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: <firstname.lastname@example.org>
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:= Assuming[Year >= 0, Simplify[Year == (Year^2)^(1/2)]] Out= True Regards, Jean-Marc