Let $V$ be a vector space and $x, w$ in $V$. Prove there is a unique vector $v$ such that: $x+v = w$.
I planned on using a proof by contradiction, that is to prove For All v in V such that x+v =w, then get a contradiction, which would then prove There exists a $v$ in $V$, such that $x+v = w$. Is this logic correct, or am I misusing quantifiers?
Well, why would you use contradiction here?
You have to show existence and uniqueness.
Let's start with existence. A good candidate vector for $v$ is of course $w - x$.
Since $w,x \in V$ and $V$ is a vector space, it holds that $w + (-x) = w - x \in V$
Now, since $x + (w-x) = w$, we proved that there is a vector $v$ such that $x + v = w$ (take $v = w-x)$
Now, let's prove uniqueness.
Suppose $x + v_1 = w$ and $x+v_2 = w$, then $x + v_1 = x+ v_2$. Add $-x$ to both sides to obtain that $v_1 = v_2 \quad \square$
Why not use contradiction? Then, you must assume that such a vector does not exist (to show existence, not uniqueness). Then you must somehow get a contradiction. The only way to reach such a contradiction, from what I can see, is to show that the vector does exist, which defeats the purpose of proof by contradiction.