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Disproving Pythagorean Theorem

  1. Sep 17, 2013 #1
    Ever since I was in grade school I have been fascinated with the idea that the Pythagorean theorem, or any other universally respected theorem, could be wrong. When I was younger I found a little proof I made to disprove it, and I came across it in an old notebook of mine. Now after taking calculus and other more advanced maths I see that Pythagoras could not be wrong, however, I am having trouble actually disproving the proof I made to disprove Pythagoras's theorem, if that makes sense. I may just be having a serious brain fart so don't kill me.

    The proof I made was this:

    Make 4 congruent right triangles with side lengths of "a" and "b" and a hypotenuse with a length "c" and put the triangles together to make a square where the hypotenuses of the triangles are on the outside of the square. it should look like this: http://4.bp.blogspot.com/-YF-2E8vTRLs/TmoSmBD65wI/AAAAAAAABX8/BPFkCMM0vGE/s1600/QST.png

    Obviously the area of the triangle is A=c^2, the Area could also be the area of each triangle A=1/2ab, there are 4 of them so it becomes A=2ab. Through substitution we get c^2=2ab.

    c^2=2ab doesn't agree with the Pythagorean theorem, the problem I have having is explaining why this is so... I cant find an error in my logic. So can you guys help me out and tell me where I went wrong?
  2. jcsd
  3. Sep 17, 2013 #2
    "c^2=2ab doesn't agree with the Pythagorean theorem"

    I agree with you up til here. So you created a square, right? Meaning that each triangle is a 45 45 90, right? So then a=b. Then c^2=a^2+b^2=a^2+a^2=2a^2=2ab. I see no reason why this contradicts the Pythagorean Theorem.
  4. Sep 17, 2013 #3
    hmm... well I do you see they don't have to be 45 45 90 triangles, it could be a rectangle too, sorry for saying square. Try an example lets say a=4 b=5 and by the Pythagorean theorem c=5 but by the formula c^2=2ab c=sqrt20..
  5. Sep 17, 2013 #4

    D H

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    This site is not the place to discuss your misunderstandings.

    Thread closed.
  6. Sep 18, 2013 #5


    Staff: Mentor

    In reference to your drawing, and working backwards, if you start with a rectangle and draw two diagonal lines, the four angles at the center will not be congruent. That means that none of the angles at the center can be 90°.

    The only way that the central angles can all be congruent (and thus 90° each) is when the rectangle is actually a square. In this case the triangles are all 45° - 45° - 90° right triangles, and the two legs are equal in length. If the square is a units on each side, each leg of the triangle that makes up a quarter of the square will be a/√2.

    No. If the legs of a right triangle are a = 4 and b = 5, then c can't be 5. This would be too short. By the Pythagorean Theorem, c = √(42 + 52) = √41 ≈ 6.4.
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