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

∴ a^{2} = 5b^{2}

This suggests that 5 is a factor of a^{2}.

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.

∴ a^{2} = 25c^{2}

∴ 25c^{2} = 5b^{2}

∴ b^{2} = 5c^{2}

This suggests that 5 is a factor of b^{2}. 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

\(\frac{1}{2}=\frac{a^2}{b^2}\)

∴ b^{2} = 2a^{2}

Hence, 2 is a factor of b^{2}.

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.

∴ b^{2} = 4c^{2}

∴ 4c^{2} = 2a^{2}

∴ a^{2} = 2c^{2}

Hence, 2 is a factor of a^{2}.

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.