Proof that the Euclidean norm is indeed a norm I apologize beforehand for this question. Its embarrassing I know. 
Anyway, here we go.
Recall:
$$ \| x \|^2_{\mathbb{R}^2} = \sum^{n}_{i = 1} x^2_{i}$$
How do we prove its a norm?
Well if its a norm it should have the three properties:


*

*Positivity $\| x \| \geq 0, \forall x \in X$ and $\| x \| = 0 $ iff $x = 0$

*Homogeneity $\| cx\| = |c| \| x \|$

*sub-additivity $ \| x + y \| \leq \| x \| + \| y \|$


1) the first is easy because its a sum of squares so its always positive and it can only be zero when x = 0 because thats the only way to make zero from the addition of a bunch of positive things.
2) Stuck in this one (homogeneity):
$$\| c x\|^2 = \sum^{n}_{i = 1} (cx_{i})^2 = \sum^{n}_{i = 1} c^2x_{i}^2 = c^2 \| x \|^2$$
which seems correct algebra but $c^2 \neq |c| $ which makes me doubt my answer. Anyone know whats going on?
3) for the 3rd one I think the way to do it is using "the extended triangle inequality", probably collecting like terms (and maybe some induction, not sure yet) and then your done. Have not tried this because I am very frustrated about not being able to get 2 yet. But I think this plan of attack for 3 should work.
 A: I am grateful for the answer to my question and the comments. However, as my title says, this is a question about Euclidean norm being a norm. So for completeness I will provide the proof for 3 (sub-additivity). My proof goes like this:
$$\| x + y\|^2 = \sum_{i=1}^{n}(x_i+y_i)^2 = \sum_{i=1}^{n}(x_i^2+y_i^2+2x_iy_i) = \sum_{i=1}^{n}x_i^2+ \sum_{i=1}^{n} y_i^2+ 2\sum_{i=1}^{n}x_iy_i $$
Thus:
$$ \| x + y\|^2 =  \| x \|^2 + \| y\|^2 + 2 x \cdot y $$
Now lets use Cauchy-Schwarz Inequality (as suggested by jimbo in the comments), we get:
$$ \| x + y\|^2 =  \| x \|^2 + \| y\|^2 + 2 x \cdot y \leq \| x \|^2 + \| y\|^2 + 2 \| x \| \| y \| = ( \| x \| + \| y \|)^2$$
which is the same as:
$$\| x + y\|^2 \leq ( \| x \| + \| y \|)^2$$
which implies:
$$ \| x + y\| \leq \| x \| + \| y \|$$
as required.
Thnx everyone! :)
A: Let $x=(x_1,x_2,\dots,x_n)\in\mathbb{R}^n$. $cx=(cx_1,cx_2,\dots,cx_n).$
The definition for the euclidean norm is the following:

$||x||=\sqrt{\sum_{i=1}^{n}x_i^2}$

For 2, $||cx||=\sqrt{\sum_{i=1}^{n}(cx_i)^2}=\sqrt{\sum_{i=1}^{n}(c^2x_i^2)}=\sqrt{c^2\sum_{i=1}^{n}x_i^2}=|c|\sqrt{\sum_{i=1}^{n}x_i^2}=c||x||$.
You can consult, for example, this article:
https://bspace.berkeley.edu/access/content/group/2fb5bd3e-8d09-40ee-a371-6cc033d854b9/ho4.pdf
