TITLE: Does this arithmetic operation have a name
QUESTION [6 upvotes]: I've came across the following product-like operation on reals
$$
  a\times b:=1 - (1-a)(1-b).
$$
This operation is commutative, associative and has $1$ as a zero element and $0$ as a unit element:
$$
  a\times 1 = 1, \qquad a\times 0  =a.
$$
Perhaps there is even a version of addition such that $a\times b$ becomes distributive. Has this operation been studied somewhere?

REPLY [3 votes]: I think it's called probabilistic or fuzzy OR.
See http://en.wikipedia.org/wiki/Fuzzy_logic
This, together with the "addition" operation$ x(+)y = (x + y - 1)$ is isomorphic to the reals with normal multiplication and addition.  The isomorphism is given by $x \to 1-x$ as mt_ says.
This operator is actually a member of a family of operators$ x + y + rxy,$ where $r$ is a non-zero complex constant.  Each of these acts like complex multiplication with the corresponding complex "addition" being $x + y + 1/r,$ and the isomorphism being $x \to \dfrac{x - 1}{r}.$  Identity is given by 0, and zero is given by $-1/r$.
These properties are straightforward to check, but I will provide proofs for two of them:
Proof of associativity:
$$x + y + z + r(xy + xz + yz) + r^2xyz \\= (x + y + rxy) + z + r(x + y + rxy)z \\= x + (y + z + ryz) + rx(y + z + ryz)$$
Proof of distributivity:
$$z + (x + y + 1/r) + rz(x + y + 1/r) \\= x + y + z + 1/r + rxz + ryz + z\\ = (x + z + rxz) + (y + z + ryz) + 1/r$$