Question

$\sqrt{2}$ is
(a) a rational number
(b) an irrational number
(c) a terminating decimal
(d) a nonterminating repeating decimal

Answer 1

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

∴ $\sqrt{2}=\frac{p}{q}$, where p and q are some integers and HCF(p, q) = 1 .... (1)
$$\begin{gathered} \Rightarrow \sqrt{2}q=p \\ \Rightarrow {\left(\sqrt{2}q\right)}^2=p^2 \\ \Rightarrow 2q^2=p^2 \end{gathered}$$
⇒ p² is divisible by 2
⇒ p is divisible by 2 .... (2)

Let p = 2m, where m is some integer.

∴ $\sqrt{2}q=p$
$$\begin{gathered} \Rightarrow \sqrt{2}q=2m \\ \Rightarrow {\left(\sqrt{2}q\right)}^2={\left(2m\right)}^2 \\ \Rightarrow 2q^2=4m^2 \\ \Rightarrow q^2=2m^2 \end{gathered}$$
⇒ q² is divisible by 2
⇒ q 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).