Question:
Consider the following two binary relations on the set \(A = \{a, b, c\}\) :
\(R_1 = \{(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)\}\) and
\(R_2 = \{(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)\}\).
Then :
Consider the following two binary relations on the set \(A = \{a, b, c\}\) :
\(R_1 = \{(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)\}\) and
\(R_2 = \{(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)\}\).
Then :
\(R_1 = \{(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)\}\) and
\(R_2 = \{(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)\}\).
Then :
Show Hint
Check symmetry by looking for reversed pairs and transitivity by looking for a chain (x,y),(y,z) whose (x,z) is missing.
Updated On: Nov 11, 2026
- both \(R_1\) and \(R_2\) are not symmetric.
- \(R_1\) is not symmetric but it is transitive.
- \(R_2\) is symmetric but it is not transitive.
- both \(R_1\) and \(R_2\) are transitive.
Show Solution
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The Correct Option is C
Solution and Explanation
R1: (b, c) is in R1 but (c, b) is not, so R1 is not symmetric. Also (b, c) and (c, a) are in R1 but (b, a) is not, so R1 is not transitive.
R2: The reverse of every pair is present: (a,b)-(b,a), (c,a)-(a,c), and (a,a), (b,b), (c,c) are self-symmetric. So R2 is symmetric. But (c, a) and (a, b) are in R2 while (c, b) is not, so R2 is not transitive.
Hence R2 is symmetric but not transitive, option (3).
Answer: (3) \(R_2\) is symmetric but it is not transitive.
R2: The reverse of every pair is present: (a,b)-(b,a), (c,a)-(a,c), and (a,a), (b,b), (c,c) are self-symmetric. So R2 is symmetric. But (c, a) and (a, b) are in R2 while (c, b) is not, so R2 is not transitive.
Hence R2 is symmetric but not transitive, option (3).
Answer: (3) \(R_2\) is symmetric but it is not transitive.
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