TITLE: Can we unify addition and multiplication into one binary operation? To what extent can we find universal binary operations?
QUESTION [72 upvotes]: The question is the extent to which we can unify addition
and multiplication, realizing them as terms in a single
underlying binary operation. I have a number of questions.

Is there a binary operation $n\star m$ on the integers
$\mathbb{Z}$ such that both addition $+$ and multiplication
$\cdot$ can be expressed as specific composition
expressions using only $\star$? A more relaxed version of
the question would allow constants into the expressions;
for example, perhaps we can arrange that $a+b=0\star(a\star
b)$ and $ab=1\star(a\star b)$.
One obvious idea is to try somehow to use a pairing
function,
so that $a\star b$ codes up both $a$ and $b$ into one
number, which then can appear as one argument to $\star$,
whose other argument will signal whether or not addition or
multiplication is desired. But there is the difficulty of
making these two tasks not interfere with one another.
Note that if we allow a trinary operation, then we can
easily do it simply by defining $\star(0,a,b)=a+b$ and
$\star(1,a,b)=ab$ and extending this arbitrarily. Can we
get rid of the need for parameters here? 
Can we prove that there is no associative such binary
operation $\star$ on the integers, from which both $+$ and
$\cdot$ are expressible by terms?
Does every ring have such a binary operation $\star$
from which both addition $+$ and multiplication $\cdot$ are
expressible as $\star$-terms? Does it matter if the ring is
finite or infinite?
Can every countable family $F$ of finitary operations on
an infinite set $Z$ be unified by a single binary operation
$\star$ on $Z$, so that every function in $F$ is the same
function as that induced by some $\star$-term?

This question arose from what I found interesting about a
recent question on
math.SE,
asked by someone who was interested in the phenomenon that
logical and and or are expressible from nand and also
from nor.

REPLY [4 votes]: Here is an easy answer to question 1.
Let $\langle x,y\rangle$ be a pairing function with $L(\langle x,y\rangle)=x$ and $R(\langle x,y\rangle)=y$.
Define:
$$\begin{align}
a \star 1 &= 2a\\
a \star 3 &= L(a)\\
a \star 5 &= R(a)\\
a \star 2b &= \langle a+b,ab\rangle
\end{align}$$
and other values of $a\star b$ equal to 1.
Then:
$$a+b = L(a\star2b) = (a\star(b\star1))\star3$$
$$ab = R(a\star2b) = (a\star(b\star1))\star5$$