Question

Examine whether the following numbers are rational or irrational.
(i) $3+\sqrt{3}$

(ii) $\sqrt{7}-2$

(iii) $\sqrt[3]{5}\times \sqrt[3]{25}$

(iv) $\sqrt{7}\times \sqrt{343}$

(v) $\sqrt{\frac{13}{117}}$

(vi) $\sqrt{8}\times \sqrt{2}$

Answer 1

(i) Let us assume, to the contrary, that $3+\sqrt{3}$ is rational.
Then, $3+\sqrt{3}=\frac{p}{q}$, where p and q are coprime and $q\ne 0$.
$$\begin{gathered} \Rightarrow \sqrt{3}=\frac{p}{q}-3 \\ \Rightarrow \sqrt{3}=\frac{p-3q}{q} \end{gathered}$$
Since, p and q are are integers.
$\Rightarrow \frac{p-3q}{q}$ is rational.
So, $\sqrt{3}$ is also rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $3+\sqrt{3}$ is rational.
Hence, $3+\sqrt{3}$ is irrational.

(ii) Let us assume, to the contrary, that $\sqrt{7}-2$ is rational.
Then, $\sqrt{7}-2=\frac{p}{q}$, where p and q are coprime and $q\ne 0$.
$$\begin{gathered} \Rightarrow \sqrt{7}=\frac{p}{q}+2 \\ \Rightarrow \sqrt{7}=\frac{p+2q}{q} \end{gathered}$$
Since, p and q are are integers.
$\Rightarrow \frac{p+2q}{q}$ is rational.
So, $\sqrt{7}$ is also rational.
But this contradicts the fact that $\sqrt{7}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $\sqrt{7}-2$ is rational.
Hence, $\sqrt{7}-2$ is irrational.

(iii) As, $\sqrt[3]{5}\times \sqrt[3]{25}$

$$\begin{gathered} =\sqrt[3]{5\times 25} \\ =\sqrt[3]{125} \\ =5,\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{integer} \end{gathered}$$
Hence, $\sqrt[3]{5}\times \sqrt[3]{25}$ is rational.

(iv) As, $\sqrt{7}\times \sqrt{343}$
$$\begin{gathered} =\sqrt{7\times 343} \\ =\sqrt{2401} \\ =49,\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{integer} \end{gathered}$$
Hence, $\sqrt{7}\times \sqrt{343}$ is rational.

(v) As, $\sqrt{\frac{13}{117}}=\sqrt{\frac{1}{9}}=\frac{1}{3},\mathrm{which}\mathrm{is}\mathrm{rational}$
Hence, $\sqrt{\frac{13}{117}}$ is rational.

(vi) As, $\sqrt{8}\times \sqrt{2}$
$$\begin{gathered} =\sqrt{8\times 2} \\ =\sqrt{16} \\ =4,\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{integer} \end{gathered}$$
Hence, $\sqrt{8}\times \sqrt{2}$ is rational.