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Rational Dependence

  1. Jul 13, 2010 #1
    Hi guys:

    I've got a problem I've been working on for some weeks and this might be the key to unlocking it.

    The question is:

    Given a vector in R^k, what is the measure of the set of vectors whose components are rationally dependent?

    Rationally dependent means for a given vector, you may find a vector with rational coefficients such that their inner product is 0.

    (1/2,1/3,1/6) is RD because of (2,3,-12), for example.
    Last edited: Jul 13, 2010
  2. jcsd
  3. Jul 13, 2010 #2


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    But this is easily generalized: every 3-tuple (p,q,r) with rational entries p,q,r is RD because
    [tex](p,q,r)\cdot (1/p,1/q,-2/r)=0[/tex]
    and 1/p,1/q,-2/r are rational. This is then easily generalized to R^n instead of R^3. Hence the set of RD vectors contains then ones with rational entries. But the latter one is already dense in R^n.
  4. Jul 13, 2010 #3
    I suppose it was late, and this meant I had to improperly state the question!

    Really, the question is does the set of RD vectors have nonzero measure over R^k, not whether they are dense or not. Of course the rationals are rationally dependent and dense, but they are a set of measure zero in R. So then the issue is whether a.a. collections of irrationals are RD, and my intuition says no.

    So RID vectors requires k-1 components to be irrational. I don't have a hold on determining exactly "how much" of these sets are RD.
    Last edited: Jul 13, 2010
  5. Jul 13, 2010 #4
    I think I understand the statement, but I'm not totally sure. I suppose that you mean what is the measure of the set of all vectors which are rationally dependent (if this is even measurable)? I've never heard of rationally dependent before...

    I think this is the answer:
    Suppose [itex]r\in\mathbb{R}^n[/itex]. Note that the measure of the set of all vectors perpendicular to [itex]r[/itex] is zero. Now, what is the measure of a countable union of sets of measure zero?

    Does that help?
  6. Jul 13, 2010 #5


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    I don't see why you're only taking a countable union. There are certainly uncountably many vectors in Rn.

    A modified argument seems to work though. Pick a vector with rational components q in Qn. Any rationally dependent vector is perpendicular to some such q, so lies in one of countably many measure zero sets
    Last edited: Jul 13, 2010
  7. Jul 13, 2010 #6
    That's not modified at all, that is the argument. I didn't want to give everything away.
  8. Jul 13, 2010 #7
    Much appreciated. I believe this solves my question.

    Don't worry about revealing the "proof", I would say that this problem is just a redefinition of a small mechanism in a larger problem, which has nothing to do with linear algebra, actually, so letting me in on the mechanism is of no great detriment to my progress.

    Thank you both!
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