MATH.2:5.1 - Absolute value fails, parity works for addition

Suppose integers are identified when they have the same absolute value. Then 1 and -1 are identified. Add the same integer 1 to both: the results are 2 and 0, which have different absolute values. This equivalence relation therefore cannot support addition inherited from integer representatives.

For a question about parity, choose a different relation: n~m when n-m is even. Reflexivity, symmetry and transitivity follow from the corresponding facts about differences divisible by 2.

If n~n' and m~m', then (n+m)-(n'+m')=(n-n')+(m-m') is even. Addition therefore preserves the relation. There are two classes:

The class of 7 plus the class of 4 is Odd. This answers the parity question using two classes. It does not determine whether the sum is 11 or another odd integer; a request for the sum itself requires the operands or more information.

Now request multiplication as well. The earlier addition argument does not settle the new operation. Here a separate calculation does:

n*m-n'*m'=(n-n')*m+n'*(m-m').

Both terms on the right are even when the corresponding inputs have the same parity. The quotient can therefore also support multiplication. The changed request led to a new compatibility argument while preserving the earlier addition result.