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Disproof to the Pythagorean Theorem?

  1. Dec 10, 2012 #1
    Ok I realize the Pythagorean Theorem is correct. I completely get that very basic concept this is just a question I have.

    On a right triangle a^2 + b^2 = c^2 with c being the hypotenuse.

    But if instead of the hypotenuse connecting the two legs you had a jagged line that went halfway up then half way to the right and then the other half to the up and the other half to the right. That line would be equal to a + b because it is going the same x distance as a and the same y distance as b.

    Now if you continued to decrease the intervals at which you switch going up and then to the right:
    You would end up with a line similar to a ton of tiny stair steps connecting the legs a and b
    This "stair" line would also be equal to a + b

    Now finally if you made the intervals in which you take these steps infinitely small (Or take the limit as it approaches zero) the "stair" line would be approximately equal to the hypotenuse and thus c = a + b

    But that is completely contradictory to the Pythagorean Theorem.

    Now I know you can approximate any function (Including a hypotenuse) with tiny step intervals because of Plank's theories on Quantum mechanics and how digital signals can perfectly recreate analog signals (Digital signals are in step form)

    I fail to see where my flaw in logic is. I know I made a mistake somewhere but I cannot see it, much appreciation if someone could point it out.
     
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  3. Dec 10, 2012 #2

    jim mcnamara

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    The Pythagorean theorem pertains to triangles constructed from three straight lines intersecting at three vertices (distinct points), right? What you describe does not meet the definition of a triangle in 2D space, using the four Euclidean axioms. The ones we all learned in high school.

    You are essentially "free" to make definitions, you just can't apply them a priori to some arbitrary theorem that is founded on a different set of definitions. First, you have to prove that the theorem also is true when used in your system of axioms.

    So, finding a counterexample (trying to disprove) some theorem using new untried definitions does not make logical sense mostly because nobody ever tried to say it was a theorem in the first place.

    This is a "semi-mathematical" way of explaining this. I am trying to communicate clearly. Let us know if you are still confused.
     
  4. Dec 10, 2012 #3

    micromass

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    You touch some deep concepts here. First there is the concept of length which is intuitively obvious to you, but which is a concept which is hard to define or grasp. There are many weird things surrounding length: for example, there exists shapes which have no length at all (= no valid length can even defined, it's not that the length is zero). Or there exist nice (=continuous) bounded curves with infinite length.

    Then there is the concept of convergence. You say that your "stair curve" approximates the hypothenuse. In fancy language, we say that the stair curve converges to the hypothenuse.

    The issue with your disproof is that the stair curve approximates the hypothenuse, but the lengths of the stair curve do not approximate the hypothenuse. It might seem obvious that the lengths need to converge as well, but apart from intuition there is no reason why we should expect this.

    So it might happen that a curve converges to another curve but the lengths of the curves do not converge. You assumed that this was true, and this was the flaw in your argument.
     
  5. Dec 10, 2012 #4

    micromass

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    If anybody has any problems figuring out what the OP means exactly, it is described here:

    images?q=tbn:ANd9GcSY4461uROBV9mzqkEH6nGG1l0V9J7y-LGcb_WMbpiraSteysNB0veFkzKIaA.jpg
     
  6. Dec 10, 2012 #5

    jim mcnamara

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    Correct me where I'm wrong -

    The Pythagorean Theorem only holds for the case where the Playfair Axiom is the usual:
    ' parallel lines that neither converge nor diverge'

    So I don't get where we can even consider curves, since the sum of the angles of a triangle in this case is 180 degrees. Proposition 1 of the Elements specifically mentions straight lines, when constructing a triangle (isoceles).

    EDIT: thanks for the picture. I was more than confused. But the "line" for the hypoteneuse violates Euclid's definition of a straight line: 'A straight line is a line which lies evenly with the points on itself.'
     
    Last edited: Dec 10, 2012
  7. Dec 10, 2012 #6
    Ok thanks you very much MicroMass and Jim

    Still a bit confused:

    Why would the lengths of the two "curves"(lets say they are curves) not converge when the two actual curves converge?

    If I had a Taylor Series Approximation of a Sin curve, the length (On the given interval of course) would converge as well as the appearance of the two curves

    So according to that logic, any quantum approximation of the hypotenuse should have the same length? Right?
     
  8. Dec 10, 2012 #7

    AlephZero

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    In a sense, you could say the theorem is a tautology, because if only holds if you define the distance between two points ##(x_1,y_1)## and ##(x_2,y_2)## to be ##\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}##.

    Of course it never occured to the ancient Greek geometers to define lengths any other way, because that is what you measure with a ruler in the "real world".

    On the other hand, it can be quite useful mathematically to define lengths in other ways, including ##|x_2-x_1| + |y_2-y_1|## as in the OP. It depends which properties of "length" are important for the math you want to do, and which are not.
     
  9. Dec 10, 2012 #8
    Ok I understand that my definition of lengths and distance may be different from Pythagorus's
    How is an approximation of the hypotenuse not equal to the hypotenuse if I take the limit as my "steps" go to infinity? According to everything I have ever been taught about approximations that should not be the case. My approximated curve should converge to the hypotenuse.
     
  10. Dec 11, 2012 #9
    The Mathematical Recreations of Lewis Carroll: Pillow Problems and a Tangled Tale (Dover Recreational Math) [Paperback] is full of such "proofs".
     
  11. Dec 11, 2012 #10

    D H

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    No, it shouldn't. Your approximation is using a different norm than the one used in the Pythagorean theorem. You're using the taxicab norm while the Pythagorean theorem uses the Euclidean norm. Of course your result isn't the same.
     
  12. Dec 11, 2012 #11

    jedishrfu

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    An interesting consequence of the OPs question is that if the hypotenuse were a sequence of steps so that the distance is a+b then this causes issues with object rotation.

    Consider a square a 5x5 which is rotated x degrees making a 3-4-5 right triangle (only now we cant say the hyp is 5) then the squares side is now (3+4)*(3+4) which has now nearly doubled in area means things are no longer invariant to rotation which probably the reason why the pythagorean theorem was written.
     
  13. Dec 11, 2012 #12
    Basically, what your calculation demonstrates is that one has to carefully define what it means for a sequence of functions to converge.

    Your sequence of functions converges to the hypotenuse in many different senses, but most importantly, it converges uniformly. That is a fairly strong type of convergence, but not strong enough to imply convergence of lengths as your calculation shows.

    An even stronger form of convergence requires that the functions f_n(x) and f_n'(x) both converge uniformly to some limit function f(x) and f'(x). Actually, what is more common is that f_n(x) and f_n'(x) converge to f(x) and f'(x) in an integral norm such as L^2. Another way of writing this is:

    [itex] \int_a^b \left(|f_n(x)-f(x)|^2+|f_n'(x)-f'(x)|^2\right)\, dx \rightarrow 0. [/itex]

    Given this kind of convergence, you can prove that the lengths of f_n converge to the lengths of f.
     
  14. Dec 11, 2012 #13

    micromass

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    I don't think the taxicab metric has anything to do with this question. The point here is that if [itex]f,f_n:[0,1]\rightarrow \mathbb{R}^2[/itex] are curves, then if

    [tex]f_n\rightarrow f[/tex]

    then that does not necessarily imply

    [tex]L(f_n)\rightarrow L(f).[/tex]

    The thing is that the length of a curve is equal to

    [tex]L(f)=\int_0^1 \sqrt{1+(f^\prime)^2}[/tex]

    So if we wanted [itex]L(f_n)\rightarrow L(f)[/itex] then we should want that the limit of integrals is the integral of the limit. This is no problem here. What is a problem is that if [itex]f_n\rightarrow f[/itex], then that does not necessarily imply [itex]f_n^\prime\rightarrow f^\prime[/itex]. It is the failure of convergence of derivatives that is the issue here.
     
  15. Dec 11, 2012 #14
    Have you actually taken the limits (it is actually two limits) as you suggest?

    If you did you might be surprised.

    That is what fractals are all about.
     
    Last edited: Dec 11, 2012
  16. Dec 11, 2012 #15

    D H

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    You're right. Just because a family of functions fn(x) approaches some other function f(x) in the sense that |fn(x)-f(x)|→0 as n→∞ does not mean that the arc length of fn(x) approaches that of f(x). Example: [itex]f_n(x) = sin(2 n \pi x)/\surd{n}[/itex] approaches [itex]f(x)=0[/itex] as [itex]n\to\infty[/itex]. Over the interval [itex]x\in(0,1)[/itex], the arc length of f(x) is simply 1, while that of fn(x) goes to infinity as n goes to infinity.
     
  17. Dec 11, 2012 #16
    The length is not converging it is remaining constant (a + b)
     
  18. Dec 11, 2012 #17

    D H

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    A series all of whose elements have the same value is a convergent series.
     
  19. Dec 11, 2012 #18
    Yes, let me state it how I intended, It is not converging to the length/limit of the hypotenuse. Each iteration of the series does not have a "delta", it is constant, most series with a limit, exhibit a property of sums approaching the limit. (I'm just not eloquent, I do know what I mean)
     
  20. Dec 11, 2012 #19
    I don't understand why you would expect a+b to approach the length of the hypotenuse, or change at all.

    No matter how tiny you make them or how many of them their are, or how smooth the drawing looks to your eye, so long as there isn't a diagonal line involved, you haven't created a triangle.

    You're basically asking why the value of [itex]\frac{1}{n} (an + bn)[/itex] never changes no matter what you stuff into n.
     
  21. Dec 12, 2012 #20
    Can't this question be generalized as: what function defines the distance between two points on a plane and what simplified version of this function arises from minimising its value ?
     
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