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Topic: Theorem Regarding Rationals
Replies: 4   Last Post: Nov 23, 2000 9:21 PM

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Hugo van der Sanden

Posts: 234
Registered: 12/8/04
Re: Theorem Regarding Rationals (stronger inequality)
Posted: Nov 23, 2000 7:11 AM
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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







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