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Proof must be integer or irrational?

  1. Sep 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose a, b ε Z. Prove that any solution to the equation x^3 +ax+b = 0 must either be an integer, or else be irrational.


    2. Relevant equations

    Not sure if this is right but x = m / n where m divides b and n divides 1

    3. The attempt at a solution

    So far i think i got x = m / n where m divides b and n divides 1 but i don't know where to go from there and i am kind of not sure if that's even right. There's probably some way to factor this right? I am trying to do this by contradiction so i assume that x = a/b where a,b belong to N


    One More problem...

    1. The problem statement, all variables and given/known data

    Prove that log2n : n ε N consists entirely of integers and irrational numbers. (it's base 2 n not log(2n))

    2. Relevant equations

    log2n = a / b. Trying to do a proof by contradiction so i started off by assuming log2n = a / b where a,b E N.

    3. The attempt at a solution

    As i said I'm doing a proof by contradiction. So far i have come up with log2n = a / b. I don't really know where to go from here. Rearranging the equation doesn't really help. Maybe use some laws for logarithms to write them differently? :/
     
  2. jcsd
  3. Sep 23, 2012 #2

    HallsofIvy

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    IF you allowed to use this, then surely you see that the only integers that divide 1 are 1 and -1! So if x is rational it is either m/1 or m/-1 for some integer m.

     
  4. Sep 23, 2012 #3
    Oh... wow okay so x could = m or -m both of which are integers. That makes sense. I don't understand how to prove that any other solution would be irrational :/
     
  5. Sep 24, 2012 #4
    what you used is called the rational root theorem, giving all possible values of rational roots, so any other roots must be .....
     
  6. Sep 24, 2012 #5

    HallsofIvy

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    That proves that any rational root must be an integer. Of course, any other roots, that is, any root that is not rational, is irrational, by definition.

    But I still wonder if you are allowed to use that. It seems too easy.
     
  7. Sep 24, 2012 #6
    My teacher told us to use that because we haven't really learned a lot yet. Thanks for your help! :)
     
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