Show that \(\sqrt{29}\) is an irrational number.
Proof:
Let \(\sqrt{29}=\frac{p}{q}\) be a number, where \(p\) and \(q\) are coprime and \(q\neq 0\).
Squaring on both sides, we get: \(29q^2=\) ......(1)
Therefore, \(p^2\) is divisible by \(29\). Hence,\(p\) can be divided by \(29\).
Substitute \(p = 29k\) in equation (1) and simplifying, we get: \(=\) \(29k^{2}\)
This means that \(q^2\) is divisible by \(29\) and hence, \(q\) is divisible by \(29\).
This implies that \(p\) and \(q\) have \(29\) as a .
And this is a contradiction to the fact that \(p\) and \(q\) are .
Hence, \(\sqrt{29}\) cannot be expressed as .
Therefore \(\sqrt{29}\) is .