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Posers and Puzzles

Posers and Puzzles

  1. 12 Jan '10 11:57
    Referring to Thread 124309 : "rational sphere" who put the simple question if a sphere made up of rational points is transparant or not...

    I say the answer is "Yes, it is totally transparant!" The following problem hints why.

    The new problem is:

    Suppose you have a sphere of radius one, a unit sphere, where every point of the sphere is the set of (x;y;z) of R^3 and every point has the distance d=1 to the center of the sphere. (That's the definition of a sphere, isn't it?)

    Question: How many of these points (x;y;z) have all of x, y, and z as rationals?

    I know six trivial points (0;0;1), (0;1;0), (1;0;0) and their negative counterparts.
    But are there more? Does it have infinitly more, or a some other number of points?

    Can you rigourously prove your answer?
  2. Standard member wolfgang59
    Infidel
    12 Jan '10 12:06
    Originally posted by FabianFnas
    Referring to Thread 124309 : "rational sphere" who put the simple question if a sphere made up of rational points is transparant or not...

    I say the answer is "Yes, it is totally transparant!" The following problem hints why.

    The new problem is:

    Suppose you have a sphere of radius one, a unit sphere, where every point of the sphere ...[text shortened]... nfinitly more, or a some other number of points?

    Can you rigourously prove your answer?
    I think rational spere (1) was talking about a solid sphere ... however.

    Your sphere has an infinite number of rational points on the surface.

    Consider the points (x, y, 0) where x and y are in the ratio of the lesser sides of a Pythagorean triangle and satisfy x^2 + y^2 = 1.

    e.g. (3/5, 4/5, 0) , (5/13, 12/13, 0) etc etc
  3. 12 Jan '10 12:17
    Originally posted by wolfgang59
    I think rational spere (1) was talking about a [b]solid sphere ... however.

    Your sphere has an infinite number of rational points on the surface.

    Consider the points (x, y, 0) where x and y are in the ratio of the lesser sides of a Pythagorean triangle and satisfy x^2 + y^2 = 1.

    e.g. (3/5, 4/5, 0) , (5/13, 12/13, 0) etc etc[/b]
    Okay, good answer.

    I thought such a 'shere', where every point within its set has a d =< 1, was called a 'ball', no?

    Alright, then if I give the further change that no coordinate can be equal on the axis, what's the answer then?

    I realize the my six points are off the question, in this new case.
  4. Standard member Palynka
    Upward Spiral
    12 Jan '10 14:55
    Originally posted by FabianFnas
    Okay, good answer.

    I thought such a 'shere', where every point within its set has a d =< 1, was called a 'ball', no?

    Alright, then if I give the further change that no coordinate can be equal on the axis, what's the answer then?

    I realize the my six points are off the question, in this new case.
    You know that we can map the sphere onto a plane, so just take the coordinates (a,b) on the plane as rational numbers and you have an infinite of (x,y,z) which are rational.
  5. 12 Jan '10 15:33
    Originally posted by Palynka
    You know that we can map the sphere onto a plane, so just take the coordinates (a,b) on the plane as rational numbers and you have an infinite of (x,y,z) which are rational.
    Give me one in the coordinates (x;y;z) where all of them are rationals and non zero. One's enough.
  6. Standard member Palynka
    Upward Spiral
    12 Jan '10 16:36
    Originally posted by FabianFnas
    Give me one in the coordinates (x;y;z) where all of them are rationals and non zero. One's enough.
    (2/3,2/3,-1/3)
  7. Standard member Palynka
    Upward Spiral
    12 Jan '10 16:41
    (4/5,1/625000000000,-3/5)
  8. Standard member wolfgang59
    Infidel
    12 Jan '10 18:34
    Originally posted by FabianFnas
    Okay, good answer.

    I thought such a 'shere', where every point within its set has a d =< 1, was called a 'ball', no?

    Alright, then if I give the further change that no coordinate can be equal on the axis, what's the answer then?

    I realize the my six points are off the question, in this new case.
    I think sphere and ball are interchangeable.

    But lets consider your hollow sphere.

    You want a coordinate on surface which is not on x=0 or y=0 or z=0.

    (3/13, 4/13, 12/13)

    I believe there are an infinite number of triples such that a^2 + b^2 + c^2 = d^2
  9. 12 Jan '10 18:39
    Originally posted by Palynka
    (2/3,2/3,-1/3)
    (4/5,1/625000000000,-3/5)
    If a point (x;y;z) is on the surface of the sphere then x^2+y^2+z^2=1.
    (2/3,2/3,-1/3) gives (2/3)^2+(2/3)^2+((-1)/3)^2 = 4/9+4/9+1/9 = 1 correct.
    but
    (4/5,1/625000000000,-3/5) gives 15/25+1/390625000000000000000000+9/25 = 0,96000000000000000000000256 not = 1 fail.

    But I asked for one point, one's enough.

    Can we from this deduce that there are infinitely many?
  10. 12 Jan '10 18:45
    Originally posted by wolfgang59
    I think sphere and ball are interchangeable.

    But lets consider your hollow sphere.

    You want a coordinate on surface which is not on x=0 or y=0 or z=0.

    (3/13, 4/13, 12/13)

    I believe there are an infinite number of triples such that a^2 + b^2 + c^2 = d^2
    Not according to my math book, but never mind. My problem was about a hollow sphere. Or else, I think the problem would be much more easy.

    (3/13, 4/13, 12/13) gives a correct result. Thanks.

    Seems that you're correct - there are probably infinite number of triplets where a, b and c belongs to Q, and each not equal to zero, and a^2+b^2+c^2=1. 1 is enough because we're talking about a unit sphere.
    But is there a proof for this?
  11. 12 Jan '10 19:02 / 1 edit
    Originally posted by FabianFnas
    Not according to my math book, but never mind. My problem was about a hollow sphere. Or else, I think the problem would be much more easy.

    (3/13, 4/13, 12/13) gives a correct result. Thanks.

    Seems that you're correct - there are probably infinite number of triplets where a, b and c belongs to Q, and each not equal to zero, and a^2+b^2+c^2=1. 1 is enough because we're talking about a unit sphere.
    But is there a proof for this?
    You may take

    a=b=2t/(t^2+2)
    c=(t^2-2)/(t^2+2)

    where t in any non-zero rational number.

    EDIT: the equation a^2+b^2+c^2=1 actually has solutions in which a, b, c are themselves squares of non-zero rational numbers. Can you find one?
  12. 12 Jan '10 19:13
    Originally posted by David113
    You may take

    a=b=2t/(t^2+2)
    c=(t^2-2)/(t^2+2)

    where t in any non-zero rational number.
    Great work. How did you came up with this formula?
  13. 12 Jan '10 21:00
    Originally posted by FabianFnas
    Great work. How did you came up with this formula?
    also, an extension of the classic pythagorean triple construction can be used to find an arbitrary number of squares that add up to a square. here is the original construction:

    a = p^2 - q^2
    b = 2pq
    c = p^2 + q^2 ---> a^2 + b^2 = c^2

    and the extension looks a little something like this:

    a1 = X1^2 + X2^2 + X3^2 + ... - Xn^2 (only the last term gets a minus sign)
    a2 = 2*X1*Xn
    a3 = 2*X2*Xn
    a4 = 2*X3*Xn
    ...
    an = 2*X[n-1]*Xn
    a(n+1) = X1^2 + X2^2 + ... + Xn^2 (all the terms get a plus sign)

    note, it's very difficult to write subscripts! except for the squares and the *'s, anything that looks like multiplication up there is meant to be a subscript (ex. a(n+1) should be a with a subscript of n+1). start small, but it is pretty easy to verify that with this construction, a1^2 + a2^2 + ... + an^2 = a(n+1)^2.
    basically all of the 2*Xi*Xn terms, when squared and added to a1^2, change their respective [-2(Xi)^2(Xn)^2] from a1^2 into a [+2(Xi)^2(Xn)^2] for a(n+1)^2.

    and so you can have an arbitrary number of squares added together to make a square. and definitely an infinite number of these "square sums" because you can choose any numbers you want for X1 through Xn. and it has an interesting implication in pedagogy that using this you can generate vectors in R^n that have all integer coordinates, and also have integer length. makes grading papers a lot easier haha

    not sure if this is well known, but was one of the proofs i did in my undergrad senior thesis... so it probably is! cheers
  14. 12 Jan '10 21:06
    Originally posted by Aetherael
    also, an extension of the classic pythagorean triple construction can be used to find an arbitrary number of squares that add up to a square. here is the original construction:

    a = p^2 - q^2
    b = 2pq
    c = p^2 + q^2 ---> a^2 + b^2 = c^2

    and the extension looks a little something like this:

    [b]a1
    = X1^2 + X2^2 + X3^2 + ... - Xn^2 (only th ...[text shortened]... t was one of the proofs i did in my undergrad senior thesis... so it probably is! cheers[/b]
    Full point! Well done!

    Thread closed with this, I suppose.
  15. Standard member Palynka
    Upward Spiral
    12 Jan '10 22:02
    Originally posted by FabianFnas
    Not according to my math book, but never mind. My problem was about a hollow sphere. Or else, I think the problem would be much more easy.

    (3/13, 4/13, 12/13) gives a correct result. Thanks.

    Seems that you're correct - there are probably infinite number of triplets where a, b and c belongs to Q, and each not equal to zero, and a^2+b^2+c^2=1. 1 is enough because we're talking about a unit sphere.
    But is there a proof for this?
    Yes, the easiest one comes from stereographic projection, like I said.