# NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.3

Exercise 1.3

## Chapter 1 Ex.1.3 Question 1

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

**Solution**

**Video 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.

## Chapter 1 Ex.1.3 Question 2

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

**Solution**

**Video 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.

## Chapter 1 Ex.1.3 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**

**Video 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.

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

*\(p\)* 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.