Is there a mathematical function such that;
f(3, 5) = 3
f(10, 2) = 2
f(14, 15) = 14
f(9, 9) = 9
It would be even more cool if there's a function that takes three (3) parameters, but that one could be solved by using recursive functionality;
f( f(3, 5), 4) = 3
$$f(x,y)=\frac{x+y-|x-y|}{2}$$
Oscar gave a nice interpretation of the above formula in his follow-up question, but I'll give a dumb derivation here for completeness.
Making use of Iversonian brackets, we have
$$\min(x,y)=x[y \geq x]+y[y < x]$$
and since $[\neg p]=1-[p]$,
$$\min(x,y)=x[y \geq x]+y(1-[y \geq x])=y-(y-x)[y-x \geq 0]$$
Now, there is the identity
$$\frac{u+|u|}{2}=u[u \geq 0]$$
and so we have
$$\min(x,y)=y+\frac{x-y-|x-y|}{2}$$
which simplifies to the desired expression.
The extension to more than two arguments is no longer as compact, though, since one now has to contend with products of Iversonian brackets ($[p \land q]=[p]\cdot[q]$).
