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Why do quaternions exist? I want to believe they exist, but all I can think of are reasons they should not exist.

These are my reasons.

  1. The quaternions are defined by the following equation:

$$i^2 = j^2 = k^2 = ijk = -1$$

There are four equalities and three variables. There should be no solution.

  1. Somehow there's a workaround for dilemma $1$. They exist. So how do you define the quantity $ijkijk$? Is it interpreted as $(ijk)(ijk)$, or $-1 \times -1$, or $1$? Or is it interpreted as $(i^2)(j^2)(k^2)$, or $-1 \times -1 \times -1$, or $-1$? $ijkijk$ can't be two values at once.

  2. I read that quaternions are not multiplication commutative. Why multiplication commutative? Why not multiplication associative or some other property?

  3. Okay, so there is a reason for number 3. Why can't I define a new set of numbers where the multiplicative identity doesn't hold? Like: $1x = x + 1?$

I've been pondering these questions for a couple of days now, so I would really appreciate an answer.

user642796
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The Turtle
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    #2 doesn't work since multiplication doesn't commute in the quaternions. $ab\ne ba$ in general for $a,b\in\mathbb H$ (where $\mathbb H$ is the set of quaternions) – Akiva Weinberger Aug 26 '15 at 22:14
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    see the part about 4 by 4 matrices with real entries https://en.wikipedia.org/wiki/Quaternion#Matrix_representations – Will Jagy Aug 26 '15 at 22:15
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    Why does the empty set exist? Once you have that plus the axioms of set theory you have everything else, including the quaternions. But it's a good question. Why does any set theoretical object exist? Only because we say so, and no logical contradiction is entailed by saying so. That's actually the reason quaternions exist. You don't need i, j, and k. You can do the same thing with 4-tuples of reals endowed with appropriate addition and multiplication rules. You get the reals from the rationals, the rationals from the integers, the integers from the naturals, & the naturals from the empty set. – user4894 Aug 26 '15 at 22:17
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    You can define a binary operation as you like and call it multiplication if you care - what properties it has, and whether they are useful properties are a different matter. Multiplication in most commonly studied algebraic structures is associative, and is often, but not always, commutative - it is associative but not commutative for the Quaternions. For the Octonions multiplication is neither commutative nor associative. Another important property is how multiplication and addition are related where both are defined - the distributive law. – Mark Bennet Aug 26 '15 at 22:20
  • Long story short: Quaternions are not real numbers, and they don't satisfy the properties you'd expect of real numbers. They're a structure called a noncommutative (or skew) field or algebra. In this particular case, the best intuition is probably to note that the unit sphere $\mathbb{H}^\times$ in the space of quaternions is $SU(2)$. It's associative but not commutative, since the same properties hold for $SU(2)$. – anomaly Aug 26 '15 at 22:25
  • In the wikipedia selection I linked earlier, the 4 by 4 matrix that $b$ is multiplying would be $i,$ the matrix multiplied by $c$ would be $j,$ finally the matrix multiplied by $d$ would be $k.$ Note that, as desired, it turns out that $ij=k,$ also $i^2 = j^2 = k^2$ are all minus the identity matrix. – Will Jagy Aug 26 '15 at 22:26
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    So, since other users answered most of the false claims you've made, I'll just add that the reason why quaternions exist, is that they add convenience for certain things. Other than that, you don't have to introduce, or accept/believe in them, for that matter. Just like complex numbers were introduced to deal with arbitrary order polynomials, namely inability to solve them operating using real numbers only, quaternions (part of the so called hypercomplex numbers) were introduced to deal with 3D solid body rotations. – Kaster Aug 26 '15 at 22:35
  • Quaternions exist because $S^3$ is parallelizable :) – johndoe Aug 26 '15 at 22:58
  • The fundamental theorem of algebra, which guarantees that a polynomial equation like $x^2 = -1$ can only have two solutions, requires commutativity. In the quaternions, there are actually infinitely many solutions to $x^2 = -1$. – pjs36 Aug 27 '15 at 02:10
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    One thing to note: mathematics lets you develop any system you please. Quaternions exist because someone drempt them up, and decided which properties they wanted applied to them. The answer to the question you didn't ask is: quaternions are important because their properties happen to have properties which are very effective at handling real-world applicable problems involving rotations. This raises them up from "mathematical oddity" to "something that actually gets used in science and engineering." – Cort Ammon Aug 27 '15 at 02:51
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    and, thus, to question 4, the answer is "you're welcome to define it as such. Nobody minds. However, that particular system may not have useful applications, so people may not elect to pay very much attention to it." – Cort Ammon Aug 27 '15 at 02:52
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    In point 1. you are giving a wrong reasoning that moreover is in the wrong context. For the latter, reasoning about numbers of equations and unknowns arises in the context of _linear_ equations, and these are not linear equations. But anyway having more equations than unknowns does not imply much if the equations are dependent, and these equations could well be dependent (I think maybe $k^2=-1$ could be deduced from the other equations, though I did not check). – Marc van Leeuwen Aug 27 '15 at 06:37
  • @johndoe: There are a lot of manifolds that are parallelizable but have no useful algebraic structure--- any closed, orientable 3-manifold, for example. It's certainly any obstruction to the existence of a field structure, but it's nowhere close to sufficient. – anomaly Aug 27 '15 at 19:15
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    I think your basic mistake is not understanding what "exist" means in math. There was a time when people said "negative numbers don't exist, because you can have 3 apples, but you can't have -3 apples". But you probably don't have a problem calling -3 a number now. There was a time when people said "the equation $x^2 = -1$ doesn't have any solutions", but then complex numbers were invented so that it does have the solutions $x = \pm i$. The $i$ $j$ and $k$ in quaternions are similar to the $i$ in complex numbers. They were *invented* to *make* quaternions exist. – alephzero Aug 27 '15 at 22:28
  • Mathematical objects in generally exist when there are no contradictions made by there existence. They don't need to correspond to anything real, physical to grant existence. Unlike in physics, in mathematics "true" ultimately means "not self-contradictant". – Vi0 Aug 28 '15 at 00:15
  • @anomaly Restricted to spheres, parallelizability IS sufficient. I am referring to the fact that $\mathbb{R}^n$ is a division algebra iff $S^{n-1}$ is parallelizable. – johndoe Aug 28 '15 at 09:32
  • $x^2 = x = 1$ shouldn't have any solutions either then? –  Aug 28 '15 at 17:19
  • Oh yes: quaternions exist! According the theorem of Wedderburn all finite field is commutative and quaternions being a non commutative field it is necessarily infinite. Indeed is the first example of non conmutative field to be known. (I used the french definition of field; for english people a field is always commutative) – Piquito Sep 01 '15 at 19:48
  • @Hurkyl: the equation $x^2=1$ where 1 is a quaternion has a non countable set of solutions (which show that if the coefficients of the equation are not in a commutative field is not true that the equation has as many solutions as its degree) – Piquito Sep 01 '15 at 19:55
  • Regarding item 4, this might be of interest: http://math.stackexchange.com/questions/259584/why-dont-we-define-imaginary-numbers-for-every-impossibility – Hans Lundmark Sep 02 '15 at 06:51
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    Re: recent edit. You had the correct terms before. Multiplication of quaternions is associative ($x(yz) = (xy)z)$, but not commutative ($xy=yx$). – user642796 Oct 03 '15 at 04:15

5 Answers5

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  1. They are not exactly variables, they are symbols. And we are just defining what happens when we apply a binary operator to them. You can always have more equations than variables, that doesn't mean you don't have a solution (even in your highschool algebra), it just means its possible there might not be a solution.

  2. You are assuming the quaternions are commutative to make your argument here. They are not. You cannot rearrange terms. ij is different than ji.

  3. Commutative means we can rearrange the terms. For example x * y = y * x. The quaternions are in fact associative which means (x * y) * z = x * (y * z). This is why it is not ambiguous to write x * y * z since the grouping doesn't matter. The quaternions are indeed associative. This is why $ijkijk$ is not ambiguous.

  4. You can define whatever kind of structure you want. The quaternions are a Group which have the identity property. However there are other algebraic structures, such as Magmas, which impose less restrictions

If you are really interested in detailed answers, check out some abstract algebra books from your local university library.

zrbecker
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  • @Danikar (from the answer) I don't go to any university yet. Perhaps there are some nice documents on the Internet that explain some detailed answers? – The Turtle Aug 26 '15 at 22:59
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    Even though you don't go to university, you can usually still get a library card to check out books. The great thing about a university library is that it is typically very underutilized by the actual students, so a lot of the books look for might be on the shelves. If you google free abstract algebra books, you should find a ton of books released by free online by professors. I can't necessarily vouch for any of them. When I was in school I tended to read a lot of Wikipedia, so that is also a very good internet resource. – zrbecker Aug 26 '15 at 23:08
  • @TheTurtle If you have a friend that goes to a University, you can ask them on your behalf to sign out books. – Jeel Shah Aug 27 '15 at 03:02
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    A mathematician who works above the ground sure calls it Lava, right? – null Aug 27 '15 at 13:03
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    I am not sure I agree on your point 4. You can not define a structure where x multiplied by the multiplicative identity produces something other than x. - So it depends on what OP means when he writes 1x=x+1. (You can of course define structures where there is no multiplicative identity, or there is a multiplicative identitity that is not the sucessor of the additive identity) – Taemyr Aug 27 '15 at 14:46
  • @Taemyr, Yes, that is kind of what I meant to say. Thanks for clarifying. – zrbecker Aug 27 '15 at 16:21
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Presumably, you accept that quadruples of real numbers exist. Like, $(3,2.4,-\pi,6)$ is a thing that exists. It's just an ordered collection of four numbers.

One way to think about the quaternions is just that they are another way of writing quadruples. So, instead of writing $(3,2.4,-\pi,6)$, we can write $3+2.4i-\pi j+6k$, and think of it as a notational difference.

Then, we define what $+$ and $\times$ means for these entities. We can define $+$ to mean anything we want, here; let's go with the following definition: $$(a,b,c,d)+(A,B,C,D):=(a+A,b+B,c+C,d+D)$$ (where $:=$ means "is defined to be equal to"). Similarly, we can define multiplication in a similar way; I won't write it out in the general case, but we can define it to be such that, in the new notation, $ijk=-1$, etc.

Then, the rest of the work is just to go through a bunch of properties, one by one, and see if they work.

  • Is addition commutative? Yes.

  • Does the distributive property hold? Yes.

  • Does every nonzero quaternion have a multiplicative inverse? Yes. (Hard to show but doable)

  • Are multiplication and addition associative? Yes.

  • Does multiplication commute? No.

  • etc.

Of course, what I haven't touched on is why they're defined this way. Or why so many of the properties above seem to coincidentally hold. But I hope that I've convinced you that these numbers exist, since you can think of $a+bi+cj+dk$ to be another way of writing the ordered quadruple $(a,b,c,d)$.

Akiva Weinberger
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Your logic for #1 is faulty. Just because you have four equations for three variables, there is nothing that concludes that there is no solution. For example, there exists a solution for the system of equations:

$$x^2=1\\ y^2=1\\ z^2=1\\ xyz=1$$

even though there are four equations for three variables.

For #2, you forgot that quaternions do not commute, thus, the expression $ijkijk$ is not equal to $iijjkk$. The expression can, however, be interpreted as$(ijk)(ijk)$ or $i(j(k(i(jk))))$ or any other combination of parentheses. No matter how you place the parentheses, the result will be $1$.

For #3: The quaternions are associative but not commutative. That is the property of the multiplication we defined.

For #4: Sure, you can define such a set. Why not? The thing is that unlike quaternions, the structure you will create will not have as nice a structure as quaternions, which are the only $4$ dimensional vector space over $\mathbb R$ on which multiplication is defined in such a way that we have a division ring.

Eric Wofsey
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5xum
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It is easy to check multiplication is not commutative from the axioms you mentioned. For instance:

From $ijk=-1$ and $k^2=-1$, one deduces $$(ijk)k=-k=(ij)k^2=-ij,\enspace\text{whence}\enspace ij=k.$$ Similarly, we have $\;jk=i,\enspace ki=j$. We deduce: $$ ji=-jk^2i=-(jk)(ki)=-ij.$$

Akiva Weinberger
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Bernard
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2

The quaternions are an example of a finite-dimensional associative real division algebra. Frobenius proved that there are only three: the reals, the complex numbers, and the quaternions (the octonions are nonassociative). So if you are inclined to believe division algebras exist, then the quaternions probably do.

On the other hand, it is a consequence of Bott's demonstration of a conjecture by Hirzebruch on the Pontrjagin classes of vector bundles over n-spheres that the sphere $S^{n-1}$ is parallelizable only for $n-1$ equal to 1, 3, or 7. So if you believe 3-spheres exist and that you can parallelize them (put an everywhere full-rank smooth vector field on them), then you should believe that $\mathbf R^4$ possesses a bilinear product operation without zero divisors, and hence that the quaternions exist. But you don't have to.

Tomo
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