Prove that \(7+3\sqrt{2}\) is an irrational.
Proof:
Assume \(7+3\sqrt{2}\) is .
If \(7+3\sqrt{2}\) is rational, it can be written as a fraction
\(\sqrt{2}=\)
Here, both and \(3q\) are integers.
Therefore, is a rational number.
This implies that \(\sqrt{2}\) is a number.
This contradicts the fact that \(\sqrt{2}\) is .
Therefore, \(7+3\sqrt{2}\) is an .
Answer variants:
\(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(= 0\)
irrational
\(\frac{p-7q}{3q}\)
\(4-3q\)
\(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q\ne 0\)
\(p-7q\)
rational
\(7p-q\)
\(\frac{7p-q}{3q}\)