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.

As $\mathbf 0 = \tuple {0, \ldots, 0}$, it follows that:


 * $\norm {\mathbf 0} = \sqrt {0^2 + \ldots + 0^2} = 0$

Suppose instead that $\norm {\mathbf v} = 0$ for some $\mathbf v = \tuple {v_1, \ldots, v_n}$ with $v_1, \ldots, v_n \in \R$.

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


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

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

Hence $\mathbf v = \mathbf 0$.

Thus holds

Let $\mathbf v = \tuple {v_1, \ldots, v_n} \in \R^n$ with $v_1, \ldots, v_n \in \R$.

Let $r \in \R$.

Then:

Thus holds.

Follows from Triangle Inequality for Vectors in Euclidean Space.

Thus holds.