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Solve

Name: Anonymous 2008-03-19 0:10

1/x = -x.  solve for x.

Name: 4tran 2008-03-19 0:50

i, -i

/thread

Name: Anonymous 2008-03-19 0:55

x^2+1 > 0 for all x in R

Name: Anonymous 2008-03-19 7:53

>>1
Impossible since it involves -, therefore i, -i.

>>3
what?

Name: Anonymous 2008-03-19 16:13

>>2
i is a cop out. Real mathematicians will tell you this equation is unsolvable.

Name: 4tran 2008-03-19 16:30

>>5
Complex analysis.

Name: Anonymous 2008-03-19 18:28

>>5
GBT middle school

Name: Anonymous 2008-03-19 19:22

-1, lol!

Name: Anonymous 2008-03-20 1:21

>>8
Fail, hard
>>7
lurk moar

Name: Anonymous 2008-03-20 1:54

-1*-1=1

Name: Anonymous 2008-03-20 2:27

>>10
yes, so?

Name: Anonymous 2008-03-20 2:40

>>9

((1)/(-1)) = ((-1)/(1)) = -1, lol!

Name: Anonymous 2008-03-20 2:43

oh, wait.  How did I not see the - in -x this whole time?  I must need reading glasses.  +-i it is, then.

Name: Anonymous 2008-03-20 2:53

>>13
, lol!

Name: Anonymous 2008-03-20 5:09

>>11
x=-1
-x=1

Name: Anonymous 2008-03-20 8:14

>>13
Give an exact figure for "i".

Name: Anonymous 2008-03-20 9:00

>>16
i is such that 0i = 1.

Name: Anonymous 2008-03-20 9:40

>>16
i*i = -1
don't be a faggot

Name: Anonymous 2008-03-20 15:19

>>16
here:
i

Name: RedCream 2008-03-21 1:53

What could ii possibly signify?

Name: 4tran 2008-03-21 2:11

>>20
= (ei*pi/2)i = e-pi/2

Name: Anonymous 2008-03-21 6:49

>>18
ok, calculate (-1)^0.5

Name: RedCream 2008-03-21 12:05

>>21
I know what it might equal, but I'm more interested in what ii can SIGNIFY.  What does it signify to take an imaginary number (essentially, a negative dimension) and then take it to an imaginary power?

Name: Anonymous 2008-03-21 15:22

>>23


The powers of real and imaginary numbers are generally defined in terms of the exponential function.

things like 2^3 have obvious meanings, but a^b for arbitrary a and b in the reals the meanings not so clear.

So using the exponential function which is quite easy to define in terms of real numbers, and it's inverse the logarithm we say that.

a^b = exp(b*log(a))

which has a definite expression in terms of real power series that we know converge (Given certain restrictions on a)


This can then be extended quite easily for complex a and b, given that we first extend our definitions of the complex exponential and logarithmic functions.

Make sense?

Name: Anonymous 2008-03-21 21:34

>>23

it works, it's consistent, it's beautiful.  gb2 high school

Name: Anonymous 2008-03-21 22:09

>>24
Wow.  Just wow.  lrn2mathematics

Name: 4tran 2008-03-21 23:58

>>23
What do you mean by "signify"?  23 = 2*2*2?

2sqrt(3) = pwnt... the best you can do is define it as limit{2x, x -> sqrt(3)}; have fun taking the 1.7 billionth power of the billionth root of 2.

In complex analysis (at least this crappy book by Brown & Churchill),
log(i) = {i*pi/2 + 2*pi*i*n | n is integer}
Log(i) = i*pi/2 (principle value)

If you notice in >>24's definition, it allows ii to be an infinite set, namely {e-pi(2n + (1/2)) | n is integer}.

Name: Anonymous 2008-03-22 0:06

>>25
shut up you queen, go to a board for people who already know everything

Name: Anonymous 2008-03-22 15:46

calculate (-1)^0.5 or you lose and are an idiot, sorry

Name: Anonymous 2008-03-22 16:15

>>29
i, faggot
btw i'm imagining plowing you in the ass right now

Name: Anonymous 2008-03-22 16:54

>>26


lrn 2 mathematics?
you trolling nigger? It makes perfect sense.

I'm sure there are other, equivalent ways of defining that shit, but this ways the way I was taught and it seems to follow quite simply from basic concepts.

Name: RedCream 2008-03-23 1:28

>>27
For me, "signify" means "forms a more comprehensive symbol".  It's an appeal to make things less abstract.  Just pushing the symbols around is not enough; in order to become functional in math, I've found it necessary to find what things signify.

I've already signified i as a "negative dimension".  Giving that definition to it other than "it's an imaginary number" simply has more significance in my mind.

Hence, when I take this negative dimension and take it to the power of a negative dimension, then what "more comprehensible" (even, "more commonly experienced") thing does that signify?

Taking any real number to the power of another rational number isn't a particularly good model, since now I'm dealing with the imaginary numbers.  They could have other properties.

Hey, I was just curious.  Hence, I wondered what other 4channers thought about it.

Name: Anonymous 2008-03-23 3:50

>>30
i is just an algebraic symbol like x and y in graphs. It's not a number. Give me the number.

Name: 4tran 2008-03-23 6:00

>>33
x, y are generally variables (ie not always the same value)
i is always the same constant (ie always the same value) (up to isomorphism with -i)

i is an imaginary number

If you're thinking about decimal expansions, then phail, since that can only represent finite real numbers.

>>32
I've generally visualized things according to their algebraic properties (ie pushing symbols around), so I can't help you with that.

Even 4 dimensions is incredibly difficult to visualize; I'm not sure how you manage to visualize negative dimensions.  Do describe.

My example was not intended to demonstrate the properties of exponentiation, but rather to demonstrate that even in a relatively simple case, intuition falls apart.  If you can find a less abstract way to visualize the sqrt(3)th power of a positive real number, I'd love to hear it.

Name: Anonymous 2008-03-23 13:01

>>33
no, you cunt
i is a complex number
this is starting to piss me off

if you're serious, i'm going to hunt you down and ram a math book up your ass.

if you're trolling, i'm going to hunt you down and rape you

so, which is it?

Name: Anonymous 2008-03-23 13:32

>>32

I've already signified i as a "negative dimension".

what?

Why not just deal with abstract concepts, it's not that hard.

Name: RedCream 2008-03-23 14:54

>>34
I discovered that i represented a negative dimensioning.  Construct a Cartesian plane, but use i for the axes.  You'll notice soon that AREAS on that plane are real numbers (i.e. the area between 0 and y +1i and x +1i is -1).

Since on the i plane (constructed of dimensionality 1 numbers), real numbers are areas (dimensionality 2), then our real numbers just sitting alone (dimensionality 0) may sit in a dimensionality number just 1 above what the i numbers do.  Hence, a dimensionality of 0-1, hence -1, hence a "negative dimension".

Thinking about this some more, I concluded that a good way to visualize this is with a black hole or other puncture through our universal fabric.  Negative dimensionality seems to represent a universe beneath or beyond ours ... and in that universe, you can use i in the same algebraic fashion as our 0-dimension numbers are.

However, I find myself on unstable ground with all this amateur crap, and when I posted about ii, I was unable to come up with a visualization.  Hence, I feel I can't signify it.

Name: Anonymous 2008-03-23 15:58

>>33
Define number, whore.

Name: Anonymous 2008-03-23 16:30

>>34
If it is a constant then you should be able to give me a figure. Even though we do not know the exact figure for Pi we know it is between 3.14159265358 and 3.14159265357. You must do the same for i.

>>35
If complex numbers are not numbers. "i" is an algebraic symbol in an equation, not a number.

Name: Anonymous 2008-03-23 16:30

>>34
>>35
Calculate i.

Name: RedCream 2008-03-23 17:48

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 = -10.5

Name: Anonymous 2008-03-23 18:00

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

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

-10.5 = -(10.5) = -1

Name: Anonymous 2008-03-23 18:57

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

Name: Anonymous 2008-03-23 18:58

>>39
complex numbers are an unordered field

Name: Anonymous 2008-03-23 19:47

>>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.

Name: Anonymous 2008-03-23 21:57

>>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

Name: Anonymous 2008-03-23 21:58

>>46
lol, quaternions.

Name: RedCream 2008-03-23 22:15

>>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.

Name: Anonymous 2008-03-23 22:58

>>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

Name: Anonymous 2008-03-23 23:32

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

Name: 4tran 2008-03-24 2:12

>>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.

Name: Anonymous 2008-03-24 2:30

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

Name: RedCream 2008-03-24 2:59

>>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.

Name: Anonymous 2008-03-24 10:34

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

Name: Anonymous 2008-03-24 11:00

>>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.

Name: RedCream 2008-03-24 19:46

>>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?

Name: Anonymous 2008-03-24 22:03

>>56
Yes.

Name: Anonymous 2008-03-24 23:41

>>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.

Name: Anonymous 2008-03-25 0:04

>>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.

Name: Anonymous 2008-03-25 2:29

>>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?

Name: 4tran 2008-03-25 2:32

>>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"?

Name: 4tran 2008-03-25 2:34

>>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.

Name: Anonymous 2008-03-25 11:33

>>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.

Name: 4tran 2008-03-25 18:50

>>63
Excellent.  Well said.
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