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]
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
>

```

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