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

• To: mathgroup at smc.vnet.net
• Subject: [mg75885] Re: [mg75812] Re: [mg75769] Re: How to get sqrt(Year^2)===Year?
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Sat, 12 May 2007 03:17:47 -0400 (EDT)
• References: <f1s1v4$fip$1@smc.vnet.net> <200705100912.FAA08020@smc.vnet.net> <200705110925.FAA05947@smc.vnet.net>

Actually, even Simplify is "stronger" than what is necessary:

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

Andrzej Kozlowski

On 11 May 2007, at 18:25, Murray Eisenberg wrote:

> "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
>

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