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.

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 $\mathbb R^n.$