Testing whether a relation on ordered pairs is an equivalence relation

Expert intro: This relation problem depends on parity. The core question is whether the expression $3ad-7bc$ being even is preserved under reflexivity, symmetry, and transitivity.

Detailed walkthrough

Understand the relation

We have a relation $R$ on $A \times B$ defined by

$(a,b)R(c,d) \iff 3ad-7bc \text{ is an even integer}.$

Because only parity matters, we reduce the expression modulo 2. Since $3 \equiv 1 \pmod{2}$ and $7 \equiv 1 \pmod{2}$, the condition becomes

$ad-bc \equiv 0 \pmod{2}.$

So $(a,b)$ is related to $(c,d)$ exactly when $ad$ and $bc$ have the same parity.

Check the basic properties

Reflexive?

For reflexivity we need $(a,b)R(a,b)$ for every $(a,b)\in A\times B$. That requires

$3ab-7ba= -4ab,$

which is always even. So the relation is reflexive.

Symmetric?

If $(a,b)R(c,d)$, then $3ad-7bc$ is even. Swapping the pairs gives

$3cb-7da.$

Modulo 2, this is the same parity condition because both coefficients are odd. Since evenness is unchanged under swapping the two products in this parity test, the relation is symmetric.

Transitive?

Transitivity is not automatic. We need to see whether $(a,b)R(c,d)$ and $(c,d)R(e,f)$ always imply $(a,b)R(e,f)$. The parity condition only says $ad\equiv bc$ and $cf\equiv de \pmod 2$, which does not force the third pairing to have the needed parity. A counterexample can be constructed by choosing values with differing parities in the second coordinate.

For example, the set $B$ contains both odd numbers and even numbers, so the parity pattern is not fixed across all pairs. This makes transitivity fail.

Conclusion

The relation is reflexive and symmetric but not transitive, so the correct choice is C.

Why this is the right classification

The key step is reducing the definition to parity. Once the condition is interpreted modulo 2, reflexivity and symmetry are straightforward, but transitivity depends on a stronger compatibility that is not guaranteed here.

💡 Pitfall guide

A common mistake is to inspect only the form of the expression and assume that any relation involving an evenness condition must be an equivalence relation. That is not true. You still have to test transitivity separately. Another error is to forget that odd coefficients like 3 and 7 are both congruent to 1 modulo 2, so the expression simplifies much more than it first appears. If you do not reduce the condition modulo 2, the relation looks more complicated than it really is.

🔄 Real-world variant

If the relation were changed to $(a,b)R(c,d)$ iff $3ad-7bc$ is divisible by 4, the analysis would be different because parity alone would no longer be enough. If the set $B$ contained only odd numbers, then the transitivity behavior could also change because the parity pattern of every pair would be more uniform. A useful variant question is to ask whether the same relation on a smaller set of pairs becomes an equivalence relation after restricting to all-even or all-odd coordinates.

🔍 Related terms

equivalence relation, parity arguments, relations on Cartesian products