I am interested in how today's professional mathematicians view the Pythagorean theorem, in terms of how the theorem fits within the axiomatic framework of mathematics. I often come across textbooks that define length by the Pythagorean theorem, so that the theorem is in essence a definition or axiom. In more modern mathematics such as linear algebra, is the Pythagorean theorem generally just used as the definition of length? Is it more conventional today to treat the Pythagorean theorem as a definition (or axiom) rather than a theorem? Are there any modern proofs of the Pythagorean theorem that don't rely on Euclidean geometry (like a proof that utilizes linear algebra/the dot product, etc.)?

  • 4
    $\begingroup$ What is modern mathematics for you? Mathematics after 1980? $\endgroup$ – Américo Tavares Aug 25 '10 at 22:48
  • $\begingroup$ I have posed the following question, now a community wiki, What is Modern Mathematics? Is this an exact concept with a clear meaning? [math.stackexchange.com/questions/3447/… $\endgroup$ – Américo Tavares Aug 27 '10 at 13:41

Length, in Euclidean geometry, is a relation, not a number. We say that two segments are of the same length if they are congruent to each other, and congruence is one of the undefined notions of Euclidean geometry. The Pythagorean theorem comes about as soon as you decide to map lengths to positive real numbers, and it does so because of how length and angles are related by the congruence and similarity axioms (e.g., how the Pythagorean theorem is proved).

Lengths, in vector spaces, can be anything that satisfies the axioms for a norm. But to have a Pythagorean theorem, you need a notion of perpendicularity of vectors, that is, an inner product, which exists only if the norm satisfies the parallelogram law. For example, the taxicab norm, which is given by $|(a,b)|=|a|+|b|$ does not come from an inner product and there is no Pythagorean theorem for that notion of length.

The natural question is why does the Pythagorean theorem in Euclidean geometry correspond to the Pythagorean theorem in a finite dimensional vector space over the reals with an inner product? (there is a theorem that states that in a finite-dimensional vector space there is only one inner product up to isomorphism, so we really can talk about THE Pythagorean theorem).

The answer is that the finite dimensional vector space over the reals with an inner product encodes the congruence and similarity axioms of Euclidean geometry. The basic exposition of (most of) this can be found in the first half of Chapter 2 of Emil Artin's wonderful book Geometric Algebra.

Take a geometry of points and lines for which the following statements are true:

  1. Any two points determine a line.
  2. Parallel postulate: for any point $p$ not on line $\ell_1$ there exists a unique line $\ell_2$ that passes through $p$ but doesn't intersect $\ell_1$.
  3. Non-triviality: there exist $3$ non-collinear points.
  4. Desargues' Theorem (the wiki article is horrible)

The first three statements allow you to define translations (transformations of the plane that send lines to parallel lines and have no fixed points), while the fourth allows you to a construct a field over which the space of translations is a $2$-dimensional vector space, and such that if you associate translations $OP$ and $OQ$ with $(0,1)$ and $(1,0)$, then any point $R$ has a translation $OR$ which can be written as $(a,b)$ with $a$ and $b$ in the field.

In this way, any affine Desarguesian plane (thing that satisfies statements 1, 2, 3 and 4) can be identified with a $2$-dimensional vector space over some field (and conversely any $2$-dimensional vector space over some field corresponds to an affine Desarguesian plane).

Alright. Now, Euclidean geometry satisfies Desargues' Theorem, so which field does it correspond to? Well, it turns out that the fact that the geometry is ordered (i.e., we have a notion of an ordering of points on a line) means that the field has to be ordered and thus is a subfield of the real numbers (this is where we need something like the continuity or Hilbert's completeness axiom which roughly state that things that should intersect do intersect, and imply that the base field is all of the real numbers).

Then the whole inner product shebang turns out to just be a codification of the various congruence axioms. The norm is given by choosing some vector to make a unit vector (segment) and then considering to which vector in the same direction other vectors (segments) are congruent to. The inner product is a matter of encoding the notion of angle between vectors as an inner product, i.e., as a function linear in both vectors, which essentially takes the notions of similarity, congruences and unit circles.

It turns out that you can do all of this in the reverse direction, and so the Euclidean plane really is a 2-d vector space over the reals with an inner product.

Addendum: a comment on symmetry groups that is too long to leave as a comment.

Every norm on a vector space has its own unit blob, that is, the set of vectors with norm less than one. Geometrically, blobs that correspond to a norm satisfy the following properties: they are convex, they are absorbing (every vector is a multiple of a vector in the blob), and they don't contain any lines through the origin. It is a theorem that any such blob corresponds to a norm.

As we know, the transformations of a vector space are linear transformations, which means that the symmetry group of an object in your vector space is going to consist of the linear transformations that bijectively send an object to itself. In other words, the group of symmetries of an object is the set of invertible linear transformations that fix that object.

Now, the blob of the Euclidean norm is the unit disk, and the unit circle is the boundary of the disk. The unit circle is the set of vectors with Euclidean norm $1$, and in general, the boundary of the unit blob for a norm will be the set of vectors with norm $1$. The group of symmetries of the boundary of the blob then will be the set of invertible linear transformations that send vectors with norm $1$ to vectors with norm $1$.

A slight caveat is that you don't want to consider just any old invertible linear transformations because when you add a norm to a vector space, you are in effect defining a topology, that is, a way to define continuity. As a result you want to work with continuous linear transformations. BUT! In a finite dimensional vector space, no matter what the norm, we have that every linear transformation is continuous.

What this means is that for any norm on, say the $2$-d vector space over the reals, the symmetry groups of the boundary of the unit blob are all invertible linear transformations of the same space. Hence, we can directly compare these groups to each other since they live in the same space (of invertible linear transformations of $\mathbb R^2$).

Example: the boundary of the unit blob for the Euclidean norm is the unit circle, i.e., all points such that $x^2 + y^2=1$. The boundary of the unit blob of the Taxicab norm is the unit diamond, i.e., all points such that $|x|+|y|=1$.

Now if you think about it, any linear transformation that sends the unit diamond to itself will also send the unit circle to itself. Hence, the symmetry group for the Euclidean norm contains the symmetry group for the Taxicab norm, and in fact it contains it properly since, for example, rotation by $45$ degrees fixes the circle, but doesn't fix the diamond.

| cite | improve this answer | |
  • $\begingroup$ Too bad I cannot upvote again, but I'm going to leave this comment for the benefit of guest: There's actually a nice systematic "dichotomous key" for determining the "point group" of geometric objects. To use some symbols, the circle has $D_{\infty}$ point group, while the diamond is a mere $D_{4}$. What do those symbols mean? Saying something has $D_{n}$ symmetry is saying that: 1. it has an n-fold center of symmetry; and 2. n lines of symmetry pass through the n-fold center of symmetry. (D by the way is an abbreviation here for "dihedral"). $\endgroup$ – J. M. isn't a mathematician Aug 26 '10 at 21:56
  • $\begingroup$ Obviously, the higher the "n", the more symmetric the object is (e.g. a hexagon is more symmetric than a square since it has more lines of symmetry). The usual circle, as you can now see, takes this to the extreme. The fact that the "circle" having this much symmetry is part and parcel with adopting the Euclidean notion of distance is quite convincing that the Euclidean norm is useful to look at. $\endgroup$ – J. M. isn't a mathematician Aug 26 '10 at 22:01
  • 1
    $\begingroup$ Thank you for the kind words, J. Mangaldan. Isn't the infinite dihedral group (semi-direct product of Z and Z/2Z) countable though and thus not the symmetry group of the circle (the symmetry group of the circle being the uncountable circle itself of course)? $\endgroup$ – Vladimir Sotirov Aug 27 '10 at 0:49
  • $\begingroup$ (i keep forgetting the reflections, circle semi-direct Z/2Z?) $\endgroup$ – Vladimir Sotirov Aug 27 '10 at 0:56

The Pythagorean theorem follows directly from the basic properties of the dot product: if $x \cdot v = 0$, then $(x + v) \cdot (x + v) = x \cdot x + v \cdot v$. There is not much to say about this proof from a modern point of view except that it does not depend on the finite-dimensionality of the underlying space, e.g. in Hilbert spaces it even generalizes to Parseval's identity. (Of course one then has to prove that dot products induce norms, but this is the standard Cauchy-Schwarz argument.)

I would like to interpret your title question more broadly. My impression is that mathematicians do not really think about the Pythagorean theorem directly (unless, perhaps, they are functional analysts!). Rather, following Klein's Erlangen philosophy, they think about the group of symmetries that the use of the Pythagorean theorem (really, dot products) implies. In the plane, this is the Euclidean group $\text{E}(2)$, in which sits the orthogonal group $\text{O}(2)$ of linear transformations preserving the dot product (rotations and reflections). The geometry determined by the Euclidean group is one where it makes sense to talk about lengths and angles and all the familiar tools of Euclidean geometry. There are other Lie groups which determine other types of geometries (the "non-Euclidean" ones).

| cite | improve this answer | |
  • $\begingroup$ I don't really understand your above proof using the dot product; if I'm not mistaken, the proof relies on the fact that the square of the dot product of a vector equals the square of the length of that vector, but how do we know that without using the Pythagorean theorem? $\endgroup$ – guest Aug 25 '10 at 0:05
  • 2
    $\begingroup$ We define the length to be the square root of the dot product of a vector with itself. This is called the norm induced by an inner product. $\endgroup$ – Qiaochu Yuan Aug 25 '10 at 0:18
  • 1
    $\begingroup$ Oh okay; now I understand your proof, but I'm still not convinced that it avoids using the Pythagorean theorem as a definition in the first place. Specifically, why is length defined as the square root of the dot product of a vector with itself? Isn't this definition of length tantamount to defining length in terms of the Pythagorean theorem? $\endgroup$ – guest Aug 25 '10 at 0:29
  • $\begingroup$ If you want to think about it that way, then yes. $\endgroup$ – Qiaochu Yuan Aug 25 '10 at 0:37
  • 1
    $\begingroup$ @guest: To say that this definition is in terms of the Pythagorean theorem is to implicitly use a basis relative to which one defines the inner product, but part of the point of using abstract inner product spaces is to do away with the need to deal with explicit bases until necessary; that's why the statement of the Pythagorean theorem I used above is basis-independent. One can define an abstract inner product space without actually writing down the "dot product" per se and then prove that orthonormal bases exist. $\endgroup$ – Qiaochu Yuan Aug 25 '10 at 0:46

Based on the previous answers, I have come to the following (possible incomplete or incorrect) conclusion: It seems that the Pythagorean theorem can be dealt with in one of two ways, either it is taken as the definition of length/distance, or it is proved by invoking the technical definitions of an inner product space. However, proving Pythagoras via the definitions of an inner product space really just shifts the question to how one motivates these definitions (namely, why does one define the length of a vector to be the square root of its dot product with itself?). Thus, one can say that the Pythagorean theorem is generally taken to be the definition of length or distance, and the only thing left to do is motivate this definition; that is, show why the Euclidean norm is such a natural system for determining length. Qiaochu has alluded to the fact that the Euclidean norm is natural because of its large symmetry group, but I'd like someone to elaborate on this a bit if possible...

| cite | improve this answer | |
  • $\begingroup$ Copying my comment: "we use here the definition of 'circle' as 'the set of all points that are a fixed distance away from a given point'. It happens that the 'circle' corresponding to the Euclidean norm (as opposed to, say, the Manhattan norm) has an infinite symmetry group (invariant under rotation by any angle with respect to its center, infinitely many axes of symmetry, etc.) which corresponds to the traditional, non-mathematician's concept of a circle." $\endgroup$ – J. M. isn't a mathematician Aug 26 '10 at 3:52
  • $\begingroup$ How does one determine the symmetry group of the "circle"? $\endgroup$ – guest Aug 26 '10 at 17:11
  • $\begingroup$ The answer too your question was too long to fit in a comment, so I added it to my original answer. $\endgroup$ – Vladimir Sotirov Aug 26 '10 at 21:20
  • $\begingroup$ The following is a comment to your response (I don't have enough reputation to comment directly on your answer): Your answer clears things up a lot, but I just want to clarify something that still confuses me. You conclude that the symmetry group of the Euclidean norm contains the symmetry group of the Taxicab norm by qualitatively imagining the unit circle and unit diamond respectively. What I want to know is whether or not there is a method for reaching this same conclusion quantitatively, without having to actually know what the "unit circle" of the different norms looks like. $\endgroup$ – guest Aug 26 '10 at 22:48
  • $\begingroup$ guest, most naively you can set up a system of equations of the form norm of Ax = norm of x (or norm of Ax = 1 where norm of x is 1), and solve, but that would be a pain. An easier way is to realize that the linear transformations are infinitely differentiable. This allows you to see that for the circle, by continuity the image of (0,1) determines all other points as a rotation or as a rotation plus flip if the determinant is negative; for the square by differentiability you have to send kinks (vertices) to kinks (vertices). Then you can solve your system of equations more easily. $\endgroup$ – Vladimir Sotirov Aug 27 '10 at 1:05

Perhaps these thoughts will shed some light on the discussion here. If one writes down a collection of axioms for Euclidean geometry, for example, Hilbert's axioms, the Pythagorean Theorem and discussion of a distance function do not explicitly appear.


A model of Euclidean geometry consists of taking points as (x,y), ordered pairs of real numbers and lines as linear equations ax + by + c = 0, (a, b, c real numbers) using the Euclidean distance function as the way to find the distance between pairs of points.

However, if one takes points as (x, y) (ordered pairs of real numbers) and lines as linear equations ax + by + c = 0 but uses |a-b| + |c- d| as the distance between the two points (a,b) and (c,d) and measures angles as one does in the Euclidean plane one gets a geometry, often called the Taxicab Plane, which obeys all of the axioms of the Euclidean Plane except for the axiom dealing with the congruence of triangles.

Details can be found in Eugene Krause's book Taxicab Geometry.

The Pyhtagorean Theorem does not hold here.

For interesting subtleties, and lots about the taxicab plane, see Dawson's article available at this site:


| cite | improve this answer | |

You don't need inner product spaces to prove the Pythagorean theorem. If abc is a right triangle with hypotenuse c, arrange four copies of it into a square whose sides are of length $a + b$. Then calculate the area of the square two different ways and equate them.

$(a+b)^2$ = ${4*ab/2} + c^2$

Expand and simplify to get the Pythagorean theorem.

| cite | improve this answer | |

It all has to do with non-euclidean geometry. The basic idea is that the distance function, or to say the same thing as a professional mathematician, the "metric", is super important in geometry. You can have any kind of distance function you like, as long as it has three key properties (look them up). This is a pretty big subject, and it all goes back to the independence of Eulcid's 5th axiom on parallel lines.

The distance between two points depends on the curvature of the underlying space, that's the key idea.

It's not possible to prove that the distance between two points is given by the Pythagorean theorem; because the actual distance between two actual points is the domain of physics, not axiomatic mathematics. The best we can do in mathematics is define what we mean by distance (with a distance function), and since it's a definition, there's nothing to prove.

The story goes that Gauss realized Euclid's 5th postulate was independent because there are many geometries where it fails, e.g. on a circle or hyperboloid. But Gauss didn't want to just tell the world he had solved the oldest problem in math, so what he did was when Riemann came to him looking for a job as a professor, he told Riemann to list 3 possible topics to give a talk on as a sort of interview for the job. So Riemann wrote two topics down, but he couldn't think of a good third topic, and he put down "foundations of geometry" even though he hadn't thought very hard about it, just because he had some nice idea's which he had not fully formulated. And since Gauss already been thinking about that, he picked the foundations of geometry; and the rest is math history. Riemann went home and came up with a theory significantly beyond what Gauss had already known.

| cite | improve this answer | |
  • $\begingroup$ I don't think the abstract definition of a metric space was particularly motivated by non-Euclidean geometry. Seeing as the definition was introduced by Frechet, my guess is it was motivated by the need to deal with function spaces in sufficient generality. $\endgroup$ – Qiaochu Yuan Aug 24 '10 at 23:40
  • $\begingroup$ although you may be correct with regard to metric spaces, the OP asked about the Pythagorean theorem, which is directly related to the non-euclidean geometry of Riemann, who introduced the well known Riemannian metric. $\endgroup$ – Matt Calhoun Aug 24 '10 at 23:47
  • $\begingroup$ So if the Pythagorean theorem is taken as the definition of length, then why is this definition so natural? Why not define length as the cube root of the cubes of the sides of a triangle or something like that? $\endgroup$ – guest Aug 25 '10 at 0:33
  • 1
    $\begingroup$ @guest: the Euclidean norm is the only norm on R^2 with an infinite symmetry group. (The symmetry group of a norm is determined by the symmetry group of its "unit circle," and the "unit circle" of the Euclidean norm is the only one that's actually a circle.) $\endgroup$ – Qiaochu Yuan Aug 25 '10 at 0:43
  • $\begingroup$ Qiaochu, what do you mean when you say that 'the "unit circle" of the Euclidean norm is the only one that's actually a circle'? (How do you that the Euclidean norm produces a circle if you haven't yet chosen the Euclidean norm as the definition of length; doesn't knowledge about a circle necessarily come from the definition of length?) $\endgroup$ – guest Aug 25 '10 at 5:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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