Proving |sv| = |s||v| and is it true for all dimensions I've been sitting here quite a while trying to figure out how to prove 
$$ |s\vec{\mathbf{v}}| = |s||\vec{\mathbf{v}}| $$
I was not told in what dimension to prove this is, so I am assuming a 3D plane. I think I have a solution but I'm not sure if my method is write. Could someone confirm I am doing it right?
So I started by setting $$\vec{\mathbf{v}} = (x, y, z)$$
$$ |\vec{\mathbf{v}}| = \sqrt{x^2 + y^2 + z^2} $$
(x, y, z) being points in a 3D graph.
Then 
$$ |s\vec{\mathbf{v}}| = \sqrt{|s|^2x^2 + |s|^2y^2 + |s|^2z^2} $$
$$ |s\vec{\mathbf{v}}| = \sqrt{|s|^2(x^2 + y^2 + z^2)} $$
$$ |s\vec{\mathbf{v}}| = \sqrt{|s|^2}  \sqrt{x^2 + y^2 + z^2} $$
$$ |s\vec{\mathbf{v}}| = |s| \sqrt{x^2 + y^2 + z^2} $$
Since
$$ |\vec{\mathbf{v}}| =  \sqrt{x^2 + y^2 + z^2} $$
Therefore
$$ |s||\vec{\mathbf{v}}| =  |s|\sqrt{x^2 + y^2 + z^2} $$
Hence
$$ |s\vec{\mathbf{v}}| = |s||\vec{\mathbf{v}}| $$
My result does get me what I was looking for but it has happened in the past that for some reason a mistake that I made got me the right answer. Also if this is true for a 3D plane, does that mean that it holds through for all dimensions?
Thanks for any and all help
 A: Your proof looks okay except that I would make the following change
$$ |s\vec{\mathbf{v}}| = \sqrt{(sx)^2 + (sy)^2 + (sz)^2} $$
$$ |s\vec{\mathbf{v}}| = \sqrt{s^2x^2 + s^2y^2 + s^2z^2} $$
$$ |s\vec{\mathbf{v}}| = \sqrt{s^2(x^2 + y^2 + z^2)} $$
$$ |s\vec{\mathbf{v}}| = \sqrt{s^2}  \sqrt{x^2 + y^2 + z^2} $$
$$ |s\vec{\mathbf{v}}| = |s| \sqrt{x^2 + y^2 + z^2} $$
Remembering that $\sqrt{s^2} = |s|$. By placing the $|s|$ within the square root in your solution, you have already to some degree presupposed that $|s\vec{\mathbf{v}}| = |s| |\vec{\mathbf{v}}|$.
The proof extends without a problem to more dimensions as long as you use a more generic formulation of the Euclidean norm:
$$|\vec{\mathbf{v}}| := \sqrt{v_1^2 + \cdots + v_n^2}$$
A: Let $\|.\|$ denote the norm induced by a scalar product $\langle.,.\rangle$, thus by definition $\|v\|:=\sqrt{\langle v,v\rangle}$. Then 
$$\|sv\|^2=\langle sv,sv\rangle=s^2\langle v,v\rangle=s^2\|v\|^2\Rightarrow\|sv\|=|s|\cdot\|v\|.$$ 
In any dimension, for any scalar product.
