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Distributivity is somehow related to counting: $a\times 1 + a\times 1 = a\times (1+1) = a\times 2$. This excludes sets of non-numeric functions from forming a ring, since if $(S,\circ)$ is a group with $S$ a set of non-numeric functions, $f\in S$; let's denote by $\square$ what would be a "multiplication", $f\circ f\circ f= 3\square f$ but $\square$ is not a internal operator because $3\not\in S$. So it seems distributivity is related to counting, and that excludes many sets from forming a ring.

A ring could have been defined as a $(R,\clubsuit,\diamondsuit)$ such that $(R,\clubsuit)$ is a group and $(R,\diamondsuit)$ is a monoid, in which case $\clubsuit$ and $\diamondsuit$ could be independent and $R$ could be a set of functions.

So is relation between distributivity and counting correct, and why is it chosen? Note: rschwieb already gave a very reasonable reason.

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  • $\begingroup$ I don't understand well your question. Are you asking about why counting is needed in ring theory? $\endgroup$ – Xam Dec 5 '16 at 17:05
  • $\begingroup$ @Xammm Let me try to reformulate: why is distributivity introduced in the definition of a ring? It seems to make the second law behave like a multiplication ($a\times 1+a\times 1=a\times (1+1) = a\times 2$), but it excludes using functions. $\endgroup$ – anderstood Dec 5 '16 at 17:08
  • $\begingroup$ That's the idea, make the second law behave like a multiplication. Mmm, why do you say it excludes to use functions? $\endgroup$ – Xam Dec 5 '16 at 17:14
  • $\begingroup$ @Xammm Because while $5+5+5=3\times 5$ involves only numbers ($\times$ is internal), $f\circ f\circ f=f^3$ involves both functions and numbers (the second operator $\bullet^\bullet$ is not internal to a set of functions). $\endgroup$ – anderstood Dec 5 '16 at 18:06
  • $\begingroup$ If $\circ$ is the second law then, $f\circ f\circ f=f^3$ means just the "cubic power" of $f$. In general, if $x\in R$, it's defined for $n\in \mathbb{N}$, $x^n=1_{R}$ if $n=0$, and $x^n=x^{n-1}\cdot x$ for $n\ge 1$, and we call $x^n$ the $n-$th power of $x$. $\endgroup$ – Xam Dec 5 '16 at 18:23
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Without a relationship between the two operations, it's very easy to get stuck with a complicated expression like, say,

$a(b+c(d+f(g+h(i+j))))$

With the law of distributivity at work, you can at least put any such expression into a sum of products.

You could ask the question "why do we require associativity for rings?" Without associativity, the number of ways to multiply elements explodes quickly: $(ab)(cd), a(b(cd)), a((bc)d)$ are just a few of the potential different values of a product of four elements. Studying such a structure is a lot harder because of this complication, and even those intrepid souls who study nonassociative rings typically still work within the constraints of a natural replacement (like the Jacobi identity, or something even weaker.)

In both cases, you can see that imposing an extra condition like this keeps the objects of interest in a workable range.

If you're interested to see a case where distributivity is partially abandoned, take a look at near-rings.

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