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I'm reading Pugh's "Real Mathematical Analysis" where C.Pugh constructs real numbers system using Dedekind cuts. Unfortunately, he omits a construction of a multiplicative inverse of a real number $x = A|B$ where $A$ and $B$ are cuts in $\mathbb{Q}$. First, some required definitions:

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

a)$A \cup B = \mathbb{Q}, A \neq \varnothing, B \neq \varnothing, A \cap B = \varnothing$

b) If $a \in A$ and $b \in B$ then $a < b$

c) A contains no largest element

A real number is a cut in $\mathbb{Q}$.

The additive inverse of a cut $x = A|B$ is defined as $(-x) = C|D$ where $C = \{r \in Q| \exists b \in B$ (not the smallest elements of $B$ ) $: r = -b \}$ and $D = \mathbb{Q} \setminus C$

Multiplication is defined for positive numbers $x > 0$. If $x = A|B, y = C|D$, then $xy = E|F$ where

$E = \{r \in \mathbb{Q}| r \leq 0$ or $\exists a \in A \ \ \exists c \in C: a > 0, c > 0, r = ac \}, F = \mathbb{Q} \setminus E$. If $x > 0, y < 0$ then we define $xy = -(x(-y))$

Now I wonder how can we define a multiplicative inverse of a positive cut $x = A|B$. Obviously, it's a cut $\frac{1}{x} = C|D$ such that $x \frac{1}{x} = 1$

So, it's such a cut $C|D$ such that for $C$ we have $\{r \in \mathbb{Q}| r \leq 0$ or $\exists a \in A \ \ \exists c \in C : a > 0, b > 0, r = ac \} = \{r \in \mathbb{Q}| r < 1 \}$

Any ideas how we can define $C$(if we define $C$, then $D = \mathbb{Q} \setminus C$)?

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Edit: Changed to correct the error modnar points out in the comments.

If $x > 0$, $$C = \{r \in \Bbb Q \mid \exists b \in B,\, rb < 1\}$$ If $x < 0$, $$C = \{r \in \Bbb Q \mid \exists b \in B,\, b < 0\text{ and } rb > 1\}$$


Original post: (has maximums when $x$ is rational)

If $x > 0$, $$C = \{ r \in \Bbb Q \mid r \le 0 \text{ or } \forall a \in A,\, ra < 1\}$$

If $x < 0$, $$C = \{ r \in \Bbb Q \mid \forall a \in A, \,ra > 1\}$$

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  • $\begingroup$ When $a$ is the multiplicative identity in $\mathbb{R}$, i.e, $a=\{r\in \mathbb{Q}\mid r<1\}$, the above $C$ becomes $a\cup \{1\}$, which has a greatest element $1$. $\endgroup$ – modnar Apr 11 at 18:23
  • $\begingroup$ @modnar - it took a while to interpret your comment, because the first time you mentioned "$a$", you actually meant "$x$", and the other two times you mentioned "$a$", you actually meant "$A$". However, you are correct, and the problem is not just with $1$, but occurs any time $x$ is rational. I've changed the answer to one version that doesn't have maximums. A different fix would have been to just add $r \ne \min B$ to each of the original set definitions, but then I'd have to point out this still works when $\min B$ doesn't exist. $\endgroup$ – Paul Sinclair Apr 11 at 21:33
  • $\begingroup$ You are right. The notation in my comment is confusing. I like your revised construction, which is quite elegant. $\endgroup$ – modnar Apr 12 at 10:31
  • $\begingroup$ I am confused. Why did you not simply take $C=\{\frac{1}{y}:x\ne y\in B\}$? $\endgroup$ – Shahab Apr 25 at 8:29
  • $\begingroup$ @Shahab - when $x < 0$, your $C$ will contain both positive elements that are $> \frac 1x$ and negative elements that are $< \frac 1x$. When $x > 0$, your $C$ will not contain $0$ itself, which the full lower Dedekind cut should include. And even for non-zero rationals, my version obviously contains every rational $<\frac 1x$, while yours requires an additional (though simple) argument to show it. And another reason is that my version doesn't require the development of division in $\Bbb Q$ before it can be introduced. $\endgroup$ – Paul Sinclair Apr 25 at 13:31
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A possible way to fix the bug mentioned under Paul's answer is $$C=\{r\in \mathbb{Q}\mid r\leq 0\ \text{or}\ (\exists s\in \mathbb{Q}\ (r<s)\ \text{and}\ (\forall a\in A\ sa<1))\}$$ for the case $x>0$. A similar trick should be able to take care of the case $x<0$, too.

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  • $\begingroup$ If you properly qualify $a$, that will work. As is, $C = \emptyset$, since for any rational $s$, there are objects $a$ for which multiplication with $s$ is defined and $sa > 1$. $\endgroup$ – Paul Sinclair Apr 11 at 21:40
  • $\begingroup$ @Paul, Indeed. $\forall a$ should be $\forall a\in A$. Thanks. $\endgroup$ – modnar Apr 12 at 10:06

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