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How is this a contradiction?

  1. Feb 21, 2008 #1
    I read this on cut-the-knot.org:

    How is this a contradiction? If [itex]p^2 = 2q^2[/itex] just says that [itex]p^2[/itex] is even, it doesn't say that 'p' is even. It is a very obvious mistake. So.. is this a stupidity on their part or something that i may have missed [or a stupidity] on my part..

    I generally wouldn't have posted this here.. but since cut-the-not.org has quality articles posted.. i would like to rethink this problem before concluding that they made a mistake.

  2. jcsd
  3. Feb 21, 2008 #2
    Is there an assumption that p and q are integers? I think that might make it work… but I agree with your objection.
    Last edited: Feb 21, 2008
  4. Feb 21, 2008 #3
    The problem assumes that p and q are integers, since it assumes that p/q is a rational number.
  5. Feb 21, 2008 #4
    Not all rational numbers are integers.


    is a rational number but not an integer.
  6. Feb 21, 2008 #5

    Gib Z

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    Slider142 never said p/q was an integer if p and q were integers, but did say p/q was rational if p and q were integers. 1 and 2 are nice integers =]

    We have that p is an integer, and that p^2 is divisible by 2. In fact, p^2, if divisible by 2, must also be divisble by 4. This is because if we express p^2 as a product of its factors, [tex]p^2 = 2^2 * k^2 * p^2 ...[/tex] where k and p are other integers/factors. All the powers must be even, in particular for 2. This is because if the power were odd, then in the factorization for p, there would be a non-integer power - ie not allowed.
  7. Feb 21, 2008 #6
    Oh, you're right. Duh. :blushing: Sorry, slider.
  8. Feb 21, 2008 #7


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    If p is odd, then p= 2n+1 for some integer n. Then p2= (2n+1)2= 4n2+ 4n+ 1= 2(2n2+ 2n)+ 1, "2 times an integer plus 1", and so odd. That is: If p is an odd integer, then so is p2". It follows that if p2 is even then p must be even.

    I would suggest that you not throw around words like "obviously wrong" and "stupidity" so easily. They tend to come back at you.
  9. Feb 21, 2008 #8
    They were talking oranges but you are talking bananas. The contradiction is that p and q were originally considered coprime!! Since if the square root of 2 = p/q where p and q are integers, we can always reduce p and q to coprime integers by dividing by the common factor.
    Last edited: Feb 21, 2008
  10. Mar 18, 2008 #9
    it's impossible two number be equal and one be even and the other be odd, and is also impossible a square of an odd/even number turn to be even/odd
  11. Mar 19, 2008 #10
    But 2*p is even, even if p is odd, so I dont know what you are trying to say.
  12. Mar 20, 2008 #11
    I am saying that if x = 2y then x is even, very simple.
  13. Mar 21, 2008 #12


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    Yes 2p is always even but that has nothing to do with what he said. He, and your original post, was talking about squares of numbers. As I showed above, the square of any even number is even, the square of any odd number is odd.

    If n2 is even, then n itself must be even. If it were odd, then n2 would have to be odd, not even.
  14. Mar 21, 2008 #13
    That is fine but Can you follow HallsofIvy's Logic and show that If x is square then y can not be odd? Thus the contradictiopn since y was assumed to be odd?
    Last edited: Mar 21, 2008
  15. Mar 21, 2008 #14
    Hi Ramsey, I am not sure what you want to know about my post...

    given x = 2y, if y is odd then x cannot be a square since x is even.

    proof: if x is a square it can be written as 4a^2, for an any a, and 4a^2 = 2y ==> 2a^2 = y ==> y is even, and this is a contradiction because we assumed that y is odd

    I am not sure if this is what you want, so I'll reproduce a proof of irrationality of 2

    supose [tex]\sqrt{2} = p/q[/tex] in the lowest terms ==> gcd(p,q) = 1

    [tex]2 = \frac{p^2}{q^2}\ ==> \ 2q^2 = p^2\ ==>[/tex] p is even (this is the conclusion about the elementary principle: two numbers cannot be equal and one be even and the other be odd, and I am not talking about q, I am talking about 2q^2, this number is even)

    as p is even, p = 2k ==> p^2 = 4k^2 ==> [tex]q^2 = 2k^2[/tex] which means that q is even also, so p/q is not in the lowest terms, and this is the contradiction, and then [tex]\sqrt{2}[/tex] cannot be expressed as a rational ==> [tex]\sqrt{2}[/tex] is irrational
    Last edited: Mar 21, 2008
  16. Mar 21, 2008 #15
    The key was the assumption that gcd(p,q) = 1 which is what Slider142 and I posted earlier.
  17. Mar 22, 2008 #16
    oh yes, ramsey, this is the key.

    sometimes I just read the first post and put some ideas, I didin't see what you guys have posted before

    I hope who have opened the thread do understand the arguments
  18. Apr 10, 2008 #17
    [itex]p^2 = 2q^2[/itex]
    so [itex]p^2[/itex] is even
    so [itex]p[/itex] is even
    so [itex]p[/itex]=0 mod 2
    so [itex]p^2[/itex]=0 mod 4
    so [itex]p^2=2q^2[/itex]=0 mod 4
    so [itex]q^2[/itex]=0 mod 2
    so [itex]q[/itex]=0 mod 2
    so q is even
    so 2|gcd(p,q)
    so gcd(p,q)>1
    Last edited: Apr 10, 2008
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