I've been looking for loop examples and stumbled upon this post. The answer describes a specific operation on $\mathbb{R} $, namely this:
$$x\ast t = \begin{cases} \ x+\frac{1}{2}t, &\text{ if }\ \frac{x}{t}\in (-\infty, -\frac{3}{2}]\cup [1,+\infty), \\\ \frac{1}{2}x+t, &\text{ if } \ \frac{x}{t}\in [-\frac{2}{3}, 1], \\\ 2x+2t,&\text{ if }\ \frac{x}{t}\in [-\frac{3}{2}, -\frac{2}{3}]. \end{cases}$$
My question is: do we implicitly define that $ x \ast 0 = x $? Division by zero is undefined and I would assume some kind of special case or at least a mentioning, especially because we are interested in a neutral element which is exactly $ 0 $.
So, is my suggestion right?
