Results 1 to 10 of 10

Math Help - This might belong in geometry, but...

  1. #1
    Junior Member
    Joined
    Feb 2007
    Posts
    70

    This might belong in geometry, but...

    The following link "disproves" the Pythagorean theorem:

    Pythagoras Disproved | Futility Closet

    The reason I'm posting it here is because it involves limits, and I have a feeling finding the flaw here is going to involve some analysis. The problem asserts that the red line converges to the hypotenuse, essentially hoping the reader will take this as obvious. I believe that this assertion is false, but I don't know how to actually prove it. Any ideas?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    The error is in the words...

    ... with each iteration, the red line more closely approximates c, but its length remains a + b...

    For the 'red line' is...

    \lim_{n \rightarrow \infty} n\cdot (\frac{a}{n} + \frac{b}{n}) = a+b

    ... whereas for the 'black line' is...

    \lim_{n \rightarrow \infty} n\cdot \sqrt{ \frac{a^{2}}{n^{2}} + \frac{b^{2}}{n^{2}}} = \sqrt{a^{2} + b^{2}} \ne a + b

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2007
    Posts
    70
    I was able to do that much, but since the claim is that this disproves the Pythagorean Theorem, I was hoping to disprove it without using the theorem itself. This step:

    <br />
\lim_{n \rightarrow \infty} n\cdot \sqrt{ \frac{a^{2}}{n^{2}} + \frac{b^{2}}{n^{2}}} = \sqrt{a^{2} + b^{2}} \ne a + b<br />

    assumes the Pythagorean Theorem, which could be seen as circular reasoning in this particular case. Is there different way to prove this wrong?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member alunw's Avatar
    Joined
    May 2009
    Posts
    188
    I don't think there is anything wrong with the claim that the red line approximates the black more closely as n increases. That is true in the perfectly respectable sense that the Hausdorff distance between the red lines and the black line tends to 0. Therefore the flaw lies in assuming that the length of the limiting curve (i.e. the black line) is necessarily the same as the limit of the sequence of lengths. In fact this is a perfectly good counter-example to show that these don't have to be equal.
    By drawing a different triangle on top of the black line, and then performing an analogous sequence of constructions we can find sequences of curves all of the same length with any length we like (so long as it is greater than c) which approximates the black line as closely as we like.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2009
    Posts
    67
    one thing about pythagoras theorem to note is that it is not really a theorem but a law on the euclidean plane. we could consider planes where pythagoras theorem does not hold and no sky would fall on our heads ... having said that the 'proof' you linked is not correct ... the length of two curves will be same in the limit if only all the intervening curves are smooth (or they have at most finitely many smooth peices)... computing lenght requires you to compute the square of the slope of the curve ..

    if you have inifinitely maney pieces then you have to take integral instead of sum. there you run into problems of convergence ...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member alunw's Avatar
    Joined
    May 2009
    Posts
    188
    Pythagoras's theorem is most certainly a theorem of Euclidean geometry, even if is only a hypothesis about the nature of actual space. In principle scientists might be one day be able to measure large real triangles accurately enough to determine that the space we live in is not actually Euclidean, but hyperbolic or spherical in nature, but if it is actually Euclidean they will never be able to prove it!
    All the red lines do consist of a finite number of smooth pieces - but the number of pieces is not bounded.
    The argument that all the red lines have the same length is a Euclidean argument, because it depends on the construction of a rectangle or parallelogram, a shape that only exists in Euclidean geometry, or if you prefer on the existence of similar triangles. So the argument breaks down in other kinds of space, though you might be able to find some other "wrong" limit for the length of red lines constructed in some analogous way.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Aug 2009
    Posts
    67
    it is not a theorem of euclidean geometry but a consequence of the way distance (or angles) is defined, which is to say the way inner product is defined. real world is neither sheprical, nor hyperbolic nor euclidean, these are spaces of constant curvature .. these are not the only kinds of spaces ... real world is more complicated than that .. and it is no speculation .. we know that it is more complicated than the three simple types. that is the core of general relativity and it has been confirmed in numerous experiments .. even something as practical as GPS/navigation depends upon relativity ... so no need of measuring 'large triangles'
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member alunw's Avatar
    Joined
    May 2009
    Posts
    188
    You are talking rubbish. Pythagoras's theorem IS a theorem of Euclidean geometry, because it can be proved based on the axioms of Euclidean geometry, though I suppose you are right that Euclidean geometry or any other geometry follows from the way distance is defined. Pythagoras is not true in other geometries even though it would be a good approximation locally, as I guess you are well aware. If it were not true I am sure it would not be proved as Euclid 1.47. I am well aware that the universe we live in might not have constant curvature, but in fact it does so far as we can tell, though that is entirely irrelevant to Pythagoras's theorem being a theorem. In common with most mathematicians I would be delighted if we lived in a universe with negative curvature since that would be more interesting, but sadly I don't think there is any evidence that we do. Perhaps you should find a physics forum to post in if you want to discuss this further. If the universe is not isotropic and homogeneous then science is a lot more difficult and can tell us nothing non-local.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Aug 2009
    Posts
    67
    i am amused at the use of words like 'rubbish' ... nevertheless ..

    it is a theorem only in the sense that zorn's lemma is a theorem of ZFC. since it is equivalent to AoC, it could well be regarded as an axiom, then AoC wud be a thoerem. so if you discount this strictly technical sense of the word 'theorem', then 'pythagoras theorem' is a result of (and equivalent) the definition of inner product in the euclidean plane.

    i will give you an exercise. if you are given a norm on R^n, can you derive the inner product from which this norm comes from ? if you do this you will understand why defining the usual norm (which is a roundabout way of assuming pythgoras) is same as defining the usual inner product.

    anyway i wont argue more on this. it is clear that you need to understand and read more carefully ... maths is not suitable for arguments, coz there is no scope of ambiguity.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member alunw's Avatar
    Joined
    May 2009
    Posts
    188
    R^n with usual norm is only one model of Euclidean geometry. Pythagoras's theorem is not an axiom of Euclidean geometry but a proper theorem of a geometry that can be axiomatised without having distance as a primitive concept. We use the usual metric because that it is the right one to make R^n a model of Euclidean geometry. When I want to do hyperbolic geometry with R^n (or rather a half space or ball within it) I use a different metric.

    Any positive definite inner product will give you Euclidean geometry - it just means your basis is not orthonormal. And not all norms come from an inner product.

    Perhaps you should take your own advice.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. inf (s) and sup (S) belong to the closure of S
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 4th 2011, 04:54 AM
  2. which two numbers do not belong in here?
    Posted in the Math Puzzles Forum
    Replies: 25
    Last Post: April 30th 2011, 07:30 AM
  3. Replies: 2
    Last Post: March 7th 2011, 08:56 AM
  4. Do the vectors (1,3,-2),(2,1,1) belong to R(T)?
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: December 13th 2009, 02:00 PM

Search Tags


/mathhelpforum @mathhelpforum