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Prove that x is irrational unless it is an integer.

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data

    This is taken from an answer book that I have. I don't understand the bolded step. Can someone explain it to me?

    Suppose x = p/q where p and q are natural numbers with no common factor. Then:

    pn/qn + an-1pn-1/qn-1 + ... + ao = 0

    and multiplying both sides by qn gives

    pn + an-1pn-1q + ... + aoqn = 0

    Now if q ≠ ±1 then q has some prime number as a factor. This prime number divides every term of the second equation other than pn, so it must divide pn also. Therefore it divides p, a contradiction. So q = ±1, which means that x is an integer.

    Once again, it's the bolded step that I don't understand. Why must it divide pn?
    Thanks in advance.
  2. jcsd
  3. Feb 12, 2012 #2


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    Looks like you're asking us to pick things up in the middle of a proof, without letting us know what it is that is being proved.

    I'm guessing that you're looking at a solution for the following.
    Prove that any root of the following polynomial of degree, n, with integer coefficients:
    xn + an-1 xn-1 + an-2 xn-2 + an-3 xn-3 + an-4 xn-4 + … + a0
    is either an integer, or the root is irrational.​
    The proof is by contradiction, and done by assuming that there is a rational, non-integer root.

    To answer your question:

    Rewrite that second equation of yours as:

    an-1pn-1q + ... + aoqn = -pn

    So, q divides the left hand side. Therefore, it must divide the right hand side.
  4. Feb 12, 2012 #3
    Thank you so much! I can't believe I missed that! Thanks!
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