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



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 $(+,\cdot,\uparrow,\uparrow,...,\uparrow).$ 

