what makes quaternion representation of rotations unique? So i'm curious about what actually makes quaternions unique.
Quaternions are 4 dimensional and multiplication is non commutative. But what part makes the solution unique? Is it because multiplication is non commutative, is it because the space is over defined, or is it because of both?
Edit 1: 
By the solution i meant a unique way to get a 3D orientation, which isn't the case for euler angles. 
 A: 
But i meant a unique solution, like when you rotate from a to b in 3 dimensions euler angles can have multiple solutions for going from a to b, while quaternions only has one 

Well... this is not true, strictly speaking. For any rigid rotation in $3$ space, infinitely many different quaternions can affect that transformation via $x\mapsto qxq^{-1}$ Just note that $qxq^{-1}=(\lambda q)x(q\lambda)^{-1}$ for any nonzero real number $\lambda$.
Then you say "ah, well I actually meant that $q$ should be a unit quaternion."  That's all well and good, it certainly narrows the field. But there are still two solutions: $q$ and $-q$.
Finally, you could further require that the scalar part of $q$ be positive. And then yes, you would finally have a unique solution most of the time.
But really, the same is true for Euler angles: after you take the time to prescribe the order in which the three transformations have to be applied, and put appropriate bounds on angles, you will have a unique solution most of the time. 
I suppose there is in a sense "less unicity" with Euler angles because they suffer the phenomenon of gimbal lock, but hopefully you get my idea that both systems of expressing rotations need constraints.
I don't understand gimbal lock well enough to explain it, but from what I've learned so far it is a topological constraint. Euler angles essentially seek a special map out of $\mathbb R^3$ into $SO(3)$, and the standard representation of rotations using unit quaternions uses instead a map out of the $3$-sphere in $\mathbb R^4$ into $SO(3)$.  Since the two domains are hooked up differently, one expresses rotations better than the other.

Is it because multiplication is non commutative, is it because the space is over defined, or is it because of both?

Remember, quaternions and Euler angles are just representing rotations. The rotations of $3$ space are going to be noncommutative no matter what, so any system representing them is going to have to be noncommutative, too. 
But your second guess that it has something to do with the space they are defined on ($\mathbb R^3$ vs $\mathbb R^4$) does hit the nail on the head, although you probably did not realize in what way it affected the representation. You can indeed say that the sets that each representation live inside are indeed responsible for the differences in each representation.
