# Proof: Show there is set of $n+1$ points in $\mathbb{R}^n$ such that distance between any two distinct points is $1$?

Argh, I hate to ask a question again so soon, especially one I feel like I should know.
Linear algebra is taking its toll, and I am not quite used to the theory side of mathematics.

Anyways, I want/need (more of a want) to show that there is a set of $n+1$ points in $\mathbb{R}^n$ such that the distance between any two distinct points is $1$.

I am also interested in proving that a set of $n+2$ points does not exist, but I imagine that should not be difficult to do after.

Thank you in advance.

PS. I really do apologize if this is a duplicate, I spent a lot of time trying to figure this out to no avail.

#### Solutions Collecting From Web of "Proof: Show there is set of $n+1$ points in $\mathbb{R}^n$ such that distance between any two distinct points is $1$?"

This approach is most reminiscent of your last question. In $\mathbb R^{n+1},$ the standard unit axis vectors such as $(1,0,0,…,0)$ are all $\sqrt 2$ apart. So, multiply by $(n+1),$ all the vectors $(n+1,0,0,…0)$ are $(n+1) \sqrt 2$ apart. These lie in the hyperplane where the sum of the coordinates is $(n+1).$ This hyperplane can be translated at then rotated so that it coincides with $\mathbb R^n,$ then shrunk by a common scalar factor so the distances are $1.$

Let me do the translation anyway: subtract $1$ from all coordinates, so the $n+1$ entries of the first vector are $(n,-1,-1,…,-1)$ and the sum is now $0.$ A normal vector to this hyperplane is $(1,1,1,…,1).$ If you can figure out how to rotate that into $(0,0,0,…,\sqrt {n+1})$ you will hhave successfully rotated all the points into the plane $x_{n+1}=0,$ which is $\mathbb R^n.$

Hint: For the first part, you can find an explicit example for any $n$. First put $n$ standard-basis unit vectors, and note the pairwise distance between any two unit vectors is the same (you will need to normalize the vectors so that the distance is $1$). Now, carefully choose the $(n+1)$th (left to you as exercise) vector.

We’ll use induction and I’ll extend the claim slightly. Specifically, what we want to show is as follows.

Show there exists a set of $n+1$ points, $x_1, x_2, \dots, x_n, x_{n+1}$ in $R^n$ such (a) $|x_i – x_j| = 1$ for all $1 \leq i <j \leq n+1$ , and (b) there exists a center point $z_n$ such that $d_n=|z_n – x_i| = |z_n – x_j|<1/\sqrt{2}$.

Basis step: It is obvious for $n=1$ that the points $x_1=0$ and $x_2 = 1$ satisfy this with $z_1 = 1/2<1/\sqrt{2}$.

Inductive step: Suppose we have the desired claim for $n$, that is the points $x_i$ for $i=1, \dots, n+1$ and $z_n$ with the desired properties. Let $e_{i}$ denote the basis vectors in $R^n$.

If we reuse the first $n+1$ points all we need to do is define a point $x_{n+2}$ such that $|x_{n+2}-x_i| = 1$ for all $0<i\leq n+1$ and a point $z_{n+1}$ such that $|z_{n+1}-x_i| = |z_{n+1}-x_j|$ for all $0<i <j\leq n+2$.

Take $x_{n+2} = z_c + h_n e_{n+1}$ where $h_n = \sqrt{1-d_n^2}$, $d_n = |z_n-x_1|$.

Also take $z_{n+1} = z_n + t_n e_{n+1}$ where $t_n = \frac{1-2d_n^2}{\sqrt{1-d_n^2}}$.

Check (a):
For $i \leq n+1$, $|x_{n+2}-x_i|^2 = |z_n – x_i|^2 + h_n^2 = d_n^2 +h_n^2 = 1$, by definition of $h_n$.

Check (b):
For $i \leq n + 1$, $d_{n+1}^2 = |z_{n+1}-x_i|^2 = d_n^2 + t_n^2$.

For $i = n+2$, $d_{n+1}^2 =$

You can actually do better: there are always $n+1$ equidistant points on an $n$-sphere. Your question is weaker because as it stands, it does not require those vectors to have the same norm.