Solve this


$\sqrt{2}$ is

(a) a rational number
(b) an irrational number
(c) a terminating decimal
(d) a nonterminating repeating decimal



Let $\sqrt{2}$ is a rational number.

$\therefore \sqrt{2}=\frac{p}{q}$, where $p$ and $q$ are some integers and $\operatorname{HCF}(p, q)=1$  ........(1)

$\Rightarrow \sqrt{2} q=p$

$\Rightarrow(\sqrt{2} q)^{2}=p^{2}$

$\Rightarrow 2 q^{2}=p^{2}$

⇒ p2 is divisible by 2
⇒ p is divisible by 2  .... (2)

Let p = 2m, where m is some integer.

$\therefore \sqrt{2} q=p$

$\Rightarrow \sqrt{2} q=2 m$

$\Rightarrow(\sqrt{2} q)^{2}=(2 m)^{2}$

$\Rightarrow 2 q^{2}=4 m^{2}$

$\Rightarrow q^{2}=2 m^{2}$

⇒ q2 is divisible by 2
⇒ is divisible by 2  .... (3)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.

Thus, $\sqrt{2}$ is an irrational number.

Hence, the correct answer is option (b).


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