Prove that \(\sqrt{5}\) is an irrational number by contradiction method.

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.

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