Re: approximation by rationals

*To*: mathgroup at smc.vnet.net*Subject*: [mg23493] Re: [mg23487] approximation by rationals*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>*Date*: Sun, 14 May 2000 16:59:57 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

If you read your own statement of the theorem the answer should become obvious. The theorem quite correctly states that if (a/b) satisfies the inequality then (a/b) is a convergent of z, but it does not state that every convergent of z satisfies the inequality. This is the difference between a necessary condition and a sufficient condition. That's all. It is in fact untrue that the necessary condition (to be a convergent) holds for Pi but not for E. It does not hold for either. Here is a proof that it also does not hold for Pi using Mathematica:: Load in the continued fractions package: In[1]:= << NumberTheory`ContinuedFractions` define the test: In[2]:= test[f_, Rational[c_, d_]] := Abs[f - c/d] < 1/(2d^2) && d >= 1 Let's take the first 30 convergents of Pi: In[3]:= conv = Convergents[Pi, 30]; Let's check how many satisfy the test: In[4]:= Length[conv1 = Select[conv, test[Pi, #] &]] Out[4]= 21 We can see the ones that do not: In[5]:= Complement[conv, conv1] // InputForm Out[5]//InputForm= {3, 333/106, 103993/33102, 208341/66317, 833719/265381, 4272943/1360120, 5706674932067741/1816491048114374, 5371151992734/1709690779483, 14885392687/4738167652} on 5/13/00 11:54 AM, Matt Herman at Henayni at hotmail.com wrote: > > 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