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.