using proof by contradiction.

Question

I am wondering whether there is another method to show that $\sqrt{118}$ is irrational. I have always been taught to use proof by contradiction for showing irrationality. Can anyone think of other methods? Please help.

Answer 1

$\sqrt{18}=3\sqrt{2}$ and $\sqrt{2}\notin \mathbb{Q}$ (for proofs of this last point not using contradiction, see wikipedia for example).

Answer 2

We see that: $$ \sqrt{18} = 3 + (\sqrt{18} - 3) = 3 + \frac{9}{3+ \sqrt{18}} = 3 + \frac{9}{3 + 3 + \frac{9}{3+ \sqrt{18}}} = 3+ \frac{9}{6 + \frac{9}{6 + \frac{9}{6 \ldots}}} $$

Hence, $\sqrt{18}$ has a continued fraction representation which is non-terminating. If it were rational, then the continued fraction would have to terminate,so that it can be evaluated. Since this is not the case, $\sqrt{18}$ is irrational.

In general: if the continued fraction representation is unending, then the number is irrational.