[srean]

No it doesn't work that way because the units of hyper-volume and length are different.

However, once you take appropriate roots of hypervolume to get same units you can safely compare. Or the otherway round take appropriate powers of length to get same units as hypervolume.

[evanb]

I agree; the fair comparison is the nth root of the hypervolume in n dimensions, (V(n))^(1/n). This monotonically decreases from n=0, which shows the counterintuitive point that people often want to make anyway: an n-sphere takes up less and less of an n-hypercube. The peak at n~=5 is illusory.

Another fair comparison is between dimension-dependent lengths is the ratio of the (hyper)volume to the surface (hyper)area V(n)/A(n). This monotonically decreases from n=1.

[cubefox]

Imagine you want to compare sizes of video game levels, where some are 2D, made from pixels, and some are 3D, made from voxels. You could stipulate that n pixels are equivalent to sqrt(n) voxels. But you could also stipulate that n pixels are equivalent to n voxels. I don't think either is more correct than the other.

[evanb]

Which is bigger: a meter, a hectare, or a liter?

[cubefox]

A hectare

[evanb]

What if the property with 1 hectare of area were 0.1m x 10^5 m? The meter is bigger than the field is wide.

Or, more to the point, suppose someone came with a different system of units and said: look, one of our standard lengths is 10^-6 of one of yours, but one of our standard areas is defined just like yours: 1 ha = (100m)^2 and (1 of our areas) = (100 of our lengths)^2.

In other words, their standard length = 10^-6m; their area = 10^-12 ha = 10^-8 m^2.

Is their standard area bigger or smaller than their standard length?

Notice that in proportion the lengths and areas are the same.

[cubefox]

With that it would probably mean the units monotonically approach 0, rather than first increasing and then decreasing. At least for whole dimensions. I'm not sure about monotonicity with fractional dimensions.