Get the complete, step-by-step math solution for: "The number of relations on the set A = \{1, 2, 3\} containing at most 6 elements including (1, 2), which are reflexive and transitive but not symmetri...". Powered by SolveForX AI math tutor.
Step 7: Evaluate cases for adding elements
Let's analyze the possible relations:
1. R1={(1,1),(2,2),(3,3),(1,2)}: Reflexive, contains (1,2), not symmetric (since (2,1)∈/R1), transitive. This is a valid relation.
2. If we add (1,3): R2={(1,1),(2,2),(3,3),(1,2),(1,3)}. Reflexive, contains (1,2), not symmetric (since (2,1)∈/R2 and (3,1)∈/R2), transitive. This is valid.
3. If we add (2,3): R3={(1,1),(2,2),(3,3),(1,2),(2,3)}. Reflexive, contains (1,2), not symmetric (since (2,1)∈/R3 and (3,2)∈/R3). Transitivity: (1,2)∈R3 and (2,3)∈R3⟹(1,3)∈R3. So, R3 must contain (1,3) to be transitive. But then it would have 6 elements: {(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}. This is R4. If we only add (2,3) and not (1,3), it's not transitive.
4. If we add (3,1): R5={(1,1),(2,2),(3,3),(1,2),(3,1)}. Transitivity: (3,1)∈R5 and (1,2)∈R5⟹(3,2)∈R5. So R5 must contain (3,2). Also, (1,2)∈R5 and (2,2)∈R5⟹(1,2)∈R5. (3,1)∈R5 and (1,1)∈R5⟹(3,1)∈R5. If (3,1) is added, (3,2) must also be added for transitivity. This would make 6 elements: {(1,1),(2,2),(3,3),(1,2),(3,1),(3,2)}. This relation is not symmetric (since (1,3)∈/R5 and (2,3)∈/R5). This is a valid relation.
5. If we add (3,2): R6={(1,1),(2,2),(3,3),(1,2),(3,2)}. Transitivity: (3,2)∈R6 and (2,2)∈R6⟹(3,2)∈R6. (1,2)∈R6 and (2,2)∈R6⟹(1,2)∈R6. No new elements are forced. This relation is not symmetric (since (2,3)∈/R6). This is a valid relation.
Let's list the valid relations with at most 6 elements:
- RA={(1,1),(2,2),(3,3),(1,2)} (4 elements)
- RB={(1,1),(2,2),(3,3),(1,2),(1,3)} (5 elements)
- RC={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)} (6 elements, forced by transitivity from (1,2) and (2,3))
- RD={(1,1),(2,2),(3,3),(1,2),(3,2)} (5 elements)
- RE={(1,1),(2,2),(3,3),(1,2),(3,1),(3,2)} (6 elements, forced by transitivity from (3,1) and (1,2))
We need to ensure no other elements are forced by transitivity that would make the relation symmetric or exceed 6 elements.
Consider RC: (1,2)∈RC and (2,3)∈RC⟹(1,3)∈RC. All pairs are present. Not symmetric because (2,1)∈/RC, (3,1)∈/RC, (3,2)∈/RC.
Consider RE: (3,1)∈RE and (1,2)∈RE⟹(3,2)∈RE. All pairs are present. Not symmetric because (1,3)∈/RE, (2,1)∈/RE, (2,3)∈/RE.
So there are 5 such relations.