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Addition/Multiplication for Dedekind cuts?

  1. Jul 12, 2011 #1
    I don't really understand the properties for adding/multiplying dedekind cuts. I get that they're closed, commutative and associative because that follows from the rational numbers (and the cut just partitions a rational number into 2 classes of rationals, plus the "cut" that only contains one number), but the other properties are confusing.

    First, I'm not really clear on how to add/multiply cuts- which happens due to them containing only rationals. I thought you take the limit of cuts (which converge to a precise limit) and this limit is the precise number that you are able to add/multiply, but it is not an actual pinpointed number and just a limit, yet limits behave just like numbers. But then I have trouble tying all of this together.

    EDIT: Okay I think I'm mixing up two ways of constructing the real numbers.

    For the identity property for addition, 0* can be chosen as all the negative rational. Say p is a point in a dedekind cut A, and q is another point such that p < q. Then p = q + [-(q-p)], where -(q-p) is a point in 0*. Thus A [itex]\subset[/itex] A + 0*, and A + 0* [itex]\subset[/itex] A due to the property of dedekind cuts to be closed downward. This makes A + 0* = A, or am i misunderstanding a concept?

    The inverse is really hard for me to imagine. A + B = 0*, so B is supposed to be a cut that gives a cut of all the negative rationals?

    Multiplication seems to be similar to addition, but the cases are divided into positive and negative. 1* should be part of the identity property.
     
    Last edited: Jul 12, 2011
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  3. Jul 12, 2011 #2

    micromass

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    Cuts have nothing to do with limits (at least not immediately). What a cut represents is all the numbers smaller than a given number. Thus the number 2 is represented by the cut

    [tex]\{x\in \mathbb{Q}~\vert~x<2\}[/tex]

    So the supremum of a cut is the number it represents. However, the supremum doesn't always exist, because we're working in the rationals here. For example, the cut

    [tex]\{x\in \mathbb{Q}~\vert~x^2<2\}[/tex]

    doesn't have a supremum. Indeed, that supremum would be [itex]\sqrt{2}[/itex] but this isn't rational!!

    This is ok.

    Do some easy examples first. You know the inverse of 2 in -2. The cut of 2 is

    [tex]A=\{x\in \mathbb{Q}~\vert~x<2\}[/tex]

    You must transform this into the cut of -2, that is

    [tex]B=\{x\in \mathbb{Q}~\vert~x<-2\}[/tex]

    The thing to look at is -B, that is

    [tex]-B=\{-x\in \mathbb{Q}~\vert~x<-2\}[/tex]

    Then you obtain all numbers larger than 2. So given a cut A, a good choice for the inverse would be

    [tex]\{-x\in \mathbb{Q}~\vert~x\notin A\}[/tex]

    However, this cut might have a largest element (which is not allowed), so we need to remove that!
     
  4. Jul 14, 2011 #3

    HallsofIvy

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    To multiply two cuts, start with "positive" cuts. A cut is positive if and only if it contains at least one rational number. If a and b are positive cuts, then ab is the set of all rational numbers, xy, where [itex]x\in a[/itex], [itex]y\in b[/itex]. Show that this is a cut. Show that if a is any cut, then a1= a. Remember that "1" is the set of all rational numbers less than 1.

    Then define: if a< 0, b> 0, ab= -(-a)(b), if a> 0, b< 0, ab= -(a)(-b), and if a< 0, b< 0, ab= (-a)(-b).
     
  5. Aug 3, 2011 #4
    Sorry to jump into this but actually I'm going through the same proof and I don't get this.

    -(q-p) in 0* I understand as some negative rational in the set of all negative rationals but the jump from there to saying A [itex]\subset[/itex] A + 0* I just can't follow and the more I think about it the worse it gets.

    I imagine A + 0* as all rationals in A [itex]\geq[/itex] 0 and A [itex]\subset[/itex] A + 0* just baffles me :(
     
  6. Aug 4, 2011 #5

    HallsofIvy

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    Then you are being too imaginative! "All rationals in A[itex]\geq[/itex] 0" is not a cut.
    (Do you see why?)

    0* is the set of all rational number less than 0- in other words the negative rationals. A+ 0* consists of all sums of rationals in A and rationals in 0*. But since all rationals in 0* are negative, we always get numbers less than the one in A.

    That is, if [itex]a\in A[/itex], [itex]o\in O*[/itex], then a- o< a. Since one of the properties of a cut is "if [itex]a\in A[/itex] and b< a, then [itex]b\in A[/itex], it follows that every member of A+ O* is in A. That is, A+ O* is a subset of A.

    Of course, it is also one of the properties of a cut that there is no largest member. If [itex]a\in A[/itex], there exist [itex]b\in A[/itex] with b> a. Then c= (a- b)< 0 so c is in O*. It follows that b+ c= b+ (a- b)= a is in A+ O*. That is, A is a subset of A+ O*: A= A+ O*.
     
  7. Aug 5, 2011 #6
    Yes, it's not closed downward.

    I think my mistake was imagining adding [itex] x \in A [/itex] and [itex] y \in B [/itex] as [itex] x_n + y_n [/itex] thus causing O* to cancel the elements of A < 0. Am I right in assuming now that it is the set of each element [itex] x \in A [/itex] added to all [itex] y \in B [/itex].

    Is it because Q is dense (and we are working with a cut) that [itex] A + O^* \subseteq A [/itex] and [itex] A \subseteq A + O^*[/itex] ? I.e., you are not losing the larger elements of A by subtracting a set of negative numbers.

    Many books (Rudin) use [itex] \subset [/itex] so I'm not sure my interpretation is correct at all.



    Thanks a lot for your help. Once I get myself muddled it's very hard to get unmuddled!
     
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