I've been playing with a computational system that represents numbers in their simple continued fraction form.(adsbygoogle = window.adsbygoogle || []).push({});

That is, CF([a_{0},a_{1}, ... , a_{n}]) =a_{0}+ [itex]\frac{1}{ a_{1}+\frac{1}{a_{2}+\ddots\frac{1}{a_{n}}} }[/itex]

Considering what types of numbers such a system can represent, the finite CF's correspond to the set of rational numbers. Therefore this set is countably infinite (as are the rationals).

Cantor showed that the real numbers, which corresponds to the set of all finite and infinite CF's, is not countably infinite. What about the quadratic surds, which correspond to infinite continued fractions with a simple repeated pattern? Surds can be represented as a finite 'rational' part and a repeat length, which seems to be countable.

A wide variety of transcendental numbers also have repeating CF representations, such as [itex]e^{1/n}[/itex] = CF([1, (n-1), 1],[0,2n,0]) = CF([1,(n-1),1,1,(3n-1),1,1,(5n-1),... ]) = [itex]1+\frac{1}{(n-1)+\frac{1}{1+\frac{1}{1+\frac{1}{3n-1+\ddots}}}}[/itex]

Consider the set of repeating CF's. { [itex] CF( [a_{0}, a_{1}, ..., a_{n} ], [v_{0},v_{1},...,v_{m}]), m<=n [/itex] }, clearly each part, being finite, is countable, is the set of pairs still countably infinite?

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# Counting continued fraction numbers

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