Everyone is familiar with distributivity of multiplication over addition of real numbers. The distributivity of two binary operations sometimes goes both ways (e.g. max and min, or for lattices in general.)

Out of curiosity, I looked at the set of real numbers for which addition distributes over multiplication. A simple computation shows that this set is

$$\{a,b,c\in \mathbb{R}\, |\, a+b+c=1\}.$$

Is this just a meaningless fluke, or is there reason to expect something like this?

  • 2
    $\begingroup$ I'm sorry; how is the "the set for which addition distributes over multiplication"? The set of all $a$ for which there exist $b$ and $c$ in $\mathbb{R}$ with $a+b+c=1$ is all of $\mathbb{R}$, and subsets of real numbers are not specified by listing three reals before the condition. I don't think you said what you meant to say. One specifies as set of real numbers by writing $$\{a\in\mathbb{R}\mid\text{condition on }a\}$$and you did not do this. $\endgroup$ – Arturo Magidin Feb 23 '11 at 1:23
  • $\begingroup$ I think he means the set of a, b, c such that a + b*c = (a+b)*(a+c). But this does not seem like an interesting condition to me. $\endgroup$ – Qiaochu Yuan Feb 23 '11 at 1:33
  • $\begingroup$ @Qiaochu: I think he means the ordered triple $(a,b,c)$ such that $a+(bc) = (a+b)(a+c)$ (which is different from "the set of real numbers). And he forgot the possibility that $a=0$... $\endgroup$ – Arturo Magidin Feb 23 '11 at 1:34
  • $\begingroup$ @Arturo Magidin: I think he means the cases where $a+(b\times c) = (a+ b)*(a+c)$, which seems to imply either that $a+b+c=1$ or that $a=0$. The same looks true for any field. $\endgroup$ – Henry Feb 23 '11 at 1:37
  • $\begingroup$ @Henry: In fact, any domain. $\endgroup$ – Arturo Magidin Feb 23 '11 at 1:38

If you mean, "for which real numbers $a$, $b$, and $c$ do we have $a+(bc) = (a+b)(a+c)$?" then you get $$a+bc = a^2 + ab+ac + bc,$$ or $a=a(a+b+c)$. This means that either $a=0$, or else $a+b+c=1$, which is probably what you were trying to say (though you forgot the possibility that $a=0$).

But this is not a "set of real numbers for which addition distributes over multiplication". Such a set would be a collection of real numbers $X$, such that for all $r,s,t\in X$ you have $r+st = (r+s)(r+t)$, which would require that given any three elements in the set, either the first one you picked is $0$, or else they add up to $1$. In particular, any number you pick, picking it three times, would have to be either $0$ or $\frac{1}{3}$; but you cannot have both $\frac{1}{3}$ and $0$, because then picking $\frac{1}{3}$ for $r$ and $s$, and $0$ for $t$, you would not have $r+(st) = (r+s)(r+t)$. So the only collections $X$ that satisfy that condition are $X=\{0\}$ and $X=\{\frac{1}{3}\}$. Not very interesting at all...

So it's probably better to think about what you are looking for as the collection of all $3$-tuples of real numbers $(a,b,c)$ such that $a+bc = (a+b)(a+c)$, which consists exactly of all $3$-tuples with either $a=0$ or $a+b+c=1$. Geometrically you get the union of two planes in $3$-space: the $yz$-plane, and the $x+y+z=1$ plane.

Nothing terribly exciting, I think, or particularly significant. But if you are interested in structures in which "addition" distributes over "multiplication" and vice-versa, then consider looking at boolean algebras, or more generally lattices, where $\wedge$ distributes over $\vee$ and vice-versa (like $\cap$ and $\cup$ do for sets, and the logical operators AND and OR do for logic).


For illustration purpose only, because it does not answer the question. However I find this example interesting and wanted to share it.

In a Boolean algebra, this short expression: $$ b+ab'=a+b $$ can be quite laborious to prove \begin{align*} b+ab'&=(a+a')b+ab' \\ &=ab+a'b+ab' \\ &=ab+ab+a'b+ab' \\ &=a(b+b')+ab+a'b \\ &=a+(a+a')b \\ &=a+b \end{align*} if you do remember that in Boolean algebra + distributes over .

Proof is in fact immediate: \begin{align*} b+ab'&=(b+a)(b+b') \\ &=a+b \end{align*}


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