Classify the rational and irrational numbers from the following :

(i) 5

(ii) $\dfrac{9}{14}$

(iii) $\sqrt{3}$

(iv) π

(v) 3.1416

(vi) $\sqrt{4}$

(vii) $-\sqrt{5}$

(viii) $\sqrt[3]{8}$

(ix) $\sqrt[3]{3}$

(x) $2\sqrt{6}$

(xi) $0.\overline{36}$

(xii) 0.202202220…

(xiii) $\dfrac{2}{\sqrt{3}}$

(xiv) $\dfrac{22}{7}$

(i) 5 can be expressed in the form $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, 5 is a rational number.

(ii) $\dfrac{9}{14}$ can be expressed in the form $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, $\dfrac{9}{14}$ is a rational number.

(iii) $\sqrt{3}$ is square root of non-perfect square i.e. 3.

Hence, $\sqrt{3}$ is an irrational number.

(iv) π is a non-terminating and non-repeating decimal.

Hence, π is a irrational number.

(v) 3.1416 is a terminating decimal, so it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, 3.1416 is a rational number.

(vi) $\sqrt{4}$ is square root of perfect square i.e. 4.

$\sqrt{4} = 2 = \dfrac{2}{1}$.

Hence, $\sqrt{4}$ is a rational number.

(vii) $-\sqrt{5}$ is square root of non-perfect square.

Hence, $-\sqrt{5}$ is an irrational number.

(viii) Given,

$\sqrt[3]{8} = 2 = \dfrac{2}{1}$.

∴ $\sqrt[3]{8}$ can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, $\sqrt[3]{8}$ is a rational number.

(ix) $\sqrt[3]{3}$ is cube root of non-perfect cube.

Hence, $\sqrt[3]{3}$ is an irrational number.

(x) $2\sqrt{6}$

Here, $\sqrt{6}$ is square root of a non-perfect square i.e. 6, thus it is an irrational number.

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

Hence, $2\sqrt{6}$ is an irrational number.

(xi) $0.\overline{36}$ is a repeating decimal.

Thus, $0.\overline{36}$ can be expressed as a fraction with an integer numerator and a non-zero integer denominator.

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

(xii) 0.2022022220… is a non-terminating and non-repeating decimal.

Hence, 0.2022022220… is an irrational number.

(xiii) $\dfrac{2}{\sqrt{3}}$.

2 is rational number and $\sqrt{3}$ is an irrational number.

Since, on dividing a rational number by irrational number the solution is always an irrational number.

Hence, $\dfrac{2}{\sqrt{3}}$ is an irrational number.

(xiv) $\dfrac{22}{7}$ can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, $\dfrac{22}{7}$ is a rational number.