What are imaginary numbers? • Hacker News Board

Re: What are imaginary numbers?

#91

Post by karpathy » Fri, Sep 21, 2012, 8:55 PM UTC

stiff wrote: post #49

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)…

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...

Re: What are imaginary numbers?

#92

Post by dllthomas » Fri, Sep 21, 2012, 8:55 PM UTC

diminish wrote: post #17

T-hawk wrote: post #4

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,…

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

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)).

Re: What are imaginary numbers?

#93

Post by alok-g » Fri, Sep 21, 2012, 9:00 PM UTC

ColinWright wrote: post #82

alok-g wrote: post #67

Earlier quoted context omitted.

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 f…

>> We have a universe of discourse that has a bunch >> of atomic things. A "set" is then a collection of things > 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. The point is that you can start with nothing, and define the set that has nothing in it. That's {}. Now we have one thing…

>> The point is that you can start with nothing, and define the set that has nothing in it. That's {}. Now we have one thing. We can define a set containing that, so we have { {} }. Now we have two things, and we can definea set containing both of them, and so on.

This helps a lot. I get it finally. :-) Thanks!

Re: What are imaginary numbers?

#94

Post by yequalsx » Fri, Sep 21, 2012, 9:03 PM UTC

esrauch wrote: post #66

yequalsx wrote: post #41

Earlier quoted context omitted.

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.

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.

Additive inverses are unique in any ring. The subtle things I skipped over because of the technicalities involved is how do I know the distributive property holds for negative integers. Indeed, what is a negative integer? How does one get them from the natural numbers? Suffice it to say that this can all be defined in a consistent, precise way and everything works out.

Re: What are imaginary numbers?

#95

Post by ColinWright » Fri, Sep 21, 2012, 9:05 PM UTC

alok-g wrote: post #85

ColinWright wrote: post #79

Earlier quoted context omitted.

> For the first set, I meant to write: > [Prepend each string with infinite zeroes] > ...000 > ...001 > ...010 ... If there are infinitely many zeros on the front, you can't actually append anything. That doesn't end up being well-defined. (Well, actually, there are transfinite ordinals, but that would confuse the issue. It's not what you mean, and it doesn't help) > Now all bit strings here have infinite length. If…

>>> [Prepend each string with infinite zeroes] > ...000 > ...001 > ...010 >> If there are infinitely many zeros on the front, you can't actually append anything. I am lost. In this first set, I do not have a decimal point anywhere. Why cannot I have an infinitely many zeros to the left of 1. It will still be just one when looked at as a number. >> Strings have a start, then they go on one place by one place As far as…

OK, so you're not talking about the usual diagonalisation. I'll try to follow what you've said and respond as I go.

  > In this first set, I do not have a decimal point anywhere.

OK, fine. But then you talked about flipping them around to come after the decimal point. When you do that you have only those strings that only have a finite number of 1s in them.

  > As far as representing it as a string, I may still start
  > from the right and work towards the left.

Yes you can, but that's not what people do when talking about Cantor and diagonalisation, so it's now completely unclear what you're talking about.

However, you start with finite strings of 0s and 1s, basically the non-negative whole numbers, represented as binary strings. Note that these are all finite, and the non-zero parts are still finite, even if you prepend an infinite number of 0s.

  > Stating the above for the decimal case, let n=1 be the
  > place right after the decimal, n=2 to the right of it, and
  > so on.

See, now you're talking about stuff after the decimal point. I'll continue ...

  > Now for place n=1, the both zero and one are covered
  > (first two cases below).

But for diagonalisation that's irrelevant. We only ask what is the first digit of the first number.

  > For n=2 place, again both zero and one are covered for
  > all possible combinations above for n=1 place

Again, irrelevant. We only ask what is the 2nd digit of the 2nd number.

  > 0.000000000000...
  > 0.100000000000...
  > 0.010000000000...
  > 0.110000000000...
  > 0.001000000000...
  > 0.101000000000...

So here if we construct the diagonal of this sequence as you've listed it we have 0.00000... Let me highlight the diagonal for you from this quoted section:

  > 0.0xxxxxxxxxxx...
  > 0.x0xxxxxxxxxx...
  > 0.xx0xxxxxxxxx...
  > 0.xxx0xxxxxxxx...
  > 0.xxxx0xxxxxxx...
  > 0.xxxxx0xxxxxx...

If we now flip this we get 0.11111....

Now observe that all of the strings you have contain only a finite number of 1s. That means that 0.11111... is not in your sequence.

  > Using the mathematical induction argument,

You haven't made an induction argument.

  > ... all combinations are covered.

For each place, both possibilities are eventually covered. but for each sequence that you give, it is eventually all zeros.

  > This must include 0.1111111111...

No, it doesn't.

  > It sits exactly where (simple) infinity sits in
  > the enumerable set {0, 1, 2, 3, ... }

Infinity does not fit in that set, and 0.1111... is not in your defined set of numbers.

Re: What are imaginary numbers?

#96

Post by ColinWright » Fri, Sep 21, 2012, 9:09 PM UTC

alok-g wrote: post #90

ColinWright wrote: post #82

Earlier quoted context omitted.

>> We have a universe of discourse that has a bunch >> of atomic things. A "set" is then a collection of things > 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. The point is that you can start with nothing, and define the set that has nothing in it. That's {}. Now we have one thing…

>> You can't say that with precision, because it is infinity you are struggling with. What you can say is that for every number n there is a cube with n on it, and that every cube has exactly one number written on it. This may be nailing it -- struggling with infinity. When you say "every number n there is a cube with n on it", what does "every" mean. Does every number include infinity? Or should not not consider inf…

  > When you say "every number n there is a cube with n on it",
  > what does "every" mean.

To be more precise, every finite number.

  > Does every number include infinity?

No, in these sorts of discussions infinity is never considered to be a number. You need explicitly to be discussing transfinite arithmetic, and we're not.

  > Or should not not consider infinity to be a number?
  > If the latter, this is probably where I went wrong.

Absolutely you should not be thinking of infinity as a number.

Re: What are imaginary numbers?

#97

Post by joe_the_user » Fri, Sep 21, 2012, 9:12 PM UTC

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)

Re: What are imaginary numbers?

#98

Post by chimeracoder » Fri, Sep 21, 2012, 9:19 PM UTC

ArbitraryLimits wrote: post #36

T-hawk wrote: post #4

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,…

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.

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.

Re: What are imaginary numbers?

#99

Post by jostmey » Fri, Sep 21, 2012, 9:46 PM UTC

jostmey wrote: post #18

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 diff…

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).

Re: What are imaginary numbers?

#100

Post by brazzy » Fri, Sep 21, 2012, 9:47 PM UTC

jostmey wrote: post #18

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 diff…

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.