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) Let

Therefore,

It is non-terminating and non-repeating

Hence is an irrational number

(ii) Let

Therefore,

It is terminating.

Hence is a rational number.

(iii) Let be the rational

Squaring on both sides

Since, x is rational

is rational

is rational

is rational

is rational

But, is irrational

So, we arrive at a contradiction.

Hence is an irrational number

(iv) Let be the rational number

Squaring on both sides, we get

Since, x is a rational number

is rational number

is rational number

is rational number

is rational number

But is an irrational number

So, we arrive at contradiction

Hence is an irrational number

(v) Let be the rational number

Squaring on both sides, we get

Now, x is rational number

is rational number

is rational number

is rational number

is rational number

But is an irrational number

So, we arrive at a contradiction

Hence is an irrational number

(vi) Let be a rational number.

Since, x is rational number,

⇒ x – 6 is a rational nu8mber

⇒is a rational number

⇒is a rational number

But we know thatis an irrational number, which is a contradiction

So is an irrational number

(vii) Let

So is a rational number

(viii) Let be rational number

Using the formula

⇒is a rational number

⇒is a rational number

But we know thatis an irrational number

So, we arrive at a contradiction

So is an irrational number.

(ix) Let $x=\sqrt{5}-2$ be the rational number

Squaring on both sides, we get

$$\begin{gathered} x=\sqrt{5}-2 \\ {x}^2={\left(\sqrt{5}-2\right)}^2 \\ {x}^2=25+4-4\sqrt{5} \\ {x}^2-29=-4\sqrt{5} \\ \frac{x^2-29}{-4}=\sqrt{5} \end{gathered}$$

Now, x is rational

$$\begin{gathered} x^2\mathrm{is}\mathrm{rational}. \\ \mathrm{So},{\mathrm{x}}^2-29\mathrm{is}\mathrm{rational} \\ \frac{{\mathrm{x}}^2-29}{-4}=\sqrt{5}\mathrm{is}\mathrm{rational}. \end{gathered}$$

But, $\sqrt{5}$ is irrational. So we arrive at contradiction

Hence $x=\sqrt{5}-2$ is an irrational number

(x) Let

It is non-terminating or non-repeating

Hence is an irrational number

(xi) Let

Hence is a rational number

(xii) Given $0.3796$.

It is terminating

Hence it is a rational number

(xiii) Given number

It is repeating

Hence it is a rational number

(xiv) Given number is

It is non-terminating or non-repeating

Hence it is an irrational number