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A Dedekind's cut question

  1. Apr 12, 2005 #1
    By Dedekind's cut we can define some irrational number, where L and R are two disjoint sets of all rational numbers.

    We know that rational and irrational numbers have different properties, that give us the ability to clearly distinguish between a rational number (expressed as a ratio between at least two integers) and an irrational numbers (cannot be expressed as a ratio between at least two integers).

    My question is this:

    Since rational and irrational numbers are clearly distinguished from each other, then what is the purpose of Dedekind’s cut, and why the word ‘cut’ is used here?

    Thank you.
     
  2. jcsd
  3. Apr 12, 2005 #2

    HallsofIvy

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    Dedinkind cuts do not give irrational numbers, nor are they used to distinguish between rational and irrational numbers.
    The purpose of Dedekind's cut is to give a rigorous definition of all real numbers as sets of rational numbers. It is not just a matter of "distinguishing between" them- in fact, since all rational numbers are real numbers, there are "rational" cuts that correspond to rational numbers.

    As far as the name is concerned, Dedekind used the term "Schnitt" (German for "cut") because, in our everyday concept of numbers, it was the boundary between L and R (the place that "cuts" between them) that determines the real number corresponding to the "cut".

    The nice thing about the Dedekind Cut definition is that it makes it easy to prove the "Least Upper Bound Property", a defining property of the real numbers (i.e.it is true in the field of real numbers but not rational numbers).

    Of course there are other ways of going from rational numbers to real numbers. You can define real numbers as equivalence classes of increasing sequences of rational numbers- making it easy to prove "monotone convergence"- or as Cauchy sequences of rational numbers-making it easy to prove the "Cauchy Criterion".
     
  4. Apr 12, 2005 #3
    Is it true that if L and R are disjoint sets of all rational numbers, there cannot be but an irrational number between them?

    And if it is true, how can we prove it?
     
    Last edited: Apr 12, 2005
  5. Apr 12, 2005 #4

    matt grime

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    Are L and R dedekind cuts still? How can you have "disjoint sets of all irrationals"? If L contains all irrationals, how can R contain any irrationals? and if we read it differently and L and R jointly contain all irrationals how can there be any irrationals "between" them? In short, it is very unclear what you're talking about. could you try starting with clear definitions of what L and R are?

    moreover dedekind cuts do not contain irrationals, they define the reals in terms of sets of rationals.
     
  6. Apr 12, 2005 #5
    Thank you Matt my mistake, the correct one is:

    Is it true that if L and R are disjoint sets of all rational numbers, there cannot be but an irrational number between them?

    L + R are sets of all Q members.

    L = {x: x^2 <2}
    R = {x: x^2 >2}

    √2 is between R and L.

    Is √2 is the result of Dedekind's cut, in this case?
     
    Last edited: Apr 12, 2005
  7. Apr 12, 2005 #6

    matt grime

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    Well, apart from the dubious nature of your mathematics, surely the answer is, in your own terms, no: put 4 instead of 2 in the brackets on the right of the < and >. Or reread the first reply to you post by Halls of Ivy.
     
  8. Apr 12, 2005 #7
    HallsofIvy wrote:
    State 1:

    If L and R are disjoint sets of all rational numbers such that any l in L < any r in R , L has no greatest member and R has no smallest member, then there must be an irrational number between L and R.

    Isn't it?


    State 2:

    If L and R are disjoint sets of all irrational numbers such that any l in L < any r in R , L has no greatest member and R has no smallest member, then there must be a rational number between L and R.

    Isn't it?


    If the answer to both states is 'yes' then I think that there is some strange situation here, because the probability to find state 2 along the real-line is greater then state 1.

    But if we define a 1-1 mapping between each arbitrary state 2 and the rational number that exists between L and R of state 2, then there are much more cases of rational numbers between irrational numbers than irrational numbers between rational numbers.
     
    Last edited: Apr 12, 2005
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