NCERT Solutions for Exercise 1.3 Class 10 Maths Chapter 1 - Real Numbers

Q1 Prove that $\sqrt 5$ is irrational.

Answer:

Let us assume $\sqrt{5}$ is rational, which means it can be written in the form $\frac{p}{q}$ where p and q are co-primes and $q\neq 0$

$\\\sqrt{5}=\frac{p}{q}$

Squaring both sides, we obtain

$\\\left ( \sqrt{5} \right )^{2}=\left (\frac{p}{q} \right )^{2}\\$

$5=\frac{p^{2}}{q^{2}}\\$

$p^{2}=5q^{2}$

From the above equation, we can see that p² is divisible by 5, therefore, p will also be divisible by 5, as 5 is a prime number. $(i)$

Therefore, p can be written as 5r

p = 5r

p² = (5r)²

5q² = 25r²

q² = 5r²

From the above equation, we can see that q² is divisible by 5, Therefore, q will also be divisible by 5 as 5 is a prime number. $(ii)$

From (i) and (ii), we can see that both p and q are divisible by 5. This implies that p and q are not co-primes. This contradiction arises because our initial assumption that $\sqrt{5}$ is rational was wrong. Hence proved that $\sqrt{5}$ is irrational.

Q2 Prove that $3 + 2 \sqrt 5$ is irrational.

Answer:

Let us assume $3 + 2 \sqrt 5$ is rational, this means it can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$\\3+2\sqrt{5}=\frac{p}{q}$

$2\sqrt{5}=\frac{p}{q}-3$

$\sqrt{5}=\frac{p-3q}{2q}$

As p and q are integers $\frac{p-3q}{2q}\\$ would be rational, which contradicts the fact that $\sqrt{5}$ is irrational. This contradiction arises because our initial assumption that $3 + 2 \sqrt 5$ is rational was wrong. Therefore $3 + 2 \sqrt 5$ is irrational.

Q3 Prove that the following are irrationals :

(i) $\frac{1}{\sqrt 2}$

Answer:

Let us assume $\frac{1}{\sqrt{2}}$ is rational, this means it can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$\frac{1}{\sqrt{2}}=\frac{p}{q}$

$\sqrt{2}=\frac{q}{p}$

Since p and q are co-prime integers $\frac{q}{p}$ will be rational, which contradicts the fact that $\sqrt{2}$ is irrational. This contradiction arises because our initial assumption that $\frac{1}{\sqrt{2}}$ is rational was wrong. Therefore $\frac{1}{\sqrt{2}}$ is irrational.

Q3 (2) Prove that the following are irrationals :

(ii) $7 \sqrt 5$

Answer:

Let us assume $7 \sqrt 5$ is rational, this means it can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$7\sqrt{5}=\frac{p}{q}$

$\sqrt{5}=\frac{p}{7q}$

As p and q are integers $\frac{p}{7q}\\$ would be rational, which contradicts the fact that $\sqrt{5}$ is irrational. This contradiction arises because our initial assumption that $7 \sqrt 5$ is rational was wrong. Therefore $7 \sqrt 5$ is irrational.

Q3 (3) Prove that the following are irrationals : $6 + \sqrt 2$

Answer:

Let us assume $6 + \sqrt 2$ is rational, this means it can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$6+\sqrt{2}=\frac{p}{q}$

$\sqrt{2}=\frac{p}{q}-6$

$\sqrt{2}=\frac{p-6q}{q}$

As p and q are integers $\frac{p-6q}{q}$ would be rational, which contradicts the fact that $\sqrt{2}$ is irrational. This contradiction arises because our initial assumption that $6 + \sqrt 2$ is rational was wrong. Therefore $6 + \sqrt 2$ is irrational.