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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.

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marked as duplicate by TomGrubb, user223391, JMoravitz, user642796 Nov 6 '16 at 5:19

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    $\begingroup$ This might be helpful: gowers.wordpress.com/2010/03/28/… $\endgroup$ – user160738 Nov 6 '16 at 1:44
  • $\begingroup$ 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. $\endgroup$ – Renan Maneli Mezabarba Nov 6 '16 at 1:48
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$\sqrt{18}=3\sqrt{2}$ and $\sqrt{2}\notin \mathbb{Q}$ (for proofs of this last point not using contradiction, see wikipedia for example).

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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.

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  • $\begingroup$ Then how come this is not irrational: math.stackexchange.com/questions/2000024/… $\endgroup$ – VadaCurry Nov 6 '16 at 1:56
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    $\begingroup$ @kunjimamu Having a non-terminating decimal form isn't necessarily the same as having a non-terminating continued fraction. Notice that this answer yields an infinitely nested fraction, and these must be irrational. The link you give only has an infinite decimal representation, which is rational if at some point, the sequence of digits becomes just a repetition of some digits. See more info here: en.wikipedia.org/wiki/Continued_fraction en.wikipedia.org/wiki/Repeating_decimal $\endgroup$ – MightyTyGuy Nov 6 '16 at 2:13
  • $\begingroup$ @MightyTyGuy Thanks for that reply. Is the above answer considered a proof? $\endgroup$ – VadaCurry Nov 6 '16 at 2:15
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    $\begingroup$ @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. $\endgroup$ – MightyTyGuy Nov 6 '16 at 2:19

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