To determine whether the relation \( R \) defined on the set \( A = \{1, 2, 3, 4, 5, 6\} \) as \( R = \{(a, b) : b = a + 1\} \) is reflexive, symmetric, or transitive, we will analyze each property step by step.
### Step 1: Define the Relation
The relation \( R \) consists of pairs \( (a, b) \) where \( b = a + 1 \). We can list the pairs based on the elements of set \( A \):
- For \( a = 1 \), \( b = 1 + 1 = 2 \) → pair \( (1, 2) \)
- For \( a = 2 \), \( b = 2 + 1 = 3 \) → pair \( (2, 3) \)
- For \( a = 3 \), \( b = 3 + 1 = 4 \) → pair \( (3, 4) \)
- For \( a = 4 \), \( b = 4 + 1 = 5 \) → pair \( (4, 5) \)
- For \( a = 5 \), \( b = 5 + 1 = 6 \) → pair \( (5, 6) \)
- For \( a = 6 \), \( b = 6 + 1 = 7 \) → pair \( (6, 7) \) (Note: 7 is not in set \( A \))
Thus, the relation \( R \) can be expressed as:
\[
R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)\}
\]
### Step 2: Check for Reflexivity
A relation is reflexive if every element \( a \in A \) is related to itself, i.e., \( (a, a) \in R \) for all \( a \).
- Check for each element:
- \( (1, 1) \notin R \)
- \( (2, 2) \notin R \)
- \( (3, 3) \notin R \)
- \( (4, 4) \notin R \)
- \( (5, 5) \notin R \)
- \( (6, 6) \notin R \)
Since none of the pairs \( (a, a) \) are present in \( R \), the relation is **not reflexive**.
### Step 3: Check for Symmetry
A relation is symmetric if for every \( (a, b) \in R \), the pair \( (b, a) \) is also in \( R \).
- Check the pairs:
- For \( (1, 2) \), \( (2, 1) \notin R \)
- For \( (2, 3) \), \( (3, 2) \notin R \)
- For \( (3, 4) \), \( (4, 3) \notin R \)
- For \( (4, 5) \), \( (5, 4) \notin R \)
- For \( (5, 6) \), \( (6, 5) \notin R \)
- For \( (6, 7) \), \( (7, 6) \notin R \)
Since for every \( (a, b) \in R \), \( (b, a) \) is not in \( R \), the relation is **not symmetric**.
### Step 4: Check for Transitivity
A relation is transitive if whenever \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \) must also be in \( R \).
- Check the pairs:
- From \( (1, 2) \) and \( (2, 3) \), we need \( (1, 3) \) → \( (1, 3) \notin R \)
- From \( (2, 3) \) and \( (3, 4) \), we need \( (2, 4) \) → \( (2, 4) \notin R \)
- From \( (3, 4) \) and \( (4, 5) \), we need \( (3, 5) \) → \( (3, 5) \notin R \)
- From \( (4, 5) \) and \( (5, 6) \), we need \( (4, 6) \) → \( (4, 6) \notin R \)
- From \( (5, 6) \) and \( (6, 7) \), we need \( (5, 7) \) → \( (5, 7) \notin R \)
Since there are pairs \( (a, b) \) and \( (b, c) \) such that \( (a, c) \) is not in \( R \), the relation is **not transitive**.
### Conclusion
The relation \( R \) is neither reflexive, nor symmetric, nor transitive.
