1`2 == 1*^-10
- To: mathgroup at smc.vnet.net
- Subject: [mg71525] 1`2 == 1*^-10
- From: "Andrew Moylan" <andrew.j.moylan at gmail.com>
- Date: Wed, 22 Nov 2006 05:22:08 -0500 (EST)
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
- Follow-Ups:
- Re: Re: 1`2 == 1*^-10
- From: "Chris Chiasson" <chris@chiasson.name>
- Re: Re: 1`2 == 1*^-10
- From: "Chris Chiasson" <chris@chiasson.name>
- Re: 1`2 == 1*^-10
- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
- Re: Re: 1`2 == 1*^-10