Griffiths Quantum Mechanics 2nd Edition: Problems 3.13

This will be a pretty short proof. We must demonstrate that two noncommuting operators cannot share a complete set of common eigenfunctions. Basically, this means that we have to prove that for any two operators, $\hat{P}$ and $\hat{Q}$ that share all of their eigenfunctions, then we must have $[\hat{P}, \ \hat{Q}]f \ = \ 0$. Despite seeming semi-intimidating, this is actually a pretty easy proof. For any $f$, we can say that it is an eigenfunction of both $\hat{P}$ and $\hat{Q}$. This means that we have:

$$\hat{Q}f \ = \ qf \ \text{and} \ \hat{P}f \ = \ pf$$
Where $p$ and $q$ are eigenvalues of the operators. We can then say that:

$$[\hat{P}, \ \hat{Q}]f \ = \ \hat{P}\hat{Q}f \ - \ \hat{Q}\hat{P}f \ = \ \hat{P}qf \ - \ \hat{Q}pf \ = \ q\hat{P}f \ - \ p\hat{Q}f \ = \ qp \ - \ pq \ = \ 0$$
And that's all there is to it (I think).