Prove that 6-√ 7 is irrational number, given that √ 7 is an irrational number.

Let us assume that \(\displaystyle 6-\sqrt{7}\) is rational \[\begin{aligned} & \therefore 6-\sqrt{7}=\frac{p}{q} ; q \\ & \Rightarrow \sqrt{7}=\frac{6 q-p}{q} \end{aligned} \] p, q are integers, \(\displaystyle \therefore 6 \mathrm{q}-\mathrm{p}\) is an integer \[\Rightarrow \frac{6 \mathrm{q}-\mathrm{p}}{\mathrm{q}} \text { is a rational number } \] \(\displaystyle \Rightarrow \sqrt{7}\) is rational number which contradicts our assumption that \(\displaystyle \sqrt{7}\) is an irrational number \[\Rightarrow 6-\sqrt{7} \text { is an irrational number } \]