Show that \(\frac{4\sqrt{17}}{5}\) is an irrational.
Proof:
Assume \(\frac{4\sqrt{17}}{5}\) is
Therefore, \(\frac{4\sqrt{17}}{5}=\)
On simplification, \(\sqrt{17}=\)
Since \(p\) and \(q\) are integers, is a number.
This implies that \(\sqrt{17}\) is a number.
But this contradicts the fact that \(\sqrt{17}\) is .
Therefore, \(\frac{4\sqrt{17}}{5}\) is .
Answer variants:
\(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q= 0\).
irrational
\(\frac{2p}{5q}\)
\(\frac{5p}{4q}\)
rational
\(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q\ne 0\).