# I.14.1 Small potential typo

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.

Hey @Kamal_Mohamed! You are correct! There’s a typo there. I will fix it!