**KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.3** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.3.

## Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.3

Question 1.

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

Solution:

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

∴ We can find integers a and b (where, b ≠ 0 and a and b are co-prime) such that \(\frac{a}{b}=\sqrt{5}\)

⇒ a = \(\sqrt{5} \cdot b\) … (1)

Squaring both sides, we have a^{2} = 5b^{2}

∴ 5 divides a^{2} ⇒ 5 divides a … (2)

∴ a = 5c, where c is an integer.

∴ Putting a = 5c in (1), we have

From (2) and (3), a and b have at least 5 as a common factor, i.e., a and b are not co-prime.

∴ Our supposition that \(\sqrt{5}\) is rational, is wrong.

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

Question 2.

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

Solution:

Let us assume that \(3+2 \sqrt{5}\) is an irrational number.

Here, p, q, ∈ z, q ≠ 0

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

∵ \(\frac{p-3 q}{2 q}\) is rational number.

But \(\sqrt{5}\) is not a rational number.

This contradicts the fact that,

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

Question 3.

Prove that the following are irrationals :

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

ii) \(7 \sqrt{5}\)

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

Solution:

i) Let \(\frac{1}{\sqrt{2}}\) is a rational number.

\(\frac{1}{\sqrt{2}}=\frac{\mathrm{p}}{\mathrm{q}}\)

\(\sqrt{2}=\frac{q}{p}\)

By Squaring on both sides,

2 × p^{2} = q^{2}

2, divides q^{2}.

∴ 2, divides q

∵ q is an even number.

Similarly ‘p’ is an even number.

∴ p and q are even numbers.

∴ Common factor of p and q is 2.

This contradicts the fact that p and q also irrational.

∴ \(\sqrt{2}\) is an irrational number.

∴ \(\frac{1}{\sqrt{2}}\) is an irrational number.

ii) Let \(7 \sqrt{5}\)is a rational number.

∴ \(7 \sqrt{5}=\frac{p}{q}\)

\(\sqrt{5}=\frac{p}{7 q}\)

Here,\(\frac{p}{7 q}\) is one rational number.

It means \(\sqrt{5}\) which is equal also a rational number.

This contradicts to the fact that \(\sqrt{5}\) is an irrational number.

This contradicts to the fact that \(7 \sqrt{5}\) is rational number.

∴ \(7 \sqrt{5}\) is a rational number.

iii) Let \(6+\sqrt{2}\) is a rational number.

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

∴ \(\sqrt{2}\) is also rational number.

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

This contradicts to the fact that \(6+\sqrt{2}\) is a rational number.

∴ \(6+\sqrt{2}\) is an irrational number.

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