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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