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MathGroup Archive 2007

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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


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