Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.4 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.4

Question 1.

Prove that the following are irrational.

(i) \(\frac{1}{\sqrt{2}}\)

Solution:

\(\frac{1}{\sqrt{2}}\)

Let us assume to the contrary that \(\frac{1}{\sqrt{2}}\) is rational. Then there exist co-prime positive integers ‘a’ and ‘b’ such that 1 a

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

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

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

Here ‘a’ and ‘b’ are integers, \(\frac{b}{a}\) is rational.

∴ \(\sqrt{2}\) is rational.

This contradicts the fact that \(\sqrt{2}\) is irrational.

So our assumption is wrong.

Hence \(\frac{1}{\sqrt{2}}\) is irrational.

(ii) \(\sqrt{3}\) + \(\sqrt{5}\)

Solution:

\(\sqrt{3}\) + \(\sqrt{5}\)

Let us assume to the contrary that \(\sqrt{3}\) + \(\sqrt{5}\) is a rational number.

Then there exist co-prime positive integers ‘a’ and ‘b’ such that

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

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

S.B.S.

This contradicts the fact that \(\sqrt{3}\) is irrational.

∴ Hence, \(\sqrt{3}\) + \(\sqrt{5}\) is irrational.

(iii) 6 + \(\sqrt{2}\)

Solution:

6 + \(\sqrt{2}\)

Let us assume on the contrary that 6 + \(\sqrt{2}\) is rational. Then there exist co-prime positive integers ‘a’ and ‘b’ such that

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

⇒ \(\sqrt{2}\) = \(\frac{a-6 b}{b}\)

\(\sqrt{2}\) is rational.

\(\frac{a-6 b}{b}\) is rational.

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

∴ 6 + \(\sqrt{2}\) is irrational.

(iv) \(\sqrt{5}\)

Solution:

\(\sqrt{5}\)

Let us assume, to the contrary that \(\sqrt{5}\) is irrational then there exist co-prime positive integers a and b such that

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

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

S.B.S. we get

(\(\sqrt{5}\) b)^{2} = (a)^{2}

5b^{2} = a^{2} ……. (1)

5 divides a^{2}.

Hence 5 divides a.

We can write a = 5c for some integer c.

Substitute a = 5c in (1) we get

5b^{2} = (5c)^{2}

5b^{2} = 25c^{2}

b^{2} = \(\frac{25 c^2}{5}\)

b^{2} = 5c^{2}

5 divides b^{2} and 5 divide b.

‘a’ and ‘b’ have atleast as a common factor.

This contradicts the fact that ‘a’ and ‘b’ have no common factor other than 1.

So our assumption is wrong.

∵ \(\sqrt{5}\) is irrational.

(v) 3 + 2\(\sqrt{5}\)

Solution:

3 + 2\(\sqrt{5}\)

Let us assume, to the contrary that 3 + 2\(\sqrt{5}\) is rational. Then there exist co-prime positive integers ‘a’ and ’b’ such that

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

2\(\sqrt{5}\) = \(\frac{a}{b}\) – 3

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

\(\sqrt{5}\) is rational.

\(\frac{a-3 b}{2 b}\) is rational.

This contradicts the fact that \(\sqrt{5}\) is irrational, so our assumption is wrong.

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

Question 2.

Prove that \(\sqrt{p}\) + \(\sqrt{q}\) is an irrational, where p, q are primes.

Solution:

Let us assume to the contrary that \(\sqrt{p}\) + \(\sqrt{q}\) is rational. Then there exist co-prime positive in-tegers ‘a’ and ‘b’.

We know that square root of any prime number is irrational, we get \(\sqrt{q}\) is rational.

This contradicts the fact that \(\sqrt{q}\) is irrational.

So our assumption is wrong.

∵ \(\sqrt{q}\) is irrational.

∵ \(\sqrt{p}\) + \(\sqrt{q}\) is irrational.