JAC Class 10 Maths Solutions Chapter 1 Real Numbers Ex 1.3

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 = \(\frac{a^2}{b^2}\)
∴ a² = 5b²
This suggests that 5 is a factor of a².
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.
∴ a² = 25c²
∴ 25c² = 5b²
∴ b² = 5c²
This suggests that 5 is a factor of b². 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
\(\frac{1}{2}=\frac{a^2}{b^2}\)
∴ b² = 2a²
Hence, 2 is a factor of b².
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.
∴ b² = 4c²
∴ 4c² = 2a²
∴ a² = 2c²
Hence, 2 is a factor of a².

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.