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Nested sequences of rational intervals

  1. Nov 18, 2008 #1
    My textbook says the following:

    For a closed interval J_n = [a_n, b_n]

    "A nested sequence of rational intervals give rise to a separation of all rational numbers into three classes (A so-called Dedekind Cut). The first class consists of the rational numbers r lying to the left of the intervals J_n for sufficiently large n, or for which r < a_n for almost all n. The second class consists of the rational numbers r contained in all intervals J_n. This class contains at most one number, since the length of the interval J_n shrinks to zero with increasing n. The third class consists of the rational numbers r for which r > b_n for almost all n. It is clear that any number of the first class is less than any of the second class, and that any number of the second class is less than any of the third class. The points a_n themselves are either in the first or second class, and the numbers b_n either in the second or third class.

    If the second class is not empty, it consists of a single rational number r. In this case the first class consists of the rational numbers less than r, the third class of the rational numbers greater than r. We say then that the nested sequence of intervals J_n represents the rational number r. For example, the nested sequence of intervals [r - 1/n, r + 1/n] represents the number r.

    If the second class is empty, then the nested sequence does not represent a rational number; these nested sequences then serve to represent irrational numbers. The individual intervals [a_n,b_n] of the sequence are for this purpose unimportant; only the separation of the rational numbers into three classes generated by this sequence is essential, telling us where the irrational number fits in among the rational ones."



    I don't understand the last paragraph. When they say "if the second class is empty, then the nested sequence does not represent a rational number", do they mean that if the nested sequences do not converge to a rational number, then it must represent an irrational number? Or do they mean that the nested sequence of intervals do not contain any point whatsoever (which I don't believe is possible if the length of each sub-interval tends to zero as n goes to infinity)?
     
  2. jcsd
  3. Nov 18, 2008 #2

    lurflurf

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    Homework Helper

    It depends how you look at it.
    Say we start with rational numbers.
    We define cuts as a way to break the rationals.
    We always get the big ones and the small ones.
    Some times a single rational stands out as separationg the two sets.
    We can think of that rational and that cut as being in related or linked.
    We can say the middle or separting class contains at most one rational.
    Sometimes no rational separates the classes
    If we are restricted to rationals the middle class is empty and we leave it at that.
    This is sometimes undesireable we want the middle class to contain a number.
    There are two ways to do this.
    One way is to just say each cut is defined by one number.
    when the one number is not rational we nave a new number we call irrational
    The set of cuts is isomorphic to a number system we then call real numbers
    Another (roundabout) method would be to first construct a new number system.
    The difference is the same as that between
    "Drats we shot and missed, quick draw a bullseye around the hole"
    or
    "Drats we shot and mised, in the future only shoot over there, where targets are so close missing is impossibel."

    "Why bother?" you say.
    Well someday you might learn something called calculus.

    In calculus one does something similar to a cut called a limit..
    The answer to some question could be a limit.
    Lets say we have a question answer is a cut.
    A non rational cut may be satisfactory
    sqrt(2) for exampe
    A non rational cut may be an step towards a rational answer
    [sqrt(2)]^2
    A non rational cut may be a good way to unstand an impossible anwer
    sqrt(2) when only rational answers are allowed might have different meaning than other impossible anwers like 11/infty or the integer between 1.1 and 1.3.
     
  4. Nov 18, 2008 #3
    Understood. Thanks for the reply, it was just the answer I was looking for :)
     
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