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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?

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    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.2011-02-23
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    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.2011-02-23
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    @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 $a=0$...2011-02-23
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    @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.2011-02-23
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    @Henry: In fact, any domain.2011-02-23
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    @Arturo: I'm unfamiliar with the term "entire ring" - could you explain what you mean?2011-02-23
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    @Zev: It just means a (not-necessarily commutative, not necessarily with identity) ring with no zero divisors. But I may have jumped the gun. If the ring is not commutative, you get $a = a^2 + ba + ac$, so $(1+b)a = a(a+c)$, and there is no obvious cancellation.2011-02-23
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    Okay, this question was hastily written, and it shows. It should say: "Given any three real numbers a,b,c st a+b+c=1, we have a+bc=(a+b)(a+c), i.e. addition distributes over multiplication. Why is this?" However, I will not dwell on this since the consensus is "no reason."2011-02-23
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    @Chester: The reason is simply "because a(a+b+c)=a if a+b+c=1".2011-02-23

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