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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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2 Answers 2

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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, 2020 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, 2020 at 21:33
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    $\begingroup$ @VoiletFlame - why go to all that trouble? Suppose $x > 0$, and pretend the negatives do not exist. $C = \{r\mid \forall a \in A, ra < 1\}$ and $E = \{ac\mid a \in A, c \in C\}$. So we see that for all $e \in E, e < 1$. Conversely, if $e \in \Bbb Q_+ < 1$ and $a > 0 \in A$, then $c = \frac ea \in \Bbb Q_+$ and $ca = e < 1$, so $c \in C$ and $ac = e \in E$. Thus $E = \{r \in \Bbb Q_+ | r < 1\}$, which means $E|F = 1$ in the reals. I'll leave how to modify the argument to include the negatives, and the corresponding argument when $x < 0$ to you. $\endgroup$
    – Paul Sinclair
    Mar 8, 2022 at 3:38
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    $\begingroup$ This is not the right place to ask questions about other definitions and approaches. To ask such questions, start your own thread about them. These are the comments on my answer here, and should only be used to address that answer. As for the definition of $C$ in my comment, I pointed out right at the start that I was going to ignore negatives, and left the matter of adjusting to include negatives to you. The negatives here (for $x > 0$ case) are just a technical addendum to both $C$ and $E$. The important action is on the positive side. $\endgroup$
    – Paul Sinclair
    Mar 8, 2022 at 13:14
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    $\begingroup$ I just said questions not directly relatd to this post need to be asked in their own thread. I meant it. There are several very strong reasons for this rule. I will not break it for you. $\endgroup$
    – Paul Sinclair
    Mar 8, 2022 at 17:23
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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, 2020 at 21:40
  • $\begingroup$ @Paul, Indeed. $\forall a$ should be $\forall a\in A$. Thanks. $\endgroup$
    – modnar
    Apr 12, 2020 at 10:06

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