What is the volume of an n-dimension cube?

Question

What is the volume of an n-dimension cube? Consider the length of each side to be $a$. How to solve this problem?

Answer 1

The "area" of this shape can be thought of as multiplying each of the side lengths together for $n$ sides. So we could say that the area is $a^n$, where $n$ is the dimension of Euclidean space you're working in and $a$ is the side length of one of the sides of the hypercube ($n$-dimensional cube)

Answer 2

Perhaps it would be easier to imagine the volume of the cube to be the "n-dimensional area". Let's consider what happens in 1D, 2D and 3D first:

n=1:

In the first dimension, the volume would be just the side length or a=$a^1$.

For n=2:

This would be the square whose side length is a and the area is $a^2$

For n=3:

This would be the cube with volume of a^3.

For n>3:

I could do an induction proof on the conclusion, however a more intuitive pattern starts to emerge:

having the dimension be $0<n\in\Bbb Z$, would most certainly mean that Volume(n-dimensional cube with side length a)=V(a,n)=$a^n$. Keep in mind that the units would be $(units)^n$ for the n-dimensional area of an n-dimensional shape with the above restrictions, The dimension should a counting

The reason I said that an induction proof would not be as good is because I am not at the level of education to do this. I know how to do a very basic induction proof, but I am unconfident for this one. I guess it would go like this:

A line has n units which I will call u which are line segments so u+u+u+...+u=nu similarly, a square has square units, but also a certain amount of squares in one line segment like row so this would be nu+nu+...nu n times getting us $n^2$ u. We have this dimension being d=2 with n d=1 shapes “within” it. number, ie a positive integer excluding zero.

The reason I said that an induction proof would not be as good is because I am not at the level of education to do this. I know how to do a very basic induction proof, but I am unconfident for this one. I guess it would go like this: A line has n units which I will call u which are line segments so A=u+u+u+...+u=nu similarly, a square has square units, but also a certain amount of squares in one line segment like row so this would be A=nu+nu+...nu n times getting us $n^2$ u. We have this dimension being d=2 with n d=1 shapes “within” it. This means that our “hypothesis” is a d=N+1 cube has n of the d=N cubes “within” it using the same dimension units. The abbreviation u represents any general unit to the n-th power.

All in all, a d=N+1 cube can be drawn by connecting the corresponding vertices of a d=N cube assuming the connecting edge length is equal to the d=N cube edge length. This means that the hypothesis is “argued out” and so is informally correct with A=($n^{N}+...+n^{N})*u$ n times getting us A=n^{N+1}*units^{N+1}.

If a unit was instead defined by,say a n-tetrahedron, these formulas would not be correct for the space as it would likely be transformed.

I am not completely sure about the zeroth dimension nor fractal/fractional dimensions nor complex dimensions nor other types of dimensions here like negative dimensions.

Here are some links to these other dimensions:

Has the notion of having a complex amount of dimensions ever been described? And what about negative dimensionality?, https://en.m.wikipedia.org/wiki/Fractal_dimension, Imaginary and Complex Dimensions

Maybe there are other dimensions?