Prove that \((\sqrt{2} + \sqrt{13})^{2}\) is an irrational number, if \(\sqrt{26}\) is an irrational number.
Proof:
Let \((\sqrt{2} + \sqrt{13})^{2}\) be .
By definition, \((\sqrt{2} + \sqrt{13})^{2}\) \(=\) , where \(p\) and \(q\) are integers.
Expand \((\sqrt{2} + \sqrt{13})^{2}\) using the identity .
Let \(a = \sqrt{2}\) and \(b = \sqrt{13}\).
\(\Rightarrow\) \((\sqrt{2} + \sqrt{13})^{2}\) \(=\) .
Thus, we have \(15 + 2 \sqrt{26}\) \(=\) .
Since, \(p\) and \(q\) are integers, \(\frac{p - 15q}{2q}\) will also be .
Therefore, \(\sqrt{26}\) is rational.
This the fact that \(\sqrt{26}\) is an .
Hence, \((\sqrt{2} + \sqrt{13})^{2}\) is an .
Hence, proved.