Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Why can Imaginary Numbers, which are called that because they don't exist, be used in mathematical proofs?

Update:

Anonymous dude,

I can't quite follow the reasoning in your second paragraph. You seem to be stating that if a field (rational numbers in this case) is proven to exist, this proves that a larger field, of which this is a subset, exists. It seems to me that any such field extension is postulated, not proven, to exist. Could you clarify this please? I seem to accept complex numbers in the way your first paragraph describes, as very useful tools in calculations but nothing more. Negative natural numbers I think of as positive natural numbers going in the opposite direction, so they cause me no problem. With rational numbers, (e.g two and three quarters) I can think of the 2 as a natural number, and the 3 in 3/4 as another natural number counting units for which the "/4" simply defines a smaller size, so they aren't a problem either. Internal consistency is not proof. Neither is precision or logical meaning.

Update 2:

Manipulating complex numbers is usually not a problem, but believing that something HAS been proven to me, when the proof includes a value which I can use, but not comprehend, is a problem. It isn't that I feel hypocritical accepting such a proof, I just question whether such things have REALLY been proven to anyone.

Update 3:

Buri,

I looked up the link you gave, and followed it to some others, but their reasoning seems to be, "Complex numbers form part of a coherent system, and are very useful in solving many problems, therefore they must exist." Thanks for the effort, anyway.

(If anyone's interested, it was Andrew Wiles so-called proof of Fermat's last theorem that brought this matter to my mind. Specifically all the complex numbers in the modular groups. I still consider the Taniyama-Shimura Conjecture to be a conjecture. Maybe I just have a closed mind as far as "i" is concerned. If you can convince me about root minus 1, then I will be unhappy, but at least then I can move on, and THAT I will appreciate, thanks)

Update 4:

anonymous dude, The definition of (x_1, y_1)*(x_2,y_2) to be (x_1 x_2 - y_1 y_2, x_1 y_2 + y_1 x_2), appears (to me) to be neither reasonable nor consistent BECAUSE an apparent proof of the existence of root -1 results from it. I don't feel comfortable with thinking of ANY numbers as being perpendicular to R. I can't make an analogy between this, and "+3-2 = 3 steps forward and 2 steps back". In the physical world, I know a third direction exists because I can observe it, but in mathematics, I can't see that just because two axes exist, it follows as the night, the day, that of necessity a third axis must also exist.

Update 5:

"This polynomial cannot be factored, precisely because there is no real number r with the property that r^2 + 1 = 0." The difficulty I have with this line of reasoning is in observing that because no real number r, satisfies this equation, r must exist, but be imaginary. Why must it exist? I am not questioning the usefulness of complex numbers in many areas. I am questioning whether we are justified in using an incomprehensible, yet definable and manipulable concept in proofs. (perhaps I should have said incomprehensible to me.)

Update 6:

In the fundamental theorem of Calculus, there is a caveat, "Provided the limit exists". Perhaps I am being overly skeptical in feeling that more branches of mathematics, including polynomials should include "Provided the value exists". I just don't know.

Update 7:

zpconn, I have to admit that I have a lot of problems with the way maths is done. Mostly along the lines of "How reasonable and sensible does a postulate have to be before it is considered axiomatic, and therefore decisive in proving a point. I'm not questioning the self-consistency of complex number construction.

Update 8:

"If you don't understand how complex numbers can be used to prove facts about real numbers, then you clearly just don't understand the proofs and have a lot more education to do in mathematics." I don't understand WHY complex numbers are trusted in proving facts about real numbers, Which is why I asked the question.

Update 9:

Steiner. Irrational numbers do not upset me because they help to make the "number line" continuous.

When mathematicians say "If This is equal to That" or "If we define This as That", and a few years later other mathematicians decide to use these statements to PROVE something, and forget about all the "If's", THAT causes me to question the validity of the proofs.

14 Answers

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  • 1 decade ago
    Favourite answer

    EDIT: Given that I have run out of space, I will leave my two constructions of the complex numbers up for reference and just keep writing over the beginning part.

    "The definition of (x_1, y_1)*(x_2,y_2) to be (x_1 x_2 - y_1 y_2, x_1 y_2 + y_1 x_2), appears (to me) to be neither reasonable nor consistent BECAUSE an apparent proof of the existence of root -1 results from it."

    Well, I have to admit that I am going to have a very hard time convincing you that the complex numbers exist if you disagree with my construction by virtue of the fact that it proves that the complex numbers exist. It is almost as if you want the non-existence of the complex numbers to be built into mathematics because you find them counter-intuitive - but if that is your desire then the burden of proof is on you to show that it CAN'T be done, not on me to show that it CAN be done. Whether you find the complex numbers "reasonable" or not, the construction I gave IS consistent and it DOES prove that there is a field extension of R which contains a square root of -1 (unless you can find a mistake?). I'm very sorry that they are not as physically intuitive to you as subtraction, but allow me to suggest that the problem lies in what physical observations you find intuitive rather than the conceptual validity of using complex numbers.

    "The difficulty I have with this line of reasoning is in observing that because no real number r, satisfies this equation, r must exist, but be imaginary. Why must it exist?"

    The second construction that I gave is very complicated, so let me distill it a little bit. First, we take a field F. Then we form the polynomial ring F[x]. Then we take an irreducible polynomial f(x) - which necessarily does not have roots in F - and form the ideal I generated by f. Then we form the quotient ring - which we can prove abstractly to actually be a field - denoted F[x]/I. In this field, f turns out to have a root. All that you need to make this construction work is an irreducible polynomial over your field, and there are many ways to show that x^2 + 1 is irreducible over R; the miraculous fact is that we only need to do this once and we have a field which contains all roots of all polynomials over it.

    END EDIT

    So first I will give a geometric construction which will be a little more intuitively appealing and equally rigorous but not really in line with the way things are done today. Given two sets S and T, there exists a third set S x T, called the product of S and T, which consists of all ordered pairs (s,t) where s is in S and t is in T. Take the set R of all real numbers, and form the set R x R (the product of R with itself). This is the set of all pairs (x, y) where x and y are real numbers. This space, called R^2, comes naturally equipped with a geometric structure: it is the ordinary Euclidean plane. It also has an algebraic structure which it inherits from R: we can define (x_1, y_1) + (x_2, y_2) = (x_1 + x_2, y_1 + y_2). This notion of addition is associative and commutative, and there is an identity element (0,0). Each element (x,y) has an additive inverse, namely (-x,-y). Thus R^2 has the structure of an "Abelian group". Let us give it a little more structure. I want to define a notion of multiplication on R^2. Define (x_1, y_1)*(x_2,y_2) to be (x_1 x_2 - y_1 y_2, x_1 y_2 + y_1 x_2). This is a perfectly reasonable and consistent notion of multiplication. One observation is that if I restrict my attention to the first coordinate, then this recovers ordinary multiplication of real numbers: (x,0)*(y,0) = (xy,0). So the real numbers with their multiplicative structure sit naturally as a subset of R^2 with its multiplicative structure. Let us dig a little deeper. Certainly this notion of multiplication is commutative and associative, and I will let you check a few more properties (it might take some time, but it's a good exercise): first, multiplication distributes over addition; second, (1,0) functions as a multiplicative identity element; and third, every (x,y) other than (0,0) has a multiplicative inverse given by (x/(x^2 + y^2), -y/(x^2 + y^2)). Once you have demonstrated these facts, you have checked off the list of axioms necessary for R^2 equipped with + and * to be a field. So let us review: we started with the set of all real numbers, we formed the product R^2 of the real numbers with itself, we defined addition on R^2, we defined multiplication on R^2, and we found that R^2 with this notion of addition and multiplication is a field. Everything so far is perfectly legal and perfectly consistent. Now let us make a quick observation: given any element (x,y) of R^2, we can write (x,y) = (x,0)*(1,0) + (y,0)*(0,1). Identifying the real numbers with elements of the form (*,0), we can write this as x + y*(0,1). So let's think about this (0,1) object, which seems to be special. One quick geometric observation is that (0,1) is sort of perpendicular to the real line; the real line is like the "x-axis", and (0,1) points along the "y-axis". But the more important observation is algebraic: with our definition, (0,1)^2 = (0,1)*(0,1) = (0*0 - 1*1, 0*1 + 1*0) = (-1,0) = -1. So low and behold, -1 has a square root in this field. In fact, R^2 with the notion of addition and multiplication that I have defined is "isomorphic" as a field to the space of complex numbers, and you can use this as a definition of the complex numbers. If you are comfortable with thinking of negative numbers as numbers that "point in the opposite direction", then maybe you are comfortable with thinking of the complex numbers as numbers which point in yet a third direction, perpendicular to the real numbers.

    So that is one way to obtain the complex numbers. But there is another way which is a little bit more natural and much more useful. First, some elementary field theory. Let F be a field; as above, this means F is a set equipped with some notion of addition and multiplication that have the following properties: addition and multiplication are commutative and associative, multiplication distributes over addition, there is an additive identity element (which we will call 0), every element a has an additive inverse, there is a multiplicative identity element (which we will call 1), and every element a except 0 has a multiplicative inverse. Given a field F, we can consider polynomials over F; these are just expressions of the form P(x) = a_n x^n + a_{n-1} x^(n-1) + .... + a_1 x + a_0 where a_0,...,a_n are all elements of F. We can then take the set of all polynomials over F; let's denote this set F[x]. One might wish to say a few words about the construction of F[x], although the way I said it can be made perfectly rigorous. But for a slightly more intrinsic approach, F[x] has a universal property among graded F-algebras. Regardless, F[x] has an additive structure (usual addition of polynomials) and a multiplicative structure (usual multiplication of polynomials). However, F[x] is not a field because most polynomials do not have multiplicative inverses (1/x is not a polynomial). A set equipped with addition and multiplication is called a ring, and so F[x] is an example of a ring. Now, some polynomials in F[x] can be factored; others cannot. Let's assume that f(x) is a polynomial which cannot be factored. We can multiply f(x) by other polynomials in F[x], and then form the set I of all products of f(x) with polynomials in F[x]. I is called an "ideal" of F[x]; an ideal is characterized by the property that the sum of any two elements in the ideal is again in the ideal and the product of any element in the ideal with an element of the ring is in the ideal. Given an ideal I, we can define an "equivalence relation" on F[x] by saying that two polynomials P(x) and Q(x) are "equivalent" if P(x) - Q(x) is an element of I. Within set theory there is a well-developed theory of equivalence relations; I do not want to go into the details. An equivalence class with respect to an equivalence relation is a collection of elements of F[x] that are all equivalent to each other; so, for example, f(x) + x and x are two polynomials in the same equivalence class because f(x) + x - x = f(x) which is an element of I and hence f(x) + x is equivalent to x. Now we can form the set of all equivalence classes of F[x], which we call the "quotient ring" of F[x] by I, denoted F[x]/I. One can show as an elementary exercise in algebra that if f(x) is an irreducible polynomial (meaning it cannot be factored) and I is the ideal generated by f(x) then F[x]/I has the structure of a field. I'm skipping over a lot of details here - like how to make sense of addition and multiplication of equivalence classes - so I will have to refer you to an algebra book. Let us just apply this to a specific example. Let F = R, the set of all real numbers, and form the ring R[x]. This is the set of all polynomials with real coefficients. Now consider the polynomial f(x) = x^2 + 1. This polynomial cannot be factored, precisely because there is no real number r with the property that r^2 + 1 = 0. So form the ideal I generated by x^2 + 1, and then form the quotient field R[x]/I. Just from abstract reasoning, we know that this is a well-defined set and it has the structure of a field; let's poke around and try to figure out what it looks like. Unfortunately I am reaching the limit of how long an answer can be in this forum, so I will skip to the punchline: if we denote the equivalence class of a polynomial P(x) by [P(x)], then we have: [x]^2 = [x^2] = [x^2 + 1 - 1] = [x^2 + 1] - [1] = [0] - [1] = -[1]. Here I am using that [x^2 + 1] = [0] because x^2 + 1 - 0 is an element of I. And thus we see that R[x]/I is a field containing R which has a square root of -1. In fact, R[x]/I is isomor

  • 1 decade ago

    (1) They do exist, as much as any real number does. The fact, however, that the basic unit is the solution to x^2 + 1 = 0 is mostly just an algebraic coincidence to the reason why they're used a lot in proof or real life (I'll get to that soon).

    (2) They do exist.

    (3) They do exist. [get the point, yet?]

    Imaginary numbers, or more generally Complex numbers, do not algebraically have to be written by using a letter i (or even j, to cover those electrical engineers). We could write i as a special coordinate pair (0,1). If we combine it with the following rules:

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

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

    Then in fact these are the complex numbers. No 'i' necessary. I didn't even have to talk about some root of a polynomial that never intersects the x-axis.

    You may say that that's a weird rule for multiplication, but actually it turns out it's quite typical of objects which are periodic in nature (such as anything that orbits another object). Now, I ask you: if it describes reality, how can we responsibly avoid using them in proof?

    Edit: By the way, if you really don't like the coordinate definition I could give an algebraically equivalent definition using 2x2 matrices of _real numbers_. What my main point is, is that we can make "Complex numbers" which have no mention of the infamous 'i'.

    Unfortunately, if you don't believe in any of the constructions of the real numbers, I can't help you there. That's a philosophical stumbling block.

  • 1 decade ago

    You've got a lot of excelent answers. I just want to point out that I think it's weird that you don't accept the existence of the complex numbers and, yet, you accept the existence of the irrationals.

    The construction of the set of the irrational numbers is maybe more abstract then the construction of the set of the complexes. You have the rationals, whose existence is clear to you. But you know there are a lot of holes on the real line. How to fill such holes? One approach is based on Cauchy sequences, you just define irrationals to be the limits of such sequences when they don't converge to a rational.

    Another approach is Dedekind's cuts, also somewhat abstract.

    Why do the complexes upset you but the irrationals do not?

    I think you're giving too much importance to the adjective imaginary. Well, the reals are so imaginary as the imaginary numbers. If this shocks you, OK, let's put it differently: the imaginary numbers are so real as the reals.

    (hope this doesn't lead you to think nothing is real...)

  • 1 decade ago

    Existence in mathematics is not as sordid an affair as metaphysical existence.

    Imaginary numbers aren't called 'imaginary' because they don't exist, but because they were thought to be contrieved conjurations. Of course geometrical basis at the time was the gold standard for proof (unlike now, thankfully) and Gauss, et al. giving complex numbers a geometrical representation (following the more arithmetical idea of an ordered pair of reals) was the final step in getting complex quantities (and hence 'imaginary' ones) accepted. You seem to never have outgrown this geometrical flavour and this may be why you have trouble believing they 'exist'.

    Mathematical existence, constitues simply that something can be imagined, and put to work in a preconceived, consistent (barring Godel's incompleteness since we will consider dicrete areas) system. Nothing less and certainly nothing more (you may be cringing now, but this is how it is nowadays- Platonism is dead, formalism lives only in spirit and in the shadows fictionalism is at play).

    And Wiles' proof certainly qualifies as a brilliant triumph in mathematics. If you can't get your head around constructions from real numbers (and hence sets, leading all the way to the modern basis of Zermaelo Frankel set theory with the axiom of choice) you'd be best left out of pure mathematics these days.

    edit- here's something really simple that might help; you agree with Dedekind's cuts and making the real line continuous, right? Now do the same with another real line perpendicular to the original et voila- complex numbers justified!

    Modern mathematics is all about definition (where all the 'if's gather), proposition and proof! Not physics or geometry.

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  • zpconn
    Lv 4
    1 decade ago

    The terminology is misleading and dangerous. The imaginary numbers form a subset of the complex numbers, and the complex numbers are defined as the field C consisting of members of the form (a,b), where a and b are real numbers, and equipped with addition and multiplication operations defined by

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

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

    It follows that every complex number (a,b) can be expanded as (a,0) * (1,0) + (b,0) * (0,1). If we write x for (x,0) and i for (0,1), this is the same as 1a + bi = a + bi. Now note the curious property that (0,1) * (0,1) = (-1,0); i.e., i^2 = -1 in the simpler notation.

    There is no valid and legitimate reason to reject this. Ordered pairs of real numbers are critical for understanding the world--they correspond to points in a plane, for example. We've done nothing here you don't already accept, but for some reason you don't want to accept the result. That's inconsistent.

    The point with all this is that the complex number admit a straightforward construction and that they exist just like any other mathematical object exists. You will find that mathematicians construct many structures like C above on the spot for use in mathematical proofs. "Mathematical existence" is not the same as metaphysical existence; the key requirement is that the system so defined be self-consistent.

    The terminology arose from mathematicians (often otherwise extremely skilled ones) who lacked the foresight to appreciate this sophisticated and subtle point.

    In many modern treatments, the complex numbers are formally constructed without needing to define them as ordered pairs of real numbers like I did above. For example, here is the standard definition: http://us.metamath.org/mpegif/mmcomplex.html

    If you're won't accept any of this, then you have some significant problems with the way all of modern mathematics is done. The above definition proceeds from the basic axioms of ZFC set theory.

    Perhaps a definition more to your liking would be as a certain subset of the set of all 2x2 real matrices. I suppose you don't have a problem with matrices? Then you don't have a problem with complex numbers--they correspond to certain 2x2 real matrices. An example: http://www.sosmath.com/matrix/complex/complex.html

    Overall, it sounds like you have a lot to learn not about complex numbers but about formal systems. Understanding the significance of the formal system in which one is working is an important but very difficult point to appreciate. Mathematics largely consists of working with different formal systems that are defined by a set of elements, a list of operations on that set, and then a list of axioms that elements in the set must obey with respect to the operations. The integers, rationals, reals, complex numbers, quaternions, octonions, sedonions, matrices, differential forms, tensors, functions, groups, rings, fields, quotient spaces, product spaces, topological spaces--well, *everything* in mathematics is set up through this procedure.

    I hope this helps.

    I'm going to respond to some of your additions to the question:

    "The definition of (x_1, y_1)*(x_2,y_2) to be (x_1 x_2 - y_1 y_2, x_1 y_2 + y_1 x_2), appears (to me) to be neither reasonable nor consistent BECAUSE an apparent proof of the existence of root -1 results from it."

    This shows you have an irrational bias against complex numbers not based in reason. Give me a proof that this definition of multiplication is inconsistent or your statement here is meaningless. Mathematics is done by proving statements, not rejecting ideas because they just don't appeal to you.

    In fact, don't waste your time: it has already been proven that the complex numbers are a perfectly consistent formal system.

    "I am not questioning the usefulness of complex numbers in many areas. I am questioning whether we are justified in using an incomprehensible, yet definable and manipulable concept in proofs. (perhaps I should have said incomprehensible to me.)"

    What's incomprehensible about complex numbers? They are easy to construct as a formal system--you've seen plenty of such constructions right here on this site. In this respect they are no different from the integers or the real numbers or the matrices or any number of familiar mathematical objects.

    "Mostly along the lines of "How reasonable and sensible does a postulate have to be before it is considered axiomatic, and therefore decisive in proving a point. I'm not questioning the self-consistency of complex number construction."

    If you don't understand how complex numbers can be used to prove facts about real numbers, then you clearly just don't understand the proofs and have a lot more education to do in mathematics.

    Here one simple example:

    Certain contour integrals in the complex plane can be used to compute difficult improper real integers because in the limit the integral along the segment of the contour off the real line goes to zero, leaving only the integral along the real line.

    As for how they can be used in the proof of Fermat's Last Theorem, why don't you go read the proof? I haven't read it and can't answer that question specifically. If it's a well-written proof and your understanding of mathematics is sufficient, it should be clear how complex numbers are used.

  • 1 decade ago

    What an excellent question, and some great debate!

    The first point I'd like to make (in response to Matt's comments) is that complex numbers and imaginary numbers are NOT the same thing. Complex numbers are made up of a "real" part and an "imaginary" part. Imaginary numbers are the "imaginary" part of that!

    I assume, initial called imaginary, because they aren't what we see so easily in life - easy to see that 1 apple + 1 apple gives us 2 apples. Easy to see that negative can mean in the opposite direction. Easy to see if we cut something in pieces we have fractions. Slightly harder to deal with irrationals, but do-able once you start getting the hang of maths.

    Now what about infinity - not really a number is it! But the concept of infinity and infinitesimals is paramount to many aspects of mathematics.

    And now to "imaginary" and "complex" numbers - we have advanced in our understanding of these but much much harder to deal with mathematically, hence not usually introduced until the final year of high school or at university level. But fundamental to many aspects of nature, science and mathematics. Personally I can really only "imagine" the concept of "imaginary" numbers and have always been fascinated by them. But I think they exist as a concept as much as infinity exists as a concept and hence we have developed some mathematics to suit.

  • Mαtt
    Lv 6
    1 decade ago

    Well they do exist! Imaginary number is a term used in high school, in college and beyond they are called complex numbers. They have applications in a wide variety of engineering, physics, and analysis. The most practical use occurs electrical engineering, without them your computer would not exist.

    The fundamental theorem of algebra tells us that x² = -4 has two solutions, just because they are complex does not mean they don't exist.

    Also all real number can be written in the form a + 0*i, so real numbers are just a subset of complex numbers.

  • 1 decade ago

    Acomplex number is made up of two parts: real and imaginary .A real number multiplied by square root of (-1) is called imaginary.

    There are many mathematical tools that have been constructed for help in analysis.whe, even -ve numbers do not exist in nature.

  • zgraf
    Lv 4
    1 decade ago

    Um yeah, what the other poster said.

    Imaginary numbers come into play in lots of physical phenomenon.

    If you listen to music for example, the analog waveform

    can be expressed as a function of time that involves complex numbers.

    For a complex signal [X(t)] = A(t) + iB(t):

    Amplitude = sqrt[A(t)*A(t) + B(t)*B(t)]

    Phase = tan-1[B(t)/A(t)]

    Although one can't hear an imaginary component of a signal,

    people can actually >> hear << differences in amplitude and phase.

  • Jeff T
    Lv 6
    1 decade ago

    The term "imaginary" is somewhat inaccurate.

    All mathematical rules are either definitions or if-then statements.

    And the imaginary numbers have a consistent, logical, set of rules. So they are useable.

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