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Is there such an operation that is distributive over addition, and is not multiplication?

Also, please no operations that are defined piece-wise, or that are trivial.

It must apply to all integers, but for all reals for example, the result does not matter.

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Do you also want a unit element for your binary operation? – Zhen Lin Dec 8 '11 at 6:20
Pardon my ignorance, but what does that mean? – soandos Dec 8 '11 at 6:21
Addition doesn't distribute over addition... – Arturo Magidin Dec 8 '11 at 6:22
@soandos: A unit element is an element $u$ such that $a\odot u = a$ for all $a$. – Arturo Magidin Dec 8 '11 at 6:22
@ArturoMagidin, fixed, thanks. No need, but it would be interesting – soandos Dec 8 '11 at 6:23
up vote 5 down vote accepted

The only such operations are those of the form $m\otimes n=mna$ for some fixed integer $a$. When $a=1$ this is ordinary multiplication, and when $a=0$ it’s trivial.

Suppose that $\otimes$ is such an operation. Then for any integer $n$ we have $$n\otimes 0=n\otimes(0+0)=(n\otimes 0)+(n\otimes 0)\,;$$ the only integer satisfying $x+x=x$ is $0$, so $n\otimes 0=0$ for every $n\in\mathbb{Z}$. Now let $a=1\otimes 1$. Then $$1\otimes 2=1\otimes(1+1)=(1\otimes 1)+(1\otimes 1)=a+a=2a\,,$$ and an easy induction shows that $1\otimes n=na$ for every positive integer $n$. We already know that $1\otimes 0=0$, so in fact $1\otimes n=na$ for every integer $n\ge 0$.

Now suppose that $n$ is a negative integer. Then $$0=1\otimes 0=1\otimes\big(n+(-n)\big)=(1\otimes n)+\big(1\otimes(-n)\big)=(1\otimes n)-na\,,\tag{1}$$ so $1\otimes n=na$, and we’ve now shown that $1\otimes n=na$ for every $n\in\mathbb{Z}$.

We can now repeat the argument to generalize the operand $1$ to any integer:

$$2\otimes n=(1+1)\otimes n=(1\otimes n)+(1\otimes n)=2an\,,$$ and another easy induction gives us $m\otimes n=mna$ for every non-negative integer $m$ and every integer $n$. Finally, the trick that I used in $(1)$ can be used again to show that $m\otimes n=mna$ for all $m,n\in\mathbb{Z}$.

Added: Because $\otimes$ is just scaled multiplication, it’s certainly both commutative and associative: $$m\otimes n=amn=anm=n\otimes m$$ and $$(k\otimes m)\otimes n=a(akm)n=ak(amn)=k\otimes(m\otimes n)$$ for any $k,m,n\in\mathbb{Z}$.

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How can I turn this into a function with conventional symbols? – soandos Dec 8 '11 at 6:40
@soandos: For each $a\in\mathbb{Z}$ you have a function $f_a$ defined by $f_a(m,n)=amn$. – Brian M. Scott Dec 8 '11 at 6:42
So thats just multiplication? – soandos Dec 8 '11 at 6:58
@soandos: It’s multiplication with an extra scaling factor. For instance, the operation $f_3$ gives you $f_3(2,5)=30$, $f_3(-4,7)=-84$, and so on. The operation $f_1$ is ordinary multiplication. – Brian M. Scott Dec 8 '11 at 7:09
@Brian It would be helpful to add a remark about associativity, which is completely absent above. – Bill Dubuque Dec 8 '11 at 15:19

The answer to your question is essentially no: every non-degenerate associative binary operation which distributes over $+$ is multiplication in some (possibly non-commutative) rng. Let us abstract the situation a little: so suppose we have an abelian group $A$, whose operation we write as $+$. As with the addition of numbers, we have a zero element, commutativity, associativity, and negative elements.

Suppose we have a binary operation $\odot$ which distributes over $+$: in terms of equations, we have

\begin{align} a \odot (b + c) & = a \odot b + a \odot c \newline (a + b) \odot c & = a \odot c + b \odot c \end{align}

The first equation tells us that $a \odot -$ is a group homomorphism $A \to A$, and the second equation tells us that the map $a \mapsto (a \odot -)$ is a group homomorphism $A \to \textrm{End}(A)$. But $\textrm{End}(A)$ is naturally a (non-commutative unital) ring, and if we have the non-degeneracy condition

$$a \odot b = 0 \text{ for all } b \text{ in } A \implies a = 0$$

then every element $a$ is mapped to a unique endomorphism $a \odot -$ in $\textrm{End}(A)$. This allows us to identify $A$ with a subset $A'$ of $\textrm{End}(A)$. The associativity condition

$$(a \odot b) \odot c = a \odot (b \odot c)$$

means we can identify $\odot$ with composition of endomorphisms, and this means that $A'$ is a subrng of $\textrm{End}(A)$. If we demand that $\odot$ have a right unit, then the non-degeneracy condition is automatically satisfied; on the other hand, if we demand that $\odot$ have a left unit and be non-degenerate, then $A'$ is even a subring of $\textrm{End}(A)$, and the left unit is also a right unit.

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