Show that \((\sqrt{5} + \sqrt{19})^2\) is an irrational number, given that \(\sqrt{95}\) is an irrational number.
Proof:
\((\sqrt{5} + \sqrt{19})^2\) \(=\) \(5 + 2 \sqrt{95} + 3\)
\(=\) \(24 + 2 \sqrt{95}\)
Let us assume that \(24 + 2 \sqrt{95}\) is a number.
\(24 + 2 \sqrt{95} =\) , where \(p\) and \(q\) are and \( q \neq 0\).
\(\sqrt{95} = \frac{p - 24q}{2q}\)
Here \(p\) and \(q\) are rationals. So, \(\frac{p - 24q}{2q}\) is a number.
This implies that \(\sqrt{95}\) is a number.
This contradicts the given that \(\sqrt{95}\) is number.
Hence, \((\sqrt{5} + \sqrt{19})^2\) is number.