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$\mathbb Q^c$ = set of irrational numbers and $\mathbb Q$ = set of rational numbers, This question is from my Mathematics assignment, and my professor specifically asked us to use continued fraction expansions to solve this. I have proved that rationals does not posses completeness axiom property, using the set $\{\lor x \in \mathbb Q, \,x> 0| x^2<2\}$ . But No idea how to prove for irrational using continued fractions

MulshxxSL
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  • $\mathbb Q^c$ = set of irrational numbers and $\mathbb Q$ = set of rational numbers – MulshxxSL Aug 27 '21 at 02:10
  • For the rational case, you can use the convergents of the continued fraction. They get closer and closer to $\sqrt{2}$ , but keep rational. For the irrational, case, I think that the fractions like $[1,2,2,2,1,1,1,1,1,1,1,1\cdots]$ do the job. They correspond to an irrational number different from $\sqrt{2}$ and get arbitary close to $\sqrt{2}$ as well , if we start with more and more twos after the first one. – Peter Aug 29 '21 at 09:41
  • Yes, Initutively we can suggest that. But I can't find a way to show that contradicts the completeness axiom using set theory. I'm not sure whether we can add continued fractions into a set and prove such – MulshxxSL Aug 29 '21 at 10:24

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Proving the continued fraction expansion of $\sqrt2$

check this out, it explores a detailed method.

You can use the convergents of the continued fraction expansion. The terms of the sequence of convergents alternate between being greater than and less than $\sqrt{2}$. If you consider the odd set of terms you can see that $\sqrt{2}$ is an upper bound. Further the absolute difference between consecutive terms of the convergents reduces and approaches 0. You can use this to show that $\sqrt{2}$ is the least upper bound of the set of odd terms. It follows that it is a set of rationals, but its supremum is irrational, hence not satisfying the completeness property.

Hope this helps :)

Sandil Ranasinghe
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