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Smallest infinity for Euclidean geometry to work

  1. Dec 20, 2014 #1
    If we choose rational numbers to represent points on a line then there will be gaps on the line and consequently the plane will be full of holes. Then we cannot say that two non-parallel line must intersect on a point (because they may meet at the gaps). So obviously we need point arranged more densely than rational number. I was wondering do we necessarily need real number line for euclidean geometry or we can stop at earlier infinity. After writing this far I realize it points to continuum hypothesis. But I don't know enough and will highly appreciate your input and comment.
    I am not a student of mathematics but dabble with its concept every now and then.
     
  2. jcsd
  3. Dec 20, 2014 #2

    Mark44

    Staff: Mentor

    What do you mean by "earlier infinity?"
     
  4. Dec 20, 2014 #3
    Now I understand i was looking for infinity that is bigger than infinity associated with rational number but smaller than that of real number. I know that it is undecidable.
    But i was originally wondering what kind of infinity is necessary for continuity in the sense of euclidean geometry.
     
  5. Dec 20, 2014 #4

    jbriggs444

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    A countable set should suffice. Model the "points" in plane geometry as ordered pairs of algebraic numbers. Then two lines defined by two pairs of "points" will always have a "point" at which they intersect.
     
  6. Dec 20, 2014 #5
    Seems to me you do need to the full set. If all you care about are lines, then you'll only need the rational numbers since if you restrict the coefficients of a line, [itex] y = ax+b[/itex] to rational numbers, you'll always get a rational solution for the intersection of two lines.

    However, what happens when I want to know the length of the diagonal of a square of side length 1? The answer is [itex] \sqrt{2}[/itex]. What about the circumference of a circle of radius 1? The answer is [itex]2\pi[/itex].

    You quickly move from rational numbers to irrational as you start looking at polygons, and then quickly to transcendental when you start looking at curves.
     
  7. Dec 20, 2014 #6

    jbriggs444

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    Even if you start looking at curve lengths, that won't take you past the computable numbers. Those are still countable.
     
  8. Dec 20, 2014 #7
    Thank you for your response. I understand your point about algebraic number. If points of the plane is defined by pair of algebraic number, then the intersection of any two lines will also be a point (I get it). But i don't get your last response about computable number. I'll really appreciate if you elaborate the point.
     
  9. Dec 21, 2014 #8

    jbriggs444

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    Computable numbers, http://en.wikipedia.org/wiki/Computable_number, are numbers that can be calculated to any desired precision by a fixed and finite program. Since there are at most countably many programs, there are at most countably many computable numbers.

    The value of a trigonometric function or inverse trigonometric function evaluated at a computable number is, if defined at all, computable. The length of a circular arc with a computable radius between two computable endpoints is, therefore, computable.
     
  10. Dec 22, 2014 #9
    What are some examples of non-computable numbers?
     
  11. Dec 22, 2014 #10

    jbriggs444

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    It is impossible to explicitly display a non-computable number. It is impossible to even present an algorithm that could produce one. After all, if such an algorithm existed, the number would have to be computable.

    One can nonetheless write down a definition for a non-computable number. For instance, http://en.wikipedia.org/wiki/Chaitin's_constant
     
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