Solution :
Suppose that \(\sqrt{5}\) is an irrational number. Then \(\sqrt{5}\) can be expressed in the form \(p\over q\) where p, q are integers and have no common factor, q \(ne\) 0.
\(\sqrt{5}\) = \(p\over q\)
Squaring both sides, we get
5 = \(p^2\over q^2\) \(\implies\) \(p^2\) = \(5q^2\) ………..(1)
\(\implies\) 5 divides \(p^2\)
\(\implies\) 5 divides p.
Let p = 5m \(\implies\) \(p^2\) = \(25m^2\) ……….(2)
Putting the value of \(p^2\) in (1), we get
\(25m^2\) = \(5q^2\) \(\implies\) \(5m^2\) = \(q^2\)
\(\implies\) 5 divides \(q^2\) \(\implies\) 5 divides q. ………(3)
Thus, from (2), 5 divides p and from (3), 5 also divides q. It means 5 is a common factor of p and q. This contradicts the supposition so there is no common factor of p and q.
Hence, \(\sqrt{5}\) is an irrational number.
