(a)OR(b)

Prove that $\displaystyle \sqrt{3}$ is an irrational number.

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

\[\begin{aligned} & \therefore \sqrt{3}=\frac{p}{q} \text {, let } p \& q \text { be co-primes and } q \neq 0 \\ & 3 q^{2}=p^{2} \Rightarrow p^{2} \text { is divisible by } 3 \Rightarrow p \text { is divisible by } 3 \\ & \Rightarrow p=3 a \text {, where ' } a \text { ' is some integer } \\ & 9 a^{2}=3 q^{2} \Rightarrow q^{2}=3 a^{2} \Rightarrow q^{2} \text { is divisible by } 3 \Rightarrow q \text { is divisible by } 3 \\ & \Rightarrow q=3 b \text {, where ' } b \text { ' is some integer } \\ & \text { (i) and (ii) leads to contradiction as ' } p \text { ' and ' } q \text { ' are co-primes. } \\ & \therefore \sqrt{3} \text { is an irrational number. } \end{aligned} \]

\(\displaystyle \mathrm{LCM}=432\)

i.e. \(\displaystyle \frac{\mathbf{4 3 2}}{\mathbf{6 0}}=7 \mathrm{~min} 12 \mathrm{sec}\).

⇒ traffic lights will change simultaneously again at $\displaystyle 7$ : $\displaystyle 7$ : $\displaystyle 12$ a.m.