10 ms·

What are imaginary numbers?

217 points by arcatek 14y ago

fchollet 14y 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?
- deleted 14y ago
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- walrus 14y ago
  Trivially, this result is clearly obvious.
  - Tipzntrix 14y 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.
- friendlyfrog 14y ago
  No it's not. i is a number such that i^2 = -1 by definition.
  - gizmo686 14y ago
    And in some contexts -1 is rotating by 180 degrees, which happens to be 290, or i*i.
T-hawk 14y 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-to-imaginary-numbers/ http://betterexplained.com/articles/a-visual-intuitive-guide... https://news.ycombinator.com/item?id=2712575 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.
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- JadeNB 14y 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 14y 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.
- kalid 14y 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).
- diminish 14y ago
  One more question; what is the "i"th root of i? Please try to use the angle metaphor :)
  - ColinWright 14y 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.
jacobolus 14y 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 http://geocalc.clas.asu.edu/pdf/OerstedMedalLecture.pdf
- tgb 14y 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.
  - jacobolus 14y 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/ http://www.geometricalgebra.net/
- jamii 14y 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 14y 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)
Jun8 14y 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 14y 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 http://www.youtube.com/watch?v=Iqf3rvg742g
yason 14y ago
The obligatory, if you're still scratching your head thinking about complex numbers: http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/ http://betterexplained.com/articles/a-visual-intuitive-guide...
jostmey 14y 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.
- yequalsx 14y ago
  EDIT: I proved the wrong thing. Correct proof is in post below.
  - alok-g 14y ago
    Can you please complete your proof? How do I go from 1.(-1) = -1 to (-1)*(-1)=(+1)?
    - yequalsx 14y ago
      Shoot. I lost sight of the goal. Sorry. a·0 = 0 for all numbers a Thus it is true that –1·0 = 0 Replace 0 on left hand side with (–1 + 1) to get –1·(–1+1) = 0 Now use the distributive property –1·(–1) + (–1)·1 = 0 Keeping in mind that 1·a = a for all numbers gives us –1(–1) + (–1) = 0 Thus –1(–1) is a number such that when added to –1 gives 0. Hence –1(–1) must be the opposite of –1 which is 1.
      - esrauch 14y ago
        You are assuming that additive inverse is unique. Also you are assuming that -1(-1) is some value at all; you could write an extremely similar proof that leaves you with 0/0 = 1 and it would be faulty to conclude that 0/0 actually is 1.
friendlyfrog 14y 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 14y ago
  You're thinking of complex numbers.
radicalbyte 14y ago
The best explanation I've seen yet is one Hobbes gave to Calvin: http://www.gocomics.com/calvinandhobbes/2009/02/18/ http://www.gocomics.com/calvinandhobbes/2009/02/18/
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luser001 14y 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-hawk 14y 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.
  - luser001 14y ago
    Ok, thx. I think I understand now.
- irishcoffee 14y ago
  * i = sqrt(-1) * i * i = i^2 = sqrt(-1) * sqrt(-1) = -1 .. i^4 = i^2 * i^2 = -1 * -1 = 1
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grandalf 14y ago
I have started reading the story of i: http://www.amazon.com/An-Imaginary-Tale-Story-square/dp/0691027951 http://www.amazon.com/An-Imaginary-Tale-Story-square/dp/0691...
alok-g 14y 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.
- nhaehnle 14y 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.
shou4577 14y 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 14y 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 14y 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 14y 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.
- gizmo686 14y ago
stiff 14y 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.
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math33 14y ago
Isn't 4 rotations 4i? Why is it i^4
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- Evbn 14y ago
  Rotation is multiplication. Translation is addition.
joe_the_user 14y 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)
- Evbn 14y 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.
- zvrba 14y 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 14y ago
Complex numbers are cool, but automatic differentiation using dual numbers is even cooler.
scotty79 14y 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.
zvrba 14y 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.

217 points by arcatek 14y 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?

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walrus 14y ago
Trivially, this result is clearly obvious.
- Tipzntrix 14y 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.
friendlyfrog 14y ago
No it's not. i is a number such that i^2 = -1 by definition.
- gizmo686 14y ago
  And in some contexts -1 is rotating by 180 degrees, which happens to be 290, or i*i.

T-hawk 14y 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-to-imaginary-numbers/ http://betterexplained.com/articles/a-visual-intuitive-guide... https://news.ycombinator.com/item?id=2712575 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.

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JadeNB 14y 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 14y 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.
kalid 14y 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).
diminish 14y ago
One more question; what is the "i"th root of i? Please try to use the angle metaphor :)
- ColinWright 14y 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.

jacobolus 14y 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 http://geocalc.clas.asu.edu/pdf/OerstedMedalLecture.pdf

tgb 14y 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.
- jacobolus 14y 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/ http://www.geometricalgebra.net/
jamii 14y 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 14y 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)

Jun8 14y 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 14y 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 http://www.youtube.com/watch?v=Iqf3rvg742g

yason 14y ago

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

jostmey 14y 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.

yequalsx 14y ago
EDIT: I proved the wrong thing. Correct proof is in post below.
- alok-g 14y ago
  Can you please complete your proof? How do I go from 1.(-1) = -1 to (-1)*(-1)=(+1)?
  - yequalsx 14y ago
    Shoot. I lost sight of the goal. Sorry. a·0 = 0 for all numbers a Thus it is true that –1·0 = 0 Replace 0 on left hand side with (–1 + 1) to get –1·(–1+1) = 0 Now use the distributive property –1·(–1) + (–1)·1 = 0 Keeping in mind that 1·a = a for all numbers gives us –1(–1) + (–1) = 0 Thus –1(–1) is a number such that when added to –1 gives 0. Hence –1(–1) must be the opposite of –1 which is 1.
    - esrauch 14y ago
      You are assuming that additive inverse is unique. Also you are assuming that -1(-1) is some value at all; you could write an extremely similar proof that leaves you with 0/0 = 1 and it would be faulty to conclude that 0/0 actually is 1.

friendlyfrog 14y 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 14y ago
You're thinking of complex numbers.

radicalbyte 14y ago

The best explanation I've seen yet is one Hobbes gave to Calvin: http://www.gocomics.com/calvinandhobbes/2009/02/18/ http://www.gocomics.com/calvinandhobbes/2009/02/18/

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luser001 14y 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-hawk 14y 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.
- luser001 14y ago
  Ok, thx. I think I understand now.
irishcoffee 14y ago
* i = sqrt(-1) * i * i = i^2 = sqrt(-1) * sqrt(-1) = -1 .. i^4 = i^2 * i^2 = -1 * -1 = 1

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grandalf 14y ago

I have started reading the story of i: http://www.amazon.com/An-Imaginary-Tale-Story-square/dp/0691027951 http://www.amazon.com/An-Imaginary-Tale-Story-square/dp/0691...

alok-g 14y 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.

nhaehnle 14y 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.

shou4577 14y 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 14y 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 14y 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 14y 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.
gizmo686 14y ago

stiff 14y 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.

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math33 14y ago

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

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Evbn 14y ago
Rotation is multiplication. Translation is addition.

joe_the_user 14y 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)

Evbn 14y 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.
zvrba 14y 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 14y ago

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

scotty79 14y 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.

zvrba 14y 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.