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Re: Is there a simple way to transform 1.1 to 11/10?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg92987] Re: Is there a simple way to transform 1.1 to 11/10?
*From*: "Szabolcs HorvÃt" <szhorvat at gmail.com>
*Date*: Wed, 22 Oct 2008 05:37:44 -0400 (EDT)
*References*: <gdkajv$4r2$1@smc.vnet.net> <48FDBF08.5040207@gmail.com>
On Tue, Oct 21, 2008 at 16:52, Alain Cochard
<alain at geophysik.uni-muenchen.de> wrote:
> Szabolcs HorvÃ¡t writes:
> > On Tue, Oct 21, 2008 at 14:49, Alain Cochard
> > <alain at geophysik.uni-muenchen.de> wrote:
> > > Szabolcs Horvat writes:
> > > > Alain Cochard wrote:
> > > > > The obvious
> > > > >
> > > > > In[1]:= x=1.1`Infinity
> > > > >
> > > > > is not syntactically correct.
> > > > >
> > > > > I understand that SetPrecision[1.1,Infinity] does not work either:
> > > > >
> > > > > In[3]:= SetPrecision[1.1,Infinity]
> > > > >
> > > > > 2476979795053773
> > > > > Out[3]= ----------------
> > > > > 2251799813685248
> > > > >
> > > > > In[4]:= N[%,20]
> > > > >
> > > > > Out[4]= 1.1000000000000000888
> > > > >
> > > > > I searched the newsgroup and thought I had the solution with Rationalize:
> > > > >
> > > > > In[5]:= Rationalize[1.1,0]
> > > > >
> > > > > 11
> > > > > Out[5]= --
> > > > > 10
> > > > >
> > > > > But
> > > > >
> > > > > In[9]:= Rationalize[1.000000001,0]
> > > > >
> > > > > 999999918
> > > > > Out[9]= ---------
> > > > > 999999917
> > > > >
> > > > > In[10]:= N[%,20]
> > > > >
> > > > > Out[10]= 1.0000000010000000830
> > > > >
> > > > > So any simple way?
> > > > >
> > > >
> > > > Hello Alain,
> > > >
> > > > Rationalize is the way to go. Floating point numbers are usually stored
> > > > in a binary (not decimal) representation on computers. 1.000000001 is
> > > > not exactly representable in binary (in the same way as 1/3 =
> > > > 0.3333333... is not exactly representable in decimal). Note that in
> > > > your example you start with a MachinePrecision number (approximately 15
> > > > digits), and then convert back to a number with 20 digits of precision.
> > > > If you start with 1.000000001`20 then everything will be fine.
> > > >
> > > > In[1]:= Rationalize[1.000000001`20, 0]
> > > > Out[1]= 1000000001/1000000000
> > > >
> > > > In[2]:= N[%, 20]
> > > > Out[2]= 1.0000000010000000000
> > > >
> > > > Another example:
> > > >
> > > > In[1]:= Rationalize[N[Sqrt[2], 30], 0]
> > > > Out[1]= 1023286908188737/723573111879672
> > > >
> > > > In[2]:= N[%, 30]
> > > > Out[2]= 1.41421356237309504880168872421
> > > >
> > > > In[3]:= % - Sqrt[2]
> > > > Out[3]= 0.*10^-30
> > >
> > > I don't really convert back to a number with 20 digits of precision.
> >
> > You used N[%,20] in your example where % was the rationalization of a
> > MachinePrecision number. If you convert it back to a MachinePrecision
> > number (simply N[%]), then you'll get the same number that you started
> > with, and not a different one.
> >
> > > It was just to convince myself that it was not 1000000001/1000000000
> >
> > Actually 1.000000001 isn't 1000000001/1000000000 either ... see below.
> >
> > > The solution you give will not be fully general (as that by bob) since
> > > you have to adapt the '20' or the '30' for each case. This is a bit
> > > painful for a given number and impossible to do in a program, it seems
> > > to me.
> > >
> >
> > No, you misunderstood me. I did not offer a solution. I just tried
> > to explain why the result returned by Rationalize[1.000000001,0] is
> > correct, and why the "problem" does not exist at all.
>
> Sorry that I misundertood you. Also, I wasn't implying the mma does
> not give the correct result.
>
> There is a problem for me because when I play with mathematical
> numbers like 1.608910743981704391 with an infinite number of zeroes
> afterwards, it is painful to manually count the figures and write it
> as 1608910743981704391/1000000000000000000
>
> > You asked how you can transform 1.1 to 11/10. This is not a
> > precisely defined question. Why do you prefer 11/10? Why not
> > 1100000000000000088/1000000000000000000, for example?
>
> Because I'd say that for pure mathematics, 11/10 is closer to 1.1 than
> 1100000000000000088/1000000000000000000 is...
But in Mathematica (and practically any other computer program) it is
really closer to 1100000000000000088/1000000000000000000. As I said,
it is represented in binary internally. And this particular number
does not have an exact binary representation, just as 1/3 or 1/7 does
not have an exact representation in decimal. So when you write 1.1,
and use the default machine precision, the computer really understands
1.0001100110011001100110011001100110011001100110011010,
which in decimal is exactly
1.100000000000000088817841970012523233890533447265625.
If you increase the precision to 20, and type 1.1`20, you'll get a
more precise (but still not exact) binary representation of 1.1:
1.000110011001100110011001100110011001100110011001100110011001100111
The corresponding decimal is not 1.1, but
1.100000000000000000008131516293641283255055896006524562835693359375.
>
> >
> > [...]
> >
> > What I meant was that you have to decide what precision to use. [...]
>
> I want infinite precision, juste like when you enter 1, Pi, or 11/10.
>
If you want infinite precision, then you have to use integers,
rational, or symbols like Pi. As I already said, floating point
numbers are always treated as inexact numbers, and their precision is
tracked by the system.
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