Question

Examine whether the following numbers are rational or irrational:

(i) $\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)$

(ii) ${\left(\sqrt{3}+2\right)}^2$

(iii) $\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$

(iv) $\sqrt{8}+4\sqrt{32}-6\sqrt{2}$

Answer 1

(i) $\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)$

$$\begin{gathered} =5^2-{\left(\sqrt{5}\right)}^2\left[\left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ =25-5 \\ =20,\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{integer} \end{gathered}$$

Hence, $\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)$ is rational.

(ii) ${\left(\sqrt{3}+2\right)}^2$

$$\begin{gathered} ={\left(\sqrt{3}\right)}^2+2^2+2\times \sqrt{3}\times 2\left[{\left(a+b\right)}^2=a^2+b^2+2ab\right] \\ =3+4+4\sqrt{3} \\ =7+4\sqrt{3} \end{gathered}$$

Since, the sum and product of rational numbers and an irrational number is always an irrational.

$\Rightarrow$$7+4\sqrt{3}$ is irrational.

Hence, ${\left(\sqrt{3}+2\right)}^2$​ is irrational.

(iii) $\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$

$$\begin{gathered} =\frac{2\sqrt{13}}{3\sqrt{13\times 4}-4\sqrt{13\times 9}} \\ =\frac{2\sqrt{13}}{\sqrt{13}\left(3\sqrt{4}-4\sqrt{9}\right)} \\ =\frac{2}{\left(3\times 2-4\times 2\right)} \end{gathered}$$

$$\begin{gathered} =\frac{2}{6-8} \\ =\frac{2}{-2} \\ =-1,\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{integer} \end{gathered}$$

Hence, $\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$ is rational.

(iv) $\sqrt{8}+4\sqrt{32}-6\sqrt{2}$

$$\begin{gathered} =2\sqrt{2}+4\times 4\sqrt{2}-6\sqrt{2} \\ =2\sqrt{2}+16\sqrt{2}-6\sqrt{2} \\ =12\sqrt{2} \end{gathered}$$

Since, the product of a rational number and an irrational number is always an irrational.

Hence, $\sqrt{8}+4\sqrt{32}-6\sqrt{2}$ is rational.