Prove that $\frac{\left(3-4\sqrt{2}\right)}{7}$ is an irrational number, given that $\sqrt{2}$ is an irrational number.

Let us assume that $\frac{\left(3-4\sqrt{2}\right)}{7}$ is a rational number.

Thus, $\frac{\left(3-4\sqrt{2}\right)}{7}$ can be represented in the form of $\frac{p}{q},$ where p and q are integers, q ≠ 0, p and q are co-prime numbers.

$$\begin{gathered} \frac{3-4\sqrt{2}}{7}=\frac{p}{q} \\ \Rightarrow 3-4\sqrt{2}=\frac{7p}{q} \\ \Rightarrow 4\sqrt{2}=3-\frac{7p}{q} \\ \Rightarrow 4\sqrt{2}=\frac{3q-7p}{q} \\ \Rightarrow \sqrt{2}=\frac{3q-7p}{4q} \\ \mathrm{Since},\frac{3q-7p}{4q}\mathrm{is}\mathrm{rational}\Rightarrow \sqrt{2}\mathrm{is}\mathrm{rational} \\ \mathrm{But},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\sqrt{2}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \\ \mathrm{Therefore},\mathrm{our}\mathrm{assumption}\mathrm{is}\mathrm{wrong}. \\ \mathrm{Hence},\frac{3-4\sqrt{2}}{7}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \end{gathered}$$