# Is there math with non-commutative multiplication of real numbers?

I'm wondering is there a math with non-commutative multiplication of real numbers. For example, we could define operator ⊗ for $$n, m ≥ 0$$:

$$n⊗ m = n\times m$$ $$n⊗ (-m) = n\times m$$ $$-n⊗ m = -(n\times m)$$ $$-n⊗ (-m) = -(n\times m)$$

Or we can choose some other rules of multiplication.

How could this math be applied?

• Those rules for multiplication of real numbers will break the distributive law. It would be hard to find an use for such a system. There are useful structures with noncommutative multiplication (matrices, quaternions) but they are essentially larger than the real numbers, which retain the commutivity. – Ethan Bolker Jan 10 '19 at 14:47
• Matrix multiplication is not commutative. – Mauro ALLEGRANZA Jan 10 '19 at 14:47
• Matrices are not real numbers – DmytroSytro Jan 10 '19 at 14:48
• @MauroALLEGRANZA: I assume what the OP means is that we define a new operator, say $\otimes$, as $n\otimes m=n\times m, n\otimes(-m)=n\times m$ etc. (for $n,m\ge 0$). – TonyK Jan 10 '19 at 14:51
• – Shufflepants Jan 10 '19 at 19:15

If I'm not mistaken one possible option is $$n\circ m=|n|m$$. Then:

Non-commutativity $$n\circ m=|n|m\not=|m|n=m\circ n$$.

(it is easy to check that it is actually associative as $$(x\circ y)\circ z=(|x|y)\circ z=||x|y|z=|xy|z.$$ whilst $$x\circ(y\circ z)=x\circ(|y|z)=|x||y|z=|xy|z$$)

• Yeah that works. In fact, $n\circ m := n\times|m|$ is what OP has defined in the question. – John Doe Jan 10 '19 at 14:57
• @JohnDoe Ah, so it is, I must admit I scanned the content of the OP after reading the title! – Lagrangian Jan 10 '19 at 15:13

I proceed assuming you wanted addition to stay the same (since you said nothing about changing it.)

If you don't care about distributivity, or you don't care about addition period, then yeah, you can take whatever function you want $$\mathbb R\times\mathbb R\to \mathbb R$$ and sometimes the order of the inputs will matter.

If you do care about addition, then you can propose whatever weird rules you want for a binary operation, but it will often be disastrous for other properties that we value about multiplication, like distributivity.

Take the second proposed axiom for example: $$n\otimes(-m)=nm$$

If we wanted distributivity, $$n\otimes m + n\otimes(-m)=n\otimes(m-m)=0$$, so that $$n\otimes(-m)=-(n\otimes m)=-nm$$. With your axiom above, we'd have $$nm=-nm$$ so that $$2nm=0$$. But this is using regular multiplication and we know that's not true in the real numbers for nonzero $$n,m$$.

I think there are some oddball binary operations on $$\mathbb R$$ that can be useful, but by and large the most-used ones are those which cooperate with addition, so that you have a ring structure.

How could this math be applied?

Try not to fall down the rabbit hole of spending time with "solutions looking for problems" and try to get into the mindset of "problems looking for solutions." Almost always (or, always?) the most fruitful mathematics are generated in the service of solving a problem, not the other way around.

BUT perhaps you meant to ask something more like this, which I think is a fair question:

What are some examples of noncommutative binary operations on the reals that have applications?

Well, now that I think about it, two come to mind:

$$a\otimes b=a/b$$ and $$a\otimes b=a-b$$. These 'have applications' but their study does not seem to go very far beyond what we already learn with regular multiplication.

• Actually, I'm looking for a way to explain to children that there are different maths by simple example without talking about complex numbers etc. So, I was thinking about how is it possible to redefine basic operators on real numbers and how it could be applied. – DmytroSytro Jan 10 '19 at 15:33
• @DmytroSytro Unsolicited advice: this doesn't seem like a very good approach for teaching children. At the very best I'd think you could consider $-, /$ as examples. If I were trying to demonstrate different types of mathematics, I would show them appropriate examples (without going too deep) from geometry, topology, derivatives of real valued functions, probability, and combinatorics. If I really had to focus on taking them out of their comfort zone with arithmetic, I'd invest in demonstrating rotations with quaternions (after demonstrating rotations with the complex numbers.) – rschwieb Jan 10 '19 at 15:50
• Your advice is good. My idea is to show them that math is not only about applying rules of a formal systems, but to show them that it's possible to create new rules and systems. – DmytroSytro Jan 10 '19 at 16:08
• @DmytroSytro I understand where you're coming from. But I would do something else than using a binary operation as an illustration. if you really wanted to go the route of applications of offbeat number systems, then there are a bunch of examples beyond complex and quaternions. Like split-complex numbers and screw theory. The history of synthetic geometry might be a good illustration of how systems change and are created. – rschwieb Jan 10 '19 at 17:04
• @DmytroSytro I don't know if children will be able to follow it, but you can show them lambda calculus, then show typed lambda calculus, then show how with curry-howard isomorphism it can be used as a replacement of classical logic for the foundation of (constructive) mathematics. – artem Jan 10 '19 at 17:11

Bit less concrete: construct a bijection of sets between your favorite non-abelian group of cardinality $$2^{\aleph_0}$$ and $$\mathbb{R}$$, “mimic” the group operation on $$\mathbb{R}$$ through this map. Example: $$GL(2,\mathbb{R})$$.

Instead of talking about truly artificial or contrived notions of number or operations on real numbers, why not at least contemplate the cliched "clock arithmetic", where (on a typical analogue wall clock) 12 + 1 = 1, so 12 behaves like 0, etc.

For that matter, I think the simplest genuine batch of "non-commutative numbers" is the Hamiltonian quaternions... which are very useful in expressing rotations, and are used for efficient rotation-in-space computations both at NASA and in 3D video games. This might agitate for figuring out a not-too-painful way of broaching the subject of quaternions... and complex numbers... rather than contriving things that are inevitably pretty bogus (and non-persuasive). After all, we don't have to give all the gruesome details...

As pointed out by several people already, your proposed $$\otimes$$ breaks distributivity. What's more, the usual multiplication $$\times$$ is almost the only binary operation which respects distributivity. In particular, there are no (continuous) non-commutative binary operations on the reals which satisfy distributivity. Here's the proof:

Consider a distributive binary operation $$\otimes$$ which is continuous in both arguments. For any $$a,b\in \mathbb{R}$$ and any integer $$n$$, we have the following:

• By repeated distributivity, $$a \otimes (nb) = a \otimes (b + b +\dots +b) = a\otimes b + a\otimes b + \dots + a\otimes b = n(a\otimes b)$$
• Then also $$a\otimes b = a \otimes \left(\frac{nb}{n}\right) = n\left(a \otimes \frac{b}{n}\right)$$, so that $$\left(a \otimes \frac{b}{n}\right) = \frac{1}{n} a\otimes b$$.
• Together the above observations show that for any rational number $$r$$ we have $$a\otimes rb = r(a\otimes b)$$.
• By continuity, for any real number $$r$$ we have $$a\otimes rb = r(a\otimes b)$$.
• The analogous observations for the left-hand argument show that for any real $$r$$ we have $$(ra)\otimes b = r(a\otimes b)$$.

Thus $$a\otimes b = ab(1\otimes 1)$$. So our binary operation $$\otimes$$ is just regular multiplication up to an overall scaling factor $$1\otimes 1$$.

So you can't keep distributivity if you want non-commutivity on real numbers. On the other hand, if you don't require distributivity, a "binary operation" is really nothing but a function of two variables, so it's kind of pointless to think about it as a "multiplication". There are lots of interesting functions of two real numbers, many of which are "non-commutative", but it's not very helpful to call them multiplication if they don't satisfy distributivity.

Summarily, I would say the answer to your question is "no, there is no useful math with non-commutative multiplication of real numbers". The story is of course quite different over other number fields or matrices. In these cases there are many interesting distributive, non-commutative binary operations.

• Actually, you can keep distributivity,without commutativity as I demonstrated in my answer. You just can't keep both distributivity and continuity without commutativity. – C Monsour Jan 11 '19 at 2:47

There's an easy way to see it's possible to do this and not only keep distributivity over addition but also to have rational numbers multiply like normal:

(1) Note that $$\Bbb{R}$$ and $$\Bbb{R}^n$$ are isomorphic as $$\Bbb{Q}$$-vector spaces (since both have algebraic dimension $$c$$) for any $$n$$, and in particular for $$n=4$$.

(2) Choose such a $$\Bbb{Q}$$-vector space isomorphism $$f$$ that takes $$1$$ to $$(1,0,0,0)$$.

(3) For real numbers $$x$$ and $$y$$, define $$xy=f^{-1}(f(x)f(y))$$ where the product $$f(x)f(y)$$ in $$\Bbb{R}^4$$ is performed by treating them as quaternions in $$\Bbb{H}$$ via $$(a,b,c,d)$$ corresponding to $$a+bi+cj+dk$$.

Now note $$x(y+z)=f^{-1}(f(x)f(y+z))=f^{-1}(f(x)(f(y)+f(z)))=f^{-1}(f(x)f(y)+f(x)f(z))=f^{-1}(f(x)f(y))+f^{-1}(f(x)f(z))=xy+xz$$ and similarly $$(x+y)z=xz+yz$$. Thus all follows because of distributivity in $$\Bbb{H}$$ and because $$f$$ and $$f^{-1}$$ preserve addition. But if $$s$$ and $$t$$ are the (necessarily irrational) real numbers that map to $$(0,1,0,0)=i$$ and to $$(0,0,1,0)=j$$, then $$st=-ts$$ in this definition of multiplication.

• However, I strongly suspect that this "construction" depends on the axiom of choice or some weaker version thereof (thinking of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ is pathological and very easily runs into such just by the smell test and I think in fact it's known that such isomorphisms require it), and thus is impossible to construct explicitly in ZFC set theory (or if it is, then at least impossible to prove that the putative explicit construction actually produces what it claims). This should be mentioned if it's true. – The_Sympathizer Jan 11 '19 at 3:21
• As basically it means that while such may "exist" in a very strict sense, we don't have to worry about them for the most part. – The_Sympathizer Jan 11 '19 at 3:22

Did someone ask for Grassman variables? Why have boring commutative multiplication where $$xy = yx$$ when you can have exciting anticommutative multiplication where $$xy = -yx$$? Exponentiation becomes super easy since $$x^n = 0$$ for $$n\ge 2$$. Differentiation and integration are now the same thing. Quantum field theory makes marginally more sense. You'll wonder why you even bothered with $$\mathbb R$$ in the first place.