1. Standard memberTheMaster37
    Kupikupopo!
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    22 Jun '08 10:05
    Originally posted by mtthw
    Probably 🙂

    I suspect proofs using the set theory definitions of cardinal numbers might be more interesting, but I don't remember much set theory!
    Using the Axioms of Zermelo and Fraenkel;

    Using S(A) to denote the sucessor of A and BU to denote the Big Union
    For all ordinals A and B and all limit-ordinals L addition is defined with the following;

    A + 0 = A
    A + S(B) = S(A+B)
    A + L = BU{ A+B | B in L}

    Using that on 1 + 2;

    1 + 2 = 1 + S(1) = S( 1 + 1 ) = S( 1 + S(0) ) = S( S( 1 + 0 ) ) = S( S( 1 ) ) = S( 2 ) = 3
  2. Joined
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    23 Jun '08 06:14
    Originally posted by joe shmo
    Is that the problem Which leads to 1 when performing the operations, If it is, then I ve read that they are offering a reward for it proof!
    i was under the impression that there was a reward for it since a while back
  3. R
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    23 Jun '08 14:00
    Originally posted by banx99
    i was under the impression that there was a reward for it since a while back
    Yes, I guess since 1937 here's a link for more info

    http://mathworld.wolfram.com/CollatzProblem.html
  4. Standard memberAThousandYoung
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    23 Jun '08 14:14
    Originally posted by Jirakon
    What about the points that define a circle with a radius of infinity?

    That's the first thing that came to mind. It would have to be infinite, since if it were a finite number of points, one could easily draw a horizontal line far above the last point. I'm pretty sure that's the solution.
    Infinities don't exist 🙁
  5. Standard memberAThousandYoung
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    23 Jun '08 14:16
    Originally posted by wolfgang59
    Just because all points have finite coordinates does not mean there is a 'y-max'

    Otherwise your "proof" could be used to prove that the set of points on x=0 is finite.
    I disagree. His proof only proves that the line x=1 does not intersect two of the points in that set.
  6. Standard memberAThousandYoung
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    23 Jun '08 14:17
    Originally posted by Palynka
    I don't see why what I said implies the set must be bounded. Take the Euclidean axis. No line that you can draw or represent as a function intersects a point of those axes only at infinity.
    But there are lines that only intersect one of the axes and not the other.
  7. Standard memberAThousandYoung
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    23 Jun '08 14:18
    Originally posted by FabianFnas
    Prove, that 1+2=3

    Is this provable? Yes it is.
    Most of math students, from lower grade of university, cannot. Can you?
    1+2 is defined as 3 is it not?
  8. Joined
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    23 Jun '08 15:07
    Line may or maynot extend to infinity. It may be a small line. So circle with Infinite radius or similar figures dont hold good IMHO.
  9. Joined
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    23 Jun '08 15:35
    Originally posted by AThousandYoung
    1+2 is defined as 3 is it not?
    No, not more then 2+3=5, because you cannot define every combinations of a, b, and c so a+b=c.

    Better to use the axioms and go from there.

    However, addition is a defined operation, is it not?
  10. Southern California
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    24 Jun '08 05:25
    Haha.
    Wanna play chess?!
  11. Standard memberAThousandYoung
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    25 Jun '08 16:55
    Originally posted by FabianFnas
    No, not more then 2+3=5, because you cannot define every combinations of a, b, and c so a+b=c.

    Better to use the axioms and go from there.

    However, addition is a defined operation, is it not?
    I would think yes more than 2+3=5, but not more than 1+3=4. 1+n gives you the number in the sequence that comes after n.
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