Prove that \(4+8\sqrt{19}\) is irrational
Proof:
Let us consider \(4+8\sqrt{19}\) is
\(4+8\sqrt{19}=\frac{p}{q}\) be a number, where \(p\) and \(q\) are and \(q\neq 0\)
\(8\sqrt{19}=\frac{p}{q}-\)
\(\sqrt{19}=\)
Since, \(p\) and \(q\) are integers, will also be number.
Therefore, \(\sqrt{19}\) is number.
This contradicts the fact that \(\sqrt{19}\) is number.
Hence, \(4+8\sqrt{19}\) is irrational number.