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I have been looking for proofs for the pythagorean theorem that don't use area calculation but calculus, complex numbers or any other interesting ways to proof it.

I would love to see any interesting proof, Shay

Martin Sleziak
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Shay Ben Moshe
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    Well, that depends on what you're willing to call the Pythagorean theorem, as well as what kind of assumptions you're willing to start with. One might argue that the machinery of certain parts of calculus and complex numbers depends on the Pythagorean theorem, so that any such proof is circular. One might argue that there are various equivalent statements of the Pythagorean theorem, and that the proofs of their equivalence are not trivial. – Qiaochu Yuan Jul 22 '11 at 19:23
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    Echoing Qiaochu: it really depends on what tools you have. For example, there is a "Pythagorean Theorem" that holds for any inner product space (in particular, for the real and complex planes): if $\langle \mathbf{x},\mathbf{y}\rangle = 0$ (if $\mathbf{x}$ and $\mathbf{y}$ are orthogonal), then $$\lVert \mathbf{x}\rVert^2 + \lVert \mathbf{y}\rVert^2 = \lVert \mathbf{x}+\mathbf{y}\rVert^2,$$using the definition $\lVert\mathbf{z}\rVert = \sqrt{\langle \mathbf{z},\mathbf{z}\rangle}$. How interesting that is, though, seems rather subjective... – Arturo Magidin Jul 22 '11 at 19:26
  • Thank you for your comment. I have seen a "proof" of the pythagorean theorem that uses complex numbers but it is circular and therefore invalid. All I am looking for is a proof that for a right-angled triangle a^2+b^2=c^2, not using geometric areas calculation. – Shay Ben Moshe Jul 22 '11 at 19:29
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    To second and amplify Qiaochu's statement, the very definition of a norm of a complex number effectively states the Pythagorean Theorem as an axiom. I suppose you could consider the multiplicativity of the norm - that is, the fact that it behaves as expected under scaling and rotations - as a 'proof' that it acts like a length, and thus as an indirect justification for the Pythagorean theorem, but that seems somewhat roundabout... – Steven Stadnicki Jul 22 '11 at 19:30
  • Arturo Magidin, I have just seen your proof. You said it yourself, it is not interesting because you have defined the distance using the result of the pythagorean theorem. – Shay Ben Moshe Jul 22 '11 at 19:30
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    @shayfalador: Again: the problem is that if you don't say what you are assuming and what you are not assuming, then your question is meaningless. "Interesting" is subjective, so it is not something that anyone but you can answer (not a good question for this site). If you don't say what you are willing to assume and what you aren't, then you won't get good answers. Even writing "$a^2+b^2=c^2$" is meaningless without you saying exactly what you are assuming and what you aren't, or what the operations mean in that context. – Arturo Magidin Jul 22 '11 at 19:33
  • I am willing to see any proof (especially if you find it interesting) that doesn't assume things such as the distance between $(a,b)$ to $(0,0)$ is $\sqrt(a^2+b^2)$. Thanks again! – Shay Ben Moshe Jul 22 '11 at 19:46
  • @shayfalador: You aren't addressing the problem. For one, if you aren't assuming anything about distances, then just what does "$a^2+b^2=c^2$" even mean? You need to say what you are assuming, not just what you are not. – Arturo Magidin Jul 22 '11 at 19:57
  • I can not be anymore specific than that. Any axiom, but those that define distance as I said, is ok. – Shay Ben Moshe Jul 22 '11 at 20:06

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Explanation in terms of linear algebra. From this blog post by Terence Tao

The statement $a^{2}+b^{2}=c^{2}$ is equivalent to the assertion that the matrices $% \begin{pmatrix} a & b \\ -b & a% \end{pmatrix}% $ and $% \begin{pmatrix} c & 0 \\ 0 & c% \end{pmatrix}% $ have the same determinant. But it is easy to see geometrically that the linear transformations associated to these matrices differ by a rotation, and the claim follows.

Américo Tavares
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There is a proof using Similar Triangles:

enter image description here

We get from $\triangle CDA$ and $\triangle ABC$ that $\displaystyle \frac{CD}{AC} = \frac{CA}{BC}$ i.e. $\displaystyle \frac{\alpha}{a} = \frac{a}{c}$ i.e. $ \alpha c = a^2$

Similarly $\displaystyle \beta c = b^2$.

Adding gives the result.

Aryabhata
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  • I know this one, it is nice, thank you. – Shay Ben Moshe Jul 22 '11 at 19:50
  • @shay: There is one which uses in-center (but uses areas) which I am guessing you haven't seen. See the last paragraph of my answer here: http://math.stackexchange.com/questions/50093/trigonometric-equality-frac1-sin-a-cos-a1-sin-a-cos-a-tan/50149#50149 – Aryabhata Jul 22 '11 at 19:52
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    I can't resist and quote Gerry Myerson on another thread: "I like to call this the one-line proof of the Pythagoras Theorem, the one line being the perpendicular to the hypotenuse." – t.b. Jul 22 '11 at 20:13
  • In this page of the 1943's book "Compêndio de Geometria" (Geometry handbook) by Diogo Pacheco de Amorim de 1943 the theorem is proved in this way. – Américo Tavares Jul 22 '11 at 20:30
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    By the way: here's my preferred way of presenting this proof. Look at this picture: the red, green and blue figures are similar "houses". Since the red and blue roofs give the green roof, the red square and the blue square must add up to the green one. QED --- Christian Blatter displayed a similar picture with the heading: "Ich bin auch ein Beweis!" ("I'm a proof, too" in German) at his fare-well lecture. [There is a joke in that title that is hard to explain to people not living in Zurich] – t.b. Jul 22 '11 at 22:18
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How about the simplest ever (I got it from a book): Imagine a right prism with the base the triangle in question A(right angle), B , C (counter clock wise). The height of the prism is arbitrary. Now fill the prism with a gas at a given pressure. On the faces, in their middle the pressure forces act (Surface * pressure, perpendicular on the face). Equate the momenta around the corner B. They should cancel (the prism does not rotate by itself). You get the Pitagora's theorem in a blink of an eye. Cheers!! PS: I wanted to upload the figure, but I am not allowed until I get more points!

Vasea Igor
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Isn't the answer is just what's in here: Proof using differentials ?

amWhy
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Let $\,\triangle ABC\,$ be a right triangle and $\,O\,$ the midpoint of hypotenuse $\,BC\,$, known to also be its circumcenter, so $\,OA = OB = OC = \frac{1}{2} BC\,$. By the median length (Apollonius') theorem:

$$ \require{cancel} \begin{align} 2\left(AB^2+BC^2\right) = BC^2 + 4 AO^2 = BC^2 + 4 \cdot \left(\frac{BC}{2}\right)^2 \;\;\iff\;\; AB^2+AC^2 = BC^2 \end{align} $$

dxiv
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An elegant proof. Consider the vector $\mathbf{c}$ in the complex plane: $\mathbf{c}= a + \mathbf{i}b$, where $a$ and $b$ are scalar values, and $a,b > 0$. Now consider the reflection of $\mathbf{c}$ in the line at an angle of 45 degrees (pi/4 radians) passing through the origin, call that $\mathbf{d}$. $\mathbf{d} = b + \mathbf{i}a$, and $\mathbf{d}$ clearly has the same length as $\mathbf{c}$.

Now take the vector product of c and d. That has the same length as the square of the length of c, and in terms of polar coordinates, it is at an angle of pi radians (by virtue of the multiplicative properties of complex numbers expressed as polar coordinates).

so $\mathbf{c\cdot d} = i(c^2) = (a+ib)\cdot (b+ia) = ab-ba+i(a^2 + b^2) = i(a^2 + b^2)$.

$c^2 = a^2 + b^2$

Air Conditioner
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Mark
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