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MathGroup Archive 2006

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Re: Re: 1`2 == 1*^-10

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71600] Re: [mg71572] Re: [mg71525] 1`2 == 1*^-10
  • From: "Chris Chiasson" <chris at chiasson.name>
  • Date: Fri, 24 Nov 2006 01:17:20 -0500 (EST)
  • References: <200611221022.FAA04444@smc.vnet.net>

One reason a person may want to work with low precision (or accuracy)
is when the given data is already of low precision (or accuracy). For
instance, I only know how tall I am to within half an inch or so. If I
were doing a calculation involving my height, I would probably enter
it as:

height=6*12+1``0

That's already down to 1.857 digits of precision.

On 11/23/06, Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
> This has come up several times before. I can't imagine why you need
> to consider numbers with two digits of precision but if you really do
> I think the best suggestion can be found here:
>
>
> http://forums.wolfram.com/mathgroup/archive/2006/Feb/msg00108.html
>
> Andrzej Kozlowski
>
>
> On 22 Nov 2006, at 19:22, Andrew Moylan wrote:
>
> > Hi all,
> >
> > Please help me understand the following behaviour, which was wrecking
> > havoc the results I get from calling the function Sort:
> >
> > Evaluating
> >   1`2 == 1*^-10
> > gives
> >   True
> >
> > Correspondingly, evaluating each of
> >   1`2 < 1*^-10
> > and
> >   1`2 > 1*^-10
> > give
> >   False
> >
> > Can anyone explain why these two numbers are declared to be equal?
> > It's
> > inconsistent with my previous understanding of how arbitrary-precision
> > numbers are interpreted in Mathematica.
> >
> > (I've resolved my sorting problem by using OrderedQ instead of Less as
> > the ordering function in Sort. But why was this necessary?)
> >
> > Cheers,
> > Andrew
> >
>
>


-- 
http://chris.chiasson.name/


  • References:
    • 1`2 == 1*^-10
      • From: "Andrew Moylan" <andrew.j.moylan@gmail.com>
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