Re: approximation by rationals

*To*: mathgroup at smc.vnet.net*Subject*: [mg23497] Re: [mg23487] approximation by rationals*From*: "Mark Harder" <harderm at ucs.orst.edu>*Date*: Sun, 14 May 2000 17:00:00 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Matt, There is no contradiction here between Mathematica's results and the theorem as you have stated it. Namely, the theorem is only a one-way implication: "conditions on a/b & z imply a/b is a convergent." This leaves room for cases where a/b is aconvergent of z and the conditions are false. What you have found in z=E & a/b=106/39 is a COUNTERexample of the CONVERSE of the theorem, which merely proves that the converse of the theorem is false. -mark -----Original Message----- From: Matt Herman <Henayni at hotmail.com> To: mathgroup at smc.vnet.net Subject: [mg23497] [mg23487] approximation by rationals > >hi, > >there is a number theory theorem which states that if |z-(a/b)|< 1/(2 >b^2), then a/b is a convergent of z. If this is true, then why when I go >into mathematica and set z=E and a/b = 106/39, which is a convergent >of E, the inequality is false? Is this a problem in mathematica's >numerical evaluation, or a problem in the theorem? Is the theorem only >valid when z is a quadratic irrational? Because the theorem works for >z=Pi. Also the stronger form of the theorem is |zb-a|<1/2b^2. This is >false when z=E. Again the same question as above. > > >Matt