Gujarat Board GSEB Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.3

Question 1.

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

Solution:

Let \(\sqrt { 5 } \) be a rational number.

Therefore we can find two integers a, b (b ≠ 0) such that \(\frac{a}{b}\) = \(\sqrt { 5 } \)

Let a and b have a common factor other than 1 then we divide both by common factor.

Then, we get \(\frac{p}{q}\) = \(\sqrt { 5 } \) …(1)

[Where p and q are co-prime]

Squaring both sides

\(\frac{p^{2}}{q^{2}}\) = 5

p^{2} = 5q^{2} …(2)

therefore p^{2} is divisible by 5, then p will be divisible by 5.

Let p = 5r Putting value of p in eqn (2) we get

25r^{2} = 5q^{2}

5r^{2} = q^{2}

This means q^{2} is divisible by 5 then q will be divisible by 5.

Which means p and q have common factor 5. We reach at the contradiction as our supposition that p and q are co-prime is wrong.

Hence \(\sqrt { 5 } \) is irrational.

Question 2.

Prove that 3 + 2 \(\sqrt { 5 } \) is irrational.(CBSE)

Solution:

Let us suppose 3 + 2\(\sqrt { 5 } \) is a rational.

Then \(\frac{a}{b}\) = 3 + 2\(\sqrt { 5 } \)

[Where a and b are integers]

⇒ \(\frac{a}{b}\) – 3 = 2\(\sqrt { 5 } \)

\(\frac{1}{2}\)[\(\frac{a}{b}\) – 3] = \(\sqrt { 3 } \)

\(\frac{1}{2}\)[\(\frac{a}{b}\) – 3] is rational as a and b are integers therefore \(\sqrt { 3 } \) should be rational. This contradicts the fact that \(\sqrt { 3 } \) is an irrational.

Hence 3 + 2\(\sqrt { 5 } \) is an irrational.

Question 3.

Prove that the following are irrationals:

- \(\frac { 1 }{ \sqrt { 2 } } \)
- 7\(\sqrt { 5 } \)
- 6 + \(\sqrt { 2 } \)

Solution:

1. Let \(\frac { 1 }{ \sqrt { 2 } } \) be a rational.

Therefore we can find two integers a, b (b ≠ 0)

Such that \(\frac { 1 }{ \sqrt { 2 } } \) = \(\frac{a}{b}\)

\(\sqrt { 2 } \) = \(\frac{b}{a}\)

\(\frac{b}{a}\) is a rational. Therefore \(\sqrt { 2 } \) will also be rational which contradicts to the fact that \(\sqrt { 2 } \) is irrational.

Hence our supposition is wrong and \(\frac { 1 }{ \sqrt { 2 } } \) is irrational.

2. Let 7\(\sqrt { 5 } \) be a rational

Therefore 7\(\sqrt { 5 } \) = \(\frac{a}{b}\)

[Where a and b are integers]

\(\sqrt { 5 } \) = \(\frac{a}{7b}\)

\(\frac{a}{7b}\) is rational as a and b are integers

Therefore \(\sqrt { 5 } \) should be rational.

This contradicts the facts that is \(\sqrt { 5 } \) irrational therefore our supposition is wrong. Hence 7\(\sqrt { 5 } \) is irrational.

3. Let 6 + \(\sqrt { 2 } \) be rational

Therefore 6 + \(\sqrt { 2 } \) = \(\frac{a}{b}\)

[where a and b are integers]

\(\sqrt { 2 } \) = \(\frac{a}{b}\) – 6

\(\frac{a}{b}\) – 6 is rational as a and b are integers therefore, \(\sqrt { 2 } \) should be rational.

This contradicts the fact that \(\sqrt { 2 } \) is irrational, therefore, our supposition is wrong.

Hence 6 + \(\sqrt { 2 } \) is irrational.