
Re: Theorem Regarding Rationals (stronger inequality)
Posted:
Nov 23, 2000 7:11 AM


Leroy Quet wrote: > > I, Leroy Quet wrote: > :It wasn't to hard to prove (by induction) that > > :for all positive rationals r, > :q(r) > s(r); > > :where, if r = m/n, GCD(m,n) = 1; q(r) = m + n; > :s(r) = sum of terms in continued fraction of r. > > Let k(r) = max(m,n). > Let j(r) = min(m,n). > (Again, r = m/n, GCD(m,n) = 1.) > > Then, for all positive rationals, r, > s(r) <= k(r). > > This is again easily proved by induction. > > Is j(r) <= s(r) > for all positive rationals, r?
No: j(7/9) = 7 > s(7/9) = 6. In general, also, s(F_n/F_{n+1}) = n, k(r) > F_n => s(r) >= n.
Hugo

