Verify that \((\sqrt{11} + \sqrt{23})^2\) is an irrational number, given that \(\sqrt{253}\) is an irrational number.
Proof:
\((\sqrt{11} + \sqrt{23})^2\) \(=\) \(11 + 2 \sqrt{253} + 23\)
\(=\) \(34 + 2 \sqrt{253}\)
Let us assume that \(34 + 2 \sqrt{253}\) is a number.
\(34 + 2 \sqrt{253} =\) , where \(p\) and \(q\) are and \( q \neq 0\).
\(\sqrt{253} = \frac{p - 34q}{2q}\)
Here \(p\) and \(q\) are rationals. So, \(\frac{p - 34q}{2q}\) is a number.
This implies that \(\sqrt{253}\) is a number.
This contradicts the given that \(\sqrt{253}\) is number.
Hence, \((\sqrt{11} + \sqrt{23})^2\) is number.