Chaos's Math Help Center

[Shivertongue he/him]

So, in short, the answer is magic.

[Eerongal he/him]

Actually, you aren't doing it right. As I show in my post, you cannot distribute square roots through multiplication in the general case. You must convert the numbers to polar form before applying the exponents. If you could do what you are suggesting, Eero, I could:

You sure? Everywhere i look up says you can.

for example: here and here and here and i could go on, but you get my point

Edit: am i maybe thinking of a rule that applies to only positive numbers or something?

[Chaos he/him]

You sure? Everywhere i look up says you can.

for example: here and here and here and i could go on, but you get my point

Edit: am i maybe thinking of a rule that applies to only positive numbers or something?

Yes, it only applies in the positive real case. It's certainly true with positive numbers, but I'm talking about the general complex case, where it is not always true. If it were, you could do this:

1 = √(-1*-1) = √(-1)√(-1) = i * i = -1.

That equality, by the way, is why mathematicians way back when didn't like using i, because it gave those nonsense answers. You can resolve the ambiguity, but you have to resort to different methods.

EDIT: In your third link, it does specify the requirement that they need to be positive. Obviously, for most people's purposes, you just use the real case, with no need for this complex arithmetic But in general, yeah, you need to go to the complex plane.

Edited March 25, 2012 by Chaos

[Alliare she/her]

Chaos, you got really upvoted because of this

I LOVE maths, and you're really good at explaining.

However, there's something that I would like to know more about. Could you explain this, please?

The reason for this is that exponents are slightly odder in the complex world' date=' because we multiply differently there (in fact, remember when I said "i is a shorthand?" It's a shorthand for complex multiplication, because it turns out there's only one way you can multiply pairs of numbers that give us a closed operation. But I digress)

[/quote']

I'm curious about the complex world, and why and how is multiplication (and, by virtue of that, exponentiation) different there. And about what 'i' really is, when you say it's a shorthand.

Edited March 25, 2012 by Alliare

[Chaos he/him]

Chaos, you got really upvoted because of this

I LOVE maths, and you're really good at explaining.

However, there's something that I would like to know more about. Could you explain this, please?

I'm curious about the complex world, and why and how is multiplication (and, by virtue of that, exponentiation) different there. And about what 'i' really is, when you say it's a shorthand.

Thanks! That means a lot Since I am going to be a graduate student in the fall, and I'll need to teach precalculus...

Your question is a very good one. To explain it, I need to talk to you a little bit about abstract algebra (which Joe here loves, so he may correct me). "Abstract algebra" sounds pretty scary, but really, at its core, we look at the properties of familiar number system--like, say, the real numbers--and we ask, "Hey, I wonder if these properties hold for different, weirder sets." An example of properties that we like would be the distributive property.

a(b+c) = ab + bc

There are a lot of properties that the integers, real numbers, etc. that have properties like this that we just take for granted. In the language of abstract algebra, those a,b, and c's don't have to be numbers. They could be a set. They could be a lot of weird things. We essentially "abstract" out what the elements of the space (integers, reals, complex numbers, sets, other things) and we say, "Well, if we define addition and multiplication appropriately, hey! Lookit that! The distributive property holds." In an abstract sense, other, weird monkey-butt objects might still have properties that we really like. So we give fancy names to things that have properties we like, like groups, rings, and fields, and then abstract algebra is about finding theorems about all groups.

Now, here's the important thing when we do this: I can't just give you a list of numbers. I also have to defined an operation to act on these numbers. Like, say, addition. Because, obviously, I can define addition a lot of stupid ways. In the language of abstract algebra, let's see how I can define a new operation, which I'll call + rather than +, because the former is our new addition. Let's say our set of numbers we are applying this operation to is the integers.

a + b = a + b.

Okay, this is dumb--I've just defined my "new" abstract operation to be the familiar addition. So for any two a and b in our set, well. You just freaking add them. But I don't have to defined them this way. You might be thinking, "Wow, we're going to define a new addition! That's going to be tough," but you have to realize mathematicians their stupidly pathological examples Take this addition, also over integers.

a + b = a + b + 1

It's a new operation, really. I'm going to add the two numbers together, and add 1 to them. I could do wackier things.

a + b = a^b

Whoa, this is addition? Well, sure, + is just a name for an operation. It does "something" on two numbers. That something could be pretty weird. (Sorry about my previous post, I didn't know I could do superscripts)

How we define our operations makes a big difference if we get things like, the distributive property to hold. In fact, for the distributive property, I also need to define a second operation, "multiplication". Let's make a multiplication right now!

a x b = 2ab

So, the regular multiplication, multiplied by 2. Do you feel powerful, defining a new freaking multiplication? Well, you can certainly write things down. As we can see, you can define a lot of operations. But in abstract algebra, we don't want just any operation; we want the operations that actually give useful properties.

Let's check the distributive property for this multiplication and the last addition that I mentioned. I bet you it's not going to work at all.

a x (b + c)

Note that I'm being very careful differentiating between my abstract operation and our "normal" operations. In this form, the distributive property would read like this: a x (b + c) = a x b + a x c. It's exactly the same as a(b+c)=ab+ac, but I'm using it with different addition and multiplication. In actual abstract algebra books, though, you have to realize that they will just the + sign down and you have to remember "Hey, this is the abstract operation". Mathematicians are lazy.

a x (b + c) = a x (b^c)

There, I did my weird addition.

Now, actually, I defined these operations on the integers. My operations don't apply on anything that isn't an integer. So I need to check that b^c is also an integer. In fact, typically the distributive property is the last property we prove for "rings"--one of the earlier properties is closure, which means that for a and b in our set, a + b is also in that set.

It's easy to see that for any integers, b^c needs not be an integer. It's certainly true in the positive case, 2^5 is sure an integer, and you won't get any decimals out. But what if I choose c to be a negative number? 2^(-1). Both are integers, and we get out 1/2, if you recall your exponent rules about negative exponents. 1/2 is certainly not an integer.

So really, without closure we can't really do too much. So I'm going to change "space" that we are working on to be the positive integers (0,1,2,3...). This way, our "distributive property check" is actually defined (I kind of like things to be defined, don't you?) But, we'll show the distributive property still fails.

a x (b^c) = 2ab^c

There, I applied our multiplication. Now, for the distributive property to hold, we need a x (b + c) = a x b + a x c. So we what we want is (2ab)^2ac

I think it's pretty clear that for that these weird-chull operations, there's no way I can doctor it so 2ab^c actually equals (2ab)^2ac.

So the distributive property fails. This means that, technically speaking, these operations don't form a "ring", but that's not really important that you know what a ring is here

Now, why did I go through this excruciatingly long exercise, instead of writing my novel like I'm supposed to today? To simply show you that there's a lot of ways to define addition and multiplication. We want to pick the "good" additions and multiplications so that we get the properties we want.

Let's finally talk about some freaking complex numbers.

Complex numbers, a + ib where a and b are real numbers, are really just a set of two real numbers. They are ordered pairs. If you've had any matrices, we can write a + ib as a row (or column) vector as so: (a, . There, that's a vector. Ordered pairs are always vectors. Mathematicians like calling this R² (instead of just "the real numbers", which they just call R), because hey, we have two dimensions of real numbers. I don't have just one real number, I have two.

So, complex addition and multiplication are a bit different from our regular addition and multiplication. Our normal real number addition just takes two numbers and adds them. In the complex world, we are adding two pairs of real numbers. Four numbers to keep track of here.

The short story is that there is only one way to define addition and multiplication to pairs of numbers in a way that gives us a "field". I'll tell you the punchline: the way we define multiplication on these pairs of numbers is actually going to correspond to using i.

Fields are really cool--the real numbers with our usual operations are a field. Basically, fields have addition, subtraction, multiplication, and division. What is a field? I don't really want to write out the definition, because really, all these abstract algebraic structures are defined by a list of properties that we want. I'm lazy, so I won't list them, but you can look on Wikipedia about it. Just skip that first intro paragraph and go down to the Definition section. I may have to write down certain properties for my proof next, but I don't want to write them all out.

Now, how would you define a + on pairs of numbers? Don't think too hard. The obvious way to do this would be to add each component of the vector:

(a, + (c, d) = (a + c, b + d)

This is exactly what you'd do in linear algebra/matrix theory. Nothing "complex" yet. But what normal vectors don't do is have a good multiplication. How the crap am I supposed to multiply vectors? If you know some physics, you may have heard of the dot product and the cross product. They are ways to "multiply" vectors. But in our multiplication, we want to multiply two pairs of numbers, and then get another pair of numbers out. Dot products give us one number back, not two, and cross products are just for three-dimensional vectors. None of our normal tricks work! Curses!

But hey, as I showed you, we can write down operations all day! Suck it, mathematicians! We're going to define our complex multiplication right now!

Let's do the obvious way.

(a, x(c, d) = (ac, bd)

It worked for addition, so hey, maybe we just multiply the components together here, too, and everything will work out! Spoiler alert: it doesn't. This multiplication will fail two very important properties of a field. Pasted from wikipedia, for a Field F:

1. There is an element, called the multiplicative identity element and denoted by 1, such that for all a in F, a · 1 = a.

Okay, nothing special here. In the real numbers, if you multiply anything by 1, you get that same thing back. 1 doesn't do anything. It's the "identity" in that you get a's "identity" back. In this definition, "1" is not necessarily the number 1. We are denoting our multiplicative identity by 1.

Our multiplicative identity is a member of our set, so that means our "identity" is another pair of numbers. For the way we defined our multiplication, it's pretty clear that (1, 1) is our identity.

(a, x(1, 1) = (a*1, b*1) and since a and b are both real numbers, that means we get (a, back. Cool, we have our identity.

However, there's another property we need for a field, and it deals with division. In the language of abstract algebra, this is the "multiplicative inverse"--it's called that because really, division isn't a new operation, it's the reverse of multiplication. (There's also the "additive inverse", which for the real numbers is just subtraction). Here's the definition:

2. For any a in F other than 0, there exists an element a^−1 in F, such that a · a^−1 = 1.

Okay, what the heck does that mean? Read a^-1 as "a inverse". For the real numbers, the inverse element is just 1/a. Because you'll get a* (1/a) = 1, and 1 is our multiplicative identity, so we're done.

At first, you think everything is fine with our multiplication, right? Because we could just pick our inverse element of (a, to be (1/a, 1/B). Then we'd just get:

(a, x(1/a, 1/B) = (a*(1/a), b*(1/B) = (1, 1)

Which is our identity! Sweet, we are done.

But wait. Not so fast. This must work for any two pairs of numbers. We have some problems right at 0, right? Now, in "pairs of numbers" land, 0 is (0, 0). So let's try to find an inverse to this number:

(0, a)

Hmmm. My inverse needs to get me (1, 1) back. Is that even possible?

(0, a)x(x, y) = (0, ay). Nope, I'm never going to get (1, 1), because that first component is always 0. Crap! That means division totally fails, because division needs to work for all pairs of numbers (except (0, 0)). This is not the multiplication we are looking for.

So, what is a multiplication that gives us the property of a field? You're not going to guess it, so I'm just going to write the answer down:

(a, x(c, d) = (ac - bd, ad + bc)

This is why I went through all that effort earlier to convince you that operations could be pretty weird, because this one is certainly weird. No one remembers this formula. But, you can verify the properties for multiplication, which I can go through for you. In this case, we our multiplicative identity is (1, 0), and we don't have the same problem as before. For this multiplication and our earlier addition, we get a field out! That's cool.

This is obviously a doctored-up example, because we would've never guessed that. And why does no one remember this formula?

Call (a, = a + ib and (c, d) = c + id.

(a, x(c, d) = (a + ib)(c + id)

Well, we can FOIL just like we always do!

(a + ib)(c + id) = ac + iad + ibc + i*i*bd

I haven't told you what i is yet. But look if we define i = √(-1), so i²= -1. Look what happens!

ac + iad + ibc + i*i*bd = ac + iad + ibc + (-1)bd. Let me group this in a more suggestive fashion:

(ac - bd) + i(ad + bc).

Hey, lookit that. Going back to our "ordered pair" notation (ac - bd) + i(ad + bc) = (ac - bd, ad + bc), which is exactly the multiplication we defined before.

What does this mean? That means multiplying complex numbers, with i being the square root of -1, it actually corresponds to this "abstract" field of pairs of numbers with a weird addition and multiplication. We can go from to the other freely. i is our shorthand for that weird multiplication in the abstract sense.

So, if it makes you feel better, complex arithmetic really is just arithmetic on pairs of numbers with addition and a weird multiplication. No square root of -1 needed for it to work! But come on, it's a lot easier to FOIL with (a+ib)(c+id). They are equivalent! So I might as well use i, because I'm lazy and life is short.

Now, okay, we defined multiplication. It's a field. But really does that multiplication correspond to, geometrically? For this, polar form is preferred. Let's let the x-axis be the real axis, and the y-axis be the imaginary axis. For a general complex number z = x + iy, we can write it like this, in polar form. We've written that complex number in terms of R, the radius from (0,0)--in complex analysis we call this the modulus of a number, and theta, the angle from the x-axis. This angle is called the argument of a number. Why these words? I don't know. In complex land mathematicians give everything a new name. It's annoying.

Multiplying two numbers actually does this: the two moduli multiply together--so the new "length" of the complex number is the two lengths multiplied together, which is kind of what you expect--but the arguments don't multiply, they add. Like, for i, which (0,1), it's radius from the origin is 1, and the angle is π/2 (in radians). Multiplying i * i, we'd sure like that to be -1. Well, i * i, we multiply our moduli (1*1 = 1), but we add our angle, π/2 + π/2 = π. If you know your unit circle, you know that we're on the unit circle with angle pi, which is exactly the point (-1,0).

Cool. In a more general sense, here's what 2+i and 3+i multiplied together looks like. I don't really like that picture, but you can kind of see that the radii are being multiplied, but the angles are adding.

Now, I don't know how much calculus you have, so I can't prove you the first and most important consequence of complex multiplication, but let's say, for kicks and giggles, that we can express a complex number like this:

z = r*e^iθ

Where r is that radius (the modulus), and θ is its angle. What happens when we multiply two z's together?

z1*z2 = r1*e^iθ1 * r2*e^iθ2 = (r1*r2)*e^{i(θ1 + θ2)}

The radii multiplied, like we want, and the angles added. Huh. This might signal to you that the exponential is a good way write a complex number in polar form. But... that's kind of weird. The exponential function is increasing. And how would we write numbers in polar form normally? If you've never seen this transformation, don't worry, but it's pretty easy to show with trigonometry that if you have (x, y) and you want to write that ordered pair in terms in of r and θ, you do this:

x = rcosθ

y = rsinθ

So, (x, y) = (rcosθ, rsinθ)

That might look familiar to you. But here, Chaos is telling you that you that you can write a complex number like this:

(x, y) = x + iy = re^iθ.

Whaaaaat?? Doesn't this mean...

(x, y) = (rcosθ, rsinθ) = rcosθ + i*rsinθ = re^iθ.

That means that...

e^iθ = cosθ + isinθ

Let's put a box around that. This is important. This is Euler's equation.

e^iθ = cosθ + isinθ

So the exponential function--this nice easy increasing function which looks like this--if you have a complex exponential... that means it is a sum of sines and cosines?

If you are staring at those graphs and wondering just what the hell happened that makes that even work at all, well. You aren't the only one. The answer is that we have this i in here--this weird multiplication--which is making things weird. Apparently, e^iπ = -1. If you just said "This post has been reported for attempting to skirt the rules", well, I introduce you to this xkcd comic: http://xkcd.com/179/

It's straightforward to prove Euler's Equation. While I kind of derived it, I assumed that you could write z = r*e^iθ to do it. I'd like to prove that I am actually allowed to do that in the first place.

To do this, I need calculus 2. Specifically, I need power series. If you've had any of that before, well, then I'll give you the quick version: if you take the power series for e^x and substitute x = iθ, that i will give you some negative terms in the power series. By separating the real and imaginary parts, you get the power series for cosine + i * the power series for sine.

If that made no sense, that's okay. You kind of need some basic calculus for you to be even convinced that works.

But there you have it, that's what i does. It's a weird multiplication which forms an algebraic field. When you look at that field, it turns out that some weird things happen, like exponentials being cosines and sines.

If you have any calculus, then I can tell you a few more weird results. Differentiation in the complex sense is a much, much stronger condition in the complex numbers than in the real numbers. For one thing, a function must satisfy the Cauchy-Riemann Equations for them to be differentiable at all.

We haven't gotten there yet in complex analysis class, but my professor keeps telling me the weirdest result: if you have one derivative defined on the complex numbers, then the function is also infinitely differentiable. One derivative gets you all of them. Apparently this has something to do with the fact you can build a power series out of any differentiable function, and we know that power series are infinitely differentiable. So... that's kind of weird, because in the real case, it's easy to write down a function that has one derivative, but you can't get a second derivative. In the complex case, if you're differentiable, you're infinitely differentiable. So in this sense, complex functions are much smoother than in the real case.

There's a lot of awesome properties of the complex numbers. For one thing, it's a generalization of the real numbers, so it's a much bigger set than the reals. It's the largest algebraic field, in fact. You get sexy, sexy results like the Fundamental Theorem of Algebra.

Is there anything that complex numbers don't give you? Well, a few things. Earlier in the thread, I explained how √(x*y) is not always √(x)√(y) in the complex sense, and since square roots are exponents, you might well suspect this is because of our weird multiplication. You need to be more precise in polar form for things to work out.

Perhaps the easiest property in the complex numbers that isn't true for the reals is that the complex numbers aren't "ordered".

In the real numbers, for two real a and b, there are three possibilities:

a = b

a < b

a > b

The real numbers are "ordered". Either you are bigger or you are smaller than a number. There's a strict ordering to this number system. That's why we can draw the real number "line". Lines have an order to them.

This isn't true in the complex numbers, because how I can determine what a complex inequality is? It's not a line, but an entire plane.

Sure, we can say that 1 + i > -1 - i, because both components in the first are bigger than the second. But let's compare 1 - i to -1 + i. Is one big than the other? Nope, not really. The complex numbers are not ordered. Though, actually, this is a consequence of them being very similar R² rather than any weirdness with i. Any vector in Rⁿ for any n obviously isn't ordered.

So in the complex world, we get a bunch of cool properties, but we have to give up some nice things that we liked in the real numbers. But the things we lose have little practical significance. In the case of being ordered, while the complex numbers are not themselves ordered, we can always take a complex number's modulus (that radius from 0), and moduli are always positive real numbers. So, if we need to order things, we can take the modulus and compare things that way.

[Alliare she/her]

... and I should be going to university or I will be really late, but just couldn't stop reading.

Honest, why can't I upvote you more than once for that post? It's a masterpiece! Some of the things you explain I had been taught before, but never in that way, never so clearly and from the basics, so I wasn't aware of what I was doing until now. Don't worry about teaching students, you're freaking awesome as a teacher

Ok, so I thin I have iot all clear (or as clear as that can be) now. Maybe I wouldn't know how to do exercices with complex numbers but the ones which are similar to the examples you've used, but I understand the theory. Thanks!

*copies the post and saves it safely in her computer*

Edited March 26, 2012 by Alliare

[Chaos he/him]

... and I should be going to university or I will be really late, but just couldn't stop reading.

Honest, why can't I upvote you more than once for that post? It's a masterpiece! Some of the things you explain I had been taught before, but never in that way, never so clearly and from the basics, so I wasn't aware of what I was doing until now. Don't worry about teaching students, you're freaking awesome as a teacher

Ok, so I thin I have iot all clear (or as clear as that can be) now. Maybe I wouldn't know how to do exercices with complex numbers but the ones which are similar to the examples you've used, but I understand the theory. Thanks!

*copies the post and saves it safely in her computer*

You're welcome to randomly pick another of post of mine and upvote it, if you so choose!

I'm really glad you like it. I wasn't sure if anything would sink in. I have a tendency to ramble a lot, and it can get a little out of hand. I was mostly doing this off the top of head (with occasional looks into my textbook), haha. This makes me extraordinarily happy that you thought it was good!

[Shivertongue he/him]

Now, I'll show you MY method of doing math. It has served me well, and I highly recommend everyone take this approach.

Step One: Identify your problem.

Step Two: Procrastinate (3-4 hours, usually playing WoW)

Step Three: Make Eric do it while you go play WoW.

Step Four: Make fun of Eric.

It's a tried and true method!

[Chaos he/him]

Now, I'll show you MY method of doing math. It has served me well, and I highly recommend everyone take this approach.

Step One: Identify your problem.

Step Two: Procrastinate (3-4 hours, usually playing WoW)

Step Three: Make Eric do it while you go play WoW.

Step Four: Make fun of Eric.

It's a tried and true method!

Step Five: Charge Will $100 per hour.

That's my favorite part.

[Shivertongue he/him]

Now, I'll show you MY method of doing math. It has served me well, and I highly recommend everyone take this approach.

Step One: Identify your problem.

Step Two: Procrastinate (3-4 hours, usually playing WoW)

Step Three: Make Eric do it while you go play WoW.

Step Four: Make fun of Eric.

It's a tried and true method!

Step Five: Charge Will $100 per hour.

That's my favorite part.

Step Six: Pay Eric with homemade boxings, then laugh.

Step Seven: Laugh at Eric more.

[Satsuoni he/him]

Um, Chaos, isn't

√(-2)√(-2)=√(2*-1)*√(2*-1) = √2 * √2 * i * i = -2, which is not the same as 2 (which is what you'd get if you said √(-2)√(-2) = √(-2*-2) = √4 = 2).

just another solution for the same problem? You could do the same trick with exponents, after all (just add 2*pi before division)

Also:

(0, a)

Hmmm. My inverse needs to get me (1, 1) back. Is that even possible?

You know, when I studied this stuff in high school and first year of University, I have always wondered what mathematicians have against the harmless infinity. It makes many calculations so much easier. Sure, it is a freaky number(inf>inf, inf=-inf, etc), but still. So there are two solutions you can define and infinity as a number (inf, 0*inf=1, 0*-inf=-1). Then your multiplication works nicely. Or you could just define all pairs containing zero to BE zero Sure, you'll get a weird 4-lobed torus-like thingie instead of plane, but who cares? At least your multiplication will be saved

Now , question:

Can a question be more about physics rather than math? On the level above quantum mechanics I don't really see the difference, so here: how does one deal with 2 wavefunctions of different, interacting (positron/electron, to be precise) functions in the same potential well? What is spinor? And what is the difference between algebra of Lie, and normal algebra, which I always assume to be the algebra of Truth?

[Chaos he/him]

Um, Chaos, isn't

just another solution for the same problem? You could do the same trick with exponents, after all (just add 2*pi before division)

With complex square roots, I am very careful in that I am taking the principal square root. Yes, there are two square roots, but I am doing the principal square root w = z^(1/2) so that w is a function, and so I can do operations like differentiation on it. I can't do that on a multivalued thing. And this way, you will always get a unique answer.

You know, when I studied this stuff in high school and first year of University, I have always wondered what mathematicians have against the harmless infinity. It makes many calculations so much easier. Sure, it is a freaky number(inf>inf, inf=-inf, etc), but still. So there are two solutions you can define and infinity as a number (inf, 0*inf=1, 0*-inf=-1). Then your multiplication works nicely. Or you could just define all pairs containing zero to BE zero Sure, you'll get a weird 4-lobed torus-like thingie instead of plane, but who cares? At least your multiplication will be saved

The thing with infinity is that it isn't a real (or complex) number. Yes, once I define the complex numbers, I will eventually make the extended complex plane which contains one extra number, infinity, but infinity is not an element of the complex numbers nor it is a real number. There is no pair of fixed real numbers (a, with the properties you describe.

We have nothing against infinity; we use it literally every day when we invoke calculus. However, it really isn't a number. Infinity is relatively meaningless outside a proper epsilon-delta definition of a limit. It's important to be rigorous with this, because otherwise you will have logical inconsistencies. This is, at least, how we do it in analysis, and that's the field I know best. I can't speak for other mathematical fields.

By the way, in the complex world, there's no such thing as negative infinity. There's only one infinity, and this is because we define it in terms of the modulus of a number.

So, anyway, it's because of mathematical rigor that we need to be very careful with infinity. Plus, I can't do proper operations, and we'll have other difficulties if we define infinity in this way.

The next suggestion you had is that all pairs of numbers containing zero to be the Zero Element. Now, there's many problems with doing this. While you could do this--as you can define things in a lot of ways, however you'd like--this is a bad way of defining this. Firstly, directly to the point at hand, the way I defined addition makes an algebraic group with respect to addition. This is very good, because the real numbers are an algebraic group with addition, too, and we always want the complex numbers to give back the real numbers and all their properties when I let the imaginary component of the complex numbers go to zero. By doing this, I wouldn't get the real numbers back, and that's a rather pointless endeavor. These group/ring/field properties are very necessary.

Specifically, the Zero Element must be unique in groups. The Zero Element has a very importantly, namely, in a group G, with operation + and Zero Element e, for all a in G, we get a+e = e+a = a. In effect, the Zero Element doesn't "do" anything under this "addition". Which is exactly what we have for the typical addition.

This for all elements in the group, including the Zero Element. So let's say we have two Zero Elements, call them, for kicks and giggles, 0 and 2.

We need 2 + 0 = 2, because 0 is a Zero Element. We need 2 + 0 = 0 + 2 = 2.

However, if 2 is also a Zero Element, that means for any a, we get a + 2 = 2 + a = a. Set a = 0, and then we get

0 + 2 = 2 + 0 = 0.

We can't have multiple Zero Elements for this property to be true. 2 must be 0 in this case (the abstract 2 and 0, I'm talking about). The Zero Element is unique.

Plus, come on. Even in non-complex land, it's very silly to say that (0,a) is "zero" for any a. Namely, let's look at some nice properties of vectors, like length. Using the distance formula, the length of (0,a) is sqrt( 0^2 + a^2) = a. I don't know about you, but I think a proper "zero vector" should have zero length. There's only one vector does this, and that's the vector with every component equal to 0.

If that wasn't the zero vector, then we've destroyed linear algebra. I kind of find the machinery of linear algebra to be kind of important, so this may not be the best zero we could choose

So you see, in defining an expansion of the real numbers, we have a couple of issues. We need the key fact that, when the "imaginary" part (which we haven't defined yet, but let's just say for now that it's another, "nonreal" component that we are adding to the usual real number) is 0, then we need to get the real numbers back. As in,

a + ib = (a,

If b = 0, then we just have the real number a. I should get the real numbers back. But, in your definition of 0, (a,0) is considered a Zero Element. So, whatever that new space is, it certainly doesn't have the properties of the real numbers, because the real numbers are an algebraic field. This may suggest to you that an "extension" of the real numbers should also be an algebraic field, and there's only one bigger field than the reals, and that is the complex numbers, as I have defined them.

Besides, even though that complex multiplication seems weird, it's certainly true that if the imaginary components are 0,

(a, 0)*(c, 0) = (a*c - 0*0, a*0 + 0*c) = (ac,0).

Which is exactly you'd want--pure real numbers multiplied by pure real numbers are real numbers.

So, long story short, I didn't want to "save" my random multiplication that I made up back there, because it turned out to be a rather poor choice of multiplications. This multiplication forms a field, and is much more useful because of it.

Now , question:

Can a question be more about physics rather than math? On the level above quantum mechanics I don't really see the difference, so here: how does one deal with 2 wavefunctions of different, interacting (positron/electron, to be precise) functions in the same potential well? What is spinor? And what is the difference between algebra of Lie, and normal algebra, which I always assume to be the algebra of Truth?

Hoo boy, I fear we may find a limit to my math knowledge eventually, haha.

We can talk physics, but realize that you can't derive physics from first principles, like I did with the complex numbers That, in my opinion, makes physics more challenging than pure mathematics.

So we have three questions here.

1. What happens when a positron and an electron are in the same potential well?

2. What is a spinor?

3. What's the difference with Lie algebra and normal algebra?

Well, for #1, the thing is... well. You aren't going to ever be able to answer this with nonrelativistic quantum mechanics. Those particles are going to very quickly annihilate into two photons, and depending how the potential well was constructed, the photons may bounce around the well, or be absorbed into the well, or escape the well.

The short answer is that you need Quantum Electrodynamics to properly make sense of this system beyond that qualitative thing I just wrote about. While I have had senior level particle physics, and have seen some of QED, I know nonrelativistic quantum mechanics much more, on a much more practical level. This is because if you think solving the Schrodinger Equation for wavefunctions is hard, the relativistic version, the Dirac Equation (well, at least, for spin 1/2 particles. The Dirac equation does not apply to spin 3/2 or others, but there are apparently other equations that work for that) is really, really hard.

Why is the Dirac equation harder? Well, it leads to your second point, about spinors. The thing is, the Dirac equation is a matrix equation, with very special, 4x4 matrices called gamma matrices in it. These gamma matrices incorporate the Pauli matrices (if you're familiar with them) and so, in a weird way, the Dirac equation has spin built right into it from the start. (Fun fact, Schrodinger actually always wanted to derive a relativistic version of the Schrodinger equation, because everyone had known about relativity for a while now. Even before Schrodinger found the equation that bears his name, he discovered what is now called the Klein-Gordon equation, which is a relativistic quantum equation, but it happens to only work for spin 0 particles. Schrodinger then later found the nonrelativistic version, which he finally decided to release because, to his surprise, it gave kinda good results)

So, what is a spinor?

Well, since the Dirac equation is a matrix equation, the "wavefunction" solutions are now a column vector with four components. So where nonrelativistic wavefunctions have a scalar amplitude, the relativistic version has four amplitude components. This makes things... a lot harder. Partial differential equations are hard enough, and now I have a vector to keep track of? As you can imagine, the Dirac equation is incredibly hard, and you kind of need to guess the form of the spinor (that is, that vector "wavefunction") and try to solve the Dirac Equation that way.

As for question #3, I can't really tell you what Lie algebra is. I may be able to in two years, when I get my Master's, but right now, I've only had one abstract algebra class, so I couldn't tell you. I direct you to Wikipedia, because honestly, that'd be the first thing I'd do to try to explain it right now

[Voidbringer]

Math thread! I am very excited about this. Math is awesome!

@Satsuoni

Firstly, a Lie Algebra is just a vector space V with an extra operation denoted [.]: VxV->V (often called the bracket or Lie bracket) which is bilinear, [x,x]= 0 for all x in V, and it satisfies the Jacobi identity (i.e. [x,[y,x]]+[y,[x,z]]+[z,[x,y]]=0). Don't worry too much about that unless you are looking to get into the theory. Based on the fact you were asking physics questions I imagine you would prefer to hear more about the applications. Just know that Lie Algebras are used in representation theory to study Lie Groups, which are special kinds of groups that are also differential manifolds (i.e. things you can do calculus on). They come up a lot in differential equations and differential geometries, more specifically you will see this used in the mathematics of string theory and twistor theory (for all you spinor fans out there). Some of the nicest examples of Lie groups are the matrix groups O(n) and SO(n). In these cases, their Lie Algebras are just their tangent spaces at their identity elements. Hope that covers at least something you may have been looking for.

@Chaos

Glad to see a fellow mathematician!

Now I have a question that you might be able to help me with since you have taken abstract algebra. I am pretty lost on this one...

Let G, H, and K, be additive abelian groups. Let f1:G -> H, f2:G -> K, g1:H -> G, and g2:K -> G be homomorphisms such that for all g in G, g1(f1(g))+ g2(f2(g)) = g, f1 o g1 = Ih, f2 o g2 = Ik, f1 o g2 = 0, and f2 o g1 = 0, where o denotes composition, Ix denotes the identity map of a set X, and 0 denotes that map that sends all elements to to the identity element of the range. Show that G is isomorphic to the direct product (sum) of H and K (HxK).

I feel like I would get this if I had some category theory under my belt but...

[Satsuoni he/him]

@Chaos

Glad to see a fellow mathematician!

Now I have a question that you might be able to help me with since you have taken abstract algebra. I am pretty lost on this one...

Let G, H, and K, be additive abelian groups. Let f1:G -> H, f2:G -> K, g1:H -> G, and g2:K -> G be homomorphisms such that for all g in G, g1(f1(g))+ g2(f2(g)) = g, f1 o g1 = Ih, f2 o g2 = Ik, f1 o g2 = 0, and f2 o g1 = 0, where o denotes composition, Ix denotes the identity map of a set X, and 0 denotes that map that sends all elements to to the identity element of the range. Show that G is isomorphic to the direct product (sum) of H and K (HxK).

I feel like I would get this if I had some category theory under my belt but...

Let me try this I am no mathematician, but I like to try. So, first, "simplify" the problem so that I could read it. There are variables (G,H,K) that have addition defined on them. There are linear-like functions (homomorphisms: f(a1+a2)=f(a1)+f(a2), for + being the abelian operation on corresponding groups, blah, blah). Let g,h,k be any values in groups. Then, what you have is:

g1(f1(g))+ g2(f2(g)) = g

f1(g1(h))=h

f2(g2(k))=k

f1(g2(k))=0 (you sure about "sends all elements to to the identity element", rather than "to zero element"? I am going to consider it zero element, h+0=0. It is weird otherwise (h+1=h))

f2(g1(h))=0

You need to prove that there are functions s(H,K)=G and s^-1(G)=(H,K), so that s(s^-1)=I, s^-1(s)=I (from wiki)

So, let us consider a (map, function, whatever) s(h,k)=g1(h)+g2(k), and s^-1(g)=(f1(g);f2(g))

then:

s(s^-1(g))=g1(f1(g))+g2(f2(g))=g

s^-1(s(h,k))=(f1(g1(h)+g2(k));f2(g1(h)+g2(k)))=(f1(g1(h))+f1(g2(k));f2(g1(h))+f2(g2(k)))=(h+0;k+0)=(h;k)

QED?

Ok, now I am confused. Either this problem did not require math above middle school level, or I have gotten something wrong. I am leaning towards the latter, so. What did I get wrong?

Also, @Chaos : Thanks for explanation! But, I am not sure Dirac equation accounts for interaction (annihilation). Does it? I was almost sure you need to go into field theory for that... If it does, that is good news

[Edit] Consistency.

Edited March 29, 2012 by Satsuoni

[Chaos he/him]

Ok, now I am confused. Either this problem did not require math above middle school level, or I have gotten something wrong. I am leaning towards the latter, so. What did I get wrong?

Also, @Chaos : Thanks for explanation! But, I am not sure Dirac equation accounts for interaction (annihilation). Does it? I was almost sure you need to go into field theory for that... If it does, that is good news

[Edit] Consistency.

Your middle school is way more awesome than mine, clearly

Voidbringer, I need to refresh myself on my abstract algebra. It was never my strongest suit. So I'll go look into it, but no promises. I kind of didn't expect so many high level math people here! I'll do my best though.

Satsuoni, the solutions of the Dirac equation certainly are what you use in Quantum Electrodynamics. I'm not familiar with the derivations for that stuff, but yes, essentially, you need the Dirac equation for that. Full QED definitely will solve your problem.

[Joe ST he/him]

Technical niggle, you misuse the term zero element where you mean identity element. The zero element is the one that sends every element to itsself. It gets confusing because the identity in the (R,+) group of addition over R, is the zero in (R,*) group of multiplication over R. So you're right in saying it's the zero element of R, but not of the addition group.

I'll come back and try your groups problem at lunch Voidbringer, seen as it sounds like I'm the most qualified in abstract algebra in this place

[Satsuoni he/him]

Technical niggle, you misuse the term zero element where you mean identity element. The zero element is the one that sends every element to itsself. It gets confusing because the identity in the (R,+) group of addition over R, is the zero in (R,*) group of multiplication over R. So you're right in saying it's the zero element of R, but not of the addition group.

I'll come back and try your groups problem at lunch Voidbringer, seen as it sounds like I'm the most qualified in abstract algebra in this place

Which one of us? Well, I assume me, since I am the layman around here. Well. OK, I see it. I used element that has (a+0=0+a=a) property, whatever its name. Then, what is the zero element of addition? Does it even exist?

So, let me get this straight. Abstract algebra is, basically, basic operations that we learn in elementary and middle school, only on arbitrary elements with arbitrary rules that fit whatever problem you are solving, covered by fancy and ambiguous terminology to hide that fact? Strictly imo, using the formulae that define the properties, instead of weird names is much more clear and reduces ambiguity and misunderstandings, but what do I know...

[Joe ST he/him]

Yes that's basically what abstract algebra is, it's defining how different actions work on sets of arbitrary elements. It get's pretty abstract pretty darn fast.

And I was talking to Eric about the zero stuff. He only knows Fields and VectorSpaces, he's a n00b

Oh, and an Algebra (in particular, a Lie Algebra) is something else entirely. Wikipedia is pretty good at explaining Abstract Algebra, once you know the terminology.

Let G, H, and K, be additive abelian groups.

Let f1:G -> H, f2:G -> K, g1:H -> G, and g2:K -> G be homomorphisms such that for all g in G,

- g1(f1(g)) + g2(f2(g)) = g, OR (f1 o g1) + (f2 o g2) = Ig,

- f1 o g1 = Ih, f2 o g2 = Ik, OR g1(f1(g)) = h, g2(f2(g)) = k

- f1 o g2 = f2 o g1 = 0

where o denotes composition, Ix denotes the identity map of a set X, and 0 denotes that map that sends all elements to to the identity element of the range.

Show that G is isomorphic to the direct product H*K

H*K is defined as {(h,k) for all h in H, k in K}

then you can define 'similar' functions, f1' f2' g1' and g2':

- f1': H*K -> H as f1'(h,k) = h

- f2': H*K -> K as f2'(h,k) = k

- g1': H -> H*K as g1'(h) = (h,e)

- g2': K -> H*K as g2'(k) = (1,k)

where 1 and e are Identity elements in H and K respectively

Then you can show that these 'similar' functions act the same as the first ones:

- f1' o g1' = Ih OR g1'(f1'(h,e)) = g1'(h) = (h,e)

- f2' o g2' = Ik OR g2'(f2'(1,k)) = g2'(k) = (1,k)

- f1' o g2' = 0' OR g2'(f1'(h,k)) = g2'(h) = (1,e)

- f2' o g1' = 0' OR g1'(f2'(h,k)) = g1'(k) = (1,e)

and

- (f1' o g1') + (f2' o g2') = i' OR g1'(f1'(h,k)) + g2'(f2'(h,k)) = (h,e) + (1,k) = (h,k)

Where 0' is the zero map and i' is the identity map over H*K.

I can't remember the correct words to use though, since it's been a year or so since I did this stuff. Basically these new 'similar' functions are enough to let you prove that G and H*K are similar, aka isomorphic. I'd suggest to not take anything I've said as rigorous proof, but I'm fairly sure that this is the way one should approach such a problem.

[Satsuoni he/him]

Ok. So, what (excepting the terminology) is wrong with my solution? It is obviously wrong, since cool, advanced group theory problem cannot be solved by such primitive methods, but I can't put my finger on what exactly is amiss (due to the lack of knowledge on said theory).

[Edit] Just noticed an edit, sorry. Wait till I read it.. done.

Now I have a question about the solution. So, first you define some new functions. Then you show that these new functions behave the same as old functions. Um. So? That does not imply that they are the same functions or that G and (H*K) are isomorphic, just that you can define the functions that act the same way on both of them, or am I misunderstanding something again? Sorry.

Edited March 29, 2012 by Satsuoni

[Voidbringer]

H*K is defined as {(h,k) for all h in H, k in K}

then you can define 'similar' functions, f1' f2' g1' and g2':

- f1': H*K -> H as f1'(h,k) = h

- f2': H*K -> K as f2'(h,k) = k

- g1': H -> H*K as g1'(h) = (h,e)

- g2': K -> H*K as g2'(k) = (1,k)

where 1 and e are Identity elements in H and K respectively

Then you can show that these 'similar' functions act the same as the first ones:

- f1' o g1' = Ih OR g1'(f1'(h,e)) = g1'(h) = (h,e)

- f2' o g2' = Ik OR g2'(f2'(1,k)) = g2'(k) = (1,k)

- f1' o g2' = 0' OR g2'(f1'(h,k)) = g2'(h) = (1,e)

- f2' o g1' = 0' OR g1'(f2'(h,k)) = g1'(k) = (1,e)

and

- (f1' o g1') + (f2' o g2') = i' OR g1'(f1'(h,k)) + g2'(f2'(h,k)) = (h,e) + (1,k) = (h,k)

Where 0' is the zero map and i' is the identity map over H*K.

I can't remember the correct words to use though, since it's been a year or so since I did this stuff. Basically these new 'similar' functions are enough to let you prove that G and H*K are similar, aka isomorphic. I'd suggest to not take anything I've said as rigorous proof, but I'm fairly sure that this is the way one should approach such a problem.

Fantastic! Looks great to me! I can definitely work out the details with that much to go on. Thanks a bunch!

[Joe ST he/him]

Ok. So, what (excepting the terminology) is wrong with my solution? It is obviously wrong, since cool, advanced group theory problem cannot be solved by such primitive methods, but I can't put my finger on what exactly is amiss (due to the lack of knowledge on said theory).

[Edit] Just noticed an edit, sorry. Wait till I read it.. done.

Now I have a question about the solution. So, first you define some new functions. Then you show that these new functions behave the same as old functions. Um. So? That does not imply that they are the same functions or that G and (H*K) are isomorphic, just that you can define the functions that act the same way on both of them, or am I misunderstanding something again? Sorry.

Well you could, instead, define a homomorphism between G and H*K that uses the predefined functions to prove that it's a bijection and thus G and H*K are isomorphic. That's possibly the best (or fastest) way to get to the answer.

[Satsuoni he/him]

Well you could, instead, define a homomorphism between G and H*K that uses the predefined functions to prove that it's a bijection and thus G and H*K are isomorphic. That's possibly the best (or fastest) way to get to the answer.

...

Ok, I'll rephrase. Is a map (f1;f2)(G->(HxK)) an isomorphism if it has an inverse (g1+g2)((HxK)->G), and is an inverse of an inverse (and vise versa) (both the map and inverse being defined over the whole set of the corresponding group)? Or are there any additional qualifiers?

[Joe ST he/him]

It should also map a all elements of G into unique elements of H*K (and |G| = |H*K|)

[Satsuoni he/him]

It should also map a all elements of G into unique elements of H*K (and |G| = |H*K|)

Um. Could you give me an example where the above does not hold while still having the mutual inverse? So, the following problem:

Give an example of two groups (A and B ), for which, there exist maps a: (A->B ) and b(B->A), such that a o b =I_A and b o a =I_B (I being an Identity map for the corresponding group) for all elements of A, B, such that either a or b does not map elements uniquely (onto corresponding group) and/or such that |A|!=|B|.

Edited March 30, 2012 by Satsuoni

[Shivertongue he/him]

Interesting new development in the math world. 2+2, it seems, now equals 16.