- #1
anb
- 3
- 0
A long time ago I took a number theory course and really enjoyed it. At one point we were shown the proof for the theorem that a number is rational if and only if it has a periodic decimal expansion. The (<=) direction is really easy if you know some Calculus, but I remember the (=>) direction having some complicated proof that I never really understood. We were never examined on it so it wasn't a big deal.
Anyways, I thought of another proof that I'm pretty sure is rigorous and more than anything I want to know if anyone has thought of it before - as far as I can tell from searching on the internet it hasn't. I'm not looking for some million dollar prize or something, as the proof isn't exactly novel, but basically it involves proving that every number with a periodic decimal expansion (not *eventually* periodic, but periodic all the way like 0.abcdefabcdef..., etc.) can be expressed as a rational number with denominator equal to 9[tex]\sum[/tex](10^n) (where the series is going from n=0 to some natural number N - sorry I don't know much about using Latex to make pretty math symbols work). In less formal terms, a number with a periodic decimal expansion can always be expressed as some rational number A/999...9 (for "N+1" number of nines in the denominator). If this seems confusing to anyone who knows a lot about number theory I could post a full proof and ask if they've seen it before, but if people have already seen it before then I'd like to know that too.
Anyways, I thought of another proof that I'm pretty sure is rigorous and more than anything I want to know if anyone has thought of it before - as far as I can tell from searching on the internet it hasn't. I'm not looking for some million dollar prize or something, as the proof isn't exactly novel, but basically it involves proving that every number with a periodic decimal expansion (not *eventually* periodic, but periodic all the way like 0.abcdefabcdef..., etc.) can be expressed as a rational number with denominator equal to 9[tex]\sum[/tex](10^n) (where the series is going from n=0 to some natural number N - sorry I don't know much about using Latex to make pretty math symbols work). In less formal terms, a number with a periodic decimal expansion can always be expressed as some rational number A/999...9 (for "N+1" number of nines in the denominator). If this seems confusing to anyone who knows a lot about number theory I could post a full proof and ask if they've seen it before, but if people have already seen it before then I'd like to know that too.