# Anyone here ever got bugged with i?

1. Jan 7, 2012

### fbs7

I know how to use i... and I know that complex numbers are mighty useful (I'm an electronic engineer, so I used them a lot)... and how they keep popping all over the place in physics. So I'm sure the algebra behind i works.

But, it never bugged anyone here that we use something that doesn't really exist?

I mean, one of the first things we learn in Logic is that from a falsehood you can derive either true statement or a false statement. So while I used i a lot, I cannot claim I can say that "i such that i2=-1" exists... just that the algebra built around it works.

It's a similar thing to this: I know that there is no ρ such that ρ = 0 and ρ = 1. But if it existed, then one could find it has such-and-such very practical properties. Perhaps it could be used as a shortcut for some mathematical calculations, even though it is impossible, doesn't exist, doesn't make sense and can never be measured.

I have been mighty bugged with i for the last 20 years: one because it works, and two because everybody love something that is impossible and illogical.

Anyone else ever felt similarly helpless with i? That we just use it because we were told to use and it works, but if we had been left to our own devices we would never think of something as crazy?

2. Jan 7, 2012

### nonequilibrium

What do you mean by existing?

Or better yet: why do you say that there is no i with i*i = -1?

Sure, there is no real number with that property, but you must realize that the term "real number" is just that: a term. The adjective "real" doesn't mean that it truly exists in some sense, whatever that means for mathematical objects.

In what sense does the "existence" of $\pi$ and i differ for you? Have you ever seen $\pi$ walking on the street? You know $\pi$ because it was defined to you. Same thing with i, one can define it. The only thing that can go wrong when defining something, is that its definition brings in a contradiction, but one can verify that the definition of i does no such thing, so all is good.

So maybe your problem is that nobody properly defined i for you? There are various ways to define it. One way is to take $\mathbb R^2$ and define an operation on it such that $(0,1) * (0,1) = (-1,0)$. One denotes $(0,1)$ as i. Anyway there are plenty of sources to turn to for a definition of i.

But I'm going to guess what your objection might be: "But whenever I take x things x times, I will never get a negative amount of things!" out of which you would conclude that if a certain x had such a property, it couldn't be real. I suppose there are several objections to this, but they all come down to: you're looking at it too naively.

First of all: now you're taking an example out of physical reality, but mathematics isn't a description of our physical reality, cause that would make it physics. Math doesn't have to conform to reality. Indeed, mathematicians long thought that it had to, but in the 19th century they realized this believe was unfounded. It grew historically out of examining the parallel postulate: it's Euclid's 5th geometric postulate that says that if given a line and a point outside of the line, there is one and only one line parallel to it (and through the point). Mathematicians tried for centuries to prove it from the other four postulates, until Beltram (1868) proved that the truth value of the parallel postulate was independent of the previous four. In other words, one could change the parallel postulate to e.g. "Given a line and a point outside of it, there is no line through it that stays parallel" without giving any contradiction with the other postulates. Reality might say otherwise, but the mathematics doesn't care. Interestingly, it was long thought that this new postulate was spielerei: meaningless mathematics, no relation to reality. But nowadays, with the universe possibly being a closed (hyper)sphere, such a postulate could actually be more physical than Euclid's.

Secondly: have you ever taken $\pi$ times something?

3. Jan 7, 2012

### Number Nine

How does it "not exist"? The complex numbers are just a normed division algebra, like the real numbers. It's kind of arbitrary of you to declare that one-dimensional division algebras exist but two-dimensional ones don't when we can clearly construct both. Some complex numbers just happen to have the property that squaring them gives you -1. Nothing unusual here.

People balked at negative numbers when they were introduced, too ("how can you have negative of something?!?"), but that doesn't change the fact that they exist in the same sense as the positive integers.

Last edited: Jan 7, 2012
4. Jan 7, 2012

### fbs7

These are great points, but $\pi$ is different. It's defined from a ratio from a geometric figure, so there's no issue about it's existence, but only that it is the same for said figures, whatever their size.

Now, the definition you referred is of course completely tight and correct, and it of course works beautifully. But then you are defining two real numbers that make a vector...

Perhaps I should just get in my head that the problem is that everybody calls it incorrectly, that is they call i "an" "imaginary" "number" when it is not one (it's two numbers), it's not imaginary (it's a pair of real numbers) and is not a number (it's a vector).

You know, it would help if Mathematicians used the correct notation... that e$\pi$.i = -1 should instead be e($\pi$,0) . (0,1) = (-1,0). Darn shorthand...

5. Jan 7, 2012

### chiro

You should realize that the development of i is quite a large and complex one.

Cardano found that certain cubic equations ended up having roots with imaginary numbers in them, and when he was a bit uncertain, he ended up plugging them into his equations and surprisingly (unsurprising for us, but back then this was like quantum mechanics), it worked very well.

Fast forward to Euler and Gauss and you get things like the fundamental theorem of algebra. It was this result that said that imaginary numbers are required as a general thing when trying to find say the roots of an equation.

Also from a geometric perspective, it turns out that i gives us a natural way to do geometry, since rotations in the plane are used by using i. Rotations, especially in many dimensions form a foundation for many things including mechanics and higher level geometry.

So at least from an algebraic and geometric sense, these things do make sense if you follow the development of these fields from say Cardano onwards (or at least Gauss and Grassman onwards).

Another thing to think of is the generalization of many functions.

For example the Gamma function generalizes the factorial function to real numbers in-between.

Consider the function (-1)^n where n is an integer. We can plot these points where they oscillate between 1 and -1 indefinitely. Now consider the analogue of what Euler did with the Gamma function. It is expected that there would be some kind of interpolation for fractional values and everything inbetween.

Euler and other people like Demoivre figured out the generalization in terms of trigonometric functions.

6. Jan 7, 2012

### fbs7

Well, as an "imaginary" "number", i doesn't really exist. The name "imaginary" acknowledges that.

From Mr. Vodka's explanation, i is a vector, or a matrix, not a number. It doesn't help when Mathematicians and Physicians use it freely as if it was a number, instead of calling it a vector. It's extremely confusing.

About negative numbers, that's another great point. I know how to define N in terms of set theory, and I guess one can define Z from N with the help of "+" and "-", and from that obtain Q with the help of "/". All that seems pretty straightforward and easily translated to everyday experiences.

But, to be honest, I have no clue how to define R. I have read about Cauchy sequences and
Dedekind cuts, and that just seems complex and out of the hat. How in the world can anyone deduce that "real numbers" have anything to do with "real world"... that is completely surprising to me. Yet we use those buggers every day - ay-ay-ay, another terrifying thing to keep me awake at night.

7. Jan 7, 2012

### Number Nine

...no.
"Imaginary" is a name. It's a sequence of phonemes/graphemes that denotes a particular mathematical construction; it has no philosophical significance. We didn't find the "natural" numbers lying around the forest somewhere...

8. Jan 8, 2012

### Jamma

Well, you seem to think that real numbers "exist" (whatever that means in your usage, I don't really see what you are getting at in your usage of the word "exist" to be honest). Well, the complex numbers are just constructed from the real numbers with some defined operations, so don't they exist too?

Let me ask you this- do vectors "exist"? Complex numbers are just vectors with a method of multiplication defined on them.

No, there is no "correct" notation. And anyway, conceptually, using i is extremely useful. It has led mathematicians to deal with things not thinking of the complex numbers as a 2 dimensional number system, but as something with one complex dimension (e.g. complex manifolds). Anyway, using i instead of (0,1) is far less cluttered, I don't see what advantage your "correct" notation has.

9. Jan 8, 2012

### HallsofIvy

Staff Emeritus
fbs7, you basic problem then is that you do not know what a "number" is- or what it means for a number to exist. If you are going to take the term "imaginary" litterally, then all numbers are "imaginary"- they are concepts not concrete objects. The complex numbers exist in every sense in which the real numbers exist.

10. Jan 8, 2012

### micromass

Staff Emeritus
Hello fbs7!

Did you know that most real numbers are in fact "undefinable"?? See http://en.wikipedia.org/wiki/Definable_real_number

That is, for most real numbers there does not exist a formula that specifies the number uniquely.

Certainly, the numbers 1,2,3,1/2,$\sqrt{2}$ and $\pi$ are definable. But these are a very special case. Most real numbers can not be defined or described.

Why do you think that such numbers exist then?? Doesn't that bug you as well? We're working with numbers that we can't even define properly. Surely, we can define the set of reals, but not all the numbers separately.

So in some sense, the number i is more real than most real numbers!! Indeed, at least we can define i properly, and at least we can work with it. It just has a nonintuitive rule that $i^2=-1$. But that just means that it can't exist in the real world. But then again, most real numbers can't exist in the real world since they can't even be defined!!!

11. Jan 8, 2012

### nonequilibrium

You're almost getting the point, but not quite. First note that the objects themselves might be what you call "vectors", but I'm defining an operation you usually don't see on what you call "vectors". The important thing is that you can define an operation on $\mathbb R^2$ by defining the "multiplication" as $(a,b) * (c,d) = (ac-bd,bc+ad)$ and the "summation" as $(a,b) + (c,d) = (a+c,b+d)$ such that the resulting "thing" has the structure of a field, which is (from wiki):
In other words, the definition of the operation just giving has e.g. the nice property that every element different from (0,0) has an inverse such that their product gives (1,0). Etc. These are of course the nice properties you know from the real numbers $\mathbb R$ and why you like to use numbers in general. (The "axioms" mentioned in the quote are things like commutativity or distributivity.)

So we have this fine structure. What next? We note that we can imbed the real numbers into this "thing" by the one-to-one correspondence $a \leftrightarrow (a,0)$ and it's crucial that this also preserves multiplication and summation operators for real numbers. As a result, we can see the real numbers as merely a subset of this larger "thing".

Now give to the object $(1,0)$ the name "1" (you can name it whatever you want, so nothing fishy going on!) and give $(0,1)$ the name "i". Consequently, any object $(a,b)$ can be written, using these two names, as $(a,b) = a(1,0) + b(0,1) = a+bi$. The summation and multiplication rules can be rewritten using these two names as:
$(a+bi) + (c+di) = (a+c) + (b+d)i$
and $(a+bi)*(c+di) = (ac-bd) + (ad+bd)i$

Remember we've done nothing fishy, right? I hope you now see why your comment
is not justified: the shorthand is entirely justified as you can make it rigorous by simply defining the two names as I did above. And you see how much easier it is to write i as opposed to (0,1), so there's no reason to not use the notation "i".

Your only worry is perhaps "okay you constructed this thing, but why call it a number?". Well: why not call it a number? Sure, what you mean by a vector, I wouldn't call a number, i.e. the object $(a,b)$ without the multiplication defined above. But with the multiplication defined above the new structure has all the properties numbers "should" have. Better yet: what do you consider to be the definition of a "number"? If it is the naive definition of "$\mathbb R$" then you're correct in saying that the "complex numbers" are not numbers, but that's not due to a bad property of the complex numbers but due to a bad property of your definition! That's like something defining "numbers" to be the rational numbers and then complaining to you that the irrational numbers cannot be numbers. A better definition for "numbers" is any kind of "thing" with the properties you usually associate to $\mathbb N$ or $\mathbb R$, correct? And as you see, this "thing" defined above satisfies these properties, and is hence called "the complex numbers" :)

PS: as you see, the thing I defined is actually the complex plane, with the real numbers being one line out of it.

12. Jan 8, 2012

### fbs7

That's a good point. N, Z, Q and R all relate quite easily to everyday experiences, so it's easy to instinctively relate to them, and measure things in the real world that correspond to them.

Similarly, one can readily associate vectors ( like 2x, 2x + 3y, 2x + 3y + 4z ) to everyday geometry, and matrixes easily translate to the real world. For both cases anyone can show real world entities that correspond to them.

But a vector or a matrix is not a number. A number is one quantity, a vector/matrix is a collection of numbers. So if one says that i is a vector, that I can take. But to say that i as number, just like 1, sqrt(2), e or π, that's not easy to take.

Let me ask the other way around... if complex numbers (a,b) are a number... then a vector (a,b,c,d,e) should be a number too... so should a 10x10x10 matrix... as should a polynomial with 200 terms (given it is equivalent to a 200-element vector)... so should rational polynomials.. and so forth

So by calling ℂ a number, one has to call a lot of other things numbers too. It doesn't make sense to me... a number is one quantity, while a 10x10 is 100 numbers, not one. So i being a set of two numbers I can understand, but it being one number, that's hard to understand.

So, is a 10x10 matrix one number or not?

13. Jan 8, 2012

### fbs7

Man, you're adding to my nightmares. But I still can be saved by the fact that anything that I measure in real world corresponds to a real number. In particular, I can take a line from here to the end of the Universe, send a rocket ship through that line, and then I can tell that at some point in time the rocket ship will reach a distance of x, for any x in R.

But where can I take a measurement in real world and get i? For example, I can measure the difference of phase between voltage and current, but I'm measuring an angle of 90 degrees... I'm not measuring i. I can imagine that angle of 90 degrees comes from two imaginary vectors that are rotating, but that's my imagination at work with a convenient shorthand. In the real world there are no two rotating vectors, just one voltage and one current, and at a given time each is one nice little number.

Last edited: Jan 8, 2012
14. Jan 8, 2012

### micromass

Staff Emeritus
To make it even worse: what would you measure in the world and get $\pi$?? A circle you could say. But do circles truly exist?? Can you actually make a truly perfect circle in this world, I think not.

All $\pi$ is, is an approximation for a real value. There are no circles in the world, but we can assume that there are. That is: if we assume something to be a circle, then we won't make much error.

So real numbers such as $\pi$ are just handy to work with. They make the calculations easier!! The same with i. It just makes things easier. That doesn't mean it actually exists.

15. Jan 8, 2012

### fbs7

Mr Vodka, and everybody else, at this point it is a good time to say I do thank everybody for the patience on explaining those things to me. I appreciate it. Everybody here is so knowledgeable and learned, and I learn a lot by reading you guys.

So, I think that's really the point: what is a number?

I understand vectors that represent ℂ have a very similar algebra as numbers, but a number is an element, while a vector is a special type of set.

The fact that ℂ has such a similar algebra as ℝ allows a very convenient shorthand notation for formulas in ℂ... but it's just a convenience to make the formulas short. When I try to measure or associate it to a physical entity in the real world, I can't. I have to measure or associate to two different things, and then use my imagination to visualize, in my head, a vector in ℂ.

16. Jan 8, 2012

### fbs7

How about my rocket ship? If I send it out on a line, and I have some sort of mathematical definition for √2 and $\pi$, then I know at some point in time my rocket ship will have reached √2 and $\pi$ meters from me, even if my measurements are precise only to a number of digits.

But if I send two ships 90 degrees apart, and one reach 1 km on the "x" direction and the other 1 km on the "y" direction, then I can't say my second ship has reached i. The square of its distance will be 1 km2, not -1 km2.

Saying the other way around, I cannot imagine any quantity in reality such that its square is -1.

17. Jan 8, 2012

### jgens

The natural numbers are defined in terms of collections of sets; that is, 0 = ∅, 1 = {∅}, ... , n = {0,...,n-1}, ... , ω = {0,1,...}. The rational numbers are defined in terms of equivalence classes of integers; that is, we define (a,b) R (c,d) if and only if ad = bc and then we define ab-1 = [(a,b)]. So if you really get down to it, these numbers are just collections of sets, which is exactly what the complex numbers are as well.

You seem to have some vague (and incorrect) notion of what a number ought to be and this is causing you problems. The simple fact is that numbers are almost always just collections of some sorts: The natural numbers are initial segments of the ordinals, the rational numbers are collections of pairs of integers, the real numbers are collections of cauchy sequences and the complex numbers are collection of pairs of real numbers. So when you break it down like this, it should be apparent that the complex numbers are numbers in the same sense that all these other number systems are.

If you assume a priori that physical space is best represented with the real line (or hyperreal line), then yes. But since every measurement we make is necessarily rational, how do we know that space is best represented with the real line and not the rational line?

18. Jan 8, 2012

### micromass

Staff Emeritus
Now you assume space to be continuous. Which it isn't.

19. Jan 8, 2012

### nonequilibrium

Very welcome :) I understand your problems and I'm convinced that once you see the picture more clearly (you're getting there!), you will not only be convinced that there is no problem, but that it's actually more beautiful than you had thought.

Okay, then I think that the last useful thing I can say is that the way you view $\mathbb C$ up until now (as a special kind of set with a special kind of operation) is correct but even more so, that you should regard $\mathbb R$ and $\mathbb N$ on a similar level. The "real numbers" or "the positive integers" have the same kind of "existence": they're special sets with special operations! I understand that intuitively they mean more, but mathematically speaking, that's how you create numbers.

How to define $\mathbb N$? Well Bertrand Russel wrote a 1000 paged book on this subject, haha (http://en.wikipedia.org/wiki/Principia_Mathematica). The reason it's more complex than defining $\mathbb C$ (which was relatively easy: I showed it to you in this thread) is because there I could use the existence of $\mathbb R$, but when you're defining $\mathbb N$ there's not much else to base it on other than abstract set theory. But let's assume that $\mathbb N$ has been defined.

How to define $\mathbb Z$? Easy enough, we first define our special set $\{ -\mathbb N \} \cup \{ 0 \} \cup \mathbb N$ and define on it the special operation that gives you what you mean by $\mathbb Z$.

How to define $\mathbb Q$? The special set is now $\mathbb Z^2$ where we define $(a,b) * (c,d) = (ac,bd)$ and $(a,b) + (c,d) = (ad+bc,bd)$. The object $(a,b)$ will be the definition of the rational number $\frac{a}{b}$, quite analogous to the complex case, huh! In this case there is a caveat: the correspondence $\frac{a}{b} \leftrightarrow (a,b)$ is not one-to-one, as e.g. $\frac{a}{b} = \frac{2a}{2b}$, but one can easily solve this with a mathematical procedure which I won't discuss now as it would take us too far out. But my point is: look, you see a rational number is by definition actually a "vector" as you call it: you could voice the same objection you had to calling complex numbers "numbers". But actually there is no problem. Whether you write a rational number as $\frac{a}{b}$ or as $(a,b)$ is irrelevant (similarly in the complex case): that's merely a name for a more abstract object; the thing which matters (and stays the same in all the different representations) is not how you write it, but how it behaves. It is due to the behavior of this "$\mathbb Z^2$ with the special operation" that we call it "the rational numbers". This is actually an important point all over mathematics that keeps recurring: it is in essence the behaviour of something that defines what it is, the rest is contingent.

One can similarly define $\mathbb R$, but it's more technical mathematically speaking, it takes a few pages and some abstract concepts, but the idea is the same as for all the other number sets.

Last edited: Jan 8, 2012
20. Jan 8, 2012

### nonequilibrium

As an aside, I'm amused to see how I am trying to persuade the OP that $\mathbb C$ is just as nice as $\mathbb R$, whereas micromass tries to say that $\mathbb R$ is just as horrible as $\mathbb C$ :p

(Micromass, note that this is in no way a critique to you, I agree with what you're saying, I just find the extreme difference in our approaches amusing.)