Is the proof of Pythagorean theorem using dot (inner) product circular?

Solution 1:

I think the question mixes two quite different concepts together: proof and motivation.

The motivation for defining the inner product, orthogonality, and length of vectors in $\mathbb R^n$ in the "usual" way (that is, $\langle x,y\rangle = x_1y_1 + x_2y_2 + \cdots + x_ny_n$) is presumably at least in part that by doing this we will be able to establish a property of $\mathbb R_n$ corresponding to the familiar Pythagorean Theorem from synthetic plane geometry. The motivation is, indeed, circular in that we get the Pythagorean Theorem as one of the results of something we set up because we wanted the Pythagorean Theorem.

But that's how many axiomatic systems are born. Someone wants to be able to work with mathematical objects in a certain way, so they come up with axioms and definitions that provide mathematical objects they can work with the way they wanted to. I would be surprised to learn that the classical axioms of Euclidean geometry (from which the original Pythagorean Theorem derives) were not created for the reason that they produced the kind of geometry that Euclid's contemporaries wanted to work with.

Proof, on the other hand, consists of starting with a given set of axioms and definitions (with emphasis on the word "given," that is, they have no prior basis other than that we want to believe them), and showing that a certain result necessarily follows from those axioms and definitions without relying on any other facts that did not derive from those axioms and definitions. In the proof of the "Pythagorean Theorem" in $\mathbb R^n,$ after the point at which the axioms were given, did any step of the proof rely on anything other than the stated axioms and definitions?

The answer to that question would depend on how the axioms were stated. If there is an axiom that says $x$ and $y$ are orthogonal if $\langle x,y\rangle = 0,$ then this fact does not logically "come from" the Pythagorean Theorem; it comes from the axioms.

Solution 2:

Let's try this on a different vector space. Here's a nice one: Let $\mathscr L = C([0,1])$ be the set of all real continuous functions defined on the interval $I = [0,1]$. If $f, g \in \mathscr L$ and $a,b \in \Bbb R$, then $h(x) := af(x) + bg(x)$ defines another continuous function on $I$, so $\scr L$ is indeed a vector space over $\Bbb R$.

Now I arbitrarly define $f \cdot g := \int_I f(x)g(x)dx$, and quickly note that this operation is commutative and $(af + bg)\cdot h = a(f\cdot h) + b(g \cdot h)$, and that $f \cdot f \ge 0$ and $f\cdot f = 0$ if and only if $f$ is the constant function $0$.

Thus we see that $f\cdot g$ acts as a dot product on $\scr L$, and so we can define $$\|f\| := \sqrt{f\cdot f}$$ and call $\|f - g\|$ the "distance from $f$ to $g$".

By the Cauchy-Schwarz inequality $$\left(\int fg\right)^2 \le \int f^2\int g^2$$ and therefore $$|f\cdot g| \le \|f\|\|g\|$$

Therefore, we can arbitrarily define for non-zero $f, g$ that $$\theta = \cos^{-1}\left(\frac{f\cdot g}{\|f\|\|g\|}\right)$$ and call $\theta$ the "angle between $f$ and $g$", and define that $f$ and $g$ are "perpendicular" when $\theta = \pi/2$. Equivalently, $f$ is perpendicular to $g$ exactly when $f \cdot g = 0$.

And now we see that a Pythagorean-like theorem holds for $\scr L$: $f$ and $g$ are perpendicular exactly when $\|f - g\|^2 = \|f\|^2 + \|g\|^2$

The point of this exercise? That the vector Pythagorean theorem is something different from the familiar Pythagorean theorem of geometry. The vector space $\scr L$ is not a plane, or space, or even $n$ dimensional space for any $n$. It is in fact an infinite dimensional vector space. I did not rely on geometric intuition to develop this. At no point did the geometric Pythagorean theorem come into play.

I did choose the definitions to follow a familiar pattern, but the point here is that I (or actually far more gifted mathematicians whose footsteps I'm aping) made those definitions by choice. They were not forced on me by the Pythagorean theorem, but rather were chosen by me exactly so that this vector Pythagorean theorem would be true.

By making these definitions, I can now start applying those old geometric intuitions to this weird set of functions that beforehand was something too esoteric to handle.

The vector Pythagorean theorem isn't a way to prove that old geometric result. It is a way to show that the old geometric result also has application in this entirely new and different arena of vector spaces.

Solution 3:

If we start from Euclidean geometry in two dimensions, then it is circular. In this settings, we define perpendicular lines as lines that, at their intersection form four equal angles. Coordinate systems are defines using a basis consisting of two perpendicular vectors of the same length.

It is that actually easy to prove that two vectors are perpendicular iff their dot product is 0 using rotational properties. We would hit a snag, however, at the claim that the two-norm equals the length, which is a direct consequence of Pythagorean theorem.

Of course, that is not the context this proof of the Pythaogrean theorem normally shows up in. It shows up when we start from a different set of definitions: We define orthogonality using dot products. We define length using 2-norm. Then we prove a Vector space Pythagorean theorem as you have shown above, without circular reasoning.

Of course, now we have two different definitions for length and perpendicularity, and two different Pythagorean theorems. It is a separate step to prove that in $\mathbb{R}^2$ and $\mathbb{R}^3$, both sets of definitions actually mean the same things, and so both Pythagorean theorems are in fact the same theorem.