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: <email@example.com> <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 >