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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1$\begingroup$ This might be helpful: gowers.wordpress.com/2010/03/28/… $\endgroup$– user160738Commented Nov 6, 2016 at 1:44
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$\begingroup$ 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 MezabarbaCommented Nov 6, 2016 at 1:48
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2 Answers
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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.
answered Nov 6, 2016 at 1:53
Sarvesh Ravichandran IyerSarvesh Ravichandran Iyer
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$\begingroup$ Then how come this is not irrational: math.stackexchange.com/questions/2000024/… $\endgroup$ Commented Nov 6, 2016 at 1:56
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1$\begingroup$ @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 $\endgroup$– TylerCommented Nov 6, 2016 at 2:13
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$\begingroup$ @MightyTyGuy Thanks for that reply. Is the above answer considered a proof? $\endgroup$ Commented Nov 6, 2016 at 2:15
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1$\begingroup$ @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. $\endgroup$– TylerCommented Nov 6, 2016 at 2:19
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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).