RE: Re: Correction re. 1`2 == 1*^-10

• To: mathgroup at smc.vnet.net
• Subject: [mg71587] RE: [mg71559] Re: [mg71526] Correction re. 1`2 == 1*^-10
• From: "Erickson Paul-CPTP18" <Paul.Erickson at Motorola.com>
• Date: Fri, 24 Nov 2006 01:17:07 -0500 (EST)

```Well since 1`2 == 0 is true as well - and 1`2 == 2 as well (implying 0 == 2, I guess  ;-)  ), I'd focus on the "`2" accuracy. It appears that "1`2 == x ; -.28 ? x ? 2.28" or 1+/-1.28. Not quiet sure what the significance is of 1.28 as an allowed approximation range, but ... Would be interesting to check the true range of `3, `4, etc. accuracy and see if there is a underlying reason.

Paul

-----Original Message-----
From: Chris Chiasson [mailto:chris at chiasson.name]
Sent: Thursday, November 23, 2006 4:41 AM
To: mathgroup at smc.vnet.net
Subject: [mg71587] [mg71559] Re: [mg71526] Correction re. 1`2 == 1*^-10

There are many very strange effects when using low precision (or
accuracy) numbers in Mathematica. I don't have a way to explain this one.

On 11/22/06, Andrew Moylan <andrew.j.moylan at gmail.com> wrote:
> In my original message (below), I wrote "I've resolved my sorting
> problem by using OrderedQ instead of Less as the ordering function in
> Sort". Instead of "OrderedQ" I should have written
> "OrderedQ[{SetPrecision[#1, Infinity], SetPrecision[#2, Infinity]}] &".
>
> I previously wrote:
>
> Hi all,
>
> 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
>
> Cheers,
> Andrew
>
>

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

```

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