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I was experimenting with binary operators, and devised a commutative operator with function $f$ such that the operator mapped $f \colon \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{R}$ in order to make up some problem (e.g. if $f(3,10)=30$ and $f(4,3)=10$, find $f(4,6)$, etc.).

My question is, are there any functions $f$ other than $a \otimes b = f(a,b) = qab+r(a+b)$ where $q,r \in \mathbb{R}$?

Also, is it always true that the operator is also distributive if $a \otimes 0 = 0$?

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up vote 4 down vote accepted

Every such $f$ has the form $f(a, b) = g(a, b) + g(b, a)$ for some arbitrary function $g : \mathbb{Z} \times \mathbb{Z} \to \mathbb{R}$. And no. For example, take $f(a, b) = a^2 b^2$.

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just a follow-up: is the function I specified the only one for which $a \otimes (b+c) = a \otimes b + a \otimes c - a \otimes 0$? (in the general form specified) – Eugene Bulkin Nov 5 '10 at 1:50
@Eugene: still no. You can take f = qab + r(a+b) + p. – Qiaochu Yuan Nov 5 '10 at 1:56
@Eugene: and anticipating a follow-up to the follow-up, these are all the solutions, which I encourage you to try to prove yourself. – Qiaochu Yuan Nov 5 '10 at 2:01

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