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Do there exist binary operators *, **, and *** on the real numbers, such that * distributes over **, ** distributes over ***, *** distributes over *, but not vice versa?

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Likely there are such over a countable, perhaps even small finite set. Do you require anything else of the operators? Gerhard "Ask Me About System Design" Paseman, 2012.09.08 –  Gerhard Paseman Sep 9 '12 at 4:11
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These conditions can be expressed by first-order formulas, hence if such operators exist on some infinite set (or on arbitrarily large finite sets), then they exist on every infinite set. –  Emil Jeřábek Sep 9 '12 at 9:27
    
What have you tried? –  Trevor Wilson Sep 9 '12 at 16:46
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1 Answer 1

I shall show that we can have commutative associative binary operations that satisfy the required properties.

Without loss of generality, we can replace $\mathbb{R}$ by any other infinite set. Consider the binary operations $+,\cdot,\uparrow$ on the interval [1,\infty] where $+,\cdot$ are just normal addition and multiplication and $x\uparrow y=\infty$ for all $x,y$. Every elementary school student should know the distributive property for addition and multiplication $(a+b)\cdot c=a\cdot c+b\cdot c$. Furthermore, we have

$$ (a\cdot b)\uparrow c=\infty=(a\uparrow c)\cdot(b\uparrow c)=\infty $$ and $$ (a\uparrow b)+c=\infty+c=\infty=(a+c)\uparrow(b+c). $$

On the other hand, the average kindergartner can verify that $(1\cdot 2)+3=5\neq 20=(1+3)\cdot(2+3)$. Therefore $+,\cdot,\uparrow$ are the required distributive operations.

One can easily modify the above example to get an algebra where all three reverse distributive properties fail. For instance take binary operations $*,**,***$ on $[1,\infty]^3$ where $$(a,b,c)*(x,y,z)=(a+x,b\cdot y,c\uparrow z)$$ $$(a,b,c)**(x,y,z)=(a\cdot x,b\uparrow y,c+z)$$ $$(a,b,c)***(x,y,z)=(a\uparrow x,b+y,c\cdot z).$$

In fact, it appears that you can get $n$ cyclic binary operations that only distribute on one side simply by letting $(+,\cdot,\uparrow,\uparrow,...,\uparrow).$

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I say instead "some desired" for "the required". I definitely like using the cube to propogate the failure of distributivity in the other direction. I suspect one can get a similar example on a three or four element algebra (perhaps even two). Gerhard "Wish I Thought Of It" Paseman, 2012.09.11 –  Gerhard Paseman Sep 11 '12 at 16:17
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