Rational and Irrational Numbers — Assertion-Reason Type Questions

Assertion (A): $-\dfrac{2}{7}$ is a rational number.

Reason (R): Any number that can be expressed in the form $\dfrac{p}{q}$ is a rational number.

1. Assertion (A) is true, Reason (R) is false.

2. Assertion (A) is false, Reason (R) is true.

3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

Any number that can be expressed in the form $\dfrac{p}{q}$ is a rational number.

The conditions are that p and q must be integers, and q must not be equal to zero.

∴ Reason (R) is false.

In the number $-\dfrac{2}{7}$, p = -2(which is an integer) and q = 7 (which is a non-zero integer).

Therefore, $-\dfrac{2}{7}$ fits the definition of a rational number.

∴ Assertion (A) is true.

Hence, option 1 is the correct option.

Assertion (A): -10 + π is an irrational number.

Reason (R): Sum of a non-zero rational number and an irrational number is an irrational number.

1. Assertion (A) is true, Reason (R) is false.

2. Assertion (A) is false, Reason (R) is true.

3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

Sum of a non-zero rational number and an irrational number is always an irrational number.

This is a fundamental property of irrational and rational numbers.

∴ Reason (R) is true.

π is an irrational number, -10 is a rational number.

So, their sum -10 + π, will be an irrational number.

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Assertion (A): $0.\overline{36}$ is an irrational number.

Reason (R): Any real number that can be expressed in the form of $\dfrac{p}{q}$ where p, q are integers, q ≠ 0 and p, q have no common factor except 1 is a rational number.

1. Assertion (A) is true, Reason (R) is false.

2. Assertion (A) is false, Reason (R) is true.

3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

Let x = $0.\overline{36}$

x = 0.363636.....      .........(1)

Multiplying both side by 100, we get

100x = 36.363636...     .........(2)

Subtracting equation (1) from equation (2), we get

⇒ 100x - x = 36.363636..... - 0.363636......

⇒ 99x = 36

⇒ x = $\dfrac{36}{99} = \dfrac{4}{11}$.

Thus, $0.\overline{36}$ is a rational number.

∴ Assertion (A) is false.

By definition,

Any real number that can be expressed in the form of $\dfrac{p}{q}$ where p, q are integers, q ≠ 0 and p, q have no common factor except 1 is a rational number.

∴ Reason (R) is true.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Assertion (A): All surds are irrational numbers.

Reason (R): All irrational numbers are surds.

1. Assertion (A) is true, Reason (R) is false.

2. Assertion (A) is false, Reason (R) is true.

3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

A surd is a number that cannot be expressed as a simple fraction and is the root of a rational number (like $\sqrt{2}, \sqrt[3]{5}$). These roots are always irrational.

∴ Assertion (A) is true.

While all surds are irrational, not all irrational numbers are surds.

For example, pi (π) is irrational number but are not roots of rational numbers and therefore are not surds.

∴ Reason (R) is false.

∴ Assertion (A) is true, Reason (R) is false.

Hence, option 1 is the correct option.

Assertion (A): The rationalising factor of 2 + $\sqrt{3}$ is 2 - $\sqrt{3}$.

Reason (R): Both 2 + $\sqrt{3}$ and 2 - $\sqrt{3}$ are surds.

1. Assertion (A) is true, Reason (R) is false.

2. Assertion (A) is false, Reason (R) is true.

3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

Multiplying 2 + $\sqrt{3}$ and 2 - $\sqrt{3}$

$\Rightarrow (2 + \sqrt{3})(2 - \sqrt{3}) \\[1em] \Rightarrow 2(2 - \sqrt{3}) + \sqrt{3}(2 - \sqrt{3}) \\[1em] \Rightarrow 4 - 2\sqrt{3} + 2\sqrt{3} - (\sqrt{3})^2 \\[1em] \Rightarrow 4 - 3 \\[1em] \Rightarrow 1.$

Since, 1 is a rational number.

Thus, we can say that the rationalising factor of 2 + $\sqrt{3}$ is 2 - $\sqrt{3}$.

∴ Assertion (A) is true.

A surd is an irrational root of a rational number.

Thus, 2 + $\sqrt{3}$ and 2 - $\sqrt{3}$ are irrational numbers containing surds $(\sqrt{3})$, but not surd itself.

∴ Reason (R) is false.

Hence, option 1 is the correct option.