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

• To: mathgroup at smc.vnet.net
• Subject: [mg71526] Correction re. 1`2 == 1*^-10
• From: "Andrew Moylan" <andrew.j.moylan at gmail.com>
• Date: Wed, 22 Nov 2006 05:22:09 -0500 (EST)

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

• Follow-Ups:
  • Re: Correction re. 1`2 == 1*^-10
    • From: "Chris Chiasson" <chris@chiasson.name>