What are imaginary numbers?

http://math.stackexchange.com/questions/199676/what-are-imaginary-numbers

by arcatek • 14 years ago

217 134 14 years ago

math.stackexchange.com

_ntka• 14 years ago

The title statement is meaningless. "Imaginary numbers" are not "multiply" by "90°". What does it even mean to "multiply" by an angle?

Correct statement: "*i is equivalent to a 90° rotation in some contexts." Which is kind of obvious to anybody who has done some maths or physics at some point?

walrus• 14 years ago

Trivially, this result is clearly obvious.

Tipzntrix• 14 years ago

I swear the only time I ever heard trivial in a lecture was once I got into university. It's a whole new ball game; high school teachers wouldn't dare say trivial just in case they had someone who was behind.

friendlyfrog3• 14 years ago

No it's not. i is a number such that i^2 = -1 by definition.

gizmo6869• 14 years ago

And in some contexts -1 is rotating by 180 degrees, which happens to be 290, or i*i.

T-hawk2• 14 years ago

Here is an even better discussion on the same topic, and the HN thread from last year: http://betterexplained.com/articles/a-visual-intuitive-guide... https://news.ycombinator.com/item?id=2712575

One great conclusion from this approach is how intuitive it becomes to understand the square root of i. I always thought you'd need another dimension to describe that, and another dimension for the square root of that unit, and so on.

But think of i as a 90° rotation, so applying it twice (squaring it) results in 180° which is -1. Then √i is a 45° rotation so applying it twice results in 90° which is i. Sure enough, this works out. The unit vector at 45° is 0.5√2 + 0.5√2 * i. Follow the rules of complex arithmetic to square that and you do indeed get i.

A rotation of -135° applied twice gives the same result. Square the unit vector of -0.5√2 + -0.5√2 * i and you also get i. We've arrived back at the axiom that all numbers have two square roots of opposite signs.

Last question: What's the cube root of i? Easy: a 30° rotation. The 30° unit vector is 0.5√3 + 0.5i, and cubing that does indeed get you i.

JadeNB• 14 years ago

> We've arrived back at the axiom that all numbers have two square roots of opposite signs.

While I like the geometric content of the rest of your post, I have to quibble on two points:

    * that a number has two (usually distinct) square roots is a theorem, not an axiom.  That is, it's not 'built in' (unlike, say, commutativity of addition), but rather can be proven on the basis of 'built-in' properties.

    * while one may technically say that i and -i have 'opposite signs', it's a good idea not to talk about the sign of a complex number at all (except possibly in the sense that sgn(z) = z/|z|).  It's better to say that the two square roots are opposite; or, if one wants to be more precise, that they are additive inverses.

jerf• 14 years ago

In higher maths, commutativity of addition can be proved from the Peano axioms without taking it as an axiom. It's a common homework problem.

kalid3• 14 years ago

Thanks for posting that :). My gut-check about whether I've intuitively understood a topic is whether I can intuit new results. I'd wager 95% of people who "learned" imaginary numbers couldn't work out the cube root of i in their head like that (I was in this camp all through high school and college, which made me realize I didn't really get them).

diminish2• 14 years ago

One more question; what is the "i"th root of i? Please try to use the angle metaphor :)

ColinWright13• 14 years ago

To answer the question, e^(it) =cos(t)+ i * sin(t). That means that when t=pi/2, cos(t)=0 and sin(t)=1, so i = e^(i * pi/2)

To get the 'i'th root we divide the exponent by i, so i^(1/i) = e^((i * pi/2)/i) which is e^(pi/2).

As for the angle metaphor, that works for multiplying by complex numbers. The usual "understanding" of exponentiation is repeated multiplication, but taking a power of "i" can't be understood as multiplying together "i" lots of the number. Similarly the "i"th root can't be understood like that.

But you knew that.

kalid3• 14 years ago

Sure! You might want to check out http://betterexplained.com/articles/intuitive-understanding-...

I like this question because it really works your intuition.

The basics: x^y means "grow at x, for y units of time". I see "2^3" as "grow at 2x for 3 units of time".

Having a base of i means your "growth" is a rotation at 90 degrees, no scaling. So i^(1/2) means a 45 degree rotation, i^3 means a 270 rotation, etc.

Raising this to the i power (or 1/i power, which is -i) means the growth that was originally purely rotational is now rotated. So instead of growing at i, you are growing at (i * 1/i = 1). So, we should expect a positive real number, greater than 1 since our growth is positive.

How long do we actually grow for? Well, the base of "i" is really e^(i* pi/2), which means "Start at 1.0 and rotate continuously for pi/2 seconds". We've now modified this to "Start at 1.0 and grow at 1.0 for pi/2 seconds", which is e^(pi/2).

So the answer is i^(1/i) = e^(pi/2) ~ 4.8

It's a bit tough with text-only, read the above article for more diagrams.

yequalsx5• 14 years ago

This is of course much trickier because to define z^w for complex numbers requires that one choose a complex logarithmic function. There are infinitely many choices for this. Everyone has decided to use the principle logarithm but it could be consistently defined using any branch cut.

dllthomas• 14 years ago

e^(pi/2)

This is is because exponentiation to an imaginary power can also be thought of as rotation. You have presumably heard e^(pii) = -1. That's because it's a half-rotation away from 1. Half of that half rotation would be i, and is sqrt(e^(pi i)), which is e^(pi/2*i), which is equivalent to (e^(pi/2))^i, and the ith root of that is obviously (e^(pi/2)).

XaspR8d• 14 years ago

On the last question, I find it really interesting to note that the nth root of i will have n solutions evenly distributed around the unit circle. (As will any unit vector.) So for the cube root of i, we'll also have -0.5√3 + 0.5i and -i as solutions, which are located at 5π/12 = 30° + 120° and 8π/12 = 30° + 2 * 120°. (On another note -i * -i * -i = -1 * -i = i. It's so simple and yet not obvious unless you do a lot of complex math...which I do not.)

Read about the roots of unity (solutions to a^n = 1) to see the same concept in action. http://en.wikipedia.org/wiki/Root_of_unity

It all comes out of the properties of rotation. We can first show that there are n nth roots of unity. It comes down to a^n = 1 -> a^n * a^n = 1 -> (a * a)^n = 1, which effectively shows that any multiple of the angle represented by a will also be a solution. If you use the full fledged exponent formulas, it's easy to see that there will be only n of these. Then if b^n = i (our case), b^n * a^n = i * 1 -> (b * a)^n = i. So adding any of the nth roots of unity to b will get a new solution for the nth root of i. Since one of the roots of unity is always 1, aka an angle of 0, we will only get n-1 new solutions from this. Hence there are n solutions. :)

EDIT: New here, didn't know the markup cues would interfere with my equations.

ArbitraryLimits• 14 years ago

Do kids not learn this in 8th grade any more? I'm seriously not trying to be snarky, I just can't think of a way to write that question that sounds unsnarky. I thought everyone learned about the polar representation of complex numbers.

citricsquid• 14 years ago

You need to differentiate between "are told" and "learn". Personally I was told a lot of things in my high school math classes, I learned very few of them.

kalid3• 14 years ago

(Author of the above article)

I learned the polar representation, yep, but didn't connect as i as an "rotation operation" that took you from one dimension to the other. For all I knew, having i be 2-dimensional was like saying "We can represent condiments by having an axis for ketchup and and an axis for mustard".

It was 2 random quantities represented as an (x,y) pair, without the notion that they were deeply connected.

chimeracoder• 14 years ago

No, they don't. If they do learn it, it's certainly not in 8th grade.

My town's school system was the second in the entire state, and they taught Algebra I (ie, factoring/FOIL) in 8th grade. I went to a well-respected private school, and they taught complex numbers in 10th grade, and polar coordinates only in the honors section of 10th/11th grade classes. For those, they taught matrix multiplication and determinants for 2x2 matrices only (ie, not 3x3 or up). This means that any real linear algebra was completely out of the picture.

This was well more than enough to score a very good score on the SAT II Math IIC exam (is this still even around?)

I was fortunate that I was able to teach myself outside of my school's curriculum (and my school also let me place out of certain classes), but overall, the standards are far below what you might hope.

jacobolus3• 14 years ago

Everyone should learn some of the tools of Geometric Algebra sometime in high school, and it would save a whole lot of confusion.

http://geocalc.clas.asu.edu/pdf/OerstedMedalLecture.pdf

tgb• 14 years ago

I like the linked paper a lot so far (still reading it) even though I disagree that everyone should learn this just so that physics majors and similar can avoid some amount of confusion.

jacobolus3• 14 years ago

Not just physics students. Anyone who has to deal with geometry should learn it. Anyone who would otherwise study trigonometry, linear algebra, vector calculus, complex analysis, projective geometry, &c. should learn it first. It’s a much better set of mathematical tools for understanding spatial relationships and transformations than the alternatives for many if not most purposes.

For anyone dealing with computer vision, graphics, or modeling type fields, I recommend the more recent book “Geometric Algebra for Computer Science”. http://www.geometricalgebra.net/

jamii• 14 years ago

I wish I had more upvotes for this. Geometric Algebra is both simpler and more general than traditional complex geometry or vector calculus and for most students is a much more intuitive way to approach complex numbers.

mej10• 14 years ago

It seems like an increasing number of people are mentioning Geometric Algebra and everyone that likes it seems to _really_ like it.

It is getting to the point where I am going to have to learn it just to see what the hype is about. (I love being able to say that there is "hype" around a field of mathematics)

Jun83• 14 years ago

I am humbled by the clarity of this answer!

The problem with textbook for all levels (other than them being outrageously expensive) is that they contain many level of abstraction, or what one may call self-censure, presumably in order not to scare kids off. One level is the general consensus of how should a topic should be taught, another layer is the author's view of how it should be taught In practice there's the third layer, where the teacher presents them in a certain way. This results in a long chain of Simon Says where the final, safe stuff that's taught, for the benefit of the students, mind you, may become very detached from the reality and excitement of the topic. Unfortunately, students always sense this and they tune it out, leading to so many people not liking math, physics, signal processing, you name it.

The joy of access to a person who is infinitely (compared to you) knowledgeable in atopic and is willing to interface you on multiple levels and telling you as it is is enormous. Best known such example, of course, is Feynman but physics SE and some other boards come close.

egiva• 14 years ago

I agree 100% that textbooks have too many levels of abstraction added over time. There's a really interesting answer given on this exact same subject by Bill Gates at the Aspen Festival - check it out here: http://www.youtube.com/watch?v=Iqf3rvg742g

yason• 14 years ago

The obligatory, if you're still scratching your head thinking about complex numbers: http://betterexplained.com/articles/a-visual-intuitive-guide...

jostmey2• 14 years ago

Why does (-1)*(-1)=(+1) ? It is arbitrary, and there really is no good reason. We could construct number lines that work differently, so that imaginary numbers never appear. Such alternative number lines would still allow us to solve the exact same physics and engineering problems. Sure, the computations would work differently, but the way we would measure and use the initial conditions in our equations would be different too. In the end, the predicted outcome would still be the same.

Imaginary numbers are an artifact of how our number line is constructed. We could construct alternative number lines where imaginary numbers do no exists. The computations involving such alternative number lines would be different, but the outcome would be the same.

yequalsx5• 14 years ago

EDIT: I proved the wrong thing. Correct proof is in post below.

alok-g14• 14 years ago

Can you please complete your proof? How do I go from 1.(-1) = -1 to (-1)*(-1)=(+1)?

friendlyfrog3• 14 years ago

It's not arbitrary. It comes directly from the axiomatics.

revelation• 14 years ago

Axiomatics are entirely arbitrary if you will. Thats the whole point: I restrict myself to some basic rules only because that allows me to show other potentially useful things, consequences and applications.

ColinWright13• 14 years ago

This sounds profound, but is wrong on so many levels. In a sense everything about mathematics is arbitrary, but there's a consistency and structure that makes such a statement unhelpful and misleading.

Consider.

If you're content with the counting numbers then we can construct the negative numbers. These have the specific property that when added to the positive number of the same size we get zero,

But most people are happy with the integers, so move on. We're reasonably happy with addition, but what is multiplication? If you think of it as repeated addition you're screwed when you want to multiply by 2 1/2. It's better to think of it as a scaling. Multiplying by 2 means that you scale things up to be twice as big. Thus 1 goes to 2, 2 goes to 4, and 5 goes to 10. Also, -1 goes to -2, -6 goes to -12, and so on.

So what do we mean when we scale by -1? We look at the sequence of scaling by 4, then by 3, then by 2, and so on, each time asking where the number 1 gets sent.

    Scale by 4 and 1 -> 4
    Scale by 3 and 1 -> 3
    Scale by 2 and 1 -> 2
    Scale by 1 and 1 -> 1
    Scale by 0 and 1 -> 0

Following this progression we see that it's natural in some sense to say that scaling by -1 means that 1goes to -1. And indeed, 2 goes to -2, and 73 goes to -73.

Scaling by -1 sends something to the same distance on the other side of zero.

So where does -1 get sent under a scaling of -1? It gets sent the same distance the other side of zero. -1 gets sent to 1.

Therefore it makes sense to say that -1 scaled by -1 is 1.

(-1) * (-1) = 1

Wecan use this to ask about the square root of -1. What geometric operation can we perform on the number line, such that doing it twice is the same as multiplying by -1?

An answer is to rotate anti-clockwise by 90 degrees. Another answer is to rotate clockwise by 90 degrees.

Pursue this, and you start to construct the Agrand diagram, and the complex numbers.

fexl• 14 years ago

Beautiful exposition. Even in the late 1700s, many mathematicians rejected the use of mere negative numbers, viewing them as anomalies which indicated that one had phrased a problem wrong to begin with. On the other hand, Euler understood everything very well and even calmly explained how to take logarithms of complex numbers, which bewildered most of his contemporaries.

Someone once said that a lot of confusion could have been avoided if, instead of the terms positive, negative, and imaginary, they had instead used the terms forward, backward, and lateral.

jostmey2• 14 years ago

Maybe I am very wrong on this issue. The arguments outlined below demonstrate that given a set of axioms or properties, there is only one consistent way to define (-1)*(-1).

brazzy2• 14 years ago

That's a mix of complete nonsense and meaningless truism.

Sure, imaginary numbers are based on the axioms for real numbers. But using other axioms ("number line" is a visualization aid appropriate for grade school, not for a serious mathematical discussion) would most definitely NOT allow us to solve exactly the same physics and engineering problems, or have the same outcomes for the same problems.

friendlyfrog3• 14 years ago

Imaginary numbers are defined. Imaginary numbers are not rotations or anything else that I keep hearing. Those are all properties of the fact that we define an imaginary number to be z = a + bi, where a and b are real numbers, and i^2 = -1. That's it.

If someone asks what is an imaginary number the correct answer is "z = a+bi, where a and b are real numbers, and i^2 = -1".

austintaylor• 14 years ago

You're thinking of complex numbers.

radicalbyte• 14 years ago

The best explanation I've seen yet is one Hobbes gave to Calvin:

http://www.gocomics.com/calvinandhobbes/2009/02/18/

luser0012• 14 years ago

I don't think I quite followed the step where he/she writes "i^4 = 1", where they are relating rotation and the natural numbers. The RHS is theoretically the concept "identity under rotation". Why should it be be the same as the natural number 1?

Maybe I missed something in the explanation. Can somebody explain?

T-hawk2• 14 years ago

There is an implicit start from the unit vector 1. i (really 1 * i) is a 90° rotation of that. i^4 is four 90° rotations. Rotate anything four times and you've rotated it 360°, ending up where you started at 1.

luser0012• 14 years ago

Ok, thx. I think I understand now.

irishcoffee• 14 years ago

* i = sqrt(-1)

* i * i = i^2 = sqrt(-1) * sqrt(-1) = -1

i^4 = i^2 * i^2 = -1 * -1 = 1

grandalf• 14 years ago

I have started reading the story of i:

http://www.amazon.com/An-Imaginary-Tale-Story-square/dp/0691...

alok-g14• 14 years ago

Can someone please also do this for:

1. Matrices, especially matrix multiplication. Unlike matrix addition, multiplication is defined in a very weird way. I think I understand where it is coming from -- defining it that way allows representing and solving linear equations. More insights, however, would help.

2. Dot and cross products. E.g., the magnitude of dot product in 3D is a.b.cos(theta), while for cross product, it is a.b.sin(theta). I never got sure what happens to other combinations like a new "vector" product whose magnitude is a.b."cos"(theta)?

3. How is a set and "belongs-to" operator defined? Most books I have come across just assume these (and later define natural numbers and addition from them).

4. Why is 0.9999... considered to be "equal" to 1. I understand them to be equal "under the limit", but not without. This seems to be in my way of understanding Cantor's infinities.

nhaehnle2• 14 years ago

I'll try to give some short pointers.

1. Matrices represent linear functions between finite-dimensional vector spaces. That is, if you have a function f from V to W that satisfies f(ax + by) = af(x) + bf(y), then there is a matrix A such that f(x) = Ax, and vice versa.

Once you understand that, try to figure out what happens to those matrices when you compose functions. In other words, when you define h(x) = g(f(x)) (assuming that f and g are linear maps that can be composed in this way), then given the matrices for g and f, what will the matrix for h look like? You will end up with exactly the rules for matrix multiplication.

The reason that you call this result "multiplication" is simply that it behaves very much like the multiplication that you are used to from the reals. In particular, you get a ring on square matrices, with (matrix) addition and multiplication that satisfy a distributive law.

2. I personally think that those angles are a bit of a red herring. In particular, the cross product generalizes in a somewhat more complicated way to higher dimensions, and then you have to talk about determinants instead of angles. That would take too much time and space to explain properly here.

3. Set and the "element-of" relation are not defined in the usual sense. They are indeed simply assumed, and you just postulate the properties that they need to satisfy, a.k.a. the axioms of set theory. It's a way of thinking that takes some getting used to, but as an analogy, try to work through Euclid. He doesn't define points or lines, either, but only postulates properties that they need to satisfy.

4. Because 0.9999... is usually interpreted as a real number, and not as an infinite sequence of characters. As a real number, 0.9999... has no meaning except as a limit, and hence they must be equal. As infinite sequences of characters, 0.9999... and 1 are of course different, but that's not how we usually interpret them.

If you think that this is in your way when understanding Cantor's infinities, perhaps you should try to use the diagonalisation argument on infinite bit strings instead of on real numbers. That way, those kinds of subtleties simply do not arise.

alok-g14• 14 years ago

Thanks!!

>> then you have to talk about determinants instead of angles

I certainly never heard this before. (I know how to calculate determinants, but never quite developed intuition around them.) Can you please say some more on this? :-)

alok-g14• 14 years ago

Here is what I am confused about with regards to diagonalisation:

Start with binary non-negative integers: 000 001 010 011 100 101 ... (goes to infinity)

This set now includes all possible bit strings of infinite length since the way these are iteratively generated includes all possibilities.

This is also an enumerable set by definition.

Let's now reverse the bits and put them after a decimal. These are just real numbers now going from zero to one. (This step is actually unnecessary I think.) .000 .100 .010 .110 .001 .101 .011 ...

This must be enumerable set too.

Using diagonalisation argument, .111111 is never to be found in this set. This is exactly where I am stuck. This number comes into the set from flipping of infinity in the original set, which includes all possible stings of infinite length.

I immediately read into your message on treating these as bit strings instead of real numbers. I still am stuck though. (Do you also see a connection to 0.99999... by the way?)

gryphonic• 14 years ago

Im going to let someone give a better response for some of these, but i will touch on a few.

2) The dot product can be thought of the projection of one vector onto another, or: If i shine a light directly at one vector with the 2nd vector in between, how long will the shadow of the 2nd vector in the system be? Do an image search for dot product to see this, and it will become clear (if you know basic trig) why the equation is ABcos(theta).

The cross product is defined as having a magnitude equal to the area of the parallelogram that the vectors create. So have both vectors start at the same point, and then mirror them to make a parallelogram (see wiki on parallelogram).

4) How are the different? If you take 1-0.9999.... = x, what is x exactly? You can't tell me the difference. This is essentially how limits work.

jacobolus3• 14 years ago

For #2 and to some extent #1, see the earlier link I posted about Geometric Algebra. For #1, you want to find a good linear algebra book or resource. I never watched them, but I’ve heard that Gil Strang’s MIT 18.06 lectures are good. http://www.youtube.com/course?list=ECE7DDD91010BC51F8

For #3, try to find Halmos’s book Naive Set Theory. It has a very nice and accessible explanations.

Jun83• 14 years ago

Here's a fantastic article explaining scalars, matrices and tensors: http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/documents/Tens...

Explanation of why dot and cross products are defined that way: http://physics.stackexchange.com/questions/14082/what-is-the...

ColinWright13• 14 years ago

I don't have time now to provide complete answers of the type you want, but I can give you a little insight.

  > Matrices, especially matrix multiplication.

Consider a transformation of space, specifically a shear, or a rotation, or an expansion (or contraction). All of these leave the origin unmoved, and a line will always become a line. They are what we call linear transformations.

So given a point (x,y,z) in 3D space, the new values of x, y, and z are obtained by equations like a * x + b * y + c * z. If you then chase through how to combine two of these transformations, one after the other, the definitions of matrix multiplication drop out.

This, by the way, also explains why you can't always divide by a matrix. Sometimes the transformation collapses the space into a lower dimensional space, and that can't be undone.

  > Dot and cross products.

For unit vectors u and v, the dot product tells you how much of u points in the direction of v. Then we want it to be true that doubling the vector length doubles the size of the product. Then the result you ask about - the magnitude of dot product in 3D is a.b.cos(theta) - automatically follows.

With the cross product, that can be defined as the area of a parallelogram that has the vectors as the side. Again, thinking about that interpretation shows how we get the answer you ask about.

  > How is a set and "belongs-to" operator defined?

Set theory is often regarded as axiomatic, and so you can go back to the axiomatic definitions. We have a universe of discourse that has a bunch of atomic things. A "set" is then a collection of things, and then each set becomes a new thing in our universe. Thus sets can contain other sets.

So a set is defined by what things are in it, so the "belongs-to" concept is fundamental to how the set in question is specified.

  > Why is 0.9999... considered to be "equal" to 1.

People trip up on this in part because they think somehow that something is moving, that 0.999... "approaches" something. But no, when we write "0.999..." we have a representation of a point on the real line. Deciding what it is that you are defining requires that you be clear and detailed about what you think 0.999... means.

So what does it mean? We drop back to saying that it's the limit of the sequence 0.9, 0.99, 0.999, 0.9999, ... so we need to ask what real number can possibly be that limit? Let's call the limit L, and ask what it might be.

Well, clearly:

   L > 0.9
   L > 0.99
   L > 0.999
   L > 0.9999

... and so on.

Now, for every number X that's less than 1, there'll be something in that sequence that will be bigger than X. That means that the limit can't be less than 1. But equally, none of those numbers is bigger than 1, so the limit can't be bigger than 1.

In fact, the smallest upper bound of { 0.9, 0.99, 0.999, 0.9999, ... } is 1.

And then, why should it not be 1? We have different representations of points elsewhere. 2/3 is also 4/6 and 6/9, and sqrt(8) is the same as 2 * sqrt(2). These are simply different ways of specifying the same place on the real line. Likewise, writing 0.9999... is just a different representation of the point more often written as 1.

You said:

  > I understand them to be equal "under the limit",
  > but not without.

You tell me what you mean by 0,9999... and then I'll answer your question in more detail.

  > This seems to be in my way of understanding Cantor's infinities.

Well, there are two different types of infinities based on set theory, but you're probably asking about cardinal infinities, the infinity of counting stuff - "How many things are in this set."

So here, I have a sack with a collection of cubes, and they're numbered from 1 onwards. Call that collection A.

Now let's take a cube and divide one face into two sections. Flip a coin, and write the result in one of the sections. in the other section, divide into two, flip a coin, write the result in one section, and then lather, rinse, repeat. We can fit infinitely many coin flips onto the cube in this way. In sack B put a cube with every possible result of this process.

I claim that the cubes from sack A and sack B cannot be paired off with nothing left over. I won't prove that here, I'm now out of time.

So, ask questions.

alok-g14• 14 years ago

Thanks for a detailed explanation. The one on matrices helps too even though others also posted the same idea.

>> We have a universe of discourse that has a bunch of atomic things. A "set" is then a collection of things

This is where I get stuck. In set theory, they define numbers (and addition) starting from a set and element-of operator. I cannot imagine a bunch of atomic things without imagining numbers/counting first. (PS: "collection" would be another word for a set, so does not help to define it.) So it seems to me that unless set and element-of are left undefined (as others here have suggested), natural numbers are more fundamental than sets.

>> 0,9999... = 1

What I am getting is that the "limit" is implicit in the statement above even though often unstated. The ellipsis is what signifies the limit there, being otherwise not mathematically defined.

>> Call that collection A

I get the precise definition of sack B. I am missing it for sac A. I understand what happens on a given cube, and that cubes in sac A are numbered 1 though infinity. Is there anything to say about the numbers written on two different cubes in sac A?

Basically what you are saying is that sac A cannot have all possible combinations like sac B is defined to have. I am not sure why that is. I see that you just added a comment on another related comment from me. I'll read that first.

shou45772• 14 years ago

I can try:

1. Matrix multiplication is indeed defined oddly. To understand why, it can help to think of matrices as "linear transformations". For example, the matrix

| 0 -1 |

| 1 0 |

corresponds to counterclockwise rotation of vectors by 90 degrees (go ahead, try multiplying the vector <1,0> by this matrix - the answer should be <0,1>). Let's call this matrix A. Also, the matrix

|-1 0 |

| 0 1 |

corresponds to reflection over the y-axis (call this matrix B). If we begin with a vector v = <1,1>, then rotate twice by 90 degrees, then reflect over the y-axis, we should get the vector <1,-1> (it helps to draw a picture, but I am limited in these comments). We can figure this out by repeatedly multiplying by matrices: Av = <-1,1>, A(Av) = <-1, -1>, and B(A(Av)) = <1,-1> (you may want to multiply these out to check, then check by evaluating the operations geometrically).

But wait - there is also a matrix which represents this transformation (that is, the transformation which is rotation counterclockwise by 90 degrees twice then reflection over the y axis). It's the matrix

| 1 0 |

| 0 -1 |

which we'll call C (go ahead and check that this matrix works the way that I just said). Since multiplying a vector by C is the same as multiplying the vector by A then A then B, we have mathematically that

Cv = B(A(Av))

Now, it sure would be nice if there was a way to "multiply" matrices together so that the product B(AA) = C. In abstract terms, we want multiplication of matrices to correspond to composition of functions, since we can use matrices to represent functions. If we define multiplication of matrices in the usual "weird" way, then this property holds!

This lets us examine composing functions by multiplying matrices, which is a very powerful tool indeed. Another interesting thing to notice is that this explains why multiplication of matrices is not commutative (that is, order matters). If we rotate counterclockwise by 90 degrees, then reflect over the y axis, that's different than reflecting over the y axis then rotating counterclockwise by 90 degrees. So BA is not equal to AB.

2. I'm not sure that I understand your question here.

3. This question is beyond my ability to answer, so I'll leave it to someone else.

4. On the one hand, there are tricks to see this (e.g. 0.333... = 1/3, so 0.9999... = 3 * 0.3333... = 3 * (1/3) = 1), but I expect that you have seen them and been unsatisfied by them (I don't believe that these are satisfying explanations, personally, because I have simply replaced the unknown fact 0.999... = 1 with the equally unfounded fact 0.333.... = (1/3)).

In order to really understand what's going on, we have to understand what we really mean by 0.999... What is meant by 0.9? This is equal to 9/10. How about 0.99? This is 9/10 + 9/100. 0.999 = 9/10 + 9/100 + 9/1000. Following this reasoning, 0.999... is equal to the infinite sum 9/10 + 9/100 + 9/1000 + ... + 9/10^k + ...

Since there are infinitely many 9s in the decimal, we must add an infinite number of fractions of the form 9/10^k.

We've opened up a can of worms here - how do we add up infinitely many things? Also, if we add up infinitely many things, how can the result possibly be finite?

A Calculus II course will answer these questions rigorously, but let me attempt to make an explanation here.

Think of it as a game of wits. You start. Choose a number less than 1 - any number at all, just so long as it is less than 1. I'll try to find some number k such that 0.9999...999 (with k 9s in it) is closer to 1 than your number.

Suppose that you pick 0.97. That's pretty close to 1, only 0.03 away. But 0.01 < 0.03, and 1 - 0.99 = 0.01, so 0.99 is closer to 1 than 0.97. I win.

You try again, this time picking 0.999878787. That's much closer to 1, but still 0.000121213. Since 0.0001 < 0.000121213, 0.9999 is closer to 1 than 0.999878787. I win again.

In fact, let's suppose that you pick the number x. Since x is less than 1, 1 - x is some positive (but very small) number. No matter how small it is, though, I can find some k such that 1/(10^k) is even smaller (this fact is called the Archimedean principle, and it is one of the fundamental theorems of the real numbers). Another way to think about it is that 1/(1-x) may be very big, but I can find some k such that 10^k is even bigger. But then 0.999...99 (with k 9s) is going to be closer to 1 than your number, and I'll win again.

Now, what does this mean? Since 0.999... is bigger than 0.999...99 (with any finite number of 9s), it is bigger than every number which is less than 1 (by the game we just played). The smallest number which is bigger than every number less than 1 is just 1. Anything smaller just won't do (since there would be numbers between it and 1). This shows that 0.999... is at least 1.

I hope that helps. It isn't as elegant as the featured answer, but maybe it can be a little bit helpful to you.

alok-g14• 14 years ago

>> 0.333... = 1/3

Indeed, I was aware of this. And as you said, it does not really answer since it just leads to the same question for 0.3333...

I understand the piece on calculus too. My maths background is mostly related to engineering mathematics, and I have generally excelled at it. I later discovered that I am not clear on some "pure" mathematics things. Hypothetically, if counting were taught to me in adulthood, I may have asked a lot of relevant questions that I did not ask in my pre-school and later lost them without knowing them. I am rediscovering some of these questions now.

esrauch3• 14 years ago

Regarding 4, don't believe anyone who claims it's an intuitive result; it is simply a properly of real numbers that you cannot have nonzero infinitesimals and that any two distinct numbers have a number between them that is not equal to either (infinitely many in fact). You can construct alternate number lines that do allow nonzero infinitesimals and then .9999... actually is not equal to 1 under that number line; this number line is a strict superset of Reals just as Reals are a strict superset of Rationals. Most of the practical properties of real numbers would still apply to this new number line; it is actually only a historical convention note that .999... is actually equal to 1.

alok-g14• 14 years ago

Do you know of something that I can read further on this? I wonder about the impact this may have on mathematics.

I did not know of this property of real numbers of nonzero infinitesimals. (Possibly there is where my confusion is starting from.)

gizmo6869• 14 years ago

There do exist number systems where .999.../=1, however, they are not a strict superset of the reals. If it were, then any operation involving only real numbers would behave identicly to the real number system.

Also, this property is not a mere convention, but rather a nessasary result of what we want the number line to be. For example, assume that X<Y. Consider Z=(X+Y)/2. Z=X/2+Y/2. X<Z<Y. I have just shown, using basic algebra, that for any 2 distinct numbers, their is a third number between them. If that were not the case, then at least one of my steps must have been invalid.

Natsu• 14 years ago

Why is 0.999... one? It's an ambiguous decimal expansion.

1 - 0.999... = 0.000... Some part of you might think that there "must" be a 1 at the end of all those zeros. The problem is that there is no end at which to put a 1.

And you'd end up with tiny holes everywhere if they're not equal. 1 = 1/3 + 1/3 + 1/3 = 0.333... + 0.333... + 0.333... = 0.999... oops. Where did our missing one go this time? Surely it's clear from this that we have nothing but 9s in that expansion and that three 3s can never be larger than 9, even if you repeat them over and over forever? And there can't be any funny business going on at the end, because infinite lists do not have ends by definition.

At some point you go through the rules and just accept that this is how they play out and that using other rules just leads to weirdness (AKA "nonstandard analysis").

shou45772• 14 years ago

I have mixed feelings about this explanation.

On the one hand, the mathematician inside of me is raging "this is neither specific, nor rigorous!" In my opinion, this definition is as close to useless as a mathematical definition can be, since the person who has this (and only this) will be hard pressed to answer any other questions involving complex numbers. For example, what is multiplication by i+1? Without further understanding, I don't think such a question can be answered.

On the other hand, the teacher in my head is celebrating. This person managed to take a concept normally feared and hated by the population at large and made it interesting and understandable. The response is well thought out, and is much more engaging to the reader than any response that I could come up with. In short, this explanation is more likely to make the person interested in mathematics, while a more technical definition would probably just make them zone out.

What do other people here think about this? Is the clarity and interest worth the loss of utility? Am I totally off base?

egsmith• 14 years ago

I would argue something is better than nothing. And if this gets people to start using the tool then maybe they will start to hit the cases where the loss of utility matters, and then this can be the source of inspiration to actually tough out the technical inspiration. Probably a long shot in general but it worked that way for me.

agumonkey• 14 years ago

What I love about it is the by-necessity introduction. Limits of previous concepts drive the next idea, I believe that it's a very efficient and general approach to communicate.

Almaviva• 14 years ago

> For example, what is multiplication by i+1? Without further understanding, I don't think such a question can be answered.

i+1 on the plane is a vector at 45 degrees and a scaling of sqrt(2). So multiplying by it scales your multiplier and rotates it 45 degrees. This is a direct extension of the intuitive understanding given.

gizmo6869• 14 years ago

As a math student I can say that if the teacher explains something in a way that

stiff• 14 years ago

A complex "number" (don't think of it as of a number! think of it like you would think of a vector, group, ring or any other abstract structure) is just an ordered pair of real numbers that behaves in a certain predefined way when being added to another complex number or multiplied by it. For an introduction, to avoid unnecessary confusion, it is best to write such "numbers" as ordered pairs using the notation: (a,b). The definitions for the operations are the following:

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

It is useful to have separate names for each part of a complex number, so the a in (a,b) is called the real part, and the b the imaginary part, but for now think about those names as completely devoid of any meaning. Now, observe that under the above definition:

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

But those are, if you consider only the real parts of the complex numbers, ordinary operations on the real numbers! An example consequence of this is that we can take some equation concerning real numbers like:

  2*x + 5 = 21

and write it down in terms of complex "numbers":

  (2,0)*x + (5,0) = (21,0)

Since as we have seen pairs of the form (a,0) behave just like real numbers, we have not changed the meaning of the equation, hence we are free to solve it using the rules of complex algebra and if we happen to arrive at another number of the form (a,0), we can take out the real part of it, plug it into the original equation in terms of real numbers and it is certain to be a valid solution.

This is one of the two properties that makes the use of complex "numbers" fruitful. At the other one we arrive if we now look at "numbers" that are NOT of the form (a,0), for example at a curious property of (0,1):

  (0,1)*(0,1) = (-1,0)

So, in the domain of complex "numbers", the "number" that corresponds to the real number -1, happens to have the equivalent of what we for real numbers call the "square root". We just talk about the "square root", but it is a different operation when we are talking about complex numbers.

Those two properties combined allowed mathematicians to tackle some problems that previously did not have a solution. One example is the problem of finding a solution to cubic equations. The math here gets more complicated, but basically it turns out that by writing cubic equations in real numbers in complex numbers instead, you can find general formulas in terms of complex numbers for finding all the possible solutions, and as we have discussed if applying such a formula in the end yields a number of the form (a,0), it is guaranteed to be a valid solution for the original real equation. Google for "cubic equations cardano" to see the details.

Now, this going back and forth between complex and real numbers is so useful, that for the purpose of brevity mathematicians sacrificed intelligibility and introduced sort of a shorthand notation of the form: a + bi, so instead of writing (0,1) as we did above, we just write i, instead of (5,0) we just write 5, and instead of (1,2) we write 1 + 2i. This is purely a trick, there is nothing magical about the "i", it is just a "dummy" variable that allows convenient carrying out of the operations with pairs described above in the manner reassembling ordinary high-school algebra we all know and love.

All this is maybe a bit elementary, but I think this is the part most people fail to understand and because of this start treating complex numbers as something mysterious. There is in fact nothing mysterious about them, you have to boil every application you see of them to the above and then you will get a clear understanding of what is happening and why they are useful. Points on the plane happen to be a model for complex numbers with rotation corresponding to multiplication and so forth, this is of course very interesting, but I feel an introduction to the topic should start with what I have just tried to explain.

Jun83• 14 years ago

"A complex "number" (don't think of it as of a number! ..."

Excellent point! Another example of misleading mathematical terminology is "random variable", which are not random or variable but instead are well-defined mappings.

psykotic3• 14 years ago

> Another example of misleading mathematical terminology is "random variable", which are not random or variable but instead are well-defined mappings.

Indeed. Kolmogorov's random variables are a special case of observables in mathematical physics.

In classical physics, the state space for a free particle is S = R^3 x R^3. As an observable we might take the particle's velocity in some particular direction, which defines a mapping f : S -> R. We want such mappings to respect the relevant structure of the state space. A classical state space usually has the structure of a smooth manifold, so the mappings should be smooth as well.

In classical stochastic physics, the states now have the structure of a probability space. We want the mappings to be measurable so we can take the preimage of a measurable set of observable values to calculate its probability. This is exactly the Kolmogorov definition of a random variable.

Quantum theory doesn't quite fit into the above scheme, but there are several ways these three cases can be unified. For example, from the viewpoint of C* algebras of observables, classical systems have commutative algebras and quantum systems have noncommutative algebras.

By the way, there is a connection to monads that I find enlightening. With monads, there is both an internal semantics and an external semantics. From the internal point of view, random variables are indeed random and variable. From the external point of view, they are neither. (This can be given a Bayesian spin by thinking of the internal and external observers as someone with respectively imperfect and perfect knowledge.) This is analogous to someone's perspective on the state monad from the internal and external points of view. From the viewpoint of someone who lives in the state monad, the same expression can return different values depending on context. But from an outsider's explicit state-passing point of view, everything is referentially transparent.

michaelochurch• 14 years ago

The way I think of "random variable" is that it's an attempt to move toward a stateless world (a la functional programming) so we can reason about it.

What we actually observe in real life is a sample from the corresponding distribution. The "random variable" concept (a function over a sample space with a probability measure) is the stateless analogue that is more usable in abstract reasoning. In probability, we don't care about the specific values the variable actually ends up producing, but all the values it can produce and how likely each is.

In the observed world, it's unknown (pre-measurement) and appears to be random. In the mathematical world, it's a function over a sample space.

It's a mechanism to reason about randomness by (though the magic of probability) taking the randomness and state out.

derleth2• 14 years ago

> don't think of it as of a number!

By the same token, you shouldn't think of a real "number" as a number, but as an infinite sequence of rational numbers. Yes, all of the reals are infinite sequences of rationals.

Also, a rational isn't a number but an ordered pair of integers. (Technically, the set of the rationals is the set of all ordered pairs of integers (excluding the ones with zero in the second position) with an equivalence class defined on them.)

The integers are also not numbers, but the naturals with a sign adjoined to them.

Finally, the naturals aren't numbers but sets of sets, with zero the empty set and all further integers built inductively from that.

So... what does it really gain you to look 'inside' the numbers you use?

btilly6• 14 years ago

One minor correction. The integers are more naturally defined as an equivalence class of pairs of natural numbers, with (a, b) equivalent to (c, d) if and only if a+d = b+c. (Your approach does not eliminate the equivalence class issue, because you have to specify that +0 and -0 are the same.)

In this formulation, you define (a, b) + (c, d) to be (a+c, b+d). And you define (a, b) * (c, d) to be (a * c + b * d, a * d + b * c). After a little work you can prove that both operations send equivalence classes to equivalence classes.

Then you can embed as follows. The natural n gets mapped to the integer (n, 0). The integer m gets mapped to the rational (m, 1). The rational r gets mapped to the real (r, r, r, r, ....). And the real x gets mapped onto the complex number x + 0i.

btilly6• 14 years ago

I saw, and experienced, this approach in school and see no value in it for improving your understanding complex numbers. The point of the approach is to teach students about abstraction and formalism, and complex numbers happen to be a convenient place to do it. But the formalism is a barrier for building up a more convenient mental model about what is really going on.

Before disputing this, in any calculation that you've ever done by hand with complex numbers, do you naturally write it as (a, b) or a + bi? I always do the latter, and it saves me both time and conceptual effort.

And a random note. If you go on past advanced Calculus, you'll encounter two subjects that take Calculus and go back to the basic foundations and build them up. The first is real analysis, for which you have to learn all of the ways that things fail to work out like you would want them to. The other is complex analysis, where you wind up learning all of the ways that everything has to work out amazingly perfectly.

The difference between the two subjects is that "differentiable" in the 2-dimensional structure of complex numbers is a far, far stronger condition than "differentiable" is for the real numbers. Indeed there actually exist functions that you can construct which are infinitely differentiable everywhere in the real numbers, but for which on no interval can you extend them to a function that is differentiable in the complex plane.

brazzy2• 14 years ago

> But the formalism is a barrier for building up a more convenient mental model about what is really going on.

Actually, I'd say that complex numbers are a wonderful opportunity for illustrating that math is nothing but a mental model, not necessarily with anything "really going on", AND at the same time incredibly useful.

IMO the best way to think about complex numbers is to start with the observation that sqrt(-1) cannot exist in the realm of real numbers, quite easily provable. Then you make the bold-assed assumption that it exists anyway and simply call it i. Then play around with it a little, and find out that adding the axiom "there is something called i, where i * i == -1" does not lead to any contradictions; instead you can work with this i just fine, and numbers of the form a+bi are closed under all your everyday operations, just by applying your everyday arithmetic rules. Neat!

Then you notice that those numbers can be interpreted as 2D vectors and some operations on them as geometrical transformations thereof. Super Neat!

Then someone notices that those numbers and the operations on them can actually be used to model certain aspects of the physical world. Holy Crap!

Math is FREAKING AWESOME!

jleader• 14 years ago

"Young man, in mathematics you don't understand things. You just get used to them."

-- John von Neumann

For what it's worth, I have sometimes manipulated complex numbers as (a, b), usually when using them as 2-D vectors. But I make no claims to being a mathematician.

kahirsch• 14 years ago

I disagree with this for several reasons.

First, I find that having more different ways to look at a problem, the better I am able to deal with it. I can look at an equation algebraically, or as a graph, e.g. I can use rectangular coordinates or polar coordinates. I can look at complex numbers as abstract entities or as points in a plane.

Second, if you look at the history of complex numbers, mathematicians were just not sure what to make of them, and had no way to have confidence that what they were doing was even consistent. Being able to interpret them as point in a plane with intuitive geometric operations gave them a huge boost.

Third, thinking of them this way led to the search for generalizations. Gauss and Hamilton tried to find a way to do arithmetic in three dimensions, or prove that it couldn't be done. Hamilton eventually found the four-dimensional quaternions. And the (ac-bd,bc+ad) definition was generalized to the Cayley-Dickson construction.

jasomill2• 14 years ago

That's not unique to complex analysis; there are smooth real functions that are nowhere real analytic. It's far easier to come up with (and understand!) examples of functions that are smooth when considered as real maps from R^2 to itself, but nowhere homomorphic when considered as complex functions of a single complex variable, though. Complex conjugation is an obvious example: as the real map (x,y) -> (x,-y), it's linear, so smooth, but only holomorphic at points where 1 = -1, so nowhere. Complex conjugation restricted to the real line does have an obvious holomorphic extension, of course!

lmkg• 14 years ago

Understanding a mathematical concept doesn't mean being able to perform computations using it. It means having an intuitive understanding for what it can represent and how to use it, and how to interpret concepts that use it.

Defining complex numbers as a bunch of arbitrary arithmetic operations on tuples lends nearly zero understanding, no matter how good you get at performing that arithmetic. Understanding complex numbers as rotations is much more useful in being able to understand, say, electromagnetism. In fact, I'll go so far as to say it makes the physics easier even if it makes the math more cumbersome.

Teaching mathematics as simple rules for manipulation of symbols is basically like telling only the punchline to a joke. Not even saying the punchline early, but just saying it on its own and providing no context. The fact that math can be (not is) just a bunch of mechanical rules is amazing when those rules can be mapped to complex and abstract phenomenon, and used to provide deeper understanding of them. The mechanical rules, in themselves, are boring. Their boringness is actually what makes them a good punchline, when contrasted with the intricacy of the systems they describe, but without that set-up, they're just boring.

aaronbrethorst• 14 years ago

Thank you for explaining in three simple paragraphs why I hated every calculus professor I ever had.

jasomill2• 14 years ago

I'd highly recommend Tristan Needham's Visual Complex Analysis to anyone interested in the subject, for exactly this reason . . . and, subsequently, Henri Cartan's Elementary Theory of Analytic Functions of One or Several Complex Variables as proof that, nevertheless, "algebraic" need not imply "boring" or "computational". As a silly example, I'll never forget Cartan's definition of "2π" as the unique positive real number such that the kernel of the homomorphism "t -> e^it" from the additive group of reals to the multiplicative group of unit-length complexes is the set of integer multiples of 2π.

pm90• 14 years ago

Maybe It's because my mind is different, but I actually found his explanation very nice. I understand that it looks like just "symbol manipulation" but that is just one way to look at it. He shows how to define complex numbers using the bare minimum assumptions. As he mentioned, the reason why complex numbers are useful is precisely because of their properties under addition, subtraction and multiplication.

Like I said, it just might be because our minds are different. But this seems like a good foundation, i.e you first show that this is all there is to it. Now, you can go ahead and explain the nice physical/geometric interpretations.

Incidentally, this is the way that most books introduce Algebra (well, mathematical algebra, with groups, rings etc, not the algebra from high school)

karpathy• 14 years ago

I was going to post something like this but you did a much better job of it.

In some sense, TLDR: "Imaginary number" is a misnomer that confuses people. 1+2i is also just a convenient but potentially confusing notation to those who don't know what it is a short-hand for.

Complex numbers are basically 2D vectors with a funny multiply operation ((a,b)(c,d) = (ac-bd,bc+ad)) and it just so happens that defining it this way leads to some interesting properties and convenient formalism when dealing with many problems.

EDIT: there is also a nice graphical way of interpreting what the operator does. It adds the angles of the original vectors (wrt x-axis) and multiplies their lengths. Better explained has a nice article on this: http://betterexplained.com/articles/intuitive-arithmetic-wit...

gizmo6869• 14 years ago

Complex numbers are not vectors. If you have (a + b) in an expression, you can do algebra with it. If you then say b=ci, (a + b) would be a complex number, but the work you just did as if it were 2 real numbers is still valid.

derleth2• 14 years ago

OK, my main point restated:

A number is whatever we want it to be, whatever it is useful to treat as a number.

That's it. That's the whole gag. Saying something isn't a number is pointless: It is if it can be, and it isn't if it's more interesting to treat it as something else.

This leads to something more fundamental:

Math is all about modeling. Math is a language for making models that are logically consistent.

Confusing a model with the thing being modeled is wrong. Saying complex numbers aren't numbers because they're rotations is going at it backwards: Complex numbers can be used to model rotations. That doesn't tie them to that one model. They can be used in other ways, too.

gizmo6869• 14 years ago

Math isn't about moddeling, moddeling is applied math.

Math is about making up rules and figuring out what stuff happens with said rules.

gizmo6869• 14 years ago

I've always viewed complex numbers in the opposite way.

Start with the set of real numbers. Then join to that set the symbol i, with the property that i*i=-1, (probably other stuff in formal math).

Then, out of convenience, treat expressions of the form a+bi as a single number.

john_gabriel• 14 years ago

Complex numbers are ill-defined and real numbers don't exist either.

http://thenewcalculus.weebly.com

math33• 14 years ago

Isn't 4 rotations 4i? Why is it i^4

Evbn3• 14 years ago

Rotation is multiplication. Translation is addition.

joe_the_user• 14 years ago

Aside from people's very worthwhile answers describing the complex number system, I think it is worth mentioning that the use of the term "imaginary" is an unfortunate historical remnant. In experience, a lot of the average student confusion comes from their trying to get their head around the naive meaning of imaginary.

Now that modern mathematics understands that all number systems are more or less games with axioms, we know that no part of a number system is really more imaginary than any other part. "Imaginary" might better be termed "augmented" - we can augment the "real" number system by adding an element "i" which we say is equal to the square root of -1.

And it is just as ironic that the "real" number field itself has perhaps as many weird elements as say the rational "imaginary numbers" (pi, e, Theta etc)

Evbn3• 14 years ago

Those are the mundane ones. Almost all real numbers (except for set of set of measure 0) cannot even be given names, and are thus beyond imagintion.

zvrba2• 14 years ago

Also ironic is that real numbers may not be "real" (i.e., exist) at all. The uncountable part which is THE part that completes rationals to continuity cannot be described or generated in any way since there are only COUNTABLY many different computer programs (or mathematical formulas).

erichocean• 14 years ago

Complex numbers are cool, but automatic differentiation using dual numbers is even cooler.

scotty79• 14 years ago

I tend to think that Real numbers is the first kind of numbers that have nothing to do with reality. There are no perfect circles, squares or triangles in reality. Real things can only look like circle if you don't look close enough. I think same goes for sinusoidal waves and everything else.

All fundamental physical laws that contain e or pi or event sqrt(2) seem suspicious to me.

zvrba2• 14 years ago

From the OP's question:

> When I tried to calculate the square root of -1 on my calculator, it gave me an error.

This was a red flag to me. This person takes the answer from a calculator as the ultimate truth, whereas it (obviously) is not. I wonder how he would have reacted had he had an advanced calculator (HP50g or some TI) and got "i" as the answer instead of an error.

> To this day I do not understand imaginary numbers. It makes no sense to me at all.

Coming from a mixed EE/CS background, I had a lot of math and math-heavy classes in the first 2.5 years (linear algebra, real, vector and complex analysis, discrete math, systems theory)..

What I observed during the 5-yr master's degree was that the people who were seeking "meaning" in math courses were the same people who 1) had the most problems with math exams, 2) had the least ability to apply the math to other domains in order to do something useful.

What worked for me was to accept that math is manipulation of abstract symbols according to some rules, combined with some ingenuity (i.e., when you have to consider external information in addition to the rules).

In the case of complex numbers, you introduce a new abstract symbol called "i" with the property that i X i=-1. Other rules that you (hopefully) learned earlier still apply. So computing (x+y)X(x+z) or (2+3i)X(2-6i) is in essence the same process except that you replace i^2 with -1 when you encounter it.

I actually enjoyed doing math on that level: as a manipulation of abstract rules and invention of new rules (theorems).

In simple EE, inducntances and capacitances are modeled as pure imaginary "resistances"). However, I somehow figured out how to make use of math (not only complex numbers) in other domains. I made use of math in electronics for drawing Bode plots, information theory, systems theory, and even economics! (In economics, I was reading a lengthy chunk of text with some summations, only to realize that, in actuality, integration was described and everything could be summarized to few sentences explaining the meaning of the involved integral. On the exam, I gave MY explanation -- based on the integral -- instead of the lecturer's, and I passed!)

To me the problem seemed to be that lecturers in other subjects resorted to awkward, special-cased explanations instead of showing us how to model the problems at hand with the math we have already learned.

What are imaginary numbers?

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