NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.3

NCERT Solutions Class 10 Maths
Chapter – 1 (Real Numbers)

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 1 Real Numbers Exercise 1.3 has been provided here to help the students in solving the questions from this exercise.

Chapter : 1 Real Numbers

Exercise – 1.3

1. Prove that √5 is irrational.
Solutions – Let √5 is a rational number.
Therefore, we can find two integers a, b (b ≠ 0) such that √5 = $\frac{a}{b}$
Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime.
a = √5b
a²= 5b²Therefore, a² is divisible by 5 and it can be said that a is divisible by 5.
Let a = 5k, where k is an integer
(5k)²= 5b²This means that b² is divisible by 5 and hence, b is divisible by 5.
b²= 5k²This implies that a and b have 5 as a common factor.
And this is a contradiction to the fact that a and b are co-prime.
Hence,√5 cannot be expressed as $\frac{p}{q}$ or it can be said that √5 is irrational.

2. Prove that 3 + 2√5 + is irrational.
Solutions – Let 3 + 2√5 is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that
3 + 2√5 = $\frac{a}{b}$
2√5 = $\frac{a}{b}$ – 3

√5 = $\frac{1}{2}$ $\left ( \frac{a}{b} -3\right )$
Since a and b are integers, $\frac{1}{2}$ $\left ( \frac{a}{b} -3\right )$ will also be rational and therefore,√5 is rational.
This contradicts the fact that √5 is irrational. Hence, our assumption that 3 + 2√5 is rational is false.
Therefore, 3 + 2√5 is irrational.

3. Prove that the following are irrationals:
(i) $\mathbf{\frac{1}{\sqrt{2}}}$
(ii) 7√5
(iii) 6 + √2

Solutions –
(i) $\mathbf{\frac{1}{\sqrt{2}}}$
Let $\frac{1}{\sqrt{2}}$ is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that

$\frac{1}{\sqrt{2}}$ = $\frac{a}{b}$

√2 = $\frac{b}{a}$

$\frac{b}{a}$ is rational as a and b are integers.
Therefore, √2 is rational which contradicts to the fact that √2 is irrational.
Hence, our assumption is false and $\frac{1}{\sqrt{2}}$ is irrational.

(ii) 7√5
Let 7√5 is a rational number.
Therefore, we can find two integers a, b (b ≠ 0) such that

7√5 = $\frac{a}{b}$ for some integers a and b

∴ √5 = $\frac{a}{7b}$

$\frac{a}{7b}$ is rational as a and b are integers.
Therefore, √5 should be rational.
This contradicts the fact that √5 is irrational.
Therefore, our assumption that 7√5 is rational is false. Hence, 7√5 is irrational.

(iii) 6 + √2
Let 6 + √2 is a rational number.
Therefore, we can find two integers a, b (b ≠ 0) such that
6 + √2 = $\frac{a}{b}$
√2 = $\frac{a}{b}$ – 6
Since a and b are integers, $\frac{a}{b}$ – 6 is also rational and hence, √2 should be rational.
This contradicts the fact that √2 is irrational.
Therefore, our assumption is false and hence, 6 + √2 is irrational.

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NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.3