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Can Someone explain Dedikind cuts?

  1. Dec 14, 2004 #1
    Can someone explain to me how dedikind cuts derive the real system from the rational number system? I understand how the cuts derive an algebra with a defined addition and multiplication, etc. How do we jump from this algebra on these cuts to assuming the existence of the irrationals? I guess I can't just make that final connection. Can someone explain this?
     
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  3. Dec 15, 2004 #2

    NateTG

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    The 'existance' of irrational numbers is related to a notion of completeness. I will describe it here as the notion that every set of real numbers that has an upper bound has a least upper bound, but there are other equivalent descriptions.

    Now, let's say that a Dedekind cut [itex]D[/itex] is a partition of the rational numbers [itex]\mathbb{Q}[/itex] into two non-empty sets [itex]B_D[/itex] and [itex]T_D[/itex] where every element of [itex]B_D[/itex] is strictly smaller than all elements of [itex]T_D[/itex].

    For two Dedekind cuts [itex]D_1,D_2[/itex] we'll say that [itex]D_1 \geq D_2[/itex] if [itex]B_{D_1} \supset B_{D_2}[/itex].

    Then if we have a non-empty set of dedekind cuts [itex]\{D_1,D_2...\}[/itex] with the upper bound [itex]D_U[/itex] (i.e. [itex]\forall D_i, B_{D_i} \subset B_{D_U}[/itex] we can take [itex]B=\bigcup B_i[/itex] and [itex]T=\mathbb{Q} - B[/itex]. It's easy to see that these two sets do form a Dedekind cut, and that this Dedekind cut is an upper bound for the Dedekind cuts in the set, and that this Dedekind cut is the smallest upper bound.

    For convenience, I will start with 'rational' Dedekind cuts [itex]D_q[/itex] where [itex]B_q = \{x \in \mathbb{Q} x < q\}[/itex]. Then, for example, it's easy to look at the set of Dedekind cuts [itex]\{D_q | q^2<2\}[/itex] (which corresponds to the set of all rational numbers whose square is less than 2). It's also clear to see that this set is bounded above by [itex]D_2[/itex] therefore, from above, there is some Dedekind cut [itex]D_{\sqrt{2}}[/itex] which is the least upper bound of that set of Dedekind cuts. You should be able to work out that it is indeed the square root of [itex]D_2[/itex].
     
  4. Dec 15, 2004 #3

    mathwonk

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    the idea is that the real numbers are points on a line, so each fixed number separates the line into two parts, the numbers to the left and the numbers to the right of that fixed number.

    then the idea is that those two parts also determine back thye original number, as the unique number in between those two subsets of the line. Finally note that since the rationals are "dense" in the real line, actually the two parts of the line are determiend by the rational points in them.

    So it goes like this: there is a one one correspondence between real numbers and subdivisions of the line into two parts, such that all points in one part are entirely to the left of all points in the other part. I.e. given a real number x, we get a subdivision of the line, into all numbers less than x and all numbers greater than or equal to x.

    Note that here the left part has no largest number, but the right part does have a smallest number namely x. Conversely any subdivision of the real line into two parts, such that all points of the left part are to the left of all points of the right part, corresponds to some unique real number. I.e. either the left part has a largest number or the right part hasa smallest number, and that largest or smallest number is the desired real number defiend by the two part subdivision.

    Now consider the same idea for rationals, we can subdivide the rationals into two parts but so that neither aprt has a largest or smallest element. e.g. take all rational less than pi and all rationals greater than pi. On the other hand some subdivisions of rationals do correspond to numbers, like dividing the rationals at 3/4. So with rationals there are more two part subdivisions than there are rationa numbers.

    Indeed each way of subdividing the rationals where there is no largest or smallest number to either part, corresponds toa real irrational number. So, starting just from the rationals, we could define the reals as all two part subdivisions of the rationals. The ones in which there is a largest or smallest element are rational numbers, and the oens with neither are irrational numbers.

    this is an enormously tedious way to define real numbers in practice, but in theory it is rather beautiful.

    by the way i see that suddenly i am a "homework helper" and "science advisor", instead of just a mouthy participant. I consider this a great honor, but it may only mean I have posted a certain large number of entries over a certain length of time. If it in fact means someone has found some of them helpful, I am grateful and gratified.
     
    Last edited: Dec 15, 2004
  5. Dec 15, 2004 #4

    Hurkyl

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    In this model, the irrationals (and rationals!) are the cuts. For example, the square root of two is the cut

    {x | x < 0 or x^2 < 2}
     
  6. Dec 15, 2004 #5
    Both of these posts help. I think what confused me is that my book started with rational Dedekind cuts, and sketched out the rest leaving it (infamously) "up to the reader". One question remains though, if the cut is rational (in other words if we start with Q and try to derive R), my book defined the cut a as by definition in the right half. Basically then I understand (from the helpful posts) that the reals are the holes in between these ribbons and say the left half has no lower bound (clearly) and no upper bound (as the cut is say at 2^1/2). We can then determine operators (+ and *) for the cuts and derive an algebra and the cuts correspond to a mapping of the real system. So 2^1/2 is represent by the dedekind cut at that irrational. But . . . if we are deriving R from Q in this way, how do you know there is only one irrational at that cut 2^1/2? How do you know there aren't two solutions to 2^1/2 at that Dedekind cut? How do you show 2^1/2 is unique?

    BTW I am taking classes in AbstractAlgebra and Group theory, then differential eq, then differential geometry, in that sequence. I am an older student, and this is for fun and education. So I am most interested in the abstract, as I already have a job, family and plenty of practicality. I just want to learn stuff I passed up on in life by running off to make cash. This is why the posts in algebra, diff geo, etc.

    Also, how do you all get equations in the posts? Is there a tutorial or web site where I can do that? It would help my homework to ask questions with the right notation.

    Thank you. You are all very helpful.
     
  7. Dec 15, 2004 #6

    Hurkyl

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    Well, how many Dedekind cuts, S, satisfy S*S = 2?

    The Dedekind cuts form a field, right? ...
     
  8. Dec 15, 2004 #7
    Hurkyl, are you saying that we need only define one solution to s^2=2? In that all cuts of s and say s' would be in the same equivalence class anyway? I guess that makes sense, finally. I think I might have this now . . .
     
  9. Dec 15, 2004 #8

    Hurkyl

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    At no point do we define a solution to x^2 = 2 -- we only define the Dedekind cuts and their arithmetic.
     
  10. Dec 15, 2004 #9
    Actually, I get it even more . . . You can define a cut C as {x in Q:f(x)<a} where x is some function of rational numbers (x,+,/,-,^). C' is {x in Q:f(x)>=a}. In rational cuts, like {x:x<3} the cut point can be set either way, here on the right half, and either half can define the cut. In irrational cuts, like {x:x*x<2}, no rational number satisfies, so the discarded C' has no lower bound, but does have a lower bound in rational cuts. But the kept cut C has no upper or lower bound in either rational or irratonal cut in this definition. So we label the set of all rational number in C as the "cut" C and label that set with the cut, a. Labeling the cut 2^1/2 just is shorthand for the equation used to define the cut and doesn't imply the existence of a rational number that actually satisfies that equation. So we don't go outside of Q to define R. So we have N, then pairs of N make Z, then pairs of Z make Q, and cuts of Q make R. Got it!
     
  11. Dec 15, 2004 #10

    Hurkyl

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    Well done!
     
  12. Dec 15, 2004 #11

    Hurkyl

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    Now that you've got it, let me confuse you again. :smile:

    In some sense, you've constructed the natural numbers, the integers, the rationals, and the reals... but in another sense it isn't appropriate to call them the natural numbers, ..., the real numbers.

    Anything that satisfies the axioms of a complete ordered field, for example, deserves to be called "the real numbers" -- the thing you constructed doesn't have an exclusive right to that title. (Of course, both will be isomorphic to each other!)

    The main point of the construction is merely to show that there is some thing that can be called the real numbers.
     
  13. Dec 15, 2004 #12

    mathwonk

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    there is a(n easy) theorem that there is only one complete (in the least upper bound sense) ordered field however, up to unique isoimorpohism, so if you think of that field as determined by its proeprties and not by the names of its elements then it is unique.
     
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