let us assume, to the contrary that \begin{align} 7\sqrt 5 \end{align} is a rational number.

\begin{align} 7\sqrt 5 = \frac{p}{q}\end{align}

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} 7\sqrt 5 &= \frac{a}{b} \\7\sqrt 5 {\text{b}} &= {\text{a}}\\\sqrt 5 &= \frac{a}{{7b}}\end{align}

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

Since, $$a,\; 7$$ and $$b$$ are integers. So, \begin{align} \frac{a}{{7b}}\end{align} is rational number and so, \begin{align} \sqrt 5 \end{align} is rational. But this contradict the fact that \begin{align} \sqrt 5 \end{align}

So, our assumption was wrong. Therefore, \begin{align} 7\sqrt 5 \end{align} is a rational number.

(iìi) \begin{align} 6{\text{ }} + \sqrt 2 \end{align}

Let us assume, to the contrary that \begin{align} 6{\text{ }} + \sqrt 2 \end{align} is a rational number.

\begin{align} 6{\text{ }} + \sqrt 2 = \frac{p}{q}\end{align}

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} 6 + \sqrt 2 &= \frac{a}{b}\\\sqrt 2 &= \frac{a}{b} - 6 \end{align}

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

Since, $$a, \,b$$ and $$6$$ are integers. So, \begin{align} \frac{a}{b} - 6\end{align} is rational number and so, \begin{align} \sqrt 2 \end{align} is also a rational number.

But this contradicts the fact that \begin{align} \sqrt 2 \end{align} is irrational. So, our assumption was wrong.

Therefore, \begin{align} 6{\text{ }} + \sqrt 2 \end{align} is a rational number.