I think it would be fun to 'prove' the Pythagorean theorem in the following way, using only a small number of axioms that relate the physical world to the world of mathematics.

Assume that the collection of lengthts in the physical world corresponds one-to-one to the set of real numbers $\mathbb R$, and that the (physical) Euclidean plane therefore should be thought of as the vector space $\mathbb R^2$, as each point in the Euclidean plane corresponds uniquely to a pair of lengths (namely vertical and horizontal). These statements can be motivated physically and we'll take them as axioms.

Now my idea is that there might also be a physical motivation for requiring that the norm on $\mathbb R^2$ (modeling the physical Euclidean plane) come from some inner product. And if we take that as an axiom, then there is an elementary result about Hilbert spaces that says that $||x+y|| = ||x||+||y||$ whenever $x,y\in\mathbb R^2$ are orthonormal, which we may interpret as the Pythagorean theorem.

So if anyone knows a physical reason that the norm (representing lengths) should come from an inner product, I'd love to hear it.

N.B. I wasn't sure if I should post this question here or in the physics community, but I think mathematicians will have thought more about the axioms of inner products and things like that, so I think this is the right place.

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    $\begingroup$ Mathematics is an artificial construction: axioms do not exist in nature. So there is little point in giving a physical interpretation to Mathematics, but it is worth to give a mathematical interpretation to reality. $\endgroup$ – Jack D'Aurizio Jun 5 '17 at 14:01
  • $\begingroup$ Once we have sound definitions of "euclidean plane", "triangle" and "area of a triangle", the Pythagorean theorem is a straightforward consequence (it is enough to have a look at the usual proofs of it). $\endgroup$ – Jack D'Aurizio Jun 5 '17 at 14:03
  • $\begingroup$ @JackD'Aurizio I'm exacty trying to avoid having to give these definitions in for instance the way Euclid and Hilbert did. $\endgroup$ – Sjorszini Jun 5 '17 at 14:10
  • $\begingroup$ @JackD'Aurizio And I'm not sure what point you're trying to make with the first comment. Of course mathematically, we can define the Euclidean plane as we want, but if we use it to model the physical world, we need some reasons to do that. $\endgroup$ – Sjorszini Jun 5 '17 at 14:11
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    $\begingroup$ I will rephrase: what is the physical reason for the quadratic reciprocity theorem to hold? I guess it is difficult to find any, and even if one is able to find a "physical reason", that is quite irrelevant. The physical world does not obey to euclidean geometry, for instance, so I think it is hard to find any physical evidence of the fifth postulate out there. $\endgroup$ – Jack D'Aurizio Jun 5 '17 at 14:17

The following "physical argument" seems the most obvious: experiment confirms that in physical space at least on our scale, the Pythagorean theorem $a^2+b^2=c^2$ for a right-angle triangle is satisfied with high precision. This is equivalent to the norm being Euclidean (which automatically implies that it comes from an inner product, by the usual polarisation formula).

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    $\begingroup$ Interesting. And the polarization identity holds on infinite dimensional Hilbert spaces $\endgroup$ – CopyPasteIt Jun 5 '17 at 17:33

The Greeks thought they were writing axioms that described the world. We know now that those are just axioms for one particular geometry.

The physical world we live in seems to be Euclidean. But that may be only because we see a very small part of it in everyday experience. Its geometry might be locally Euclidean but hyperbolic or elliptic on a large scale.

Locally, we can measure lengths and angles and check that the Pythagorean Theorem is true to within measurement error. Mathematically, the Pythagorean theorem is essentially equivalent to having lengths and angles determined by an inner product.

Globally the question is open. A search for is the physical world euclidean find many interesying links. Here's one:


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  • $\begingroup$ This has little to do with my question. I'm trying to figure out a reason for having the norm (i.e., distances) on what we intuitively think of as the plane (think of a flat sheet of paper) come from an inner product. $\endgroup$ – Sjorszini Jun 6 '17 at 17:16
  • $\begingroup$ If I understand your quesyion I think @MichailKatz 's answer (and the third paragraph of mine) provides an answer. $\endgroup$ – Ethan Bolker Jun 6 '17 at 17:42

The length of a 'vector' on the real line is easy to understand, and while tinkering you notice that it can be expressed as

$|x| = \sqrt {x^2}$

How can we extend this to the plane? On the plane you now have 'vectors with angles' to contend with, whenever you are not on the $x$ or $y$ axis.

Experimenting further, you notice that multiplication has some properties on the real line, and using the notation $<x,y> = xy$, you find that

  1. $<u+v,w>=<u,w>+<v,w>$

  2. $<\alpha v,w>= \alpha <v,w>$

  3. $<v,w>=<w,v>$

  4. $<v,v>$ is non-negative and equal to $0$ if and only if $v=0$.

You are excited now since the absolute value of any number is $x$ is $\sqrt {<x,x>}$.

There is only one function, called the inner product function

<.,.> : $\mathbb R^2 \times \mathbb R^2 \to \mathbb R$

satisfying (1) thru (4) and the following 'physical' requirements:

(5)' If $e_1 = (1,0)$ then $<e_1,e_1> = 1$

(6)' If $e_2 = (0,1)$ then $<e_2,e_2> = 1$

(7)' $<e_1,e_2> = 0$

so that, with (2), (5)' and (6)' you have the length of any vectors 'without' angles. The only immediate motivation for (7)' is to be able to solve this:

Exercise: Show that the length of $e_1 + e_2$, $\sqrt {<e_1+e_2,e_1+e_2>}$ is equal to $\sqrt 2$.

But, if your inner product is to 'understand angles', maybe it only needs to 'identify' just one orthogonal pair of vectors.

Using your new tool, you find that the length of any vector

$\sqrt {<v,v>}$.

is given by the Pythagorean theorem - you don't have to prove it.

You've read some of Euclid's Elements and figure this is good enough for you!

Imagine your thrill when you learn that there is a geometric interpretation of the inner product - the $cos(\theta)$ comes into play.

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  • $\begingroup$ I don't see where the motivation for generalizing the points 1-4 comes from. We could try to generalize any other property of multiplation of real numbers, but my question is why we need these specific ones. $\endgroup$ – Sjorszini Jun 6 '17 at 17:08
  • $\begingroup$ Yes, the main thing you want to extend is the 'cool' formula for the absolute value. When you go to 2-space, you have to find a way to 'capture' angle info in the inner product. I have a another answer that might provide better motivation. The 'measuring device' has to spit out a number for any two vectors. $\endgroup$ – CopyPasteIt Jun 6 '17 at 17:39

The OP asks for a

physical reason that the norm (representing lengths) should come from an inner product

Ans: If you believe in the physicality of angles and the concept of two vectors being orthogonal to each other, it is how you want go - welcome to Euclidean Space. For 2-dim space, get out that right-angle Cartesian coordinate system graph paper and go to work!

In another answer I motivated this somewhat, but what really drives the point home is

The Gram–Schmidt process. Let

(1) ${\displaystyle \mathrm {proj} _{\mathbf {u} }\,(\mathbf {v} )={\langle \mathbf {v} ,\mathbf {u} \rangle \over \langle \mathbf {u} ,\mathbf {u} \rangle }{\mathbf {u} }}$

be your starting point, and you'll be forced to discover the standard inner product on $\mathbb R^2$. Before having its definition, you will find that you want:

If $e_1 = (1,0)$ then $<e_1,e_1> = 1$

If $e_2 = (0,1)$ then $<e_2,e_2> = 1$

$<e_1,e_2> = 0$

and that $\sqrt {<v,v>}$ better represent the length of a vector $v$. The one thing that needs work to get it to all fall out is the commutativity , $<u,v>=<v,u>$, but working with (1) you construct the isosceles triangle:

enter image description here

and you want to show that the red line segments are of equal length.

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  • $\begingroup$ I like your idea about starting with projections. But I'll need to think for a bit about how to go about it exactly, because the way you define a projection here requires that you already have some kind of inner product. $\endgroup$ – Sjorszini Jun 7 '17 at 20:06
  • $\begingroup$ When you set up the coordinate Cartesian Plane you want to use algebra to demonstrate geometric concepts. You can accept Pythagorean theorem to take the length of vector, but you are still have to handle angles. You are searching for a cool way to get both together without having to analyze a circle. $\endgroup$ – CopyPasteIt Jun 7 '17 at 22:40
  • $\begingroup$ By the way, if you like working in the complex plane, you can study the circle and get your angles another way (Euler's formula, etc.). And you will quickly come to the conclusion that you want Pythagorean's theorem; see math.stackexchange.com/a/2225716/432081 $\endgroup$ – CopyPasteIt Jun 7 '17 at 22:46
  • $\begingroup$ Motivation: Project $v = (\sqrt(2)/2,\sqrt(2)/2)$ onto both $(1,0)$ and $(0,1)$. Then, using geometry and pictures, project $(1,0)$ onto $v$. etc. $\endgroup$ – CopyPasteIt Jun 7 '17 at 23:47

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