# using proof by contradiction. [duplicate]

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.

• This might be helpful: gowers.wordpress.com/2010/03/28/… Commented Nov 6, 2016 at 1:44
• Interesting question! The only definition of irrational number that I know is "not rational", so I think that proofs by contradiction are the natural approach. Commented Nov 6, 2016 at 1:48

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.
• @kunjimamu Yes. There's a theorem that states that the value of an infinite continued fraction must be irrational. Here, the author explicitly constructs an infinite continued fraction representation of $\sqrt{18}$, and so that number must be irrational. Commented Nov 6, 2016 at 2:19
$\sqrt{18}=3\sqrt{2}$ and $\sqrt{2}\notin \mathbb{Q}$ (for proofs of this last point not using contradiction, see wikipedia for example).