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Question about rationals on the real line.

  1. Mar 8, 2012 #1
    I have seen how one maps the rationals to the naturals by using prime bases. So it is quite clear that the rationals are countable. But it seems strange to me that between any 2 reals there is a rational. So it seems like there would be just as many rationals as reals. This probably has a simple answer but it is not clear to me. Any help will be appreciated.
     
  2. jcsd
  3. Mar 8, 2012 #2
    I think the statement says something exists, but not that it is the only thing in there.
     
  4. Mar 8, 2012 #3

    Office_Shredder

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    A lot of your pairs of real numbers will have the same rational number. For example for each positive plus negative real number you might pick zero.
     
  5. Mar 8, 2012 #4
    What do you mean by " For example for each positive plus negative real number you might pick
    zero "
    Ok lets look at the irrationals and the rationals. Between any two irrationals there is a rational.
     
  6. Mar 8, 2012 #5
    Yes, this is a bit counterintuitive. But say you could go through all the pairs of irrationals; and for each pair, pick a rational that's between them. If you could do this, you'd end up picking the same rational many times. For example you'd find 0 between, say, -pi and pi. And you might also pick 0 as the rational between -sqrt(2) and sqrt(2).

    Since the rationals are countable and the irrationals uncountable, it MUST be the case that we will always end up re-using some rationals a lot of times as "the rational between two irrationals."

    What's confusing here is that the rationals are equinumerous to the natural numbers; but they have very different order properties. The natural numbers can be well-ordered: there's a first one, and a second one, and a third one, etc.

    The rationals could in theory be well-ordered too. But in their usual order, they are not well ordered. Between every rational there's another rational. That's a very different order property than the natural numbers.
     
  7. Mar 8, 2012 #6
    Suppose that your two reals are very close, so they are equal up to the 100th decimal digit. They differ in the subsequent digits; for example, for one of them the 101th digit could be 3 and, for the other, 7. You could find a rational number in-between, say, the one whose 101th digit is 6 and all subsequent digits are zero.

    But, of course, that is not the only rational you can imagine. You could choose the one whose digits from the 101th position are 51234 and then all zeros. In fact, there is an infinitude of rationals in between.

    Now, how many irrationals do you think you could find between the same two numbers? How many non-repeatable sequences of digits can you form after the 101th digit? Another whole infinitude of them.

    So, if between your two reals there are infinite rationals, as well as infinite irrationals, what is the argument, again, to say that there are just as many rationals as irrationals?

    What I try to say boils down to realizing that, if your two reals are a,b, with a<b, there is a bijection between the interval [a,b] and the interval [0,1], namely the map f:[a,b]->[0,1] defined as f(x) = (x-a)/(b-a). As SteveL27 points out, the rationals may be countable but they do not line up in the ordinary lesser-to-greater order, so there is no such thing as two contiguous rationals, let alone two contiguous reals.
     
    Last edited: Mar 8, 2012
  8. Mar 8, 2012 #7
    @ Dodo : your saying that there are no 2 rationals right next to each other. and the same for the reals. Ok Its starting to make a little more sense.
     
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