Get the complete, step-by-step math solution for: "Let A=\{1{,}2{,}3\}. Then show that the number of relations containing (1{,}2) and (2{,}3) which are reflexive and transitive but not symmetric is thr...". Powered by SolveForX AI math tutor.
Step 5: Construct Relations and Check Conditions
Let R0={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}. We need to add elements to R0 such that the relation remains reflexive and transitive, but becomes non-symmetric. The pairs that could potentially be added are (2,1), (3,1), (3,2).
1. R0: Reflexive and transitive. Not symmetric because (1,2)∈R0 but (2,1)∈/R0. This is one such relation.
2. Consider adding (2,1): If (2,1) is added, then (1,2)∈R and (2,1)∈R. Also (2,3)∈R and (3,1)∈R (from R0). Transitivity would require (1,1), (2,2), (3,3) (already there). Also (2,3) and (3,1) implies (2,1) (already added). (1,2) and (2,3) implies (1,3) (already there). (2,1) and (1,3) implies (2,3) (already there). If (2,1) is added, then (1,2) and (2,1) are present, making it symmetric for these pairs. To maintain non-symmetry, we must ensure (3,1) or (3,2) are not added if their symmetric counterparts are not. If we add (2,1), then (1,2) and (2,1) are present. If we don't add (3,1) and (3,2), then (1,3) is present but (3,1) is not, so it's not symmetric. This forms a valid relation.
3. Consider adding (3,1): R1=R0∪{(3,1)}. Reflexive and transitive. Not symmetric because (1,2)∈R1 but (2,1)∈/R1. This is a second relation.
4. Consider adding (3,2): R2=R0∪{(3,2)}. Reflexive and transitive. Not symmetric because (1,2)∈R2 but (2,1)∈/R2. This is a third relation.
5. Consider adding (2,1) and (3,1): R3=R0∪{(2,1),(3,1)}. Transitivity: (3,1)∈R3 and (1,2)∈R3⟹(3,2)∈R3. So we must add (3,2). Then R3=R0∪{(2,1),(3,1),(3,2)}. Now, (1,2) and (2,1) are present. (1,3) is present, (3,1) is present. (2,3) is present, (3,2) is present. This relation is symmetric. So this is not a valid relation.
6. Consider adding (2,1) and (3,2): R4=R0∪{(2,1),(3,2)}. Transitivity: (3,2)∈R4 and (2,1)∈R4⟹(3,1)∈R4. So we must add (3,1). This again leads to a symmetric relation.
7. Consider adding (3,1) and (3,2): R5=R0∪{(3,1),(3,2)}. Reflexive and transitive. Not symmetric because (1,2)∈R5 but (2,1)∈/R5. This is a fourth relation.
Let's re-evaluate the possible relations systematically:
Base relation Rbase={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}.
Possible pairs to add (excluding those that would make it symmetric for (1,2), (2,3), (1,3)): (2,1), (3,1), (3,2).
Case 1: No additional pairs. R1=Rbase.
- Reflexive: Yes.
- Transitive: Yes (checked in step 4).
- Symmetric: No, because (1,2)∈R1 but (2,1)∈/R1. (Valid)
Case 2: Add (2,1). R2=Rbase∪{(2,1)}.
- Reflexive: Yes.
- Transitivity check: (2,1)∈R2 and (1,3)∈R2⟹(2,3)∈R2 (already present). (3,1) is not present, so (3,2) is not forced. (1,2) and (2,1) are present. (1,3) is present, but (3,1) is not. So it is transitive.
- Symmetric: No, because (1,3)∈R2 but (3,1)∈/R2. (Valid)
Case 3: Add (3,1). R3=Rbase∪{(3,1)}.
- Reflexive: Yes.
- Transitivity check: (3,1)∈R3 and (1,2)∈R3⟹(3,2)∈R3. So we must add (3,2).
- This leads to R3′=Rbase∪{(3,1),(3,2)}.
- Symmetric: No, because (1,2)∈R3′ but (2,1)∈/R3′. (Valid)
Case 4: Add (3,2). R4=Rbase∪{(3,2)}.
- Reflexive: Yes.
- Transitivity check: (1,2)∈R4 and (2,3)∈R4⟹(1,3)∈R4. (3,2)∈R4 and (2,3)∈R4 does not force anything new. (1,3) is present, but (3,1) is not. (2,3) is present, but (3,2) is present. (1,2) is present, but (2,1) is not. So it is transitive.
- Symmetric: No, because (1,2)∈R4 but (2,1)∈/R4. (Valid)
Let's list the relations that satisfy the conditions:
1. RA={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}
- Reflexive: Yes
- Transitive: Yes
- Symmetric: No (e.g., (1,2)∈RA but (2,1)∈/RA)
2. RB={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(2,1)}
- Reflexive: Yes
- Transitive: Yes (e.g., (2,1)∈RB and (1,3)∈RB⟹(2,3)∈RB, which is present)
- Symmetric: No (e.g., (1,3)∈RB but (3,1)∈/RB)
3. RC={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(3,1),(3,2)}
- Reflexive: Yes
- Transitive: Yes (e.g., (3,1)∈RC and (1,2)∈RC⟹(3,2)∈RC, which is present)
- Symmetric: No (e.g., (1,2)∈RC but (2,1)∈/RC)
These are the three relations that satisfy all conditions.