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.