Very mean squareroot!

sanctus · #1 (**permalink**) 01-25-2008, 05:46 AM

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

Qfwfq · #2 (**permalink**) 01-25-2008, 07:16 AM

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).

freeztar · #3 (**permalink**) 01-25-2008, 08:30 AM

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

Qfwfq · #4 (**permalink**) 01-26-2008, 12:45 AM

Quote:

Originally Posted by freeztar

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.

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

freeztar · #5 (**permalink**) 01-26-2008, 01:14 AM

Quote:

Originally Posted by Qfwfq

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

I see your point.

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

modest · #6 (**permalink**) 01-26-2008, 02:56 AM

Quote:

Originally Posted by freeztar

(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.

CraigD · #7 (**permalink**) 01-26-2008, 09:17 AM

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.

Qfwfq · #8 (**permalink**) 01-28-2008, 03:09 AM

Quote:

Originally Posted by CraigD

$sqrt{x}$ 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 $z^2=\alpha$ .

sanctus · #9 (**permalink**) 01-28-2008, 10:39 PM

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.

Qfwfq · #10 (**permalink**) 01-29-2008, 12:23 AM

Quote:

Originally Posted by sanctus

it implies that the definition of the imaginary unit is not unique.

Well, that doesn't really matter, does it?

Hypography?

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