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Derive that \(\sqrt{31}\) is an irrational number.
Proof:
Let \(\sqrt{31}=\frac{p}{q}\) be a number, where \(p\) and \(q\) are coprime and \(q\neq 0\).
Squaring on both sides, we get: \(31q^2=\) ......(1)
Therefore, \(p^2\) is divisible by \(31\). Hence,\(p\) can be divided by \(31\).
Substitute \(p = 31k\) in equation (1) and simplifying, we get: \(=\) \(31k^{2}\)
This means that \(q^2\) is divisible by \(31\) and hence, \(q\) is divisible by \(31\).
This implies that \(p\) and \(q\) have \(31\) as a .
And this is a contradiction to the fact that \(p\) and \(q\) are .
Hence, \(\sqrt{31}\) cannot be expressed as .
Therefore, \(\sqrt{31}\) is .
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