Talk:Finite Simple Continued Fraction has Rational Value

I am not sure what you are looking for here as an improvement. Personally, I would argue that the rational numbers are a field, so in particular, they are closed under addition and each element has a multiplicative inverse. But to formalize this, I think there is no way around an inductive proof. I will however fix an error which I just spotted. KarlFrei (talk) 07:08, 1 October 2018 (EDT)


 * Indeed, apart from tidying it up a bit (which I've done), I can't see any immediately obvious improvements myself. The "Improve" tag has been removed, and if anyone does have a proof which they think is "better", they are welcome to add it as a second proof. --prime mover (talk) 08:46, 1 October 2018 (EDT)