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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: [mg75909] Re: How to get sqrt(Year^2)===Year?
  • From: Hatto von Aquitanien <abbot at AugiaDives.hre>
  • Date: Sun, 13 May 2007 05:45:21 -0400 (EDT)
  • References: <f1s1v4$fip$1@smc.vnet.net> <f1upol$8hr$1@smc.vnet.net>

dh wrote:

> 
> 
> Hi,
> 
> a square root has actually two values but by choosing a branch cut, the
> 
> function is made unique.

Kronecker was right.

> Therefore, the Sqrt of a real number is by definition positive. 

Hu?  Do you mean the square root of a positive number is by definition real?

> Therefore, Sqrt[x^2] is not equal to x if x<0. 

Wait for the coffee to kick in before you post.
 
> However, if you tell Mathematica that Year is positive you will get what
> you
> 
> want. E.g.
> 
> Simplify[x==Sqrt[x^2],x>0]
> 
> hope this helps, Daniel

Well, it helps in this circumstance.  I believe Convert will do the same,
but I'm not sure if it works in all circumstances.  I do wish there were
some way for Mathematica to know that something such as a physical constant
is real and positive.  Perhaps doing that under all circumstances would
turn into a major performance hit, but it would be nice in circumstances in
which it is inappropriate to consider cases of x < 0, or x not real in
Sqrt[x].
 
> 
> 
> 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?

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