(a) Prove that √ 3 is an irrational number.

Let \(\displaystyle \sqrt{3}\) be a rational number.

\(\displaystyle \therefore \sqrt{3}=\frac{\mathrm{p}}{\mathrm{q}}\), where \(\displaystyle \mathrm{q} \neq 0\) and p & q are coprime.

\(\displaystyle 3 \mathrm{q}^{2}=\mathrm{p}^{2} \Rightarrow \mathrm{p}^{2}\) is divisible by $\displaystyle 3$

⇒ p is divisible by $\displaystyle 3$----- (i)

\(\displaystyle \Rightarrow \mathrm{p}=3 \mathrm{a}\), where 'a' is a postive integer

\(\displaystyle 9 \mathrm{a}^{2}=3 \mathrm{q}^{2} \Rightarrow \mathrm{q}^{2}=3 \mathrm{a}^{2} \Rightarrow \mathrm{q}^{2}\) is divisible by $\displaystyle 3$

⇒ q is divisible by $\displaystyle 3$ ----- (ii)

(i) and (ii) leads to contradiction as 'p' and 'q' are coprime.

\(\displaystyle \therefore \sqrt{3}\) is an irrational number.

\(\displaystyle (\sqrt{2}+\sqrt{3})^{2}=2+3+2 \sqrt{6}=5+2 \sqrt{6}\)

Let us assume, to the contrary, that \(\displaystyle 5+2 \sqrt{6}\) is rational

\(\displaystyle \therefore 5+2 \sqrt{6}=\frac{\mathrm{a}}{\mathrm{b}} ; \mathrm{a}, \mathrm{b}\) are integers, \(\displaystyle \mathrm{b} \neq 0\)

\(\displaystyle \therefore \sqrt{6}=\frac{\mathrm{a}-5 \mathrm{~b}}{2 \mathrm{~b}}\)

RHS is a rational number, whereas LHS is an irrational number.

∴ Our assumption is wrong.

\(\displaystyle \Rightarrow 5+2 \sqrt{6}=(\sqrt{2}+\sqrt{3})^{2}\) is an irrational number