Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hey guys I have a quiz soon and I really dont know how to prove this question. I tried my best but it is not working. Please help out with anything or hints.

Let $(A'_1, A'_2)$ be a Dedekind cut of ${\mathbb{Q}}$ that represents the same real number as $(A_1, A_2)$. Let $C'_1 = A'_1 + B_1$ and $C'_2 = {\mathbb{Q}} \diagdown C'_1$.

Prove that $(C'_1, C'_2)$ represents the same real number as $(C_1, C_2)$.

By Dedekind cut, it means a pair $(A,B)$ such that:

(i) $A,B\neq\emptyset,$

(ii) $A\cap B=\emptyset,$

(iii) $A\cup B=\Bbb Q,$

(iv) $y\in A$ whenever $y\in\Bbb Q$ and there is some $x\in A$ with $y<x,$ and

(v) $A$ has no greatest element.

share|cite|improve this question
What is $C_1$ and $C_2$? – Pedro Tamaroff Feb 26 '13 at 17:50
@Peter: Presumably $C_1=A_1+B_1$ and $C_2=\Bbb Q\setminus C_1$. – Brian M. Scott Feb 26 '13 at 17:52
Do you know when two Dedekind cuts represent the same real number? You’ve not given us your definition of Dedekind cut $-$ there are different ones in use $-$ but if it’s what I think, this happens only when the cut represents a rational. For example, $0$ is represented by both $\langle(\leftarrow,0),[0,\to)\rangle$ and $\langle(\leftarrow,0],(0,\to)\rangle$. – Brian M. Scott Feb 26 '13 at 17:54
i will fix the question again – MathGeek Feb 26 '13 at 17:56
No, that’s not what I meant. Prove that if $\langle A_1,A_2\rangle$ and $\langle A_1',A_2'\rangle$ represent the same real, then $A_1=A_1'$ and $A_2=A_2'$. Then the result that you want is trivial, because $A_1'+B_1=A_1+B_1$. – Brian M. Scott Feb 26 '13 at 18:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.