What is the length of the diagonal of a tesseract (four-dimensional cube) with side length $a$? What is the length of the diagonal of a tesseract (four-dimensional cube) with side length $a$?
Additional info:
- Diagonal of a line: $a$
- Diagonal of a square: $\sqrt{2a^2}$
- Diagonal of a cube: $\sqrt{3a^2}$
- Diagonal of a tesseract: ??
 A: I believe that by "diagonal" you mean "diameter", which is a more general term from geometry which means "the greatest distance that two points on the shape can reach from each other".  In the standard dimensions that we can visualize, these would be the distances between the vertices of the square and the cube.  
In multiple dimensions the idea still generalizes, so the answer, in short, is $\sqrt{n*a^2}$, where $n$ is the number of dimensions.  But why?
Let's assume that one corner of the (hyper)-cube, which in four dimensions is known as a tesseract, is at the origin, (0,0,...,0).  Then the corners can all be given coordinates. For example, one corner is at (a,0,...,0), and the furthest corner from the origin would be at (a,a,...,a).  The distance from this corner to the origin, then, is given by:
$$D = \sqrt{a^2 + a^2 + ... + a^2} = \sqrt{n*a^2}$$
A: The tesseract is definitely a mathematical object.
Let us take an Euclidean affine space of dimension 4 to define it, for example the canonical space $E=\mathbb{R}^4$. Let $a>0$. We define the "tesseract $T_a$ of edge a" as the convex envelope of $\{(\mp\frac a2,\mp\frac a2,\mp\frac a2,\mp\frac a2)\}$. $E$ is also a metric space and we can therefore define the diameter of $T_a$ : here it is $\sup_{x,y \in T_a} \|x-y\|=\sqrt{a²+a²+a²+a²}=\sqrt{4a²}=2a$.
We can now define, if we wish, a "diagonal of $T_2$" as a segment $[xy]\subset T_2$ such that $\|x-y\|=4$ . For example, $[(-1,-1,-1,-1)(1,1,1,1)]$ is a diagonal of $T_2$...
