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

Question

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?

Answer 1

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 $x\circ(y\circ z)=x\circ(|y|z)=|x||y|z=|xy|z$)

Answer 2

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.

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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.

Answer 3

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: $GL(2,\mathbb{R})$.

Answer 4

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...

Answer 5

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.

Answer 6

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.

Answer 7

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.