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MathGroup Archive 2006

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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71572] Re: [mg71525] 1`2 == 1*^-10
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 23 Nov 2006 05:41:44 -0500 (EST)
  • References: <200611221022.FAA04444@smc.vnet.net>

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
>


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