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I'm currently working through Pugh's Analysis (for fun.) Currently, I'm working on the following problem:

A multiplicative inverse of a nonzero cut $x=A|B$ is a cut $y=C|D$ such that $x*y=1^*$. If $x>0^*$, what are $C$ and $D$? If $x<0^*$, what are they? Prove that $x$ uniquely determines $y$.

For the first two, I have respectively, $$y=\{r\in\mathbb{Q}: r<0 \text{ or } \exists b\in B \text{ is not the least element of } B \text{ } r=\frac{1}{b}\}|\text{ The rest of }\mathbb{Q}$$ $$y=\{r\in\mathbb{Q}: \exists b\in B \text{ is not the least element of } B \text { and } b< 0 \space r=\frac{1}{b}\}|\text{ The rest of }\mathbb{Q}$$

Then, I have that for positive $x$, assume that $x$ does not uniquely determine $y$ and there exists $z$ such that $x*z=1$ and $y\neq z$ it follows that $x*z=x*y$ and hence $y=z$. This feels right, but I kind of feel like I'm implicitly using division in the last step, which I probably shouldn't be. Am I missing something?

  • Based on your definition of $y$ it is obvious that $y$ is uniquely determined by $x$ (because the definition of $y$ is crucially dependent on $A$ and $B$). – Paramanand Singh May 06 '15 at 04:20
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    If you want to follow your own approach, you need to show that distributive law holds and then $x(y - z) = 0$ and since $x \neq 0$, we have $y - z = 0$ and then $y = z$. So two ideas are needed 1) distributive law 2) product of two non-zero cuts is non-zero. – Paramanand Singh May 06 '15 at 04:23

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