I don't give a rat's ass if there's a proof or not connecting the two. I'm not a mathematician, but it makes sense from a calculus perspective that the number of dimensions ends up as a divisor.
He asks how to visualize the area of a triangle by bounding it in a rectangle, where it becomes obvious that the triangle takes up half the area.
He then poses the question about the volume of a pyramid where you can use a similar technique.
https://www.maa.org/external_archive/devlin/LockhartsLament....
From your corner draw lines to the four ceiling corners -- a cube diagonal, two face diagonals and a cube edge. This is the pyramid. Now draw lines to the corners of one of the walls opposite you -- it's the same pyramid, but on its side. And lastly draw lines to the other wall opposite you -- again the same pyramid. You have now covered the entire cube with three identical pyramids, so the volume of the pyramid is a third of the volume of the cube. The interesting part is that the bases of these pyramids are on the three dimensions, giving the intuition that the /3 is due to the dimension. This proof might even generalize in higher dimensions.
Edit: I think this is also the gist of the top answer in TFA, but it's weirdly formulated imo.
EDIT: got around to reading TFA and it's also the top answer, so there's a nice visualization.
Diving deeper into the proof, say the hypercube is a unit hypercube, spanning from (0, ..n, 0) to (1, ..n, 1) -- where "..n" means "a sequence of length n". Then:
Each corner is a point (b_1, ..n, b_n) where each b_i is 0 or 1.
You're located at (0, ..n, 0).
The hypercube has 2n faces, which can be represented as a pair (i, b) where 0<=i<n and b is 0 or 1. The face (i, b) touches the 2(n-1) corners whose i'th coordinate is equal to b.
You touch half of those faces: specifically the faces (i, 0) for each 0<=i<n.
You can draw a pyramid to the other half of the faces, since you don't touch them. These pyramids must all have the same volume, by the symmetry of coordinate permutations, which is an operation that preserves volume.
TFA - What does TFA stand for? The Free Dictionary
Life is too short for proofs. - Gilbert Strang (in one of his lectures)
Beware of bugs in the above code; I have only proved it correct, not tried it.
Donald Knuth
Edsger W. Dijkstra
CONTEXT:
Argument three is based on the constructive approach to the problem of program correctness. Today a usual technique is to make a program and then to test it. But: program testing can be a very effective way to show the presence of bugs, but is hopelessly inadequate for showing their absence. The only effective way to raise the confidence level of a program significantly is to give a convincing proof of its correctness. But one should not first make the program and then prove its correctness, because then the requirement of providing the proof would only increase the poor programmer’s burden. On the contrary: the programmer should let correctness proof and program grow hand in hand. Argument three is essentially based on the following observation. If one first asks oneself what the structure of a convincing proof would be and, having found this, then constructs a program satisfying this proof’s requirements, then these correctness concerns turn out to be a very effective heuristic guidance. By definition this approach is only applicable when we restrict ourselves to intellectually manageable programs, but it provides us with effective means for finding a satisfactory one among these.
[0] Edsger Dijkstra - Turing Award Lecture - The Humble Programmer - 1972
https://www.cs.utexas.edu/~EWD/transcriptions/EWD03xx/EWD340...
Proof: Take height 1 for simplicity of notation. Then we get the volume as the integral over crosssections (Fubini). Each cross section is rescaled by a factor (1-z), where z is the height. Rescaling changes the cross sections measure by (1-z)^(n-1). Integrate that to get 1/n. Done.
Edit: apparently it does for pyramids, so it should for cones as well. General formula for volume of n-dimensional pyramid is A*h/n where A is the volume of the base.
If the 4th dimension is time, does that mean a quarter of the "space" is spread out over time? What does that even mean? Anyone familiar with n-dimensional space able to weigh in?
When you look at time as the zeroth dimension rather than the fourth, it should be obvious that the fourth dimension is simply another spatial dimension. Think about how a two-dimensional plane is a cross section of a three-dimensional object, and line would be a cross section of a plane, and a (zero-dimensional) point is a cross section of a line. So too can our three-dimensional universe be looked as a cross section of a theoretical four-dimensional existence. A three-dimensional object that moves in the fourth dimension would simply cease to exist in our universe, and a four-dimensional object that moves in the fourth dimension would change which 3D cross section is visible to us.
When you think about the nth dimension as a projection of the n+1th dimension, it begins to make a lot more sense. That doesn’t mean you can necessarily visualize the 4th spatial dimension directly, but you can at least visualize how the 3rd dimension can be a projection of the 4th, just like how we can easily see that the 2nd dimension as a projection of the third.
It will generalize to any shape consisting of stacked rescaled slices of some other shape where the scaling factor increases linearly.
Of course one can fill in why in this case it happens to be the dimension in the denominator—several sibling posts, particularly gniv's (https://news.ycombinator.com/item?id=36562093), have done so nicely—but I was responding specifically to the disclaimer:
> I don't give a rat's ass if there's a proof or not connecting the two. I'm not a mathematician, but it makes sense from a calculus perspective that the number of dimensions ends up as a divisor.
which seemed to suggest that it should be obvious without mathematical reasoning.
A 2D projection of a cone isn't intuitively 1/3 of a cylinder with the same dimensions.