(A) Prove that 5-2 √ 3 is an irrational number. It is given that √ 3 is an irrational number.

Assuming \(\displaystyle 5-2 \sqrt{3}\) to be a rational number.

Let \(\displaystyle 5-2 \sqrt{3}=\frac{a}{b}\) where \(\displaystyle a\) and \(\displaystyle b\) are integers & \(\displaystyle b \neq 0\)

\[\Rightarrow \sqrt{3}=\frac{5 b-a}{2 b} \]

Here RHS is rational but LHS is irrational.

Therefore our assumption is wrong.

Hence, \(\displaystyle 5-2 \sqrt{3}\) is an irrational number.

\[\begin{aligned} 5 \times 11 \times 17+3 \times 11 & =11 \times(5 \times 17+3) \\ & =11 \times 88 \text { or } 11 \times 11 \times 2^{3} \end{aligned} \]

It means the number can be expressed as a product of two factors other than $\displaystyle 1$, therefore the given number is a composite number.