Algebraic structure for subtraction limited by 0 from below.

Let's assume an algebraic structure with elements from non-negative real numbers with the operation $x - y := max(x - y, 0)$.

It fails at least 2 out of 3 group definition properties:

1. Associativity: $(5 - 6) - 2 \neq 5 - (6 - 2)$
2. Identity: $x = x - 0 \neq 0 - x = 0, \forall x>0$ (i.e. there's only right identity)
3. Inverse element equals to itself (unless one requires two-side identity to exist)

This algebraic structure is at least magma (i.e. it's closed under the operation), can it be further sharpened to a less general algebraic structure?

Solution 1:

I will denote your operation by $\ominus$ so that we do not mix it with the ordinary minus sign. Following this chart that you posted in the comment section, and because the operation is not associative and does not admit a unit element as you proved, we are led to check whether $(\mathbb R_+,\ominus)$ is a quasigroup. That is, given any $a,b \in \mathbb R_+$, can we find a unique $x\in \mathbb R_{+}$ such that $a\ominus x = b$ (left divisibility) and a unique $y\in \mathbb R_+$ such that $y\ominus a = b$ (right divisibility).

Let us check left divisibility first. We have $a\ominus x = \max(a-x,0)$. In particular, because $x$ is non-negative, we see that $a\ominus x \leq a$. Therefore, if $b > a$ then the equation $a\ominus x = b$ does not have any solution $x\in \mathbb R_+$. It follows that left divisibility fails in general.

At this stage, we already understand that $(\mathbb R_+,\ominus)$ is not a quasigroup, and according to the previous chart it can not be anything else than a magma.

For sake of completeness, let us now check right divisibility. We have $y\ominus a = \max(y-a,0)$. It is not difficult to find out that $y = a+b \in \mathbb R_+$ is a solution, but is it unique ? Let us distinguish a few cases.

• Case $a=b=0$ : In this case, the equation we need to solve is $\max(y,0) = 0$. There is only one solution in $\mathbb R_+$, that is $y = 0 = a+b$.

• Case $b=0$ and $a > 0$ : Notice that if $y$ is any number satisfying $0\leq y < a$, we have $y\ominus a = 0$. Therefore, any such $y$ is a solution. There is an infinite number of them.

• Case $a,b > 0$ : Let $y$ be a solution of $y\ominus a = b$. In particular, $\max(y-a,0) > 0$, so we must have $y-a > 0$ and $\max(y-a,0) = y-a$. The equation becomes $y-a = b$ so that $y = a+b$.

Thus, in general, the solution $y$ fails to be unique. Therefore right divisibility fails too.