Examine, whether the following numbers are rational or irrational:

Solution:

(i) Let $x=\sqrt{7}$

Therefore,

$x=2.645751311064 \ldots$

It is non-terminating and non-repeating

Hence $\sqrt{7}$ is an irrational number

(ii) Let $x=\sqrt{4}$

Therefore,

$x=2$

It is terminating.

Hence $\sqrt{4}$ is a rational number.

(iii) Let $x=2+\sqrt{3}$ be the rational

Squaring on both sides

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

$\Rightarrow x^{2}=4+3+4 \sqrt{3}$

$\Rightarrow x^{2}=7+4 \sqrt{3}$

$\Rightarrow x^{2}-7=4 \sqrt{3}$

$\Rightarrow \frac{x^{2}-7}{4}=\sqrt{3}$

Since, x is rational 

$\Rightarrow x^{2}$ is rational

$\Rightarrow x^{2}-7$ is rational

$\Rightarrow \frac{x^{2}-7}{4}$ is rational

$\Rightarrow \sqrt{3}$ is rational

But, $\sqrt{3}$ is irrational

So, we arrive at a contradiction.

Hence $2+\sqrt{3}$ is an irrational number

(iv) Let $x=\sqrt{3}+\sqrt{2}$ be the rational number

Squaring on both sides, we get

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

$\Rightarrow x^{2}=3+2+2 \sqrt{6}$

$\Rightarrow x^{2}=5+2 \sqrt{6}$

$\Rightarrow x^{2}-5=2 \sqrt{6}$

$\Rightarrow \frac{x^{2}-5}{2}=\sqrt{6}$

Since, x is a rational number

$\Rightarrow x^{2}$ is rational number

$\Rightarrow x^{2}-5$ is rational number

$\Rightarrow \frac{x^{2}-5}{2}$ is rational number

$\Rightarrow \sqrt{6}$ is rational number

But $\sqrt{6}$ is an irrational number

So, we arrive at contradiction

Hence $\sqrt{3}+\sqrt{2}$ is an irrational number

(v) Let $x=\sqrt{3}+\sqrt{5}$ be the rational number

Squaring on both sides, we get

$\Rightarrow x^{2}=(\sqrt{3}+\sqrt{5})^{2}$

$\Rightarrow x^{2}=3+5+2 \sqrt{15}$

$\Rightarrow x^{2}=8+2 \sqrt{15}$

$\Rightarrow x^{2}-8=2 \sqrt{15}$

$\Rightarrow \frac{x^{2}-8}{2}=\sqrt{15}$

Now, x is rational number

$\Rightarrow x^{2}$ is rational number

$\Rightarrow x^{2}-8$ is rational number

$\Rightarrow \frac{x^{2}-8}{2}$ is rational number

$\Rightarrow \sqrt{15}$ is rational number

But $\sqrt{15}$ is an irrational number

So, we arrive at a contradiction

Hence $\sqrt{3}+\sqrt{5}$ is an irrational number

(vi) Let $x=(\sqrt{2}-2)^{2}$ be a rational number.

$x=(\sqrt{2}-2)^{2}$

$\Rightarrow x=2+4-4 \sqrt{2}$

$\Rightarrow x=6-4 \sqrt{2}$

$\Rightarrow \frac{x-6}{-4}=\sqrt{2}$

Since, x is rational number,

$\Rightarrow x-6$ is a rational nu8mber

$\Rightarrow \frac{x-6}{-4}$ is a rational number

$\Rightarrow \sqrt{2}$ is a rational number

But we know that $\sqrt{2}$ is an irrational number, which is a contradiction

So $(\sqrt{2}-\sqrt{2})^{2}$ is an irrational number

(vii) Let $x=(2-\sqrt{2})(2+\sqrt{2})$

$\Rightarrow x=(2)^{2}-(\sqrt{2})^{2} \quad\left\{\operatorname{As}(a+b)(a-b)=a^{2}-b^{2}\right\}$

$\Rightarrow x=4-2$

$\Rightarrow x=2$

So $(2-\sqrt{2})(2+\sqrt{2})$ is a rational number

(viii) Let $x=(\sqrt{2}+\sqrt{3})^{2}$ be rational number

Using the formula $(a+b)^{2}=a^{2}+b^{2}+2 a b$

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

$\Rightarrow x=2+3+2 \sqrt{6}$

$\Rightarrow x=5+2 \sqrt{6}$

$\Rightarrow \frac{x-5}{2}$ is a rational number

$\Rightarrow \sqrt{6}$ is a rational number

But we know that $\sqrt{6}$ is an irrational number

So, we arrive at a contradiction

So $(\sqrt{2}+\sqrt{3})^{2}$ is an irrational number.

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

Squaring on both sides, we get

$x=\sqrt{5}-2$

$x^{2}=(\sqrt{5}-2)^{2}$

$x^{2}=25+4-4 \sqrt{5}$

$x^{2}-29=-4 \sqrt{5}$

$\frac{x^{2}-29}{-4}=\sqrt{5}$

Now, x is rational

$x^{2}$ is rational.

So, $x^{2}-29$ is rational

$\frac{x^{2}-29}{-4}=\sqrt{5}$ is rational.

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

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

(x) Let 

$x=\sqrt{23}$

$x=4.79583 \ldots$

It is non-terminating or non-repeating

Hence $\sqrt{23}$ is an irrational number

(xi) Let $x=\sqrt{225}$

$\Rightarrow x=15$

Hence $\sqrt{225}$ is a rational number

(xii) Given $0.3796$.

It is terminating 

Hence it is a rational number

(xiii) Given number $x=7.478478 \ldots$

$\Rightarrow x=7 . \overline{478}$

It is repeating

Hence it is a rational number

(xiv) Given number is $x=1.1010010001 \ldots$

It is non-terminating or non-repeating

Hence it is an irrational number