State in each case, whether true or false :

(i) $\sqrt{2} + \sqrt{3} = \sqrt{5}$

(ii) $2\sqrt{4} + 2 = 6$

(iii) $3\sqrt{7} - 2\sqrt{7} = \sqrt{7}$

(iv) $\dfrac{2}{7}$ is an irrational number.

(v) $\dfrac{5}{11}$ is a rational number.

(vi) All rational numbers are real numbers.

(vii) All real numbers are rational numbers.

(viii) Some real numbers are rational numbers.

(i) $\sqrt{2}$ = 1.41, $\sqrt{3}$ = 1.732 and $\sqrt{5}$ = 2.24

$\sqrt{2} + \sqrt{3}$ = 3.14, which is not equal to 2.24.

$\therefore \sqrt{2} + \sqrt{3} \ne \sqrt{5}$

Hence, above statement is false.

(ii) Given,

$2\sqrt{4} + 2 = 6$

Solving, L.H.S. :

$\Rightarrow 2\sqrt{4} + 2 \\[1em] \Rightarrow 2 \times 2 + 2 \\[1em] \Rightarrow 6.$

Since, L.H.S. = R.H.S.

Hence, above statement is true.

(iii) Given,

$3\sqrt{7} - 2\sqrt{7} = \sqrt{7}$

Solving L.H.S. :

$\Rightarrow 3\sqrt{7} - 2\sqrt{7} \\[1em] \Rightarrow \sqrt{7}(3 - 2) \\[1em] \Rightarrow \sqrt{7} \times 1 \\[1em] \Rightarrow \sqrt{7}.$

Since, L.H.S. = R.H.S.

Hence, above statement is true.

(iv) Since, in $\dfrac{2}{7}$ denominator is not equal to zero.

2 and 7 have no common factor.

∴ $\dfrac{2}{7}$ is rational number.

Hence, above statement is false.

(v) Since, in $\dfrac{5}{11}$ denominator is not equal to zero.

5 and 11 have no common factor.

∴ $\dfrac{5}{11}$ is rational number.

Hence, above statement is true.

(vi) Both, rational and irrational numbers are real numbers.

Hence, above statement is true.

(vii) Real numbers are both rational as well as irrational number.

Hence, above statement is false.

(viii) Some rational numbers are also real numbers.

Hence, above statement is true.