MATH.16:5.1 - Two classifications, then one shared integer

Two separately chosen integers have been classified. The first report gives its remainder modulo 2; the second gives its remainder modulo 4. The task is to combine the reports so that both can be recovered, and to add combined reports by adding their corresponding remainders. The two reports should determine the complete combined result.

Let A={0,1}, with addition reduced modulo 2, and B={0,1,2,3}, with addition reduced modulo 4. Adding the two reported numbers into one number loses recovery: reports (0,1) and (1,0) both give 1. Keeping a pair supplies both projections and requires no additional choice. All eight pairs are possible because the original integers may be chosen separately.

Now both reports must describe the same integer. The pair (0,1) fails: an integer with remainder 1 modulo 4 is odd. The modulo-4 report determines parity through t(b)=b modulo 2. Set s to the identity on A and construct the compatible pairs:

Q={(0,0),(1,1),(0,2),(1,3)}.

Every member comes from an integer. Addition stays within Q: (1,1)+(1,3)=(0,0). A report determines the original integer only modulo 4.

Additional structure – groups. With the stated modular additions, A and B are groups. The identity on A and the parity map t preserve addition, so Q with componentwise addition is also their pullback in groups.

The projection (a,b) -> b has inverse b -> (t(b),b). If the next calculation is easier with one modulo-4 value, MATH.7 transports it through these maps. If the next question is which integers may be identified while retaining both reports, MATH.2 instead constructs their quotient modulo 4.

Returning to functions on the underlying sets, a proposed update exposes another choice. Incrementing the modulo-4 component alone sends (0,0) to (0,1), outside Q. Incrementing both components gives (1,1) and preserves agreement for every pair in Q. The intended change to the underlying integer determines which component updates belong together.

Additional structure – groups. The joint increment is not a homomorphism: it takes the identity (0,0) to (1,1). It is suitable for updating the represented integer, but fails a requirement to preserve the group operation.