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Re: Re: How to get sqrt(Year^2)===Year?

• To: mathgroup at smc.vnet.net
• Subject: [mg75812] Re: [mg75769] Re: How to get sqrt(Year^2)===Year?
• From: Murray Eisenberg <murray at math.umass.edu>
• Date: Fri, 11 May 2007 05:25:22 -0400 (EDT)
• Organization: Mathematics & Statistics, Univ. of Mass./Amherst
• References: <f1s1v4$fip$1@smc.vnet.net> <200705100912.FAA08020@smc.vnet.net>
• Reply-to: murray at math.umass.edu

"Full" in FullSimplify is unnecessary, and Element[Year, Reals] is 
redundant here:

   Simplify[(Year^2)^(1/2),Year>=0]
Year

Jens-Peer Kuska wrote:
> Hi,
> 
> FullSimplify[Year == (Year^2)^(1/2), Element[Year, Reals] && Year > 0]
> 
> gives True
> 
> Regards
>    Jens
> Hatto von Aquitanien wrote:
>> If I have some expression which takes the square root of a square, it
>> evaluates leaving the whole square root expression unchanged.  
>>
>> Year == (Year^2)^(1/2)
>>
>> just returns what I entered when evaluated.
>>
>> Year == Year
>>
>> evaluates to True.
>>
>> How do I persuade Mathematica to evaluate the first expression completely?
> 

-- 
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305

• Follow-Ups:
  • Re: Re: Re: How to get sqrt(Year^2)===Year?
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>

• References:
  • Re: How to get sqrt(Year^2)===Year?
    • From: Jens-Peer Kuska <kuska@informatik.uni-leipzig.de>