Prove that $\displaystyle \left(5 \sqrt{3}+\frac{2}{3}\right)$ is an irrational number given that $\displaystyle \sqrt{3}$ is an irrational number.

Let \(\displaystyle 5 \sqrt{3}+\frac{2}{3}\) be a rational number. \(\displaystyle \therefore 5 \sqrt{3}+\frac{2}{3}=\frac{\mathrm{a}}{\mathrm{b}}\) where a and b are integers and \(\displaystyle \mathrm{b} \neq 0\). \(\displaystyle 5 \sqrt{3}=\frac{\mathrm{a}}{\mathrm{b}}-\frac{2}{3}\) \(\displaystyle \sqrt{3}=\frac{3 \mathrm{a}-2 \mathrm{~b}}{15 \mathrm{~b}}\) 3a - 2b and 15b are integers. ∴ RHS is rational. But LHS \(\displaystyle =\sqrt{3}\) is an irrational number which is contradiction to our supposition. Hence \(\displaystyle 5 \sqrt{3}+\frac{2}{3}\) is an irrational number.