Talk:Construction of real numbers

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The definition of Dedekind cuts on this page seems incorrect. According to the page, a partiton of an ordered field, (A,B), is a Dedekind cut, where A and B are non-empty sets, such that A is closed downwards and B is closed upwards. Then, A and B intersect at a point - ie, they do not form a partition.

The page discusses a non-closed addition, where (\{a\le 0\}, \{b>0\})+(\{a<0\}, \{b\ge 0\}) is not a Dedekind cut. But of course, neither (\{a\le 0\}, \{b>0\}) nor (\{a<0\}, \{b\ge 0\}) is a Dedekind cut by this page's definition, which requires both sets that make the cut to be closed.

No, this is wrong. The definition does not require each set to be topologically closed. You're confusing topologically closed with downward- or upward-closed. (Walt, it wasn't me who wrote these first comments.) Revolver 02:03, 16 Dec 2004 (UTC)
Oops, sorry about that. I thought this was all one big comment by you. -- Walt Pohl 05:12, 16 Dec 2004 (UTC)

The definition I'm familiar with defines A as an open set, extended downwards. B as closed and extended upwards. The least upper bound of A is the number defined by the cut. See [1] for a reference to one such formulation.

Other references define them as a non-empty, downwards extended set, that is bounded-above, and doesn't contain a greatest element. The the least-upper bound of this set defines the number at the cut. See [2] for a reference to one such formulation.

Please see Talk:Dedekind cut Revolver 16:30, 15 Dec 2004 (UTC)

The definition on this page only requires that A be "closed downward", not closed. Likewise for B. -- Walt Pohl
Please read the comments at Talk:Dedekind cut. This is precisely the problem. Unless A is required to be open (equiv, B is required to be closed), addition is not a closed operation. Read the comment in the article -- it is absolutely correct. We need a definition where the operation makes sense! The best way around this for the particular case of Q seems to be to required that A be open. More general constructions in general posets may handle the situation differently. Revolver 02:01, 16 Dec 2004 (UTC)

[edit] The Dedekind cut arithmetics is erroneous

Note, that if you define (A_x,B_x)\, as the cut corresponding to 2-\sqrt2\,, and (A_y,B_y)\, as corresponding to 2+\sqrt2\,, and apply the rules as written, then neither (A_{\rm sum},B_{\rm sum})\,, nor (A_{\rm prod},B_{\rm prod})\, is a Dedekind cut. The former is not even a partition, since 4 \notin A_{\rm sum} \cup B_{\rm sum}; and the latter is a partition but not a correct cut, since 2 is the greatest element of A_{\rm prod}\,.

The older constructions were based only on the set of positive rational numbers. The cuts may be defined for the whole set \Bbb{Q}\,; but in that case the rules for the arithmetic get immensely more complex than the thing exhibited at the moment. (The same goes for any other ordered field.) The rules for arithmetics now are not completely adapted to this, even in spite of the 'cheating' at multiplication by making a restriction to nonnegative cuts. My personal suggestion is to make another approach. I've made a longer suggestion for this at de:Diskussion:Dedekindscher_Schnitt, which I prefer not to repeat here. JoergenB 01:16, 10 October 2006 (UTC)

I suggest we only do it for positive rationals, as that will give an idea of what is going on, (and anyone who can understand it for positives, can work out the extension.. its just tedious..). any ideas? --TM-77 14:23, 28 October 2006 (UTC)

[edit] Construction by decimal expansions

We can take the infinite decimal expansion to be the definition of a real number, considering expansions like 0.9999... and 1.0000... to be equivalent, and define the arithmetical operations formally.

I don't know whether such a definition would work; it seems reasonable. I think it should be pointed out that we could use (any base? any integer as a base?). In any case surely it doesn't need to have anything to do with the number 10. Brianjd | Why restrict HTML? | 11:46, 2005 Apr 24 (UTC)

Doing that is the easy part. The hard part is to prove that the newly created field is complete, that is any Cauchy sequence converges. Oleg Alexandrov 15:14, 24 Apr 2005 (UTC)
maybe you guys can tell me how to properly define the sum of 0.44444444 and 0.66666666 (I mean the infinite decimals, of course)....in such a way that your definition applies to any two decimal expansions...--345Kai 20:40, 3 October 2006 (UTC)
0.444... + 0.666... = (0.4 + 0.6) + (0.04 + 0.06) + (0.004 + 0.006) + ... = 1 + .1 + .01 + ... = 1.111... --Ian Maxwell 04:15, 14 October 2006 (UTC)
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