# Sum of two Dedekind cuts is well-defined

Can you help me with this?

Let $x=A \mid B$ and $x'=A' \mid B'$ be cuts in $\mathbb Q$. It is defined

$$x+x'=(A+A') \mid \text{ rest of } \mathbb Q$$

Show that although $B+B'$ is disjoint from $A+A'$, it may happen in degenerate cases that $\mathbb Q$ is not the union of $A+A'$ and $B+B'$.

EDIT: As the comment below asked, I'll include the definition of a cut in $\mathbb Q$.

A cut in $\mathbb Q$ is a pair of subsets $A,B$ of $\mathbb Q$ such that

(a) $A\cup B=\mathbb Q$, $A\neq \emptyset$, $B\neq \emptyset$, $A\cap B=\emptyset$

(b) if $a\in A$ and $b\in B$ then $a<b$.

(c) $A$ contains no largest element.

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You might want to include the definition of "cuts" as well. Such low level details might depend crucially on the exact definitions (and could even vary from one book to another). –  Srivatsan Jan 3 '12 at 15:46
This appears to be homework. Please read the faq, add the homework tag, and edit the question to include what you have done so far and where you are stuck. When you do, I will remove my downvote. –  Nate Eldredge Jan 3 '12 at 16:23
sorry if I don't know the rules of here since I'm new. I've added the homework tag, but since I reached nothing in solving it, I prefered not to touch the questions statement. please tell me if there is some other thing I must do. –  Goodarz Mehr Jan 3 '12 at 16:53

You can simply take the cut corresponding to $\pm\sqrt2$, i.e.
$\newcommand{\Q}{\mathbb{Q}}A=\{x\in\Q; x<\sqrt{2}\}$ and $B=\{x\in\Q; x>\sqrt{2}\}$;
$A'=\{x\in\Q; x<-\sqrt{2}\}$ and $B'=\{x\in\Q; x>-\sqrt{2}\}$.
Clearly $A+A'=\{x\in\Q; x<0\}$ and $B+B'=\{x\in\Q; x>0\}$.
If you prefer to define the cut without any reference to real numbers you can get rid of $\sqrt{2}$ easily. E.g. $A=\{x\in\Q; x\le 0 \lor x^2<2\}$.