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Proving that \sqrt{p} is irrational

  1. Nov 27, 2013 #1
    I'm aware of the standard proof.

    What I'm wondering is why we can't just do the following. Given, I haven't slept well and I'm currently out of caffeine, so this one might be trivial for you guys.

    Suppose, by way of contradiction, that ##\sqrt{p}=\frac{m}{n}##, for ##m,n\in\mathbb{Z}## coprime. Then, ##p=\frac{m^2}{n^2}##. Because ##p## is an integer, ##\frac{m^2}{n^2}## must be as well. However, because ##m## and ##n## are coprime, so are ##m^2## and ##n^2##. Thus, ##n^2=1## is necessary for ##\frac{m^2}{n^2}## to be an integer. But that means ##m^2=p##, and ##p## is prime. Thus, a contradiction is met and we see that ##\sqrt{p}## is irrational.

    Is this valid? I'll probably figure out my mistake (if I made one) by the time I get back with adequate caffeination, but until then, I'd like to make sure I figure it out.

    Thank you.

    Edit: Gee, I guess it might be important to mention that ##p## is prime. :facepalm:
     
    Last edited: Nov 27, 2013
  2. jcsd
  3. Nov 27, 2013 #2
    In my opinion, it's valid.
     
  4. Nov 27, 2013 #3

    jbunniii

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    It looks OK to me.
     
  5. Nov 27, 2013 #4

    Curious3141

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    Completely valid. The last step basically assumes that a prime cannot be a perfect square, which is true and fairly obvious. But if you really want to make the proof completely obvious, you can state: ##m^2 = p##. Hence ##m|p## (m divides p), which is a contradiction.
     
  6. Nov 28, 2013 #5

    PeroK

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    Actually, it doesn't matter that p is prime. It's valid for any +ve integer: the square root is either an integer or irrational; it cannot be a proper rational.
     
  7. Nov 28, 2013 #6
    Thank you, all.
     
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