True/false: Let $v,w \in \mathbb{R}^2$. If $v \perp w$, then we have that $\left \| v+w \right \|= \left \| v \right \|+\left \| w \right \|$ 
True/false: Let $v,w \in \mathbb{R}^2$. If $v \perp w$, then we have
  that $\left \| v+w \right \|= \left \| v \right \|+\left \| w \right
\|$

I think actually it doesn't work because we have the triangle inequality saying
$$\left \| v+w \right \| \leq \left \| v \right \|+\left \| w \right \|$$
But because $v$ and $w$ are orthogonal to each other, we have that
$$\left \langle v,w \right \rangle=0$$
So the statement should be true because we have that $0=0$?
 A: $$\left \| v+w \right \|^2=(v+w)\cdot(v+w)=\left \| v\right \|^2+\left \| w \right \|^2+2v\cdot w$$
Once $v\perp w$ then $v\cdot w=0$ and then we have
$$\left \| v+w \right \|^2=\left \| v\right \|^2+\left \| w \right \|^2\le (\left \| v\right \|+\left \| w \right \|)^2$$
A: If $v\perp w$, then you easily see that
$$
\|v\|^2+\|w\|^2=\|v+w\|^2
$$
(Pythagoras’ theorem). If also $\|v\|+\|w\|=\|v+w\|$ holds, we get
$$
\|v\|^2+\|w\|^2=\|v+w\|^2=\|v\|^2+2\|v\|\,\|w\|+\|w\|^2
$$
hence
$$
\|v\|\,\|w\|=0
$$
So the relation only holds when one of the vectors is $0$.
A: If it's true, you would have prove it for all $v,w \in \mathbb{R}^2$ such that $v \perp w$. 

But to prove it's false, it suffices to show a single example for which the hypothesis is true but the conclusion is false.

The point here is that the variables $v,w$ are unquantified, so the default quantifier is "for all".

In other words, the statement 
$\qquad$"Let $v,w \in \mathbb{R}^2$. If $v \perp w$, then we have
that $\left \| v+w \right \|= \left \| v \right \|+\left \| w \right \|$."
has the same truth value as the statement
$\qquad$"For all $v,w \in \mathbb{R}^2$, if $v \perp w$, then $\left \| v+w \right \|= \left \| v \right \|+\left \| w \right \|$."

So it makes sense to test the claim against an example, to see if it has any chance of being true.

For a simple test case, let $v = (1,0)$ and $w = (0,1)$.

Then $v,w \in \mathbb{R}^2$ and $v \perp w$, so the hypothesis is satisfied.

But $v+w = (1,1)$, so we have
$$\left \| v \right \|=1$$
$$\left \| w \right \|=1$$
$$\left \| v+w \right \|=\sqrt{2}$$
and hence
$$\left \| v+w \right \| < \left \| v \right \| + \left \| w \right \|$$
Thus, for this example, the conclusion fails.
It follows that the given statement is false (since it's not always true).
