In the solution for I.14.1 (in the repo, it is found in `I.14.2/challengeDescription.md`

):

There may be some variation as well; for example, to create

\frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)

the Z can be applied to either qubit to obtain the same result.

I believe the + sign should be a -. I tried to reason about it, and I wasn’t able to see how we can reach |+> by applying a Z to either qubit.