Prove that $\displaystyle \left(4 \sqrt{2}+\frac{5}{3}\right)$ is an irrational number given that $\displaystyle \sqrt{2}$ is an irrational number.

Let \(\displaystyle 4 \sqrt{2}+\frac{5}{3}\) be a rational number. \(\displaystyle \therefore 4 \sqrt{2}+\frac{5}{3}=\frac{\mathrm{a}}{\mathrm{b}}\) where a and b are integers and \(\displaystyle \mathrm{b} \neq 0\) \(\displaystyle 4 \sqrt{2}=\frac{\mathrm{a}}{\mathrm{b}}-\frac{5}{3}\) \(\displaystyle \sqrt{2}=\frac{3 \mathrm{a}-5 \mathrm{~b}}{12 \mathrm{~b}}\) 3a - 5b and 12b are integers. ∴ RHS is rational. But LHS \(\displaystyle =\sqrt{2}\) is an irrational number which is contradiction to our supposition. Hence \(\displaystyle 4 \sqrt{2}+\frac{5}{3}\) is an irrational number.