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

*To*: mathgroup at smc.vnet.net*Subject*: [mg75770] Re: How to get sqrt(Year^2)===Year?*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Thu, 10 May 2007 05:13:30 -0400 (EDT)*Organization*: The Open University, Milton Keynes, UK*References*: <f1s1v4$fip$1@smc.vnet.net>

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? If a >= 0 then sqrt(a^2) = a is always true. Otherwise, the identity may not hold (i.e. sqrt((-1-i)^2) != -1-i). In[1]:= Assuming[Year >= 0, Simplify[Year == (Year^2)^(1/2)]] Out[1]= True Regards, Jean-Marc