Rational & Irrational Numbers — Assertion-Reason Type Questions

Assertion (A): The number obtained on rationalizing the denominator of $\dfrac{1}{\sqrt{5} - 2}$ is $2 + \sqrt{5}$.

Reason (R): If the product of two irrational numbers is rational, then each one is called the rationalizing factor of the other.

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false.

Answer

Given,

$\dfrac{1}{\sqrt{5} - 2}$

Rationalizing the denominator of $\dfrac{1}{\sqrt{5} - 2}$,

$\Rightarrow \dfrac{1}{\sqrt{5} - 2} \times \dfrac{\sqrt{5} + 2}{\sqrt{5} + 2} \\[1em] \Rightarrow \dfrac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2}\\[1em] \Rightarrow \dfrac{\sqrt{5} + 2}{5 - 4} \\[1em] \Rightarrow \dfrac{\sqrt{5} + 2}{1} \\[1em] \Rightarrow 2 + \sqrt{5}$

∴ Assertion (A) is true.

We know that,

If the product of two irrational numbers is rational, then each one is called the rationalizing factor of the other.

∴ Reason (R) is true.

Hence, Option 3 is the correct option.

Assertion (A): Each of the numbers $\sqrt[3]{2}, \sqrt[3]{3}, \sqrt[3]{4}, \sqrt[3]{5}, \sqrt[3]{6}, \sqrt[3]{7}$ is irrational.

Reason (R): The cube roots of all natural numbers is irrational.

1. A is true, R is false

2. Both A and R are true

3. A is false, R is true

4. Both A and R are false.

Answer

$\sqrt[3]{2}, \sqrt[3]{3}, \sqrt[3]{4}, \sqrt[3]{5}, \sqrt[3]{6}, \sqrt[3]{7}$ are irrational, because these are cube roots of not perfect cubes.

∴ Assertion (A) is true.

The cube roots of all perfect cubes are rational. Thus, we cannot say that the cube roots of all natural numbers is irrational.

∴ Reason (R) is false.

Hence, Option 1 is the correct option.