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Confirmation of irrational proof

  1. Aug 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove ##\sqrt n## is irrational

    2. Relevant equations



    3. The attempt at a solution



    Assume ## p^2/q^2 = n ## is an irreducible fraction.
    If ##p^2 = nq^2##, then q is a multiple of n. Call this ##p' = nq##
    substituting this for our original equation. We get ##p'^2 = nq^2##
    Implies ##(nq)^2 = nq^2## thus our original assumption is false.

    Is this horse crap? Don't just give it away if it is wrong please just indicate where I went wrong.
     
  2. jcsd
  3. Aug 10, 2013 #2

    micromass

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    Where does this proof fail if ##n=4##?
     
  4. Aug 10, 2013 #3

    HallsofIvy

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    Does it bother you at all that the statement you claim to have proved is NOT true? What if, say, n= 4? What happens to your proof in that case?

    It is, of course, true if you add the requirement that n is NOT a "perfect square". How have you used that requirement in your proof?
     
  5. Aug 10, 2013 #4
    Would if fail with this assumption right off the bat?

    Assume ##p/q = \sqrt4 ## is an irreducible fraction.
    Because ## p/q = 2 ##
     
  6. Aug 10, 2013 #5

    micromass

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    OK, so your proof does fail for some ##n##. Your proof should make this clear.
     
  7. Aug 10, 2013 #6
    Yeah, I should of just included the condition because it was given in the book just excluded it in my post.
    Thanks for the help
     
  8. Aug 11, 2013 #7

    Ray Vickson

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    Your argument is incomplete. Essentially, you claim that if p^2 is divisible by n then p itself is also divisible by n. You cannot just say it---you need to prove it.

    BTW: it is not always true for all integers n, so something about n needs to be specified.
     
  9. Aug 12, 2013 #8
    Yeah, I fooled myself. Let's see if I did it again. What about:

    ##p/q = \sqrt n ##, Assume ## (p,q) = 1 ## and ##n ## is not a perfect square.

    Squaring, we have, ## p^2 = nq^2 ##


    If ## n=/= 1## There exists a prime integer ## k## such that ## k/n##

    Substitution gives:

    ## p^2 = (n/k)q^2. ## Implies that ## kp^2 = nq^2 ## and since we know that that k divides n, ##p^2 ## and ## q^2 ## share a common factor.Thus, our assumption ## (p,q) = 1 ## is false so ## \sqrt n ## is irrational.


    What do you think Ray trick myself again?
     
  10. Aug 12, 2013 #9
    This is crap. I need to think
     
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