ac.commutative algebra - Bizarre operation on polynomials

I came across something similar in the context of extending boolean functions to real arguments. I thought it was pretty amusing, so let me share it here.

If we understand 1 to be "true" and 0 to be false, one way to define "x and y" is as xy. Similarly, "x or y" can be defined as x + y - xy, and "not x" can be 1 - x. There are good reasons to prefer these definitions over others. For example, if x and y are the probabilities of independent events, then xy, x + y - xy, etc. are respectively the probabilities of the conjunction, disjunction, etc. This gives a systematic way to extend boolean functions ${mathrm{false, true}}^nto{mathrm{false, true}}$ to polynomials $[0,1]^nto[0,1]$. These polynomials satisfy some (but relatively few) of the nice properties of their boolean counterparts, like de Morgan's law: x + y - xy = 1-(1-x)(1-y).

On the other hand, we could treat 0 as "true" and infinity as "false" and try to define boolean functions on the nonnegative real line $[0,infty]$. It seems we can begin by taking our polynomials defined above, expressed appropriately, and interchanging addition with multiplication and subtraction with division!

Conjunction: $xcdot y to x + y$

Disjunction: $(x + y) - (xcdot y) to (xcdot y) / (x + y)$

Negation: $1 - x to 1 / x$

Amusingly, these substitutions preserve de Morgan's law: $xy/(x + y) = 1/(1/x + 1/y)$. I don't think you can run with this all the way to the finish line. For example, the polynomial for exclusive-or is x + y - 2xy, but I don't see an easy way to express that to make the substitution go through. However, I do believe we have the following:

For every boolean function ${mathrm{false, true}}^nto{mathrm{false, true}}$, there is a polynomial extension $f:[0,1]^nto[0,1]$ and a rational extension $g:[0,infty]^nto [0,infty]$ such that, expressed appropriately, $f$ and $g$ are obtained from one another by the addition/multiplication interchange described above.

For example, for exclusive-or, we have $(x + y - xy)(1 - xy) to xy/(x+y) + 1/(x+y) = (1 + xy)/(x + y)$. However, $(x + y - xy)(1-xy) = x + y - 2xy + xy(1-x)(1-y) neq x + y - 2xy$, so we used a "noncanonical" polynomial. There are other examples where we can use the canonical polynomial, though. For example, for the 3-ary majority function, we have

$$xy + xz + yz - 2xyz to (x + y)(x + z)(y + z)/(x + y + z)^2.$$

I know this is not exactly what you asked about, since it involves subtraction and it's really an operation on expressions, not functions, but I hope it's in the right spirit.

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