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No Divisors Between an Integer and Twice it

  1. Nov 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Suppose that ##x \in \mathbb{N}## is such that ##n < x < 2n##, where ##n## is another natural number. I want to conclude that it is impossible for ##x## to divide ##2n##. Also, is there a way to naturally generalize this to all integers?
    2. Relevant equations


    3. The attempt at a solution
    Since ##x \in (n,2n)##< then ##x = n + \ell## for some natural number ##\ell \in (0,n)##. Now, suppose the contrary, that ##x## divides ##2n##. Then ##2n = k(n + \ell)## for some ##k \in \mathbb{Z}##. Immediately we can say that ##k## must be a positive integer. Rearranging the equation gives

    ##n(2-k) = k(n+\ell)##

    If ##k## is either ##1## or ##2##, then we get ##n = \ell## or ##2 \ell = 0##, both of which are contradictions. If ##k > 2##, then the LHS will be negative but the RHS will be positive, a contradiction.

    How do I make this a little more rigorous?
     
    Last edited: Nov 4, 2016
  2. jcsd
  3. Nov 4, 2016 #2

    fresh_42

    Staff: Mentor

    ##2n = k(n + \ell) \nRightarrow n(2-k) = k(n+\ell)## and I don't see why you may assume ##k \in \{1,2\}##.
    However, I don't think ##\ell## is needed at all. What do you get by simply assuming ##xk=2n##?
     
  4. Nov 4, 2016 #3
    Whoops! Rearranging the equation should actually give ##n(2-k) = k \ell##. And I am not not assuming that ##k## always takes the values ##1## and ##2##: I was just considering special cases.

    To answer your question, I don't know what ##xk=2n## implies.
     
  5. Nov 4, 2016 #4

    fresh_42

    Staff: Mentor

    I meant, if you substituted ##x## by ##\frac{2n}{k}## in your assumption ##n<x<2n##, you could divide the whole thing by ##n## and see where it get's you to.
     
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