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Real positive mappings

  1. Jul 9, 2008 #1
    Prove that for any
    [tex]f:\mathbb{R}\rightarrow\left(0,\infty \right)[/tex]
    there is a rational number q and an irrational number i such that
    [tex]f(q)f(i) > \left( \frac{q-i}{2} \right)^2[/tex]
    (Or if this is false find a counterexample)

    2. Relevant equations

    3. The attempt at a solution
    The best idea to me seems to be to look for a contradiction, that is assume
    [tex]f(q)f(i) \leq \left( \frac{q-i}{2} \right)^2[/tex]
    for all rational q and irrational i and try to come up with a contradiction.

    The first approach I tried was to fix a q, say q=0, then for every epsilon greater than zero there exists an irrational number i such that
    [tex]f(i)\leq \epsilon[/tex]
    (since every neighbourhood of a real number contains a rational number)
    similarly there exists a rational number q such that
    [tex]f(q)\leq \epsilon[/tex]
    So both these conditions are fine on their own, so I need to think of some way to combine them.

    Define a sequence of irrational numbers, i_0(n) by
    [tex] f(i_0(n)) \leq 1/n[/tex]
    Then define interlaced sequences of rational and irrational numbers:
    [tex]f(q_k(n)) \leq f(i_{k-1}(n))[/tex]
    [tex]f(i_k(n)) \leq f(q_k(n))[/tex]
    But I haven't found anything useful to do with these sequences - they could be quite trivial, e.g.f(q_k(n))=f(i_k(n)) for all k,n and q_k(n)=q_l(n) for all n, l and k similarly for the irrationals.

    So I'm stuck, I've got to think of some way to impose both conditions (the one on the rationals and the one of the irrationals) simultaneously (no single sequence will do the trick), but I can't think of a way to do it.

    The only other way I can think of trying to approach it is with some abstract formalism (e.g. topology), but since f isn't even continuous topology won't do. Maybe an argument based on cardinality could be feasible, but I can't think of one.
    Last edited: Jul 9, 2008
  2. jcsd

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