Bihar Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.3

Bihar Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.3 Textbook Questions and Answers.

BSEB Bihar Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.3

Question 1.
Prove that \(\sqrt{5}\) is irrational. [CBSE, Delhi 2009]
Solution:
Let us assume, to the contrary, that \(\sqrt{5}\) is rational.
Squaring on both sides, we get
5 = \(\frac{a^{2}}{b^{2}}\) or 5b² = a² ………………. (1)
This shows that a² is divisible by 5.
It follows that a is divisible by 5. …………………. (2)
So, a = 5m for some integer m.
Substituting a = 5m in (1), we get
5b² = (5m)² = 25m²
or b² = 5m² or b² is divisible by 5
and hence b is divisible by 5.
From (2) and (3), we can conclude that 5 is a common factor of both a and b.
But this contradicts our supposition that a and b are coprime.
Hence, \(\sqrt{5}\) is irrational.

Question 2.
Prove that 3 + 2\(\sqrt{5}\) is irrational.
Solution:
Let us assume, to the contrary, that 3 + 2\(\sqrt{5}\) is a rational number.
Now, let 3 + 2\(\sqrt{5}\) = \(\frac{a}{b}\), where a and b are coprime and b ≠ 0.
So, 2\(\sqrt{5}\) = \(\frac{a}{b}\) – 3 or \(\sqrt{5}\) = \(\frac{a}{2b}\) – \(\frac{3}{2}\)
Since a and b are integers, therefore
\(\frac{a}{2b}\) – \(\frac{3}{2}\) is a rational number.
∴ \(\sqrt{5}\) is an irrational number.
But \(\sqrt{5}\) is an irrational number.
This shows that our assumption is incorrect.
So, 3 + 2\(\sqrt{5}\) is an irrational number.

Question 3.
Prove that the following are irrationals:

1. \(\frac{1}{\sqrt{2}}\)
2. 7\(\sqrt{5}\)
3. 6 + \(\sqrt{2}\)

Solution:
1. Let us assume, to the contrary, that \(\frac{1}{\sqrt{2}}\) is rational.
That is, we can find co-prime integers p and q (≠ 0) such that

Since p and q are integers, \(\frac{2p}{q}\) is rational, and so \(\sqrt{2}\) is rational.
But this contradicts the fact that \(\sqrt{2}\) is irrational.
So, we conclude that \(\frac{1}{\sqrt{2}}\) is irrational.

2. Let us assume, to the contrary, that 7\(\sqrt{5}\) is rational.
That is, we can find co-prime integers p and q (≠ 0) such that 7\(\sqrt{5}\) = \(\frac{p}{q}\).
So, \(\sqrt{5}\) = \(\frac{p}{7q}\).
Since p and q are integers, \(\frac{p}{7q}\) is rational and so is \(\sqrt{5}\).
But this contradicts the fact that \(\sqrt{5}\) is irrational.
So, we conclude that 7\(\sqrt{5}\) is irrational.

3. Let us assume, to the contrary, that 6 + \(\sqrt{2}\) is rational.
That is, we can find integers p and q (≠ 0) such that
6 + \(\sqrt{2}\) = \(\frac{p}{q}\) or 6 – \(\frac{p}{q}\) = \(\sqrt{2}\)
or \(\sqrt{2}\) = 6 – \(\frac{p}{q}\)
Since p and q are integers, we get 6 – \(\frac{p}{q}\) is rational, and so \(\sqrt{2}\) is rational.
But this contradicts the fact that \(\sqrt{2}\) is irrational.
So, we conclude that 6 + \(\sqrt{2}\) is irrational.