Prove that \((\sqrt{11} + \sqrt{19})^2\) is an irrational number, given that \(\sqrt{209}\) is an irrational number.
Proof:
\((\sqrt{11} + \sqrt{19})^2\) \(=\) \(11 + 2 \sqrt{209} + 19\)
\(=\) \(30 + 2 \sqrt{209}\)
Let us assume that \(30 + 2 \sqrt{209}\) is a number.
\(30 + 2 \sqrt{209} =\) , where \(p\) and \(q\) are and \( q \neq 0\).
\(\sqrt{209} = \)
Here \(p\) and \(q\) are rationals. So, is a number.
This implies that \(\sqrt{209}\) is a number.
This contradicts the given that \(\sqrt{209}\) is number.
Hence, \((\sqrt{11} + \sqrt{19})^2\) is number.
Answer variants:
irrational
\(\frac{q}{p}\)
\(\frac{2p}{q-30p}\)
integers
\(\frac{p-30q}{2q}\)
\(\frac{q-30p}{2q}\)
whole number
rational
\(\frac{p}{q}\)