Prove that \((\sqrt{2} + \sqrt{19})^2\) is an irrational number, given that \(\sqrt{38}\) is an irrational number.
Proof:
\((\sqrt{2} + \sqrt{19})^2\) \(=\) \(2 + 2 \sqrt{38} + 3\)
\(=\) \(21 + 2 \sqrt{38}\)
Let us assume that \(21 + 2 \sqrt{38}\) is a number.
\(21 + 2 \sqrt{38} =\) , where \(p\) and \(q\) are and \( q \neq 0\).
\(\sqrt{38} = \frac{p - 21q}{2q}\)
Here \(p\) and \(q\) are rationals. So, \(\frac{p - 21q}{2q}\) is a number.
This implies that \(\sqrt{38}\) is a number.
This contradicts the given that \(\sqrt{38}\) is number.
Hence, \((\sqrt{2} + \sqrt{19})^2\) is number.