Re: Re: Re: How to get sqrt(Year^2)===Year?
Actually, even Simplify is "stronger" than what is necessary:
Refine[(Year^2)^(1/2), Year >= 0]
On 11 May 2007, at 18:25, Murray Eisenberg wrote:
> "Full" in FullSimplify is unnecessary, and Element[Year, Reals] is
> redundant here:
> Jens-Peer Kuska wrote:
>> FullSimplify[Year == (Year^2)^(1/2), Element[Year, Reals] && Year
>> > 0]
>> gives True
>> 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
> 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
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