How is the reflexive property demonstrated for the relation a mod $3 = b$ mod 3?

The naive condition $r_1s_2 = s_1r_2$ is insufficient because in a general commutative ring, elements can be annihilated by nonzero elements of $S$. The refined condition requires the existence of some $t \in S$ such that $t(r_1s_2 - s_1r_2) = 0$, which accounts for zero-divisor phenomena and ensures the equivalence relation is well-defined.

Conditions: $R$ is a general commutative ring (not necessarily an integral domain).; $S \subset R$ is multiplicatively closed.

No, the numerical demonstration using specific numbers like 10, 7, and 13 is merely an illustrative example, not a rigorous formal proof applicable to all integers.

Conditions: The demonstration uses specific numerical values.; A formal proof would require generalizing the property for all integers.

Transitivity is shown by demonstrating that if the first number has the same remainder modulo 3 as the second, and the second has the same remainder as the third, then the first has the same remainder as the third.

Conditions: The relation is equality of remainders modulo 3.; The property requires that if x is related to y and y is related to z, then x is related to z.

The symmetric property is demonstrated by showing that if one number has the same remainder modulo 3 as another, then the second number also has the same remainder modulo 3 as the first.

Conditions: The relation is equality of remainders modulo 3.; The property requires that if x is related to y, then y is related to x.

Ordinary equality on numbers is an equivalence relation because it satisfies the three required properties: reflexivity, symmetry, and transitivity.

Conditions: Equality is viewed as a relation on numbers.; The relation must be checked against the three properties of an equivalence relation.