State, giving reason, wether the given number is rational or irrational:

(i) $(3 + \sqrt{5})$

(ii) $(-1 + \sqrt{3})$

(iii) $5\sqrt{6}$

(iv) $-\sqrt{7}$

(v) $\dfrac{\sqrt{6}}{4}$

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

(vii) $(3 + \sqrt{3}) (3 - \sqrt{3})$

(i) Given,

$(3 + \sqrt{5})$

3 is a rational number as it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{5}$ is an irrational number as it is a square root of a non-perfect square i.e. 5.

The sum of a rational number and an irrational number is always irrational.

Hence, $(3 + \sqrt{5})$ is a irrational number.

(ii) Given,

$(-1 + \sqrt{3})$

-1 is a rational number as it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{3}$ is an irrational number as it is a square root of a non-perfect square i.e. 3.

The sum of a rational number and an irrational number is always irrational.

Hence, $(-1 + \sqrt{3})$ is a irrational number.

(iii) Given,

$5\sqrt{6}$

5 is a rational number as it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

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

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

Hence, $5\sqrt{6}$ is an irrational number.

(iv) Given,

$\sqrt{7}$, is an irrational number as it is a square root of a non-perfect square i.e. 7.

∴ $-\sqrt{7}$ is an irrational number.

Hence, $-\sqrt{7}$ is an irrational number.

(v) Given,

$\dfrac{\sqrt{6}}{4}$

4 is a rational number as it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

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

The division of a rational number and an irrational number is always irrational.

Hence, $\dfrac{\sqrt{6}}{4}$ is an irrational number.

(vi) Given,

$\dfrac{3}{\sqrt{2}}$

3 is a rational number as it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{2}$ is an irrational number as it is a square root of a non-perfect square i.e. 2.

The division of a rational number and an irrational number is always irrational.

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

(vii) Given,

$\Rightarrow (3 + \sqrt{3}) (3 - \sqrt{3}) \\[1em] \Rightarrow (3)^2- (\sqrt{3})^2 \\[1em] \Rightarrow 9 - 3 \\[1em] \Rightarrow 3$

3 is a rational number as it can be expressed in the form of $\dfrac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, $(3 + \sqrt{3})(3 - \sqrt{3})$ is a rational number.