Of n-dimensions and Euclidean distances

This is more of an excercise in writing formulae on my blog than actually doing some math. But then again, who needs a reason to write equations?

Well anyway, there was this doubt that a friend asked me, which was put to the students by a prof in his class. We know that the Euclidean distance, between two points and in n-dimensions, is given by:

We also know that for any three points in 2-dimensional space, , , and , the Triangle Inequality holds, namely,

This is a well known result in Euclidean Geometry, and I'm loathe to go into the proof of this here. Now the question, was to find three points , and in 3-dimensional space, which violate the Triangle Inequality!

At first glance, one will easily see that even in 3-dimensional space, the three points will still form a triangle, and so the Triangle Inequality should hold. But how do we prove it? This is where geometric visualization aided me.

You see, in 3-d space, one has a lot more freedom to twist and turn shapes. Maybe it has to do with the degrees of freedom, which I never understood clearly. In 2-d, all that one can do with a shape, say a square, is to translate, rotate, or scale it. Put it in more layman terms, you can put the square at any place on the paper, turn the paper whichever way, and even shrink or expand the paper. Translation is possible in two directions, and rotation in only one ('coz , which basically says that rotating anticlockwise is the same as rotating clockwise through a larger angle). All this, while still retaining the properties of the square. In 3-d, on the other hand, all that one can do is still essentially the same, translate, rotate, or scale it. But here, you can translate or rotate in 3 different directions (unlike 2-d, rotation is possible in 3 directions, since rotations in different directions are not commutative).

I'm pretty sure all this previous paragraph would give a sore headache, to any layman. What I've put forward, in not-so-informal terms, is the justification for the following transformations:

Translate the origin to , thus making , and .
Rotate the axes, such that lies on the x-axis, thus making , and .
Rotate the axes, such that lies on the xy-plane, thus making .

All these transformations are perfectly legitimate, in the sense that the relative positions of the three points do not change. But by doing this, we notice that we've transformed our axes such that all three points now lie on the same plane (the xy-plane here). The truth is that the three points always lay on the same plane, ('coz any three points in 3-d lie on a plane) but we've shifted our axes such that this plane became our xy-plane. In any case, I claim that these three points are the same as three points in 2-d space, since I've reduced the component of all of them to zero. And we know that the Triangle Inequality will hold for any three points in 2-d space.

Thus we see, that the Triangle Inequality should indeed hold for any set of three points in 3-d space as well. And so the original question now takes a further leap: Will the Triangle Inequality hold for any set of three points in n-dimensional space? With the help of reasoning similar to the one above, I claim yes!

Key concept used: Any set of n points in n-dimensional space lie on the same (n-1)-dimensional hyperplane. Is this really true? you tell me :P