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

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

\[\begin{aligned} & \therefore \sqrt{3}=\frac{\mathrm{p}}{\mathrm{q}}, \text { where } \mathrm{q} \neq 0 \text { and let } \mathrm{p} \text { \& } \mathrm{q} \text { be coprimes. } \\ & \Rightarrow 3 \mathrm{q}^{2}=\mathrm{p}^{2} \\ & \Rightarrow \mathrm{p}^{2} \text { is divisible by } 3 . \\ & \Rightarrow \mathrm{p} \text { is divisible by } 3 . \quad \text { ----- (1) } \end{aligned} \]

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

\[\begin{aligned} & \therefore 9 \mathrm{a}^{2}=3 \mathrm{q}^{2} \\ & \Rightarrow \mathrm{q}^{2}=3 \mathrm{a}^{2} \\ & \Rightarrow \mathrm{q}^{2} \text { is divisible by } 3 \\ & \Rightarrow \mathrm{q} \text { is divisible by } 3 \quad \text { ----- (2) } \end{aligned} \]

∴ $\displaystyle 3$ divides both p & q.

(1) and ($\displaystyle 2$) leads to contradiction as p and q are coprimes.

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

True,

\[\because 2 \times 3 \times 5 \times 7+7=7 \times(2 \times 3 \times 5+1) \text { has more than two factors. } \]

False,

\[\because 2 \times 3 \times 5 \times 7+1=211 \text { has only two factors. } \]