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Math magic trick

  1. Mar 26, 2003 #1

    ahrkron

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    This is related to the Banach-Tarski paradox. One of the sites mentioned in that thread has a beautiful "experiment" on equidcomposability: Let A be a unit circle, and B be a unit circle with one point missing. It turns out that you can divide B into two subsets, rotate one of them, and then put the pieces back in such a way that you obtain a full circle.

    I just couldn't resist posting this.
     
  2. jcsd
  3. Mar 27, 2003 #2
    Since it's close to 3am for me, I may be suffering from a case of LOSM (Lack Of Sleep Moronism), but...
    Isn't there 2*pi radians in a circle? And if my brain is still working correctly, that would mean there are only 6 points that are an integer number of radians clockwise around a circle.
    Unless of course you keep making revolutions around the circle, but isn't that cheating?
    And if you are going to cheat this way, why bother with set V since it would then be empty if you include everything in set U?
    Also, if you keep going around the circle, eventually you'll get to the point that was missing, therefore there will be one integer that doesn't get included in set U and due to the irrationality of Pi preventing two such points from coinciding, won't you still be missing a point after rotating the set?

    Sorry, but this sounds like the 'ol 2=1 "proof". Please explain why better if this is true.
     
  4. Mar 27, 2003 #3

    Hurkyl

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    U doesn't include everything. There are only a countable number of points in U (one for every integer) and there are an uncountable number of points on the circle (one for every real number in [0, 2pi))


    It's a slick proof... an easier way to see what's happening is to remove the geometric part and just look at the set of numbers {1, 2, 3, ...} if you subtract 1 from everything in it you get {0, 1, 2, 3, ...}, so yo don't "lose" any points, but you gain a point. The geometric argument carefully puts the missing point exactly where the gained point would be after subtracting 1.

    Hurkyl
     
  5. Mar 27, 2003 #4
    I think you just restated my point.
    I presume we can agree that set U must be either finite or infinate. My thoughts go, if set U is finite, you can't make a circle with an infinite number of points, and if set U is infinite it must include everything. Maybe the term "countably infinate" is confusing me; my understanding is that the terms "infinite" means "infinite" and "countably" means integers only. So I think that an infinite set is still infinite even though it doesn't include all real numbers. Then again the concept of infinity is certainly not "real" in a numerical sense.

    Yes, it's a bit easier to comprehend using numbers. I'm starting to understand the rational behind it, but I think my main confusion is in the definition of set U (and set V). Let me define the missing point in set B to be located at the point 2 radians clockwise from x. Thus set U includes {1, 3, 4, ...} since the almost-circle set B doesn't include that point. Then when you rotate set U, or subtract 1 if you prefer, you end up with {0, 2, 3, ...}.

    I think the "trick" is that after rotating set U, the missing point is now situated at a place where set V contains a point, thus completing the circle. Is that correct? But I don't understand how set V can contain anything.

    Maybe answering the following question would help:
    What is the highest integer in set U, or how many times are you allowed to go around the "circle" counting integer radians?

    This "paradox" is very counter-intuitive, hence my miscomprehension. Keep in mind that the highest math I'm familiar with is basic analytical geometery/calculus up to differential equations (including partials) and basic matrix theory. I've never heard of the term "equidecomposability" until now, or if I have, I promptly forgot it.

    Since both ahrkron and hurkyl are implying it is true, I want to accept it, but I'd like to understand it before I accept it. Do you have a link to a good proof of this?
     
  6. Mar 27, 2003 #5

    Hurkyl

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    U is infinite; it uses all positive integers.

    However, let me give a couple proofs that the set U cannot be everything.


    Proof 1: the set of points in U has zero length

    Choose any number s between 0 and the circumference of the circle.

    We can interpret the set U as being a list of points in the circle. Now, I will choose a system of circular arcs that cover the entire set U. Let the n-th arc be any circular arc containing the n-th point in the list and having length s/2^n

    Now, the set of arcs I chose covers the entire set U (because for every n, the n-th point is in the n-th arc). However, the total length of the arcs is s/2 + s/4 + s/8 + s/16 + ... = s. Notice that this proof does not depend on the value of s, we can choose any positive value for s... however since the arcs cover U, the overall length of U must be less than the total length of the arcs... so for every choice of s the length of U must be less than s!!! The only possible choice for the length of U is zero, so the set U has zero measure... but the circle's circumference is not zero, so the set U cannot possibly be the entire circle, or even a significant portion of it!


    Proof 2: some specific points not in U.

    Label the points of the circle by their angle from the x-axis. Each point, of course, can have many labels differing by some multiple of 2 pi.

    The definition of U says that all points with angle measure equivalent to a positive integer are in U. Consider the point P with angle measure pi / n where n is a positive integer. Suppose P is in U. Since both pi / n and some positive integer m are labels for P, we have:

    pi / n - m = 2 pi k for some integer k.

    Rewriting this equation gives:

    pi (1 / n - 2 k) = m
    pi = m / (1 / n - 2 k)

    However, the right hand side is a rational number while the left hand side is irrational! Therefore, any point with label pi / n for some positive integer n cannot be in U. This same argument can be extended to prove a lot more points are not in U.

    Proof 3: The set of points in the circle is larger than the set of points in U


    The points on the circle can be put into 1-1 correspondence with an uncountably infinite set, for example the real numbers in the interval [0, 2 pi). The set of points in U can be put into 1-1 correspondance with a countably infinite set, the positive integers. Because uncountable sets are larger than countable sets, U cannot contain the whole circle.


    Do any of those proofs make sense?

    Hurkyl
     
  7. Mar 27, 2003 #6
    So was J-man's statement here correct, the way you described things in your last post Hurkyl makes it sound like it is.
     
  8. Mar 27, 2003 #7
    That's really cute... and it doesn't even use Choice! I wonder how far you can go like that? Similar 'paradoxes' that don't use AC, I mean.
     
  9. Mar 28, 2003 #8
    the 'trick' is about geometric points having by definition no dimension.
    so..if you you take a point with no dimension off B its not a circle with a hole in it.
    in order to get a hole into the circle, the hole's dimension must be defined (not measured in points but valid measure).

    if a beam crosses a circle, you don't get a hole in the circle, not even a missing point by taking the beam away. the beam and the circle share their crossing points (which anyway have no dimension), so if you take the beam away, there's still the circles point.

    no trick.
    just a question of definition.

    ..neat one, though.
     
    Last edited by a moderator: Mar 28, 2003
  10. Mar 28, 2003 #9

    Hurkyl

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    Points have "dimension zero", not "no dimension"... but dimension is a red herring, we can talk about whether the circle is missing a point without even mentioning the idea of dimension.

    Hurkyl
     
  11. Mar 29, 2003 #10
    Can this problem be treated by a Cauchy integral?
     
  12. Mar 30, 2003 #11
    I think it is not of utmost importance that this circle be defined presicely.
    Lets make the missing point lie at (1,0)

    In vector form circle C(t)

    C(t) = ( Cos(t), Sin(t) ) 0 < t < 2&pi;

    Note 0 < t < 2&pi;

    NOT

    0 <= t <= 2&pi;
     
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