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Math Help - [SOLVED] Help with powers and roots

  1. #1
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    [SOLVED] Help with powers and roots

    There's something in my notes that says p^2 is divisible by 3 so p is divisible by 9.
    I don't understand this...

    If p^2 / 3
    shouldn't
    p be divisible by 3^(1/2)?
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  2. #2
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    Quote Originally Posted by Noxide View Post
    There's something in my notes that says p^2 is divisible by 3 so p is divisible by 9. Mr F says: This statement is false. A simple counterexample is p = 6. Speak to your teacher.

    I don't understand this...

    If p^2 / 3
    shouldn't
    p be divisible by 3^(1/2)?
    A correct statement is if p^2 is divisible by 3 then p is divisible by 3.
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  3. #3
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    Is that still true in the context of q^2 = (p^2)/3?
    So could I say that p is divisible by 3 and also that q is divisible by 3?


    I'm asking this because the weird phrasing of P^2 / 3 so p is divisible by 3 came up in my notes and when i looked it up on the internet it also came up here:

    http://mathforum.org/library/drmath/view/52619.html
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  4. #4
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    Quote Originally Posted by Noxide View Post
    Is that still true in the context of q^2 = (p^2)/3?
    So could I say that p is divisible by 3 and also that q is divisible by 3?


    I'm asking this because the weird phrasing of P^2 / 3 so p is divisible by 3 came up in my notes and when i looked it up on the internet it also came up here:

    Math Forum - Ask Dr. Math
    An exact quote of that link is
    [snip]
    p and q are integers
    [snip]

    3 q^2 = p^2

    So p^2 is divisible by 3. That means that p must be as well, so p^2 is
    divisible by nine.
    which is very different to what you've said in this thread. I already told you that
    if p^2 is divisible by 3 then p is divisible by 3.
    .

    The proof given in the link is very clear. I don't know exactly what your question or problem is.
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  5. #5
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    Haha, now it's a problem.
    The problem is that I CANT READ!

    Thanks for you help!
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