The alphabet consists of just two characters, $0$ and $1$. How do I go about proving that it's undecidable?
I was thinking of reducing the general case to binary form meaning if the alphabet has more than two characters encode them into binary. I don't really know how to go about proving it though.
Any help would be appreciated. Thanks
The standard proof of undecidability of PCP (see e.g. Hopcroft, Motwani, Ullman's "Introduction to Automata Theory, Languages, and Computation" section 9.4 in the third edition, or any other standard text in the area) is to set the pairs up so one of the branches is the starting configuration of a TM followed by some marker, while the other is just the marker. There are pairs which just reproduce the symbols of the first on the second (so they match), with special pairs that then give the effect of the move of the TM (second string gets the "before move" configuration around the state to match the same position on the first, first one gets "after move" configuration). If the TM accepts, another bunch of pairs takes over to level off both strings. The proof isn't hard, just very long-winded (there are many cases to consider, and make sure the construction can't derail).
Doing it for a binary alphabet is certainly possible (can code any alphabet you wish and states of the TM in binary), but doing it directly is just too much hassle.