Question

Examine, whether the following numbers are rational or irrational:

(i) $\sqrt{7}$ (ii) $\sqrt{4}$

(iii) $2+\sqrt{3}$ (iv) $\sqrt{3}+\sqrt{2}$

(v) $\sqrt{3}+\sqrt{5}$ (vi) $(\sqrt{2}-2{)}^2$

(vii) $(2-\sqrt{2})(2+\sqrt{2})$ (viii) $\sqrt{2}+\sqrt{3}{)}^2$

(ix) $\sqrt{5}-2$ (x) $\sqrt{23}$

(xi) $\sqrt{225}$ (xii) 0.3796

(xiii) 7.478478..... (xiv) 1.101001000100001.....

Answer 1

(i) $\sqrt{7}$

It is an irrational as 7 is not a perfect square.

(ii) $\sqrt{4}$

It is a rational number as 4 is a perfect square of 2.

(iii) $2+\sqrt{3}$

It is an irrational number as sum of a rational number and an irrational number is also an irrational number.

(iv) $\sqrt{3}+\sqrt{2}$

Irrational as sum of two irrational numbers is also an irrational number.

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

is an irrational number as sum of two irrational numbers is also an irrational.

(vi) $(\sqrt{2}-2{)}^2$ = 2 + 4 + 2 $\sqrt{2}$ $\times$ 2 = 6 + 4 $\sqrt{2}$

{$\because (a-b{)}^2=a^2+b^2-2ab)$}

it is an irrational number as sum of a rational and an irrational number is an irrational number.

(vii) $(2-\sqrt{2})(2+\sqrt{2})$ = $(2{)}^2-(\sqrt{2}{)}^2$
$\because (a+b)(a-b)=a^2-b^2=4-2=2$
$Whicharationalnumber$

(viii) $\sqrt{2}+\sqrt{3}{)}^2=2+3+2\sqrt{2}\sqrt{3}$

{ $\because (a+b{)}^2=a^2+b^2+2ab$}

= 5 + 2 $\sqrt{6}$

Which is an irrational number as sum of a rational and an irrational number is an irrational number.

(ix) $\sqrt{5}-2$ is an irrational number as difference of an irrational number and a rational number is also an irrational number.

(x) $\sqrt{23}$ is an irrational number as $\sqrt{23}$ is not a perfect square.

(xi) $\sqrt{225}$ = 15

(xii) 0.3796 is a rational number its decimal is terminating.

(xiii) 7.478478..... = 7. $\overline{478}$

Which is non-terminating recurring decimal. Therefore it is a rational number.

(xiv) 1.101001000100001.....

It is an irrational number as its decimal is non-terminating non-recurring decimal.