Solve

1/x = -x.  solve for x.

To stop the foolishness being promoted by >>39 and >>40, I'll change my terms to:

1i¹ⁱ

... since i is just a marker that the numbers we're talking about are imaginary.  Other than their imaginary character, they possess the characteristics of the real numbers.

After all, we avoid the use of other sorts of markers, like the base10 notation (where π ≈ 3.14159₁₀).  When we say i, we really mean 1i.

Also, why do we so duly accept the combined terminology of xi+y?  That's like adding a line and an area.  They're just not in the same classification, so to speak.  (Pardon my amateur use of terminology.)  I think that the xi+y notation is better represented as a vector, having two parts that are not directly related to each other, yet together do properly describe one object.

>>40
i = √-1 = -1^0.5

>>39
"42" is an algebraic symbol in an equation, not a number.

>>41
I think you mean (-1)^0.5.

-1^0.5 = -(1^0.5) = -1

>>39
>>40
you be trollin
you gonna get raped

>>39
complex numbers are an unordered field

>>37

why does negative area imply negative dimension?

That just doesn't follow. Dimension is really defined, and this is very loose here, as the amount of information needed to signify the structure. Areas require two "linearly independant" elements to make any sense.

You ever do any linear algebra?

The complex plane has dimension two, surely you can agree to that?

The elements 1 and i are linearly indepedent and spanning.

The real number line has dimension one.

So surely the imaginary number line, by your same argument has dimension 1 as well.


The complex plane is merely a way to visualise algebra on ordered pairs of numbers with multiplication and addition defined in a certain way it doesn't "signify" anything in the real world. As I said before, if you want to do any even quasi-serious maths you need to stop trying to anchor stuff in experience and deal with the abstract.

>>41
it pretty much is a vector, much like
(1, 2, 3) being the same vector as i1 + 2j + 3k
where i, j and k are the unit vectors in the x, y, z directions

>>46
lol, quaternions.

>>45
It's not a negative area.  It's just an area on the imaginary Cartesian plane.  Areas have a dimensionality of 2.  The Cartesian plane model assigns numbers to a dimensionality of 1.  Since those numbers are based upon i, and the areas are real, then it only stands to reason that that relation could mean that imaginary numbers have a dimensionality of -1.

As for linear algebra, my formal math education stopped before that point.  So I follow along with discussions of algebra, the calculus, and statistics, but little beyond that range.

As for your use of the "complex plane" ... you seem to be using another model other than the imaginary Cartesian one I used.  I'm not talking about xi+y.  I'm talking about a Cartesian plane where BOTH axes are based upon i.  In the real world, the Cartesian plane used BOTH axes as real numbers, so that seemed a valid move.  And then I discovered that on such a plane, real numbers became AREAS, compared to the imaginary ones which were LINES.

Unless you can point out how forming a Cartesian plane with imaginary numbers along BOTH axes is somehow invalid, I think my point stands about the relation I found.

>>48
not even a vector space
i*i=-1 which is real and hence not an element of your plane
hence your plane does not satisfy one of the requirements of a vector space
also no multiplicative inverse

>>47
not really trying to go there

>>49
Too late. You've awoken the spirit of W.R. Hamilton.

>>37
Interesting, but I don't quite follow.  Don't real numbers have dimensionality 1?

Your visualization with a black hole is quite interesting, but I should point out that "real" black holes require at least 5 dimensions for proper embedding.

>>39
The set of complex numbers do not form an ordered field.  You cannot place bounds on i.

>>41
They are indeed often represented as a vector, with the imaginary axis orthogonal to the real axis.

If you really want, you can say xi + y1, where "1" is the real multiplicative identity.

>>49
RedCream's definition satisfy the requirements of a vector space.  The whole thing may not form a field, but it is still a vector space.  The "i*i = -1" example only demonstrates a bad way to define an inner product space.  The set of continuous functions form a vector space, and the inner product of any 2 functions results in a real number, which is not in the original vector space.  Similarly, the notion of multiplicative inverse is moot.

>>51
you're right
forgot that vector spaces only need scalar multiplication
back to linear algebra I go

>>49
i²=-1 is a straightforward geometric operation on the imaginary Cartesian plane as I gave it.  Why is such an operation invalid?

Remember, the plane is defined only by two axes, and each axis is defined as a +/- rational range of i.  This naturally results in the minimum unit of a line (a dimension 1 number).  But it also results in another unit of dimension 2.  Are we supposed to ignore areas on the Cartesian plane?  We used those things in Calculus all the time with no such basic conflict.

Sorry, but I can't see where I've gone wrong with this construction.  If i is truly just an imaginary form of a number line, then all other constructions of it should behave in the same way.

>>53
like I said above, back to linear algebra I go

>>53

You're not understanding what dimension means at all, I think that's your problem.

Just as you say "real numbers correspond to areas" in your complex plane, real numbers correspond to area in the normal cartesian plane, just you have negative areas instead of positive, and that pretty trivially happens because of you multiplying each axis by sqrt(-1).

You're calling an area something of dimension two, but it's not got dimension two if you just specify the magnitude of the area, it's like a vector, it's magnitude still has dimension one.

I can't really think of a clearer way to tell you that you're wrong.

>>55
OK, I can see that there's a fundamental confusion of terms at the heart of my prior understanding.

Still, am I to conclude that there's nothing particularly significant in making a geometrical construct as I gave it?  We seem to have come up with the concept of i since otherwise we'd just have to stare at the impossibility of "what's the square root of a negative number".  The concept of i serves a lot of purpose, and in doing so, seems to have all the validity of the real number system we're based in.  So when I extend a geometrical argument, what breaks?

Take this for instance:  You have a physical object in the i-verse, and it's measured linearly in units of i.  But this object seems to have real numbers wherever there's an area on it.  What's wrong with reaching that sort of conclusion?  Did I err in even supposing an i-verse?

>>56
Yes.

>>56
I think imaginary numbers only make sense when used together with the reals. After all, the complex numbers are defined as an completion of the reals.

>>58
Well, not necessarily. Pure imaginary numbers are a set in their own right. Not to be insulting, but saying that is like saying irrationals only make sense when used with rationals since, together, they form the real numbers.

>>59
Is the set of purely imaginary numbers useful for anything? I guess it's basically just an isomorphism of the real line, isn't it?

>>56
I don't see anything wrong with your assessment, though I'm not so sure about your claim of "negative dimension".

A use for your system would be another way of imagining flat Minkowski space.  The 3 spatial dimensions are as usual, but the time dimension is measured in imaginary units.  In such a case, we can use the usual pythagorean theorem to calculate "4 distances".  A real distance implies causal disconnection, and an imaginary distance implies one event happened before another (0 distance are events that are light like separated).

In mathematics, you can assume pretty much anything you want.  If your assumptions lead to a contradiction, then everything falls apart.  Until it reaches a contradiction, we can't know for sure.

>>58
I've heard of the reals being a completion of the rationals, but not of the complex numbers being a completion of the reals.  I guess we can call it "algebraic completion"?

>>60
By themselves, no.  It's sort of an isomorphism.  The problem is that this set is not closed under multiplication, so it's not a field.

>>56

It obviously has real numbers for an area. In the normal cartesian plane we have real numbers for areas, you've just mapped everything from this plane, to a plane where the axis have been multplied through by i.

It trivially follows that areas in this plane, are just the additive inverse of the areas in the other plane.

Do you know what a jacobian is?

You've basically described a transformation x->u=ix and y->v=iy

the jacobian of this transformation is what areas are scaled by and is the determinant of the jacobian matrix

( du/dx  dv/dx)
(             )  at least I hope I'm remembering this shit right 
(du/dy   dv/dy)

anyway, that's pretty obviously  i*i=-1.

Hence the negative area thing makes perfect mathematical sense.

Again, not sure how more succinctly to describe this.

>>63
Excellent.  Well said.