3 Tutor System
Starting just at 265/hour

# 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.