Definition of Dedekind Cut Addition I'm currently reading through Pugh's Analysis (for fun.) I'm currently stuck on this question:
Given two cuts $x=A|B$ and $y = A'|B'$, then $x+y=(A+A')|\text{rest of $\mathbb Q$}$, show that although $B+B'$ is disjoint from $A+A'$, their union might not be $\mathbb Q$.
I understand this question is basically asking why cut addition isn't defined to be $x+y=(A+A')|(B+B')$. I keep trying to find a degenerate case, but can't seem to really get started. Showing that they're disjoint is easy enough, but I haven't really made any progress from there. Hints would be appreciated.
 A: Consider the case where $A|B$ represents an irrational and $A'|B'$ its negative. Is $0\in(A+A')\cup (B+B')$?
A: I think this is where Hardy's Pure Math goes beyond the modern textbooks which describe the theory of Dedekind's cut. Hardy knows that a student will face this exact problem when additon of cuts $A\mid B$ and $A'\mid B'$ is defined by $A + A'\mid B + B'$.
Let $a$ denote a general member of $A$ and $a'$ a general member of $A'$. Similar notation for $b, b'$ corresponding to $B, B'$. Also $c$ is general member of $A + A'$ and $c'$ the general member of $B + B'$.
Thus $ c = a + a', c' = b + b'$. I quote Hardy verbatim (except for the notation used here):

"There can not be more than one rational number which does not belong either to $A + A'$  or to $B + B'$. For suppose there were two, say $r$ and $s$, and let $s$ be the greater. Then both $r$ and $s$ must be greater than every member $c$ of $A + A'$ and less than every member $c'$ of $B + B'$; and so $c' - c$ can not be less than $s - r$. But $$c' - c = b + b' - (a + a') = (b - a) + (b' - a')$$ and we can choose $a, b, a', b'$ so that both $(b - a)$ and $(b' - a')$ are as small as we like; and this plainly contradicts our hypothesis.
If every rational number belongs to $A + A'$ or to $B + B'$ then the classes $A + A'$, $B + B'$ form a Dedekind cut of the rational numbers, that is to say, a number $\gamma$. If there is one which does not, we add it to $B + B'$. We now have a Dedekind cut, which clearly must be rational, since it corresponds to the least member of $B + B'$."

Thus in the above argument by Hardy we see that the exceptional case occurs when we are adding two numbers which are irrational and their sum turns out to be rational. This is what is given as an actual example in the other accepted answer.
