Euclidean Space is Normed Vector Space

Theorem
The Euclidean norm is a norm on the Euclidean space $\R^n$.

Proof
We prove the norm axioms.

Positive definiteness
As $\mathbf 0 = \left({0, \ldots, 0}\right)$, it follows that:


 * $\left\Vert{ \mathbf 0 }\right\Vert = \sqrt{ 0^2 + \ldots + 0^2 } = 0$

Suppose instead that $\left\Vert{\mathbf v}\right\Vert = 0$ for some $\mathbf v = \left({v_1, \ldots, v_n}\right)$ with $v_1, \ldots, v_n \in \R$.

As $\displaystyle \sqrt{\sum_{i \mathop = 1}^n v_n^2 } = 0$, it follows from squaring both sides that:


 * $\displaystyle \sum_{i \mathop = 1}^n v_n^2 = 0$

From Even Power is Non-Negative, it follows that for all $i$: $v_i = 0$.

Hence $\mathbf v = \mathbf 0$.

Triangle Inequality
Follows from Triangle Inequality: Vectors in Euclidean Space.