Prove that root6+root5 is irrational

Lets assume √6 + √5 is a rational number.

√6 + √5 = $\frac{p}{q}$

$$\begin{gathered} Then ther exists co-prime positive integers p and q such that \\ \sqrt{6}+\sqrt{5}=\frac{p}{q} \\ \Rightarrow \frac{p}{q}-\sqrt{5}=\sqrt{6} \\ squaring both sides, we get, \\ \Rightarrow {\left(\frac{p}{q}-\sqrt{5}\right)}^2={\sqrt{6}}^2 \\ \Rightarrow \frac{p^2}{q^2}+5-2\sqrt{5}\frac{p}{q}=6 \\ \Rightarrow \frac{p^2}{q^2}+5-6=2\sqrt{5}\frac{p}{q} \\ \Rightarrow \frac{p^2}{q^2}-1=2\sqrt{5}\frac{p}{q} \\ \Rightarrow \left(\frac{p^2-q^2}{q^2}\right)\times \frac{q}{2p}=\sqrt{5} \\ \Rightarrow \left(\frac{p^2-q^2}{2qp}\right)=\sqrt{5} \\ \Rightarrow \sqrt{5 }is a rational number. \{because pq are integers and \frac{p^2-q^2}{2qp} is rational\} \\ this contradicts the fact \sqrt{5}is irrational. \\ So, our assumption is incorrect. \\ thus, \sqrt{6 }+\sqrt{5 } is a irrational. \\ Regards! \end{gathered}$$