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Prove √ is not a rational number

  1. Sep 13, 2012 #1
    1. The problem statement, all variables and given/known data

    "Prove m√n is not a rational number for any natural numbers with n,m > 1, where n is not an mth power"

    2. Relevant equations

    Natural numbers for us start at 1.
    Since we know n is not an mth power, then n [itex]\neq[/itex] km for an arbitrary integer k.

    3. The attempt at a solution

    I believe this would follow a similar argument for the classic proof that √2 is irrational.

    We set up the contradiction that m√n is actually rational. Then am=nbm for some a,b [itex]\in[/itex] [itex]Z[/itex], b [itex]\neq[/itex] 0, a,b are irreducible.
    Then am is a multiple of n, so a is a multiple of n.

    I'm assuming I need to prove this as a lemma.
    I've attempted proving this by modular arithmetic.

    Prove by contrapositive (If a is not a multiple of n, then am is not a multiple of n). But then there would be n-1 cases to check (infinitely many), so I follow with induction.
    Base case: n=2, a [itex]\equiv[/itex] 1 (mod 2), then a2 [itex]\equiv[/itex] 12 = 1 [itex]\equiv[/itex] 1 (mod 2).
    Base case checks out, so we can assume the inductive hypothesis that this is true for all n.
    To show show n+1, the (mod n) turns into (mod n+1), so how do I use the inductive hypothesis on this?
    This is the point where I get lost. Any suggestions?
     
  2. jcsd
  3. Sep 13, 2012 #2

    jgens

    User Avatar
    Gold Member

    You can do this proof much more easily just by counting prime factors.
     
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