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Prove Square Root of 15 is Irrational

  1. Aug 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove Square Root of 15 is Irrational


    3. The attempt at a solution

    Here's what I have. I believe it's valid, but I want confirmation.

    As usual, for contradiction, assume 15.5=p/q, where p,q are coprime integers and q is non-zero.

    Thus, 15q2 = 5*3*q2 = p2

    Since 5 and 3 are prime, they must divide p. However, since the lcm(5,3) = 15, it must be the case that 15 divides p. Thus, p=15k for some k.

    Then 15q2 = 15*15*k2, so q2=15k2. By the same argument, this implies 15 divides q.

    However, we have reached a contradiction, since we assumed that p and q were coprime.

    Is this a valid argument?
     
  2. jcsd
  3. Aug 23, 2012 #2
    If I understand the arguement correctly, it's saying...the square root of any prime number is necessarily irrational by definition of prime. Since sqrt(x*y)=sqrt(x)*sqrt(y), and since 5 and 3 are both prime, the square root of their product must be irrational. Makes sense to me.
     
  4. Aug 23, 2012 #3
    Hmm, well one of my main questions is if it's valid to say that 15 divides p and q, with the lcm argument. Does that park make sense?
     
  5. Aug 23, 2012 #4
    I think what you're saying is analogous to the arguement I suggested. In fancy mathematical jargon, there's an isomorphism between our arguements.
     
  6. Aug 23, 2012 #5

    micromass

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    Yes, this is valid.

    So you're saying that the product of irrational numbers is irrational?? Then what about [itex]\sqrt{2}\sqrt{2}[/itex]?
     
  7. Aug 23, 2012 #6
    Hahaha, thanks for your help. The only reason I was confirming is that my professor essentially said what you said, and we're supposed to give another type of argument..

    I just wanted to make sure that if 3 divides p and 5 divides p, 15 divides p. Lol, making sure I'm not making things up!
     
  8. Aug 23, 2012 #7

    Bacle2

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    Yes. Try to do it separately for 3 and for 5, to show that 3|p and 5|p . Then

    p=3t , p=5s , so 5s=3t . But, as you said , since gcd(3,5)=1, we must have 3|s, so

    s=3q , so p=5(3q).

    A more general result is that the square root of a rational number x is rational iff x is a perfect square.
     
  9. Aug 23, 2012 #8
    Thanks again!
     
  10. Aug 23, 2012 #9

    Lol, good point! :D
     
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