HBSE 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3

Haryana State Board HBSE 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.3

Question 1.
Prove that √5 is an irrational.
Solution :
Let us assume, that √5 is a rational. It can be expressed in the form of \(\frac{a}{b}\), where a and b are coprime positive integers and b ≠ 0.
∴ √5 = \(\frac{a}{b}\) (Where a and b are coprime ∴ HCF of a and b is 1)
Squaring on both sides
5 = \(\frac{a^2}{b^2}\)
5b² = a² ………….(i)
Therefore, 5 divides a². It follows that 5 divides a . [By theorem 1.3]
Let a = 5c and put this value in equation (i)
[Where c is any positive integer]
5b² = (5c)²
5b² = 25c²
⇒ \(\frac{25}{5}\) c² = b²
⇒ 5c² = b² ……….(ii)
It means b² is divisible by 5. It follows that b is divisible by 5 [By theorem 1.3]
From equations (i) and (ii) we say that 5 is a common factor of both a and b. But this contradicts the fact that a and b are coprime so, they have no common factor. So, our assumption that √5 is a rational number is wrong.
Therefore, √5 is an irrational number.

Question 2.
Prove that 3 + 2√5 is an irrational.
Solution :
Let us assume 3 + 2√5 is a rational. It can be expressed in the form of \(\frac{a}{b}\), where a and b are coprime positive integers and b ≠ 0.
∴ 3 + 2√5 = \(\frac{a}{b}\)
\(\frac{a}{b}\) – 3 = 2√5
\(\frac{a-3 b}{b}\) = 2√5
\(\frac{a-3 b}{2 b}\) = √5
\(\frac{a-3 b}{2 b}\) = rational
(∵ a and b are positive integers)
So, from equation (i) √5 is a rational.
But this contradicts the fact √5 is an irrational. So, our assumption that 3 + 2√5 is a rational, is wrong.
Hence, 3 + 2√5 is an irrational number.

Question 3.
Prove that the following are irrationals:
(i) \(\frac{1}{\sqrt{2}}\)
(ii) 7√5
(iii) 6 + √2
Solution:
(i) Let us assume, that \(\frac{1}{\sqrt{2}}\) is a rational. It can be expressed in the form of \(\frac{a}{b}\), where a and b are coprime positive integer.
∴ \(\frac{1}{\sqrt{2}}=\frac{a}{b}\)
(Where HCF of a and b is 1 and b ≠ 0)
\(\frac{1}{2}=\frac{a^2}{b^2}\) (Squaring both sides)
⇒ b² = 2a² …………….(i)
It means b² is divisible by 2. It follows that b, is divisible by 2 [By theorem 1.3]
Let b = 2c (Where c is any positive integer).
And put b = 2c in equation (i)
(2c)² = 2a²
⇒ 4c² = a²
⇒ 2c² = a² …………….(ii)
It means a² is divisible by 2. It follows that a, is divisible by 2. [By theorem 1.3]
From (i) and (ii) we say that 2 is the common factor of a and b. But this contradicts the fact that a and b are coprime. So, our assumption that \(\frac{1}{\sqrt{2}}\) is a rational, is wrong.
Hence, \(\frac{1}{\sqrt{2}}\) is an irrational number.

(ii) Let us assume, that 7√5 is a rational.
It can be expressed in the form of \(\frac{a}{b}\), where a and b are coprime positive integer and b ≠ 0.
∴ 7√5 = \(\frac{a}{b}\) (Where HCF of a and b is 1)
√5 = \(\frac{a}{7 b}\)
∵ a and b are positive integers.
∴ \(\frac{a}{7 b}\) is a rational.
Therefore, √5 is a rational.
But this contradicts the fact that √5 is an irrational. So, our assumption that 7√5 is a srational, is wrong.
Hence, 7√5 is an irrational number.

(iii) Let us assume, that 6 + √2 is a rational.
It can be expressed in the form of \(\frac{a}{b}\), where a and b are coprime positive integer and b ≠ 0.
∴ 6 + √2 = \(\frac{a}{b}\) (Where HCF’ of a and b is 1)
⇒ √2 = \(\frac{a}{b}\) – 6
⇒ √2 = \(\frac{a-6 b}{b}\)
∵ a and b are positive integers.
∴ \(\frac{a-6 b}{b}\) is a rational.
Therefore, √2 is a rational.
But this contradicts the fact that √2 is an irrational. So, our assumption that 6 + √2 is a rational, is wrong
Hence, 6 + √2 is an irrational number.