# Prove that the following are irrationals :

Question : Prove that the following are irrationals :

(i)  $$1\over \sqrt{2}$$

(ii)  $$7\sqrt{5}$$

(iii)  $$6 + \sqrt{2}$$

Solution :

(i)  Let us assume, to the contrary, that $$1\over \sqrt{2}$$ is rational. That is, we can find co-prime integers p and q (q $$\ne$$ 0) such that

$$1\over \sqrt{2}$$ = $$p\over q$$  or   $$1\times \sqrt{2}\over \sqrt{2}\times \sqrt{2}$$ = $$p\over q$$ or  $$\sqrt{2}\over 2$$ = $$p\over q$$

or  $$\sqrt{2}$$ = $$2p\over q$$

Since p and q are integers $$2p\over q$$ is rational, and so $$\sqrt{2}$$ is rational.

But this contradicts the fact that $$\sqrt{2}$$ is irrational.

so, we conclude that $$1\over \sqrt{2}$$ is an irrational.

(ii)  Let us assume, to the contrary, that $$7\sqrt{5}$$ is rational. That is, we can find co-prime integers p and q (q $$\ne$$ 0) such that $$7\sqrt{5}$$ = $$p\over q$$.

So,  $$\sqrt{5}$$ = $$p\over 7q$$

Since p and q are integers, $$p\over 7q$$ is rational and so is $$\sqrt{5}$$.

But this contradicts the fact that $$\sqrt{5}$$ is irrational. So, we conclude that $$7\sqrt{5}$$ is an irrational.

(iii)  Let us assume, to the contrary, that $$6 + \sqrt{2}$$ is rational. That is, we can find co-prime integers p and q (q $$\ne$$ 0) such that

$$6 + \sqrt{2}$$ = $$p\over q$$  or   6 – $$p\over q$$ = $$\sqrt{2}$$

or  $$\sqrt{2}$$ = 6 – $$p\over q$$

Since p and q are integers, 6 – $$p\over q$$ is rational, and so $$\sqrt{2}$$ is rational.

But this contradicts the fact that $$\sqrt{2}$$ is irrational.

so, we conclude that $$6 + \sqrt{2}$$ is an irrational.