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Re: with machine-precision input

  • To: mathgroup at
  • Subject: [mg128403] Re: with machine-precision input
  • From: Bill Rowe <readnews at>
  • Date: Tue, 16 Oct 2012 03:22:56 -0400 (EDT)
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On 10/14/12 at 11:40 PM, fateman at (Richard Fateman)

>try this on Mathematica 8:

>q = 429515858585961022071539/6922263581864661506963;

>Note that q was chosen so that
>Rationalize[N[q,45],0] == q.

Not here:

In[1]:= q = 429515858585961022071539/6922263581864661506963;
Rationalize[N[q, 45], 0] == q

Out[2]= False


In[4]:= Rationalize[N[q, 45]] == q

Out[4]= True

>Peculiarly, then SetPrecision[N[q,45],100]-N[q,100] is not zero but
>about -3.9E-59

Why would you expect this to be zero? Per the documentation:

SetPrecision will first expose any hidden extra digits in the
internal binary representation of a number, and, only after
these are exhausted, add trailing zeros.

So unless q can be exactly expressed with 45 digits there will
be a non-zero difference. Here:

In[4]:= SetPrecision[N[q, 45], 100] - N[q, 100]

Out[4]= 3.3475286706596422492*10^-78

but note:

In[5]:= SetPrecision[N[q, 45], 100] - N[q, 45]

Out[5]= 0.*10^-44

as it should.

And here is:

In[6]:= $Version

Out[6]= 8.0 for Mac OS X x86 (64-bit) (October 5, 2011)

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