Diameter of N-Cube

Theorem
Let $Q_n = \left[ {c-R\,.\,.\,c+R } \right]^n$ be an $n$-cube in Euclidean $n$-Space equipped with the usual metric.

Then the diameter of $Q_n$ is given by:


 * $\operatorname{diam}\left({Q_n}\right) = 2 R \sqrt{n}$

Corollary
The diameter of $Q_n$ is the length of some diagonal of $Q_n$

Proof
Write:


 * $Q_n = \displaystyle \prod_{i \mathop = 1}^n \left[{c-R \,.\,.\, c+R}\right]_i$

Let $x, y \in Q_n$

By the definition of the usual metric, the distance between any two points $x$ and $y$ is given by:


 * $\displaystyle \Vert y - x \Vert = \left({\sum_{i \mathop = 1}^n \left({y_i - x_i}\right)^2}\right)^{1 / 2}$

By the monotonicity of positive powers this sum is maximal when each summand is maximal.

Consider $x_i, y_i$ in the $i$th interval:


 * $\left[{c-R \,.\,.\, c+R}\right]_i$

To maximize $|y_i - x_i|$, take $x_i = \min \left[{c-R \,.\,.\, c+R}\right]_i$ and $y_i = \max \left[{c-R \,.\,.\, c+R}\right]_i$

Then $|y_i - x_i| = \left \vert {c+R-(c-R)} \right \vert = 2R$.

By the definition of an $n$-cube, each interval is of the same length.

Thus:

Proof of Corollary
To minimize the sum in question, we chose each coordinate $y_i$, $x_i$ of $x$ and $y$ to be endpoints.

Thus any $x, y$ so chosen is a vertex, by the definition of vertex.

Certainly $x \ne y$ because were they equal, the distance between them would be zero, and the sum would not be maximal.

The result follows from the definition of a diagonal.