# GSEB Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.3

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
p2 = 5q2                            …(2)
therefore p2 is divisible by 5, then p will be divisible by 5.
Let p = 5r Putting value of p in eqn (2) we get
25r2 = 5q2
5r2 = q2
This means q2 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:

1. $$\frac { 1 }{ \sqrt { 2 } }$$
2. 7$$\sqrt { 5 }$$
3. 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.