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>