Very mean squareroot!

[sanctus]

Have you already noticed that

[math]\sqrt{-x}\cdot\sqrt{-y}=\pm\sqrt{xy}[/math]

 

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

 

1)

[math]\sqrt{-x}\cdot\sqrt{-y}=\sqrt{(-1)\cdot x \cdot(-1)\cdot y}= \sqrt{xy}[/math]

Since -1*-1=+1

 

2)

[math]\sqrt{-x}\cdot\sqrt{-y}= i\cdot\sqrt{x}\cdot i\cdot\sqrt{y}=i^2\sqrt{xy}=-\sqrt{xy}[/math]

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]

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]

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

[Qfwfq]

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 [imath]a^2[/imath]? Think carefully of what "square root" actually means.

 

[math]a^2=a\cdot a=-a\cdot -a=(-1)a\cdot(-1)a=(-1)(-1)a\cdot a=(+1)a\cdot a[/math]

 

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

 

[math]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[/math].....:doh:

[freeztar]

[math]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[/math].....:doh:

 

I see your point.

:)

 

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

[modest]

 

(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]

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.

 

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

 

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

 

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]

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

[math](ab)^{\frac12}=(a)^{\frac12}(b)^{\frac12}=(-a\cdot-b)^{\frac12}=(-a)^{\frac12}(-b)^{\frac12}[/math]

 

:evil:

 

A square root of [imath]\alpha[/imath] is a root of the equation [imath]z^2=\alpha[/imath].

[sanctus]

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)

[math] \sqrt{-x}\cdot\sqrt{-y}=\sqrt{(-1)\cdot x \cdot(-1)\cdot y}= \sqrt{1}\cdot\sqrt{xy}=\pm \sqrt{xy}[/math] since [imath]\sqrt{1}=\pm 1[/imath]

 

2)

[math]\sqrt{-x}\cdot\sqrt{-y}= \pm i\cdot\sqrt{x}\cdot \pm i\cdot\sqrt{y}=\mp i^2\sqrt{xy}=\pm\sqrt{xy}[/math]

Since [imath]\sqrt{-1}=\pm i[/imath], but this is strange as it implies that the definition of the imaginary unit is not unique.

[Qfwfq]

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

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

[sanctus]

Why doesn't it matter? I mean it should be astated in a definition if it is not unique (would have saved me half an hour), so to me it would have matered very much!

 

But I agree in practictal things it doesn't seem to matter...usually :)

[Qfwfq]

Why doesn't it matter?

In the strictest sense there is only one imaginary unit and the other is its additive inverse.

 

However, because the constructive definition of the unit doesn't distinguish between them, it's true that you could exchange them; this is an automorphism of [imath]\mathbb{C}[/imath] and, unlike changing sign in [imath]\mathbb{R}[/imath], not just of the additive group but of the actual field. This automorphism is, of course, complex conjugation.

 

In any case, good question. :)