9
$\begingroup$

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.

$\endgroup$

closed as unclear what you're asking by Brandon Carter, John Gowers, Macavity, Michael Albanese, Daniel W. Farlow Mar 28 '16 at 20:21

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 12
    $\begingroup$ Well...what does the theorem even say without a geometric representation? $\endgroup$ – lulu Mar 28 '16 at 17:50
  • 2
    $\begingroup$ Did you check the 116 proofs proposed on this site? $\endgroup$ – J.-E. Pin Mar 28 '16 at 17:51
  • 2
    $\begingroup$ 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. $\endgroup$ – Ian Mar 28 '16 at 17:53
  • 1
    $\begingroup$ Possible duplicate of Proof of Pythagorean theorem without using geometry for a high school student? $\endgroup$ – John Gowers Mar 28 '16 at 17:57
  • 2
    $\begingroup$ 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. $\endgroup$ – Doug M Mar 28 '16 at 17:59
17
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ How do you derive the inner product definition? It appears only to be a postulate. $\endgroup$ – Carpenter Oct 1 '16 at 5:32
  • $\begingroup$ 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 $\endgroup$ – Timothy Jan 22 at 4:14
  • $\begingroup$ 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. $\endgroup$ – Timothy Jan 22 at 4:16
1
$\begingroup$

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).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.