Prove that \((\sqrt{29} + \sqrt{31})^2\) is an irrational number, given that \(\sqrt{899}\) is an irrational number.
Proof:
\((\sqrt{29} + \sqrt{31})^2\) \(=\) \(29 + 2 \sqrt{899} + 31\)
\(=\) \(60 + 2 \sqrt{899}\)
Let us assume that \(60 + 2 \sqrt{899}\) is a number.
\(60 + 2 \sqrt{899} =\) , where \(p\) and \(q\) are and \( q \neq 0\).
\(\sqrt{899} = \frac{p - 60q}{2q}\)
Here \(p\) and \(q\) are rationals. So, \(\frac{p - 60q}{2q}\) is a number.
This implies that \(\sqrt{899}\) is a number.
This contradicts the given that \(\sqrt{899}\) is number.
Hence, \((\sqrt{29} + \sqrt{31})^2\) is number.