On the train home, I thought I would try to prove $\pi$ is irrational. I needed a definition, so I used:

$\pi$ is the area of the unit circle.

But what is a circle?

A circle is the set of tuples $(x,y)$ satisfying "$x^2 + y^2 \leq 1$".

That seems a little awkward - is there a more natural way to say it?

The standard distance metric on $\mathbb{R}^2$ is "$\sqrt{(x1-x2)^2 + (y1-y2)^2}$". A circle is the unit ball of this metric, and $\pi$ is its area.

But why do we use this metric in the first place? Why not use the Taxicab metric - it's linear so it'd seem to satisfy even more nice mathematical properties.

Because of the Pythagorean theorem.

And what property of $\mathbb{R}^2$ lets us prove the Pythagorean theorem?

This one took a little more thought. I think it comes about because we want to associate lengths to lines that are equivalent under translations and rotations. Translations seem pretty natural - but I'm having trouble defining rotations without being circular:

  • If you multiply a coordinate tuple by the appropriate matrix, you get a rotation - but the matrix's entries involve $\sin$ and $\cos$, which I would define as the coordinates on the unit circle(which is circular).

  • I can define the set of all infinite lines that go through the origin. Except for the vertical line, you might assume they all have the form $L_m = \{(x,y)|y = mx\}$ for some $m \in \mathbb{R}$. Given an $m \in \mathbb{R}$, I can't find an obvious way to get the point on the unit circle corresponding to it.

  • The complex number route just gives $\sin$ and $\cos$. Circular, again.

  • We can determine whether two vectors are orthogonal using the dot product. That reduces defining the entire unit circle to defining just a single quadrant(if we also assume the length of $-1 * x$ is the same as $x$).

  • A quick Google search shows circles were a primitive notion in Euclid's Elements.

  • There is another question/answer given here, but I was still left confused as to what a rotation actually is in $\mathbb{R}^2$: Why is the Euclidean metric the natural choice?

So my questions are:

  1. Out of all the possible metrics, why choose the Euclidean metric as the natural one? I think it's rotation that's the root cause, but that might end up being a red herring.
  2. How do we define a natural equivalence of points up to rotation from first principles?
  3. This is related to #1. If we define the set of infinite lines through the origin, you can choose a representative element from each one and form a perimeter curve(assuming a continuous choice function). What properties does the circle have that would make it a natural choice?
  • 6
    $\begingroup$ I must point out that the taxicab metric is not linear, it is only piecewise linear. So it doesn't satisfy as many nice mathematical properties as you might hope. $\endgroup$
    – user856
    Oct 26, 2012 at 6:26
  • $\begingroup$ See en.wikipedia.org/wiki/Parallelogram_law, section 2. $\endgroup$
    – wj32
    Oct 26, 2012 at 12:48
  • 2
    $\begingroup$ "$\pi$ is the area of the unit circle" Better to say: "$\pi$ is the area of the unit disk." The circle is better defined as the set of tuples $(x,y)$ satisfying $x^2+y^2=1$. $\endgroup$ Mar 10, 2014 at 0:49

6 Answers 6


The Euclidean metric is special because it comes from what is called an inner product, and up to scaling it is the only metric that does so. This allows you to talk about angles between vectors in a sensible way, which you cannot do with other metrics.

So really we don't choose to use the Euclidean metric, so much as we choose to use the dot product (the only inner product on $\mathbb R^2$, up to scaling) and we get the Euclidean metric as a result.

  • 3
    $\begingroup$ Up to scale and rotation. ($ Q D Q^T $ is positive definite for orthogonal $ Q $ and positive diagonal $ D $). $\endgroup$
    – Ian
    Jan 31, 2015 at 15:14
  • $\begingroup$ Why is it not possible to talk about angles between vectors in a sensible way with other metrics? Because the SAS Postulate isn't satisfied in the resulting geometries? $\endgroup$ Jul 4, 2017 at 11:03

The obvious reason why the euclidean metric is standard is that this is how distances appear to behave in nature. I would think in any universe the "Standard" metric aliens would talk about (if they talk about such a concept) is the one that most obviously conforms to actual physical distance in their universe. There are of course nice mathematical reasons for the euclidean metric.

  • 2
    $\begingroup$ Some physics may suggest that space is Euclidean only in certain scales. At very large or very small scales it may not be. If so, you must assume that those aliens are about the same size as us. $\endgroup$
    – GEdgar
    Feb 4, 2013 at 17:08

The ancient architects needed the right angle and different constructions where used. In time Pythagoras' theorem become obvious by experiments and the only theory satisfying that condition is Euclidean geometry.


The so-called Euclidean metric is used to make $\mathbb R^2$ a model for the Euclidean plane. As described in Euclid's Elements.

The isometries of this metric are the congruences of Euclidean geometry. In particular: rotations are isometries, as you say in 1. [In addition, translations and reflections are isometries.]


It is a bit philosophical, I guess, but when you don't want measure distances, but in stead measure "orders of magnitudes of distances", then a non-archimedean distance is better.


We use 2-norm simply because it works i.e. it satisfies properties of Euclidean space.

  • No dimension is more important than other (e.g. Chebyshev distance doesn't qualify).

  • $\mathbb{R}^2$ is a real coordinate space. So integers have to work too. For positive integers $x, y, z$ $ x^n+y^n=z^n$ doesn't work for $n > 2$. [1]

  • There is no reference point. E.g. distance using 1-norm depends of which vector you choose as an axis

  • When you rotate distance doesn't change, hence the unit circle.

[1] https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem

  • $\begingroup$ I don't understand your point about Fermat's Last Theorem. Do you mean that if $x,y \in \mathbb{Z^n}$, then their distance $d(x,y) \in \mathbb{Z}$ as well? That's false even for the Euclidean distance. $\endgroup$ Apr 7, 2019 at 19:54

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