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Old 01-25-2008   #1 (permalink)
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Very mean squareroot!

Have you already noticed that
\sqrt{-x}\cdot\sqrt{-y}=\pm\sqrt{xy}

Thing is that for half an hour I searched which of the two reasoponings is wrong:

1)
\sqrt{-x}\cdot\sqrt{-y}=\sqrt{(-1)\cdot x \cdot(-1)\cdot y}= \sqrt{xy}
Since -1*-1=+1

2)
\sqrt{-x}\cdot\sqrt{-y}= i\cdot\sqrt{x}\cdot i\cdot\sqrt{y}=i^2\sqrt{xy}=-\sqrt{xy}
Since i^2=-1

It took me as already said more than 3 min to figure it out, hope to you it isn't obvious at first read why both versions are right (and incomplete)...just so that I don't feel too stupid


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Last edited by sanctus; 01-28-2008 at 11:37 PM..
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Old 01-25-2008   #2 (permalink)
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Lightbulb Re: Very mean squareroot!

I'm just replying on the fly but I'd say it's a matter of considering both roots of a number vs. only considering the principal root (of a positive real).


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Old 01-25-2008   #3 (permalink)
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Re: Very mean squareroot!

That is a good question. I'd be more inclined to go with the second scenario you presented.


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Old 01-26-2008   #4 (permalink)
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Lightbulb Re: Very mean squareroot!

Quote:
Originally Posted by freeztar View Post
I'd be more inclined to go with the second scenario...
I'd be more inclined to confirm my on-the-fly reply.

What is the square root of a^2? Think carefully of what "square root" actually means.

a^2=a\cdot a=-a\cdot -a=(-1)a\cdot(-1)a=(-1)(-1)a\cdot a=(+1)a\cdot a

Try taking "the" square root of each one and pondering it over.

a^4=a\cdot a\cdot a\cdot a=(-1)a\cdot(-1)a\cdot(-1)a\cdot(-1)a=a\cdot a\cdot(-1)a\cdot(-1)a.....


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Old 01-26-2008   #5 (permalink)
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Re: Very mean squareroot!

Quote:
Originally Posted by Qfwfq View Post
a^4=a cdot a cdot a cdot a=(-1)a cdot(-1)a cdot(-1)a cdot(-1)a=a cdot a cdot(-1)a cdot(-1)a.....
I see your point.


(what's up with the quoted [math]?)


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Last edited by freeztar; 01-26-2008 at 02:20 AM.. Reason: latex issue
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Old 01-26-2008   #6 (permalink)
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Re: Very mean squareroot!

Quote:
Originally Posted by freeztar View Post

(what's up with the quoted [math]?)
I've noticed that as well, it strips the backslashes. We would have to ask Alex what's up.

-modest

PS - I think latex is handled beautifully on hypography and I realize how much work would have to go into that.


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Old 01-26-2008   #7 (permalink)
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Post Arithmatic expression vs. set description connotation of the square root

I think we’ve had essentially this same discussion before.

The root of the dilemma, I think, is that the square root glyph can be interpreted several ways.

\sqrt{x} can be considered simple equivalent to x^{\frac12}, e^{\frac12 \ln x}, or various infinite series expressions, in which case its sign is positive.

Or, \sqrt{x} can be considered the description of a set, \{ a : x = a \cdot a \}, which is equivalent to \{ -\sqrt{x}, \sqrt{x} \}.

As long as writer and reader are aware of which connotation is appropriate for the use to which the glyph is being put, there should be no problem.


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Old 01-28-2008   #8 (permalink)
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Re: Arithmatic expression vs. set description connotation of the square root

Quote:
Originally Posted by CraigD View Post
sqrt{x} can be considered simple equivalent to x^{frac12}, e^{frac12 ln x}, or various infinite series expressions, in which case its sign is positive.
(ab)^{\frac12}=(a)^{\frac12}(b)^{\frac12}=(-a\cdot-b)^{\frac12}=(-a)^{\frac12}(-b)^{\frac12}



A square root of \alpha is a root of the equation z^2=\alpha.


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Old 01-28-2008   #9 (permalink)
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Re: Arithmatic expression vs. set description connotation of the square root

First of all it took me 30min and not 3min to figure it out.
Second I agree with you, the way I saw it is:
1)
\sqrt{-x}\cdot\sqrt{-y}=\sqrt{(-1)\cdot x \cdot(-1)\cdot y}= \sqrt{1}\cdot\sqrt{xy}=\pm \sqrt{xy} since \sqrt{1}=\pm 1

2)
\sqrt{-x}\cdot\sqrt{-y}= \pm i\cdot\sqrt{x}\cdot \pm i\cdot\sqrt{y}=\mp i^2\sqrt{xy}=\pm\sqrt{xy}
Since \sqrt{-1}=\pm i, but this is strange as it implies that the definition of the imaginary unit is not unique.


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Old 01-29-2008   #10 (permalink)
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Re: Arithmatic expression vs. set description connotation of the square root

Quote:
Originally Posted by sanctus View Post
it implies that the definition of the imaginary unit is not unique.
Well, that doesn't really matter, does it?


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