Prove that \(\sqrt{5}\) is irrational.

Let us suppose that \(\sqrt{5}\) is rational,
So there should be two co- prime integers (p and q, where q \(\ne\) 0 ) such that:

\(\sqrt{5} = {{p}\over{q}} \)
\(=>\sqrt{5} × q = p \)

On squaring both sides we'll get:
\(=>5q^2 = p^2 \) ............................(i)

From equation (i) it can be clearly seen that 5 is a factor of \(p^2\)

Therefore, 5 is also factor of p.
Let p = 5r, where r is an integer.

On substituting the value of p in equation (i),
\(=>5q^2 = 25r^2 \)
\(=>q^2 = 5r^2 \).............................(ii)

From equation (ii) it can be clearly seen that 5 is a factor of \(q^2\)

Therefore, 5 is also factor of q.
Which means 5 is a common factor of p and q which contradicts our assumption,

that p and q are co-prime and also that \(\sqrt{5}\) is rational.

Hence it is proved that \(\sqrt{5}\) is irrational.