$\mathbb Q^c$ = 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
2026-03-31 11:09:47.1774955387
Use the continued fraction expansion of $\sqrt{2}$ and show that $\mathbb Q$ and $\mathbb Q^c$ does not possess the Completeness Axiom Property.
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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 :)