CBSE Class 10 Chapter 1 Real Number Exercise 1.2 Solutions

Q1.

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

Ans.

To prove that \(\sqrt {5}\) is irrational, we can use proof by contradiction. Assume that \(\sqrt {5}\) is a rational number. This means it can be expressed as a fraction in the simplest form of \(\frac {a}{b}\), where both \(a\) and \(b\) are integers with no common factors other than \(1\) (i.e., \(a\) and \(b\) are coprime). \(\sqrt {5} = \frac {a}{b}\) let's square both sides of this equation: \( \left(\sqrt {5}\right)^{2} = \left(\frac {a}{b}\right)^{2}\) \(5 = \frac {\left(a^{2}\right)}{\left(b^{2}\right)}\) By rearrange the equation: \(5b^{2} = a^{2}\) As per above equation, we can see that \( a^{2}\) must be a multiple of \(5\) because \(5\) is on the left side. This implies that \(a\) must also be a multiple of \(5\), because if \(a\) were not a multiple of \(5\), its square wouldn't be. So, we can write a as \(5k\), where \(k\) is an integer. \( a = 5k\) Substitute this back into the equation: \(5b^{2}= (5k)^{2}\) \(5b^{2} = 25k^{2}\) Now, divide both sides by 5: \(b^{2}= 5k^{2}\) This means that b2 must also be a multiple of 5, which implies that b must be a multiple of 5. Now, we have shown that both a and b are multiples of 5. However, we assumed earlier that a and b are coprime, meaning they have no common factors other than 1. This is a contradiction because if both a and b are multiples of 5, they do have a common factor, namely 5. Since our assumption that \(\sqrt {5}\) is rational leads to a contradiction, we conclude that \(\sqrt {5}\) cannot be rational. Therefore, \(\sqrt {5}\) is irrational.

Q2.

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

Ans.

To prove that \(3 + 2\sqrt{5}\) is irrational, we can use a proof by contradiction. Assume that \(3 + 2\sqrt{5}\) is rational, which means it can be expressed as: \(3 + 2\sqrt{5} = \frac {a}{b}\) where "\(a\)" and "\(b\)" are two integers with no common factors (other than 1) and "\(b\)" is not equal to zero. Manipulate the equation to isolate \(\sqrt {5}\): \(2\sqrt {5} = \left(\frac {a}{b}\right)-3\) \(\sqrt {5} =\frac {\left[\left(\frac {a}{b}\right)-3 \right]}{2}\) Now, we'll square both sides of the equation to get rid of the square root: \(5 = \frac {\left[\left(\frac {a}{b}\right) -3\right]^2}{4}\) Next, multiply both sides by 4 to clear the fraction: \(20 = \left(\frac {a}{b}-3\right)^2\) Now, we have: \(20=\frac {\left(a-3b\right)^2}{b^2}\) We can see that \(\left(a-3b\right)^2\) is a perfect square because it equals 20 times \(b^2\) This is an integer. Therefore, \(\left(a-3b\right)^2\) must also be an integer. Now, consider the left-hand side of the equation 20. It is a positive integer. However, the square of any rational number is also rational. This means that \(\left(a-3b\right)^2\) is a rational number since it's the square of a rational number. So, we have a contradiction: \(\left(a-3b\right)^2\) is both an integer (as 20 is an integer) and a rational number. This is not possible because an integer cannot be both rational and not rational. Therefore, our initial assumption that \(3 + 2\sqrt {5}\) is rational must be false. Hence, we have proved by contradiction that \(3 + 2\sqrt {5}\) is irrational. Other Short Method : Assume \(3 + 2\sqrt {5}\) is rational. Then, \(\sqrt {5} = \left(3 + 2\sqrt {5}\right) - 3\) is also rational. However, \(sqrt {5}\) is known to be irrational, which contradicts our assumption. Therefore, \(3 + 2\sqrt {5}\) must also be irrational.

Q3.	Prove that the following are irrationals : (i) \(\frac {1}{\sqrt {2}}\)          (ii) \(7\sqrt {2}\)               (iii)  \(6+\sqrt {2}\)
Ans.
(i)	Claim: \( \frac{1}{\sqrt{2}} \) is irrational. Proof by Contradiction: Step 1: Assume for contradiction that \( \frac{1}{\sqrt{2}} \) is rational. So, we can write this number as: \( \frac{1}{\sqrt{2}} = \frac{a}{b} \) (where \( a \) and \( b \) are two coprime integers and \( b \) is not equal to zero). Step 2: Multiply both sides of the equation by \( \sqrt{2} \): \(\frac{1}{\sqrt{2}} \times \sqrt{2} = \left(\frac{a}{b}\right) \times \sqrt{2}\) \(1 = \left(\frac{a}{b}\right) \times \sqrt{2}\) Step 3: Now, divide both sides by \( b \): \(\frac{1}{b} = \left(\frac{a}{b}\right) \times \frac{\sqrt{2}}{b}\) \(b = a\sqrt{2} \quad \text{or} \quad \frac{b}{a} = \sqrt{2}\) Here, because \( a \) and \( b \) are integers, so \( \frac{a}{b} \) are rational numbers, so \( \sqrt{2} \) should be a rational number. But \( \sqrt{2} \) is irrational number. It makes a contradiction. Hence \( \frac{1}{\sqrt{2}} \) is irrational.
(ii)	Claim: \( 7\sqrt{2} \) is irrational. Proof by Contradiction: Step 1: Assume for contradiction that \( 7\sqrt{2} \) is rational. So, we can write this number as: \( 7\sqrt{2} = \frac{a}{b} \) (where \( a \) and \( b \) are two coprime integers and \( b \) is not equal to zero). Step 2: Divide both sides of the equation by 7: \(\frac{7\sqrt{2}}{7} = \frac{\frac{a}{b}}{7}\) \(\sqrt{2} = \frac{a}{7b}\) Now \( 7b \) and \( a \) are integers, so \( \frac{a}{7b} \) is a rational number. But \( \sqrt{2} \) is irrational. It leads to a contradiction and makes the initial assumption false. Hence \( 7\sqrt{2} \) is irrational.
(iii)	Claim: \(6+\sqrt{2}\) is irrational. Proof by Contradiction: Step 1: Assume for contradiction that \(6+\sqrt{2}\) is rational. So, we can write this number as: \(6+\sqrt{2} = \frac{a}{b}\) (where \(a\) and \(b\) are two coprime integers and \(b\) is not equal to zero). Step 2: Rearrange the equation to isolate \(\sqrt{2}\): \(\sqrt{2} = \frac{a}{b} - 6 \quad \text{or} \quad \sqrt{2} = \frac{a - 6b}{b}\) Here, because \(a\) and \(b\) are integers, so \(\frac{a - 6b}{b}\) is a rational number, so \(\sqrt{2}\) should be a rational number. But \(\sqrt{2}\) is irrational. It leads to a contradiction. Hence \(6+\sqrt{2}\) is irrational.