What is the sidelength of the largest square centered at 0 that contains no point with all integer coordinates? (You are allowed to rotate the square any way you want). It’s not hard to see the answer is sqrt(2).

How about the side length of the largest n-dimensional cube centered at 0 that contains no point with integer coordinates? My knee-jerk reaction was sqrt(n), but the 2-dimensional case is completely misleading (because in two dimensions a cube and a cross-polytope are the same thing). In fact it’s no more than C*sqrt(log(n)), for a relatively small constant C which is independent of n. An in fact, there is a conjecture that it’s even less: that the side length of a cube that contains no integer coordinates is bounded above by a constant independent of dimension.

]]>No offence taken :-).

The calculation you show for the “area” of the circle is the “area” of the circle IF it is projected onto a 2d plane (pi*r^2). For three dimension it would be (4/3 * pi *r ^3). Where as the “area” that you calculate of the outer box is indeed 2^9 since you are calculating it in 9 dimensions. Please refer to the formula about how to calculate “area” or “volume” of a sphere in 9-dimension.

Besides, the easy way to see is that sqrt(9)-1=3-1=2. So the circle has a radius 2 and its diameter is 4. But each side of the box is still of unit 2. It is easy to see now that the sphere will leak out of the box since a sphere with diameter 4 can not fit inside a box where every side is at most 2.

]]>I am a high school student, so perhaps this is just my limited understanding of the subject, but something doesn’t make sense to me here.

For the 2-dimensional model, we had an inner circle with a radius of sqrt(2) – 1 and an outer square area of 2^2.

For the 3-dimensional model, we had an inner circle with a radius of sqrt(3) – 1 and an outer box volume of 2^3.

So, for the N-dimensional model, we had an inner circle with a radius of sqrt(n) – 1, so wouldn’t we have an outer box volume (or whatever the shape would be called in other dimensions) of 2^n.

If that part is true then the example of N=9 (or this model in the 9th dimension) would be:

sqrt(9) -1

3-1

2

area of the circle:

(3.14)(2)^2

12.56

area of the outer box:

2^9

512

Which makes more sense to me than having the inner circle leak out of the box. Of course, I could be wrong (I’ve been wrong once or twice) and if I am please explain it to me.

P.S. No offence intended by showing all of my steps like that. it is sort of habit for me anymore

Thanks, Krys

]]>Your argument is more philosophical and it seems you do not understand dimensionality much. Since you seems to be too disturbed about the terms area ( which is nothing but volume in 2 dimension ) and volume. Let us define a measure m.

A sphere in N-dimension is defined with (p,r) where p is a point in N-dimensional space and r is the radius. Now, a sphere in N-dimension is nothing but a set of points which are AT MOST r distance from p in any direction. Now, this measure m is nothing but the integration of the (N-1)-dimensional manifold that wraps the sphere in question. Remember that integration gives area under a curve ( measure m under 1D manifold in a 2D plane ). I have given the formula on how to calculate volume or measure of a sphere in N-dimension. I request you to read more about manifolds and higher dimensions before you quibble over terminology such as “area” (volume in 2D), “length/periphery” (volume in 1D) and volume.

Regards,

]]>Similarly for three dimensions we put a sphere of maximum volume(not area) in the central gap

for the fourth dimension we will maximize some other metric lets say x. (what area is in 2 dimensions and volume in three dimensions, x is the the equivalent in 4 dimensions) not volume.

Your argument based on volume is hence incorrect.

To answer the queston,

note that we use area in 2 dimensions

and volume in 3 dimensions

now imagine if we are unware of the third dimension and live in a world of two dimensions

we would argue similarly that the area(surface area) of higher dimension cubes would soon become too big that it will overshoot the box. But it is not the case because we use volume(which we dont know about as we live in 2 dimensional world) to see if the sphere would fit in the space

]]>I was talking about the N-dimensional space so the dimension of the sphere (manifold) is N-1. For example on a plane (2D) when you draw a circle, the circle itself is a 1D manifold embedded in a 2D space. If you do not know what a manifold is, please refer to wikipedia or wolfram mathworld. ]]>