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, > > 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/