(A) Prove that √ 2 is an irrational number.

Let \(\displaystyle \sqrt{2}\) be a rational number.

\(\displaystyle \therefore \sqrt{2}=\frac{\mathbf{p}}{\mathbf{q}}\), where \(\displaystyle q \neq 0\) and let \(\displaystyle p \& q\) be co-primes.

\(\displaystyle 2 q^{2}=p^{2} \Rightarrow p^{2}\) is divisible by \(\displaystyle 2 \Rightarrow p\) is divisible by $\displaystyle 2$ ----- (i)

\(\displaystyle \Rightarrow p=2 a\), where ' \(\displaystyle a\) ' is some integer

\(\displaystyle 4 a^{2}=2 q^{2} \Rightarrow q^{2}=2 a^{2} \Rightarrow q^{2}\) is divisible by \(\displaystyle 2 \Rightarrow q\) is divisible by $\displaystyle 2$ ----- (ii)

(i) and (ii) leads to contradiction as ' \(\displaystyle p\) ' and ' \(\displaystyle q\) ' are co-primes.

\(\displaystyle \therefore \sqrt{2}\) is an irrational number.

\[\begin{aligned} & p=x^{2} y^{3}, q=x y^{4}, r=x^{5} y^{2} \\ & \operatorname{HCF}(p, q, r)=x y^{2} \\ & \operatorname{LCM}(p, q, r)=x^{5} y^{4} \\ & \operatorname{HCF} \times \operatorname{LCM}=x^{6} y^{6} \\ & p \times q \times r=x^{8} y^{9} \\ & \Rightarrow \operatorname{HCF}(p, q, r) \times \operatorname{LCM}(p, q, r) \neq p \times q \times r \end{aligned} \]