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Integer and Rational Number Subtleties in an Algebra Problem

  1. Mar 16, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##S = \{\frac{1}{n} + \mathbb{Z} ~|~ n \in \mathbb{N} \}##. I am trying to show that ##f : \mathbb{N} \rightarrow S## defined by ##f(n) = \frac{1}{n} + \mathbb{Z}## is a bijection. Surjectivity is trivial, but injectivity is a little more involved.

    2. Relevant equations


    3. The attempt at a solution

    Suppose that ##f(n) = f(m)##. Then ##\frac{1}{n} + \mathbb{Z} = \frac{1}{m} + \mathbb{Z}## or ##\frac{1}{n} - \frac{1}{m} \in \mathbb{Z}##. This implies ##\frac{m-n}{nm}## is an integer. If ##m \neq n##, then the fraction is not zero but a ratio of two nonzero integers. However, if ##\frac{p}{q}## is ##\frac{m-n}{nm}## in reduced form, how do I know that ##q = 1## can't be true; how do I know that ##m \neq n## and ##p=1## can't both be true?
     
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  3. Mar 16, 2017 #2

    fresh_42

    Staff: Mentor

    If ##\frac{m-n}{nm}## is a natural number, which you can assume since both are and one has to be larger than the other (or equal), what does it mean that ##nm\,\vert \,(m-n)\, ##? Just try some values to see what I mean.
     
  4. Mar 18, 2017 #3

    pasmith

    User Avatar
    Homework Helper

    Suppose [itex]n[/itex] and [itex]m[/itex] are positive integers such that [tex]\frac1n - \frac1m = p \in \mathbb{Z}.[/tex] Now [tex]
    0 < n = \frac{m}{1 + pm}[/tex] and as [itex]m > 0[/itex] we have that [itex]1 + pm > 0[/itex] and so [itex]p > -\frac1m[/itex]. However in fact [itex]n \geq 1[/itex], so we can get an upper bound on [itex]p[/itex] as well.
     
  5. Mar 18, 2017 #3

    pasmith

    User Avatar
    Homework Helper

    Suppose [itex]n[/itex] and [itex]m[/itex] are positive integers such that [tex]\frac1n - \frac1m = p \in \mathbb{Z}.[/tex] Now [tex]
    0 < n = \frac{m}{1 + pm}[/tex] and as [itex]m > 0[/itex] we have that [itex]1 + pm > 0[/itex] and so [itex]p > -\frac1m[/itex]. However in fact [itex]n \geq 1[/itex], so we can get an upper bound on [itex]p[/itex] as well.
     
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