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

*To*: mathgroup at smc.vnet.net*Subject*: [mg75759] Re: How to get sqrt(Year^2)===Year?*From*: dimitris <dimmechan at yahoo.com>*Date*: Thu, 10 May 2007 05:07:41 -0400 (EDT)*References*: <f1s1v4$fip$1@smc.vnet.net>

I guess you are interested in real arguments. Then the simplification is valid only for nonegative arguments and Mathematica of course knows this. In[68]:= oo == Sqrt[oo^2] (Simplify[%, #1[oo, 0]] & ) /@ {Greater, Equal, Less} Out[68]= oo == Sqrt[oo^2] Out[69]= {True, True, False} Dimitris =CF/=C7 Hatto von Aquitanien =DD=E3=F1=E1=F8=E5: > 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? > -- > http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updat= ed > http://911research.wtc7.net > http://vehme.blogspot.com > Virtus Tutissima Cassis