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3|n^2 implies 3|n

  1. Mar 25, 2012 #1
    I'm completely lost on this one. I need this to be able to solve that the square root of three is irrational. So it's a proof within a proof, but I like this way best. Please help me out.
    This is what I need to know how to prove.

    3|n^2 implies 3|n where n is some integer.
  2. jcsd
  3. Mar 25, 2012 #2


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    3 is a prime number. If 3 were not a prime factor of n, it could not be a factor of n2.
  4. Mar 26, 2012 #3
    I think I understand what you're saying and that looks like a contrapositive proof, but you don't actually prove it. Could you elaborate?
  5. Mar 26, 2012 #4
    Suppose :

    [tex]n \equiv a \pmod 3 ~\text{and}~ a \neq 0[/tex]
    then :
    [tex]n^2 \equiv a^2 \pmod 3 ~\text{and}~ a^2 \neq 0[/tex]
    hence :
    [tex] 3 \nmid n^2[/tex]
    contradiction .

  6. Mar 26, 2012 #5
    I think the Fundamental Theorem of Arithmetic (uniqueness up to order of prime factorizations) might help?
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