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

Jharkhand Board JAC Class 10 Maths Solutions Chapter 1 Real Numbers Ex 1.3 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 1 Real Numbers Exercise 1.3

Question 1.
Prove that $$\sqrt{5}$$ is irrational.
Solution:
Let us assume that $$\sqrt{5}$$ is rational.
So, we can find co-prime integers a and b such that $$\sqrt{5}$$ = $$\frac{a}{b}$$
Then, squaring on both the sides, we get
5 = $$\frac{a^2}{b^2}$$
∴ a2 = 5b2
This suggests that 5 is a factor of a2.
Since 5 is a prime number, by theorem 1.3, 5 is also a factor of a.
Let a = 5c, where c is an integer.
∴ a2 = 25c2
∴ 25c2 = 5b2
∴ b2 = 5c2
This suggests that 5 is a factor of b2. Since 5 is a prime number, by theorem 1.3, 5 is also a factor of b.
Thus, a and b have a common factor 5.
But, this contradicts the fact that a and b are co-prime.
Hence, our assumption that $$\sqrt{5}$$ is rational is incorrect.
So, we conclude that $$\sqrt{5}$$ is irrational.

Question 2.
Prove that 3 + 2$$\sqrt{5}$$ is irrational.
Solution:
Suppose 3 + 2$$\sqrt{5}$$ is rational.
Then, 3 + 2$$\sqrt{5}$$ = $$\frac{a}{b}$$; where a and b are co-prime integers.
Now, 3 + 2$$\sqrt{5}$$ = $$\frac{a}{b}$$
∴ 2$$\sqrt{5}$$ = $$\frac{a}{b}$$ – 3
∴ 2$$\sqrt{5}$$ = $$\frac{a-3 b}{b}$$
∴ $$\sqrt{5}$$ = $$\frac{a-3 b}{2 b}$$
As a and b are integers, $$\frac{a-3 b}{2 b}$$ is a rational number and so is $$\sqrt{5}$$.
This contradicts the fact that $$\sqrt{5}$$ is irrational.
Hence, our assumption is incorrect.
So, we conclude that 3 + 2$$\sqrt{5}$$ is irrational.

Question 3.
Prove that the following are irrationals:
1. $$\frac{1}{\sqrt{2}}$$
2. 7$$\sqrt{5}$$
3. 6 + $$\sqrt{2}$$
Solution:
1. $$\frac{1}{\sqrt{2}}$$
Suppose $$\frac{1}{\sqrt{2}}$$ is rational.
Then, $$\frac{1}{\sqrt{2}}=\frac{a}{b}$$ where a and b are co-prime integers.
Squaring on both the sides, we get
$$\frac{1}{2}=\frac{a^2}{b^2}$$
∴ b2 = 2a2
Hence, 2 is a factor of b2.
Since 2 is a prime number, by theorem 1.3, 2 is also a factor of b.
Let b = 2c, where c is an integer.
∴ b2 = 4c2
∴ 4c2 = 2a2
∴ a2 = 2c2
Hence, 2 is a factor of a2.

Since 2 is a prime number, by theorem 1.3, 2 is also a factor of a.
Thus, a and b have a common factor 2. which contradicts our assumption that a and b are co-prime integers.
Hence, our assumption that $$\frac{1}{\sqrt{2}}$$ is rational is incorrect.
So, we conclude that $$\frac{1}{\sqrt{2}}$$ is irrational.

2. 7$$\sqrt{5}$$
Suppose 7$$\sqrt{5}$$ is rational.
Then, 7$$\sqrt{5}$$ = $$\frac{a}{b}$$ where a and b are co-prime integers.
This gives, $$\sqrt{5}$$ = $$\frac{a}{7 b}$$
As a and b are integers, $$\frac{a}{7 b}$$ is a rational number and so is $$\sqrt{5}$$.
This contradicts the fact that $$\sqrt{5}$$ is irrational.
Hence, our assumption is incorrect.
So, we conclude that 7$$\sqrt{5}$$ is irrational.

3. 6 + $$\sqrt{2}$$
Suppose 6 + $$\sqrt{2}$$ is rational.
Then, 6 + $$\sqrt{2}$$ = $$\frac{a}{b}$$ where a and b are co-prime integers.
This gives, $$\sqrt{2}$$ = $$\frac{a}{b}$$ – 6
As a and b are integers, $$\frac{a}{b}$$ – 6 is a rational number and so is $$\sqrt{2}$$.
This contradicts the fact that $$\sqrt{2}$$ is irrational.
Hence, our assumption is incorrect. So, we conclude that 6 + $$\sqrt{2}$$ is irrational.