Get the complete, step-by-step math solution for: "Let A=\{1{,}2{,}3\}. The number of relations on A, containing (1{,}2) and (2{,}3), which are reflexive and transitive but not symmetric, is:". Powered by SolveForX AI math tutor.
Step 7: Analyze adding pairs and their implications
We systematically consider adding the remaining pairs: (2,1), (3,1), and (3,2).
1. If we add (2,1) to R1, the relation R2={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(2,1)} is reflexive and transitive. It is not symmetric because (1,3) is present but (3,1) is not.
2. If we add (3,1) to R1, for transitivity, we must also add (3,2) (from (3,1) and (1,2)). The relation R3={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(3,1),(3,2)} is reflexive and transitive. It is not symmetric because (1,2) is present but (2,1) is not.
3. If we add (3,2) to R1, the relation R4={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(3,2)} is reflexive and transitive. It is not symmetric because (1,2) is present but (2,1) is not, and (1,3) is present but (3,1) is not.
We can also add (2,1) and (3,2) to R1. This requires adding (3,1) for transitivity. The resulting relation is R5={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(2,1),(3,2),(3,1)}. This relation is reflexive and transitive. It is not symmetric because (1,2) and (2,1) are present, (2,3) and (3,2) are present, (1,3) and (3,1) are present. Wait, this relation IS symmetric. So this case is not valid.
Let's re-evaluate. The pairs that can be added are (2,1), (3,1), (3,2).
Consider the minimal relation R1={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}. This is one such relation.
Now, we can add elements to R1 as long as the properties are maintained and it remains non-symmetric.
Possible relations:
1. R1={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)} (Not symmetric, reflexive, transitive)
2. R1∪{(2,1)}={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(2,1)}. This is reflexive and transitive. It is not symmetric because (1,3) is in it but (3,1) is not. (2nd relation)
3. R1∪{(3,2)}={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(3,2)}. This is reflexive and transitive. It is not symmetric because (1,2) is in it but (2,1) is not. (3rd relation)
4. R1∪{(2,1),(3,2)}={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(2,1),(3,2)}. This is reflexive and transitive. It is not symmetric because (1,3) is in it but (3,1) is not. (4th relation)
What about adding (3,1)? If (3,1) is added, then for transitivity, (3,1) and (1,2)⟹(3,2) must be added. Also (3,1) and (1,3)⟹(3,3) (already present). So if (3,1) is added, (3,2) must also be added.
Let's consider the pairs that would make the relation symmetric: (2,1),(3,2),(3,1).
We need to ensure that at least one of (2,1),(3,2),(3,1) does not have its symmetric counterpart in the relation, or that if it does, another pair exists without its symmetric counterpart.
Let Rbase={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}. This is reflexive and transitive, and not symmetric.
We can add any subset of the remaining pairs (2,1),(3,1),(3,2) as long as transitivity is maintained and the relation is not symmetric.
Let's list all possible reflexive and transitive relations containing (1,2) and (2,3):
1. R1={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)} (Not symmetric)
2. R2=R1∪{(2,1)} (Not symmetric, since (1,3) is in it but (3,1) is not)
3. R3=R1∪{(3,2)} (Not symmetric, since (1,2) is in it but (2,1) is not)
4. R4=R1∪{(2,1),(3,2)} (Not symmetric, since (1,3) is in it but (3,1) is not)
5. R5=R1∪{(3,1),(3,2)} (Adding (3,1) implies adding (3,2) for transitivity with (1,2)). This relation is not symmetric because (1,2) is in it but (2,1) is not.
6. R6=R1∪{(2,1),(3,1),(3,2)} (Adding (3,1) implies adding (3,2) for transitivity with (1,2)). This relation is symmetric because all pairs (a,b) have (b,a) present. For example, (1,2) and (2,1), (2,3) and (3,2), (1,3) and (3,1). So this relation is symmetric and must be excluded.
So, there are 5 such relations.