Proof of vector norm formula

Question

I was trying to find something that proves the (Euclidean) vector magnitude formula for $3+$ dimensional vectors. $$\mid x|=\sqrt{x_1^2+x_2^2+\cdots+x_n^2}.$$

It seems easy to prove by induction if we can show that in an $N$ dimensional space any point in $N-1$ dimension, the origin and a point in the $N$^th dimension line forms a right triangle.

This is clearly true up to $3$ dimensions but I don't know how to prove this for $4$ (or higher) dimensions.

How would you prove/justify the Euclidean norm formula?

Answer 1

The first three dimensions are easy to visualize, but after that, it gets to be rather brain-breaking to try, so I wouldn't. Adding new dimensions is somewhat like moving from real to complex numbers--we declare that there is some new structure related to the old one and satisfying certain properties (and may even go so far as to construct it set-theoretically), but we don't really worry about such a structure existing in the "real world." In this case, we simply declare that the $(n+1)$th dimension should be a dimension that is orthogonal to each of the first $n$ dimensions, and add on as many as we like in this way.

Answer 2

The following is a right triangle: $$ \text{vertices}:\quad (\ \underbrace{0,0,0,\ldots,0}_{n\text{ components}}, 0 ),\quad (x_1,x_2,\ldots,x_n, 0),\quad (x_1,\ldots,x_n,x_{n+1}) $$

The vector from the second point above to the third is $(0,0,\ldots,0,x_{n+1})$ and that makes a right angle with the vector $(x_1,\ldots,x_n,0)$.

The induction hypothesis says that in $(n-1)$-dimensional space, the distance from the first point above to the second is $\sqrt{x_1^2+\cdots+x_n^2}$. The length of the vector $(0,\ldots,0,x_{n+1})$ is $|x_{n+1}|$. So you have $$ \sqrt{\sqrt{x_1^2+\cdots+x_n^2}^2 + |x_{n+1}|^2 } = \sqrt{x_1^2+\cdots+x_n^2+x_{n+1}^2}. $$