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Is there a purely algebraic proof for the Pythagorean theorem that doesn't rely on a geometric representation? Just algebra/calculus. I want to TRULY understand the WHY of how it is true. I know it works and I know the geometric proofs.

pancini
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Carpenter
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    Well...what does the theorem even say without a geometric representation? – lulu Mar 28 '16 at 17:50
  • The thing is : what is your definition of "space" and "distance", then ? – Captain Lama Mar 28 '16 at 17:50
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    Did you check the 116 proofs proposed on this site? – J.-E. Pin Mar 28 '16 at 17:51
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    When you've stripped out the geometry, it is not so clear that the algebraic fact, which follows straightforwardly from the definition of an inner product, a norm induced by an inner product, and orthogonality, has any geometric significance. Indeed this proof just goes as follows: assume $x,y$ are orthogonal vectors, then $| x+y |^2=\langle x+y,x+y \rangle = \langle x,x \rangle + \langle y,y \rangle + \langle x,y \rangle + \langle y,x \rangle = | x |^2 + | y |^2 + 0 + 0$ $= | x |^2 + | y |^2$ if $x,y$ are orthogonal. – Ian Mar 28 '16 at 17:53
  • The "distance" would involve the units of a graph. A graphed circle might be involved in an algebraic proof. I don't want to accept Pythagorean theorem like a physical law, I want to actually understand why it's the case. – Carpenter Mar 28 '16 at 17:53
  • Because it's a theorem about the 3 sides of a right triangle, it's impossible to provide a proof without using geometry at all, although maybe a proof that uses geometry "behind the scenes" can be done. Maybe this is what you're looking for? http://www.mathsisfun.com/geometry/pythagorean-theorem-proof.html –  Mar 28 '16 at 17:53
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    Possible duplicate of Proof of Pythagorean theorem without using geometry for a high school student? – John Gowers Mar 28 '16 at 17:57
  • Thanks J-E Pin for the link. – Carpenter Mar 28 '16 at 17:58
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    My question to you is why would a purely algebraic proof be more satisfying or convincing than a geometric proof? Generally, geometric proofs are a bit more intuitive, and for most of the history of mathematics, considered to be more rigorous. – Doug M Mar 28 '16 at 17:59
  • In response to "I don't want to accept Pythagorean theorem like a physical law," the Pythagorean Theorem is essentially essentially a physical law about flat, two-dimensional space. By making certain geometric assumptions about this flat space (i.e., Euclid's five axioms) we can derive consequences, like the Pythagorean Theorem, geometrically. Captain Llama's answer is essentially the only way to escape making geometric assumptions, by making algebraic assumptions (you'd make geometric definitions instead) and building geometry from algebra. – pjs36 Mar 28 '16 at 18:06
  • @Donkey_2009 The proposed "duplicate" appears not to be about the Pythagorean Theorem. The asker of that question seems to have conflated the theorem with the existence of Pythagorean triples of integers, for which one does not need to be concerned with right angles at all. I see you knew that already; but why should we suppose that this asker has made the same error? – David K Mar 28 '16 at 18:21
  • Maybe these could be relevant for the question: http://math.stackexchange.com/questions/733874/which-axioms-are-behind-the-pythagorean-theorem http://math.stackexchange.com/questions/23074/the-pythagorean-theorem-and-hilbert-axioms http://math.stackexchange.com/questions/675522/whats-the-intuition-behind-pythagoras-theorem/676731#676731 – Marco Disce Mar 28 '16 at 19:05
  • I'm not sure what you're trying to ask. The question was closed so I made a pending edit and am wondering if it makes it better by your standards. Maybe if you accept it and read my answer at https://www.quora.com/Mathematics-What-is-a-purely-rigorous-abstract-proof-of-the-Pythagorean-theorem/answer/Timothy-Bahry then from that figure out what I didn't know you meant to ask and tell me what you're trying to ask, I might be able to further improve it. Maybe what you want to ask will turn out to be such that an answer like my Quora answer answers your question. – Timothy Dec 03 '18 at 05:10
  • I'm not sure what you were trying to ask. Maybe you meant to say "rigorous proof." A rigorous proof of a mathematical statement is probably a more complete formal proof or a written proof from which experts can figure out a more complete formal proof. Did you mean to ask something like "I know how to deduce the distance formula from the following axioms" 1. For any real numbers x, y, z, w, d((x, y), (x + z, y + w)) = d((0, 0), (z, w)) 2. For any real number x, y, d((0, 0), (x, y)) is a nonnegative real number. 3. For any nonnegative real number x, d((0, 0), (x, 0)) = x 4. For any real numbers – Timothy Jan 22 '19 at 02:37
  • x, y, d((0, 0), (x, -y)) = d((0, 0), (x, y)). 5. For any real numbers x, y, z, w, d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y)) $\times$ d((0, 0), (z, w)). How do I show that a function from $(\mathbb{R}^2)^2$ to $\mathbb{R}$ satisfying those conditions even exists? All a proof by similar triangles does is show that any function that satisfying those conditions must be the function d((x, y), (z, w)) = $\sqrt{(z - x)^2 + (w - y)^2}$ It doesn't show that that function actually does satisfy those conditions." Maybe something like this answers you question. "Conditions 2-5 hold because – Timothy Jan 22 '19 at 02:48
  • $d((0, 0), (xz - yw, xw + yz)) = \sqrt{(xz - yw)^2 + (xw + yz)^2} = \sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = \sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = \sqrt{(x^2 + y^2)(z^2 + w^2)} = \sqrt{x^2 + y^2}\sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$. Now it's straight forward to show that condition 1 holds also." I guess other people can't figure out what you're asking because they just unconsciously use the assumption d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y)) × d((0, 0), (z, w)) which it turns out can be proven entirely using linear algebra after you define – Timothy Jan 22 '19 at 02:54
  • distance to be the function it was deduced from the 5 conditions to be. I once had the same question except in my head I instead used something more like d(x, y) to mean d((0, 0), (x, y)) replacing the notations in conditions 2-5 and omitting condition 1 because once you prove it satisfies the 4 conditions, it's so easy to define in terms of the function from $\mathbb{R}^2$ to $\mathbb{R}$, a function from $(\mathbb{R}^2)^2$ to $\mathbb{R}$ and show that it satisfies all 5 conditions but I figured out the answer to my own question. This question is not a duplicate of – Timothy Jan 22 '19 at 03:04
  • https://math.stackexchange.com/questions/1381903/proof-of-pythagorean-theorem-without-using-geometry-for-a-high-school-student like one of the comments under body of the question suggested. The body of that question was actually a totally different question about the infinitude of Pythagorean triples and maybe the author of that question thought that once you prove the infinitude of Pythagorean triples, you can show that the Pythagorean theorem holds for any member of $\mathbb{Z}^2$ whose coordinates are the first 2 terms in a triple or that you could assume all 5 conditions and then deduce – Timothy Jan 22 '19 at 03:11
  • the Pythagorean theorem from the fact that all complex numbers have a square root in the complex number system. That question was probably a question no one finds useful where as this one can still be improved into a really good question. According to https://meta.stackexchange.com/questions/217754/stance-on-repost-question-instead-of-editing-to-invalidate-answers/217757#217757, since fixing up your question like that would probably invalidate the answers, it's probably better for you to ask another question as a fixed up version of this question linking this question. – Timothy Jan 22 '19 at 03:17

2 Answers2

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The "modern" approach is this : first we define the field $\mathbb{R}$ (for instance, it's the only totally ordered field with the supremum property).

Then we define what a $\mathbb{R}$-vector space is : it's an abelian group with an external action of $\mathbb{R}$ satifying some axioms.

Then there is a notion of dimension : we can define a vector space of dimension $2$.

The notion of Euclidean distance is obtained by defining what an inner product is : it's a symmetric bilinear form such that $\langle x,x\rangle>0$ if $x\neq 0$. The distance is then $||x-y||$ with $||x||=\sqrt{\langle x,x\rangle}$. We also have the notion of orthogonality from this inner product.

Well, once you did that, then the Pythagorean theorem is a triviality : $||x-y||^2 = \langle x-y,x-y \rangle = \langle x,x \rangle - 2\langle x,y \rangle + \langle y,y \rangle = ||x||^2 + ||y||^2$ (assuming of course that $\langle x,y\rangle=0$, the orthogonality hypothesis).

Of course all the work went into the definitions, which is contrary to the basic approach of geometry which deduces properties of distance from a set of axioms (generally ill-defined, but it can be made precise with a little work).

The interesting thing about this modern approach is that algebraic structures come before geometric content. This is powerful because algebraic structures have enough rigidity. For instance, if you start with a set of points and lines satisfying some incidence axioms, it's very hard to define what it means that it has a certain dimension. But if you have a vector space structure, then it's easy.

Of course it can be a little disappointing beacause it feels like we "cheated" : we made the theorem obvious by somewhat putting it in the definitions. But on the other hand, it's very clear and precise : can you properly define what distance or an angle is using "high school geometry" ? Not so easy. Even in Euclide's Elements, this is kind of put under the rug as "primitive notions". This approach makes everyting perfectly well-defined and easy to work with.

Captain Lama
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    How do you derive the inner product definition? It appears only to be a postulate. – Carpenter Oct 01 '16 at 05:32
  • Were you treating $x$ and $y$ like points in $\mathbb{R}^2$ and defining that for any points in $\mathbb{R}^2$, $x$ and $y$, $\langle x, y \rangle$ means the dot product of $x$ and $y$? If so, I think I actually understand your proof. I independently thought of that proof long before I understood what you were saying in this one and wrote my form of it in the comments under the body of the question. From reading the comment to this answer, it seems that the OP, Carpenter doesn't find the answer good enough and to them, it was proven that that definition satisfies certain assumptions but wasn't – Timothy Jan 22 '19 at 04:14
  • proven that anything satisfying those assumptions must be that function. I think most people go the other way and assume those assumptions and deduce the distance formula but don't check that that function actually satisfies them. – Timothy Jan 22 '19 at 04:16
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A "proof" of the Pythagorean Theorem depends on some kind of definitions of:

  • right angle
  • length/area
  • stright line

The axioms of Euclid are not completely formalized but we have other formal axiomatic systems that mimic euclidean axioms and definitions of these notions (for example the Hilbert's axioms) so that we can derive Pythagorean theorem there. A formal proof with these axiomatic systems wouldn't require any reference to pictures in principle.

Proving Pytagorean Theorem in completely different context such as analytic geometry (or"calculus") could be possibly trivial or meaningless depending on what definition of "right angle" we are going to consider. For example it would be trivial if you define a right angle with the scalar product and the distance with $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$, but you could try with different ones and the proof of the theorem could get more and more complicated depending on which definition you want to take (you could want to define areas with Peano Jordan's measure for example).

Marco Disce
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