Test the three properties on R={(1,1)} over A={1,2,3}: reflexivity fails (missing (2,2),(3,3)), while symmetry and transitivity hold vacuously/trivially.
Let A={1,2,3} and R={(1,1)}.
Step 1 — Test REFLEXIVITY.
Definition: R is reflexive if (a,a)∈R for every a∈A.
Here A has three elements, so reflexivity demands all three of:
(1,1)∈R✓,(2,2)∈R?,(3,3)∈R?
But R contains only (1,1). Both (2,2) and (3,3) are missing.
⇒R is NOT reflexive.
This single fact eliminates options (A) and (B), since both assert reflexivity.
Step 2 — Test SYMMETRY.
Definition: R is symmetric if, whenever (a,b)∈R, we also have (b,a)∈R.
We must check this for every pair actually in R. There is only one such pair: (1,1).
- Take (a,b)=(1,1). Its reverse is (b,a)=(1,1). Is (1,1)∈R? Yes. ✓
Every pair in R satisfies the requirement, so
⇒R IS symmetric.
(A useful principle: any relation consisting only of pairs of the form (a,a) is automatically symmetric, because such a pair is its own reverse.)
Step 3 — Test TRANSITIVITY.
Definition: R is transitive if, whenever (a,b)∈R and (b,c)∈R, we also have (a,c)∈R.
Look for all chains inside R. The only element is (1,1), so the only possible chain is:
- (a,b)=(1,1) and (b,c)=(1,1) ⇒ we need (a,c)=(1,1)∈R. It is. ✓
There is no other combination to check, so the condition holds throughout:
⇒R IS transitive. …