Can someone please help me to prove what is the sum of two rational numbers (is rational obviously but why) with dedekind cuts, what is the sum of rational and an irrational and the sum of two irrational numbers also with the definition of dedekinds cuts?

I give the definition of cut dedekind

(a) Cut $A\neq \emptyset$ and $A\neq\Bbb Q$.

(b) If $x\in A$ and $y \in\Bbb Q$ and $y<x$, then $y\in A$.

(c) $A$ contains no largest number, that is if $x \in A$ then there is $y\in A$ so that $x<y$

  • $\begingroup$ Please help, I'm so lost... $\endgroup$ – art Mar 5 '13 at 17:34
  • $\begingroup$ Have you been given a definition for the sum of two Dedekind cuts, or are you supposed to come up with one for yourself? $\endgroup$ – Brian M. Scott Mar 5 '13 at 17:38
  • $\begingroup$ sorry Brian M. Scott i will be more exactly $\endgroup$ – art Mar 5 '13 at 17:47
  • $\begingroup$ cut dedekind of rationals is a subset A of Q such that a) A≠∅ and A≠Q. (b) If x∈A and y∈Q and y<x, then y∈A. (c) A contains no largest number, that is if x∈A then there is y∈A so that x<y $\endgroup$ – art Mar 5 '13 at 17:48
  • $\begingroup$ didi i helped you if i understand your question? $\endgroup$ – art Mar 5 '13 at 17:48

I’ll show that the sum of two rational cuts is rational. Suppose that $A$ and $B$ are rational cuts; then there are $a,b\in\Bbb Q$ such that $A=\{q\in\Bbb Q:q<a\}$ and $B=\{q\in\Bbb Q:q<b\}$. By definition $$A+B=\{p+q:p\in A\text{ and }q\in B\}\;;$$ I’ll prove that $A+B=\{q\in\Bbb Q:q<a+b\}$, the Dedekind cut corresponding to the rational number $a+b$.

Suppose that $r\in A+B$. Then there are $p\in A$ and $q\in B$ such that $r=p+q$. Since $p\in A$, we know that $p<a$, and similarly, since $q\in B$, we know that $q<b$, so $r=p+q<a+b$. This shows that $A+B\subseteq\{q\in\Bbb Q:q<a+b\}$.

Now suppose that $r\in\Bbb Q$ and $r<a+b$. Let $d=\frac12(a+b-r)>0$, and note that $d$ is rational. Let $p=a-d$ and $q=b-d$; then $p,q\in\Bbb Q$. Moreover $p<a$ (since $d>0$), so $p\in A$, and $q<b$, so $q\in B$, and $$p+q=(a-d)+(b-d)=a+b-2d=a+b-(a+b-r)=r\;.$$ Thus, $r\in A+B$, and we’ve shown that $\{q\in\Bbb q:q<a+b\}\subseteq A+B$. Putting the pieces together, we conclude that $A+B=\{q\in\Bbb Q:q<a+b\}$. $\dashv$

Here I didn’t have to prove directly that $A+B$ was a Dedekind cut, because I could show that it was equal to something known to be a Dedekind cut. You should try to prove that for all Dedekind cuts $A$ and $B$, $A+B=\{p+q:p\in A\text{ and }q\in B\}$ is a Dedekind cut.

  • It’s easy to show that $A+B\ne\varnothing$.
  • Suppose that $r\in A+B$, $s\in\Bbb Q$, and $s\le r$; you need to show that $s\in A+B$. You know that $r=p+q$ for some $p\in A$ and $q\in B$. Let $d=\frac12(r-s)$, and consider the rational numbers $p-d$ and $q-d$.
  • Suppose that $r\in A+B$; you need to show that there is an $s\in A+B$ such that $r<s$. Start by writing $r=p+q$ for some $p\in A$ and $q\in B$, and use the fact that $A$ and $B$ have no largest elements.

Once you’ve done that, you can show that if $A$ is a rational cut and $B$ is an irrational cut, then $A+B$ is irrational. You already know that $A+B$ is a Dedekind cut, so you just have to show that it’s not a rational Dedekind cut: for each $r\in\Bbb Q$, $A+B\ne\{q\in\Bbb Q:q<r\}$. This can be done by contradiction: show that if $A+B=\{q\in\Bbb Q:q<r\}$ for some $r\in\Bbb Q$, then the cut $B$ is rational.

When both $A$ and $B$ are irrational cuts, $A+B$ can be either rational or irrational, depending on exactly which cuts $A$ and $B$ are; you can’t prove any general conclusion here.

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    $\begingroup$ THANK THANK YOU AGAIN!!good ideas! $\endgroup$ – art Mar 5 '13 at 18:48
  • $\begingroup$ @art: You’re very welcome. $\endgroup$ – Brian M. Scott Mar 5 '13 at 18:48

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