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MathGroup Archive 2000

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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



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