Identify the following as rational or irrational numbers.

Solution:

(i) Given number is $x=\sqrt{4}$

$x=2$, which is a rational number

(ii) Given number is $3 \sqrt{18}$

$\Rightarrow 3 \sqrt{18}=3 \sqrt{3 \times 3 \times 2}$

$\Rightarrow 3 \sqrt{18}=3 \times 3 \sqrt{2}$

$\Rightarrow 3 \sqrt{18}=3 \times 3 \sqrt{2}$

$\Rightarrow 3 \sqrt{18}=18 \sqrt{2}$

$3 \sqrt{18}=3 \sqrt{3 \times 3 \times 2}=3 \times 3 \sqrt{2}=9 \sqrt{2}$

So it is an irrational number

(iii) Given number is $\sqrt{1.44}$

Now we have to check whether it is rational or irrational

$\Rightarrow \sqrt{1.44}=\sqrt{\frac{144}{100}}$

$\Rightarrow \sqrt{1.44}=\frac{\sqrt{144}}{\sqrt{100}}$

$\Rightarrow \sqrt{1.44}=\frac{12}{10}$

$\Rightarrow \sqrt{1.44}=\frac{6}{5}$

$\Rightarrow \sqrt{1.44}=1.2$

So it is a rational

(iv) Given that $\sqrt{\frac{9}{27}}$

Now we have to check whether it is rational or irrational

$\Rightarrow \sqrt{\frac{9}{27}}=\frac{\sqrt{9}}{\sqrt{27}}$

$\Rightarrow \sqrt{\frac{9}{27}}=\frac{3}{\sqrt{3 \times 3 \times 3}}$

$\Rightarrow \sqrt{\frac{9}{27}}=\frac{3}{3 \sqrt{3}}$

$\Rightarrow \sqrt{\frac{9}{27}}=\frac{1}{\sqrt{3}}$

So it is an irrational number

(v) Given that $-\sqrt{64}$

Now we have to check whether it is rational or irrational

Since, $-\sqrt{64}=-8$

So it is a rational number

(vi) Given that $\sqrt{100}$

Now we have to check whether it is rational or irrational

Since, $\sqrt{100}=10$

So it is rational number