Interesting reults related to $a+b-ab$

Abstract Algebra

Consider a unital ring $R$ with identity $1$. If $a\in R$ has a right inverse $b$, then writing $a=1-z$ and $b=1-w$, we have $$1=ab=(1-z)(1-w)=1-z-w-zw.$$ The condition on $z$ and $w$ is thus $$z+w-zw=0.$$ Since this condition does not involve the identity, we can use it for an arbitrary ring.

Definitions: Let $R$ be an arbitrary ring. An element $z\in R$ is called right (left) quasi-regular if there exists an element $w$ in $R$ such that $z+w-zw=0$ $(z+w-wz=0)$. The element $w$ is called a right (left) quasi-inverse of $z$.
Let $R$ be a unital ring with identity $1$. Then $z\in R$ is called right (left) quasi-regular if  $1-z$ has a right (left) inverse.
If $z$ is both left and right quasi-regular, then $z$ is quasi-regular.

Let $R$ be a commutative ring and define a binary composition (known as the circle composition) in $R$ by $$a\cdot b=a+b-ab.$$ One can check that $(R,\cdot)$ is associative and forms a semigroup. Here in $(R,\cdot)$, the identity is not $1$, but $0$: $a\cdot 0=0=0\cdot a$. The set of quasi-regular elements $Q$ are the elements $a$ s.t. $a\cdot b=0$ and $b\cdot a=0$, namely they are the units of $(R,\cdot)$. We know that the set of units forms a group. It follows that the set of quasi-regular elements together with the circle composition $(Q,\cdot)$ is a group.

Claim: The mapping $$\phi:Q\to U(R)\\ a\mapsto 1-a$$ is an isomorphism of $Q$ onto $U(R)$. [$U(R)$ denotes the group of units of $R$.]
Proof: It is clear that the map is a bijection with inverse $a\mapsto 1-a$. ($1-a\mapsto 1-(1-a)=a$)
$\phi(a\cdot b)=\phi(a+b-ab)=1-a-b+ab=(1-a)(1-b)=\phi(a)\phi(b)$. $\Box$

Claim: Any nilpotent element is quasi-regular.
Proof: If $r^n=0$, and $s=1+r+r^2+\cdots+r^{n-1}$, then $(1-r)s=s(1-r)=1$, so nilpotent elements are quasi-regular. $\Box$

Corollary: If $r$ is nilpotent, then both $1+r$ and $1-r$ are units. [For $1+r$, take $s=\sum_{i=0}^{n-1} (-1)^i r^i$]

Logic

Consider a truth function s.t. $T(A)=1$ if the statement $A$ is true; $T(A)=0$ if $A$ is false. To generalise this, let $T(A)=a$, $T(B)=b$. $$T(A\cap B)=a+b-ab\\ T(A\cup B)=ab$$ The use of $T$ allows one to reduce logical problems to algebraic equations. For more information, one can refer to a text on Boolean algebra.

There are similar results of characteristic functions and sets.
$$\chi_{A\cup B}=\chi_A+\chi_B-\chi_{A\cap B}\\ \chi_{A\cap B}=\chi_A\chi_B\\ \chi_{A\triangle B}=\chi_A+\chi_B-2\chi_{A\cap B}\\ \\ |A\cup B|=|A|+|B|-|A\cap B|$$

More to explore:
idempotents
orthogonal idempotents
Jacobson radical