$x,y$ are perpendicular if and only if $x\cdot y=0$. Now, $||x+y||^2=(x+y)\cdot (x+y)=(x\cdot x)+(x\cdot y)+(y\cdot x)+(y\cdot y)$. The middle two terms are zero if and only if $x,y$ are perpendicular. So, $||x+y||^2=(x\cdot x)+(y\cdot y)=||x||^2+||y||^2$ if and only if $x,y$ are perpendicular. ( I copied this)

I think this argument is circular because the property

$x\cdot y=0 $ implies $x$ and $y$ are perpendicular

comes from the Pythagorean theorem.

Oh, it just came to mind that the property could be derived from the law of cosines. The law of cosines can be proved without the Pythagorean theorem, right, so the proof isn't circular?

Another question: If the property comes from the Pythagorean theorem or cosine law, then how does the dot product give a condition for orthogonality for higher dimensions?

Edit: The following quote by Poincare hepled me regarding the question:

Mathematics is the art of giving the same name to different things.

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    $\begingroup$ Usually, the standard way to define angles in vector spaces is via inner product. So, no. The argument is not circular. $x \dot y = 0$ iff $x \perp y$ is pretty much the definition of $x$ and $y$ being perpendicular. $\endgroup$ Feb 14 '19 at 17:58
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    $\begingroup$ @MilanStojanovic Can you define what is real world orthogonality? $\endgroup$ Feb 14 '19 at 18:16
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    $\begingroup$ the Euclidean norm $\|(x,y)\|^2 = (x,y)\cdot(x,y) = x^2 + y^2$ was defined in light of the knowledge of the Pythagorean theorem. So, yes, I would say it is circular to prove the Pythagorean theorem using the inner product. $\endgroup$
    – Doug M
    Feb 14 '19 at 18:38
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    $\begingroup$ @DougM But isn't that like cheating? I mean in almost every axiomatic system, our axioms come from some well-known facts that somebody has purified them among many other facts. Does this mean if we try to prove these basic facts using our axioms, our reasoning is circular? I disagree. We should take our axioms as something like God given facts, regardless of why they make sense. $\endgroup$ Feb 14 '19 at 18:43
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    $\begingroup$ The Euclidean Norm is not an axiom, it is a definition. A definition that was chosen because it logically follows from the Pythagorean theorem, which is a direct consequence of the axioms of Euclidean geometry. We could define a non-Euclidean norm, and then we should not expect the Pythagorean theorem to hold as used above. $\endgroup$
    – Doug M
    Feb 14 '19 at 19:33

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.

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    $\begingroup$ Thank God finally someone said what I had on my mind! I was trying to say this but I didn't know how to put it in words in a good way. You said exactly what I wanted to say but I didn't know how to say it. Thanks. +1 $\endgroup$ Feb 14 '19 at 22:57
  • $\begingroup$ I suppose it is a matter of interpretation, but Euclid's axioms were chosen because the Greeks thought them "obviously true", not as something that they could pick and choose to get the geometry they wanted. For them, math and physics were the same thing, and the axioms were the basis of their understanding of nature. $\endgroup$ Feb 15 '19 at 3:47
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    $\begingroup$ @PaulSinclair The question is whether the Greeks already "knew" that Euclidean geometry was correct before those axioms were chosen. It is precisely because it was so "obviously true" and so inextricable from their view of the shape of space that I doubt the abstraction was the first form of their geometry. Reducing the axioms to just a handful may have been (for them) much like reducing matter to a handful of elementary particles was to us (except that their reduction was understood to be perfect, and ours is not). $\endgroup$
    – David K
    Feb 15 '19 at 6:06
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    $\begingroup$ The predecessor of Greek geometry was the geometry of the Egyptians and Babylonians, which were collections of rules and formulas (algorithms, really) that were apparently obtained from experience, rather than logic, and were occasionally flawed. As I understand it, Thales of Miletus travelled and learned these rules, but realized some of them could be obtained logically from simpler assumptions - such as the theorem that vertical angles are equal, which is attributed to him. He brought the idea home, where others expanded on it haphazardly until Euclid collected it into a coherent system. $\endgroup$ Feb 15 '19 at 16:58

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.

  • $\begingroup$ What is your opinion however to the main question : Is the proof circular or not ? $\endgroup$
    – Peter
    Feb 16 '19 at 8:40
  • $\begingroup$ @Peter - please read the last sentence of my post. $\endgroup$ Feb 16 '19 at 14:32

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.


Remember that $\mathbb{R}^n$ as a vector space over $\mathbb{R}$ with standard addition and scalar multiplication isn't rich enough to do geometry. When we talk about geometry, particularly the Euclidean geometry, we want to study the transformations that keep some properties like length and angles preserved, namely isometries. Most of our theorems will be about how two angles, lengths or congruent circles and we define two shapes to be congruent if and only if they can be transformed into each other by one of these 'transformations' (isometries). Viewing geometry from this point of view, i.e. through its group of transformations, is related to what is called the Erlangen program.

Anyway, it turns out that both of these properties, i.e. length and angle, can be packaged into one algebraic concept called the inner product (which I'm going to define below). More precisely, if $\vec{x} \in \mathbb{R}^n$, then

$$\| \vec{x} \| = \sqrt{\langle \vec{x}, \vec{x} \rangle}$$ $$\angle (\vec{x}, \vec{y}) = \arccos\bigg(\frac{\langle \vec{x},\vec{y}\rangle}{\sqrt{\langle \vec{x},\vec{x}\rangle} \sqrt{\langle\vec{y},\vec{y}\rangle}}\bigg)$$

The last definition makes sense because of the Cauchy-Schwartz inequality which is true for inner products: $$|\langle \vec{x},\vec{y} \rangle|^2 \leq \langle\vec{x},\vec{y} \rangle \langle \vec{y}, \vec{y}\rangle$$

And hence, there exists a unique angle between any two vectors.

Now, by definition, an inner product is a bilinear form $\langle \cdot, \cdot \rangle: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ such that

  1. $\langle \vec{x}, \vec{x} \rangle \geq 0$ and $\langle \vec{x}, \vec{x} \rangle = 0$ if and only if $\vec{x}=\vec{0}$
  2. $\langle \vec{x}, \vec{y} \rangle = \langle \vec{y}, \vec{x} \rangle$
  3. $\langle \overrightarrow{\alpha x+\beta z}, \vec{y} \rangle = \alpha\langle \vec{x}, \vec{y} \rangle + \beta\langle \vec{z}, \vec{y} \rangle$

Again, by definition two vectors are perpendicular if and only if their inner product is zero. Accepting these definitions, the Pythagorean theorem can now be proven without any circular reasoning.

Now I understand that you feel kind of uncomfortable because you say that the standard inner product on $\mathbb{R}^n$ comes from the Pythagorean theorem for $n=2,3$. And I understand that some people here interpret this as circular reasoning. However, I have two issues with this interpretation:

  1. our definition for the standard inner product on $\mathbb{R}^n$ for $n>3$ doesn't come from any previously known fact. Does this mean that our definition for $n>3$ is less valid than for $n=2,3$ or we should give up on higher dimensions?

  2. It happens times and times in physics, as well as mathematics, that a new theory is based on some known facts in earlier theories as observations for a general pattern. Then to show that our new theory is better, they try to show that the results predicted by the previous theory can be obtained in the new framework as well. Should we say that it's circular reasoning? So, if someone applies a stronger theory to prove something that has motivated it in the earlier theory, is it circular reasoning? It seems that many people here think so. Even though I don't challenge it, I don't like this interpretation. I believe that once we have a new theory with its own definitions, which can be taught to a person without any prior knowledge of the earlier facts, it is not unreasonable to focus our attention only on our current framework and forget about its past like it never existed. Meanwhile, we can use our new theory to prove the results proven in our earlier theory.

  • 4
    $\begingroup$ @MilanStojanovic I have an objection to that kind of thinking but I don't know how to put it in words properly. This is where we need a logician like Asaf Karagila. I think you are talking about two different systems. The law of cosines is a theorem in the Euclidean geometry with a different set of axioms. Here, we are working in another system. The law of cosines is no longer a theorem. It's nothing here. All we have here is a vector space equipped with an inner product. Nothing more, nothing less. $\endgroup$ Feb 14 '19 at 18:48
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    $\begingroup$ @downvoter You're welcome to leave a comment and explain why you think this answer is not helpful. $\endgroup$ Feb 14 '19 at 19:16
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    $\begingroup$ You say "we do not need to think in terms of Euclidean geometry anymore." But the OP is asking about the Pythagorean Theoreom which is a theorem of Euclidean geometry. In other words you have not really engaged with the question raised by OP, you have just said that it is not an interesting question. I think a lot of people would disagree. $\endgroup$
    – Jbag1212
    Feb 14 '19 at 20:10
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    $\begingroup$ @Jbag1212 No. All I'm saying is that the statement $\|x+y\|^2 = \|x\|^2+\|y\|^2$ can be proven using the tools we have developed. Now, obviously, the tools we have developed come from some well-known facts in the Euclidean geometry when $n=2,3$. But that's it. Our tools are stronger than the Euclidean geometry in the sense that they're generalized to $n>3$ very naturally while the Euclidean geometry fails to do so. So, even though the Euclidean geometry was developed earlier, it seems reasonable to stick to our new, more powerful system. $\endgroup$ Feb 14 '19 at 20:17
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    $\begingroup$ "Now, obviously, the tools we have developed come from some well-known facts in the Euclidean geometry" - that's the point though. These tools were developed using well-known facts in Euclidean geometry. So to prove the Pythagorean Theorem from these tools is "circular" if these tools were motivated by the Pythagorean Theorem in the first place. As far as I am aware, the dot product is not the only reasonable definition of inner product available for $\mathbb{R}^n.$ $\endgroup$
    – Jbag1212
    Feb 14 '19 at 20:21

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