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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: [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



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