My understanding behind motivation of additive inverse of a cut set is as follows :
For example, for the rational number 2 the inverse is -2. Now 2 is represented by the set of rational numbers less than it and -2 is represented by the set of rational numbers less than it. So, if the cut set $\alpha$ represents a rational number then the inverse of $\alpha$ is the set $\{-p-r : p\notin \alpha , r >0\}$. But if the cut set does not represent a rational number then is the above definition is correct ? I think we will miss the first rational number which does not belong to $\alpha$ intuitively. Should not the set $\{-p : p\notin \alpha \}$ be the inverse now ? Confused.
There is no "first rational number that does not belong to $\alpha$," except when $\alpha$ represents that rational number.
If there is a "first rational," $r$, not in $\alpha$, that would mean that for any $r'<r$, $r'\in\alpha.$ Since $r\notin\alpha$, that means that $\alpha$ is the cut for $r$.
The entire motivation for the odd definition of $-\alpha$ is the case where the cut $\alpha$ does represent a rational number. When $\alpha$ represents an irrational, we can just write: $$-\alpha = \{r:-r\not\in\alpha\}$$
This doesn't work when $\alpha$ represents a rational number $q,$ because, under this (flawed) definition, we'd have $-q\in-\alpha$, and it is bigger than all other elements of $-\alpha$.