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Prove, if p and q are distinct prime numbers

  1. Jan 27, 2010 #1
    Prove, if p and q are distinct prime numbers, then sqrt(p/q) is irrational. I know how to prove that if p is a distinct prime number, then sqrt(p) is irrational. From there let sqrt(p) = q/r and then prove but for this I'm stuck. Do we let sqrt(p/q) = (a/b)/(c/d).Thanks.
     
    Last edited by a moderator: Jan 7, 2014
  2. jcsd
  3. Jan 27, 2010 #2
    Re: proof

    Assume not and that sqrt(p/q) is equal to a/b, for integers a, b and in lowest terms..
    Then you can say a2q = b2p. This implies that p divides a2q, but q is prime, so p divides a2. Since p is prime and divides a2=(a)(a), we know it divides a. So write a as a multiple of p and write out your equation and rearrange to isolate b2 (or b). Can you find a contradiction here?
     
  4. Jan 27, 2010 #3
    Re: proof

    So from a2q = b2p which implies that p divides a2. I understand up to there but when you say write a as a multiple of p I get confused. Could you elaborate a little bit more on that? Thanks.
     
  5. Jan 27, 2010 #4
    Re: proof

    If p divides a2 then p divides a (why? There is the famous fact that if p is prime and p divides ab then p divides a or p divides b).
    So since p divides a, we can write out a=pc for some integer c.
    Now, back to the original equation a2q = b2p
    a=pc implies that p2c2q = b2p
    Now, isolate b2 in this equation. Can you say anything about p dividing b here?
     
  6. Jan 27, 2010 #5
    Re: proof

    Okay now I understand what you meant by multiple but where are you getting that famous fact from? Okay so now if we isolate b2 in this equation we will have
    b2 = pc2q and by taking the square root of both sides we'll have
    b = sqrt(pc2q) = csqrt(pq).

    Why would p divide b?
     
  7. Jan 27, 2010 #6
    Re: proof

    You have correctly isolated b2 but you do not need to reduce it.
    b2 = pc2q means that b2 is a multiple of p, which means that p divides b2.
    This implies that p divides b.
    Now p divides a AND p divides b, contradicting that a/b was in lowest terms.

    The famous fact (that prime p divides ab implies p divides a or p divides b) will be in the first few chapters of any number theory book. Here is a link: http://en.wikipedia.org/wiki/Euclid's_lemma. This fact is very important in number theory and even in group theory & ring theory so I would advise that you understand it.
     
  8. Jan 27, 2010 #7
    Re: proof

    The reason your saying b2 is a multiple of p is because c/d = k and if we isolate c we get c = dk. So in this c is b2, d is p and k is c2q right?

    also how do go from p dividing b2 to p dividing b? Thanks.
     
  9. Jan 27, 2010 #8
    Re: proof

    You yourself found the equation b2 = pc2q, which is the same thing as b2 = p(c2q), which means that b2 is equal to p times something. It doesn't get much clearer than that.

    b2 = bb and using the famous fact we see that p divides b.
     
  10. Jan 27, 2010 #9
    Re: proof

    Okay for the first part kind of what I was saying that c = dk. It's like saying 12 = 6*k where k = 2 right? I know it's pretty clear but I just want to make sure I fully digest this stuff.

    Okay now for p divides b2 because according to Euclid's lemma it must divide one of the factors right? Thanks. I really appreciate your help.
     
  11. Jan 27, 2010 #10
    Re: proof

    Yup.

    Yes. p is prime and divides bb so by the lemma it divides b.
     
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