Question:
Given sets $ A $ (odd positive integers ≤ 20) and $ B $ (positive integers ≤ 30 divisible by 5), and relations:
Detailed Explanation of the Solution 📝
Let’s break down the reasoning for both statements about the relations $ R_1 $ and $ R_2 $ step by step!
1. Understanding the Sets
Set $ A $: Odd positive integers ≤ 20
$ A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} $
Set $ B $: Positive integers ≤ 30 divisible by 5
$ B = {5, 10, 15, 20, 25, 30} $
2. Relation $ R_1 $: $ a $ is a factor of $ b $ ($ a \mid b $)
Transitivity Check
Definition: A relation $ R $ is transitive if whenever $ (a, b) \in R $ and $ (b, c) \in R $, then $ (a, c) \in R $.
Here: $ (a, b) \in R_1 $ means $ a \mid b $ (i.e., $ b $ is divisible by $ a $).
Transitive Property for Divisibility:
If $ a \mid b $ and $ b \mid c $, then $ a \mid c $.
Example: If $ 5 \mid 15 $ and $ 15 \mid 30 $, then $ 5 \mid 30 $.
Conclusion:
Since divisibility is always transitive, $ R_1 $ is transitive! ✅
3. Relation $ R_2 $: $ (a + b) \mod 15 = 0 $
Symmetry Check
Definition: A relation $ R $ is symmetric if whenever $ (a, b) \in R $, then $ (b, a) \in R $.
Here: $ (a, b) \in R_2 $ means $ (a + b) $ is a multiple of 15.
Test for Symmetry:
$ (a, b) \in R_2 \implies (a + b) \mod 15 = 0 $.
$ (b, a) \in R_2 $ if $ (b + a) \mod 15 = 0 $.
Since addition is commutative, $ a + b = b + a $, so at first glance, it seems symmetric.
BUT: The important detail is that $ a \in A $, $ b \in B $, and the relation is only defined for pairs where the first element is from $ A $ and the second from $ B $.
If $ (a, b) \in R_2 $, $ a \in A, b \in B $, but $ (b, a) $ may not be in $ R_2 $ because $ b $ may not be in $ A $ and $ a $ may not be in $ B $.
Example:
Let $ a = 5 \in A $, $ b = 10 \in B $.
$ (5, 10) \in R_2 $ if $ 5 + 10 = 15 $, which is divisible by 15.
But $ (10, 5) $ requires $ 10 \in A $ and $ 5 \in B $. But $ 10 \notin A $ (since $ A $ only has odd numbers), so $ (10, 5) $ is not even a valid pair in $ R_2 $.
Conclusion:
$ R_2 $ is not symmetric because the relation is not defined for all possible swaps of $ a $ and $ b $ between the sets. ✅
Summary Table
Statement
True/False
Reason
$ R_1 $ is transitive
✅ True
Divisibility is always transitive.
$ R_2 $ is not symmetric
✅ True
Swapping $ a $ and $ b $ may produce an invalid pair due to set membership.
Key Takeaways
Transitivity: For divisibility relations, if $ a \mid b $ and $ b \mid c $, then $ a \mid c $.
Symmetry: For relations with different sets, swapping elements may not always result in a valid pair, so symmetry can fail even if the operation itself (like addition) is symmetric.
