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

*To*: mathgroup at smc.vnet.net*Subject*: [mg71559] Re: [mg71526] Correction re. 1`2 == 1*^-10*From*: "Chris Chiasson" <chris at chiasson.name>*Date*: Thu, 23 Nov 2006 05:41:29 -0500 (EST)*References*: <200611221022.FAA04447@smc.vnet.net>

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/

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