Cyclic Distribution on the reals? Do there exist binary operators *, **, and *** on the real numbers, such that * distributes over **, ** distributes over ***, *** distributes over *, but not vice versa?
 A: 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 $(+,\cdot,\uparrow,\uparrow,...,\uparrow).$
