Proof

The definitions involve implicitly the choices of representatives, so we need to check that the definitions do not depend on these choices.

Suppose a' + I = a + I and b' + I = b + I. Then a' = a + r and b' = b + s for some r, s I. Then both (a' + b') - (a + b) = r + s and a'b' - ab = as + rb + rs clearly belong to I. We conclude that (a' + b') + I = (a + b) + I and a'b' + I = ab + I, so that addition and multiplication are well defined.

It remains to check the definitions of the ring axioms. These are routine checks and are left to the reader.