iProblem

[srimukh]

Was just playing around with the equations, when I found something strange.

Suppose there is a number (1-x)^0.5, then pull out a -1 out of it so that it becomes i(x-1)^0.5

Now again, pull out a -1 from the root so that the term inside the root is reverted back to the original,

So, it becomes (i^2)(1-x)^0.5

(1-x)^0.5 => -(1-x)^0.5

 

So, what does it mean? How is it possible to turn a positive real number to a negative real number without following any sort of symmetric algebraic algorithm?

[mooeypoo]

I'm not much of a mathematician, but isn't it implied, whenever there is an even root, that the solution is either positive or negative?

 

As in, sqrt(4) is both -2 and +2... so it's really +/- sqrt(4).

 

So it seems that all you did was pluck out a possible solution out of the available ones anyways.

 

Mathematicians, is this right?

[srimukh]

I'm not much of a mathematician, but isn't it implied, whenever there is an even root, that the solution is either positive or negative?

 

As in, sqrt(4) is both -2 and +2... so it's really +/- sqrt(4).

 

So it seems that all you did was pluck out a possible solution out of the available ones anyways.

 

Mathematicians, is this right?

But why the equality? Roots of 4 aren't equal.

[mooeypoo]

They're both equal to the square root of four.

 

The trick you made doesn't say that one root equals the other. It says that the square root is equal the negative square root. The square root is BOTH negative and positive, so a part of it is also equal to the negative square root.

 

Again, I'm not entirely sure I'm right here, we should ask one of the math experts to join in, probably.

 

 

The only problem I can see with my own reasoning is that if you start your "trick" with +sqrt() instead of just sqrt (hence, you CHOSE the positive result first) then it makes a little less sense. So I'm not entirely sure.

 

I am pretty sure this is the sort of "trick" that lies on the inconsistency of those roots. For that matter, if I follow your logic, I could, theoretically, say that since

[math]\sqrt{4}=\pm 2[/math]

Then I can rewrite to:

[math]\sqrt{4}=-2=2[/math]

Which, since it's an equality, I can get rid of the first part and just write:[math]-2=2[/math] which obviously makes no sense.

 

I *think* the issue arises out of the roots having two possible values but the key word here is *OR*. It's EITHER positive 2 OR negative 2. Not both together. If that makes any sense? Two options for the solution. Either this, or that. Hence, equating the two up once you isolated a particular solution makes no sense.

 

Now, uhm, I'm hoping a mathematician can hop in and explain this whole thing, I'm totally not sure i'm not talking out of my rear orifice.

 

~moo

[Xittenn]

But why the equality? Roots of 4 aren't equal.

 

I didn't quite understand this?

 

Just to talk because I like to do so ....

 

If you have an equation [math] f(x) = x^2 [/math] then the relation [math] F [/math] is a function such that for each [math] x \in dom F [/math] there is only one [math] y [/math] such that [math] xFy [/math].

 

So squaring [math] (-3)^2 = 9 [/math] and also [math] (3)^2 = 9 [/math] and each are ordered pairs such that [math] \left \langle x, F(x) \right \rangle \in F [/math] and as such the function is not single rooted and is not one-to-one or an injection.

 

Because the function F is not single rooted the inverse of the function [math] F [/math] is not a function but a relation as it is a mapping from a single value to multiple values.

 

Because of this finding the root of a number does not find a single valued answer but finds a set of possible answers. In reality it should not be written that [math] F^{-1}(1) = \pm(1) [/math] but should preferentially be [math] F^{ -1 } \left [ \left \{ 1 \right \} \right ] = \left \{ -1, 1 \right \} [/math] and where negation of the set reveals that [math] \left \{ -1, 1 \right \} = - \left \{ -1, 1 \right \} [/math] in fact holds because [math] \left \{ -1, 1 \right \} = \left \{ -1, 1 \right \} [/math] by the Axiom of Extensionality.

 

but again I didn't understand what it was you were saying in your second post as noted above .... :/

 

And \llbracket \rrbracket is not functioning for image brackets, what is the proper latex ??

Edited March 19, 2011 by Xittenn

[the tree]

Guys, the problem is a lot simpler than you're making it out to be.

Suppose there is a number (1-x)^0.5, then pull out a -1 out of it so that it becomes i(x-1)^0.5

This step: [imath]\sqrt{(-1)(x-1)}=\sqrt{(-1)}\sqrt{(x-1)}[/imath] just isn't right.

 

In general [imath]\sqrt{ab}=\sqrt{a}\sqrt{b}[/imath] is only for positive reals: try [imath]a=b=-1[/imath] for instance.

[Shadow]

I made a similar mistake way back.

[the tree]

Yeah, I think it's quite a common mistake as most people have been using [imath]\sqrt{ab}=\sqrt{a}\sqrt{b}[/imath] for long a time before they are introduced to complex numbers.

[DrRocket]

Was just playing around with the equations, when I found something strange.

Suppose there is a number (1-x)^0.5, then pull out a -1 out of it so that it becomes i(x-1)^0.5

Now again, pull out a -1 from the root so that the term inside the root is reverted back to the original,

So, it becomes (i^2)(1-x)^0.5

(1-x)^0.5 => -(1-x)^0.5

 

So, what does it mean? How is it possible to turn a positive real number to a negative real number without following any sort of symmetric algebraic algorithm?

 

The problem arises with what one means by [math]z^{\frac {1}{2}}[/math]. the convention that you learn in high school algebra is that it is the positive square root if [math]z>0[/math].

 

In elementary calculus you learn that [math]x^y = e^{y\ log \ x}[/math] when [math] x>0 [/math]

 

This also works for complex numbers [math]z^y = e^{y\ log \ z}[/math], but one must be very careful with what is meant by [math]log \ z [/math]. [math] log \ z = log (|z|) + i \ Arg (z) [/math] but [math] Arg (z)[/math] is only defined modulo [math]2 \pi [/math] and as one winds around the origin [math]Arg[/math] is either ambiguous or discontinuous. Mathematicians handle this by either "picking a branch of the logarithm' or by working on a Riemann surface instead of the complex plane.

 

In "picking a branch of the logarithm" one deletes from consideration some infinite line segment starting at the origin -- deleting the negative real axis corresponds to the usual convention of elementary algebra and calculus.

 

The net result is that the simple aglebraic rules, like [math] x^a x^b = x^{a+b}[/math], [math](xy)^a = x^ay^a[/math] , [math](x^a)^b = x^{ab}[/math] do not hold in general when dealing with complex numbers.

Edited March 20, 2011 by DrRocket