Partial problem set 10 - 1 2 3 4 5 6 A A A A A A Reflexive let a a a = 0 Since 0 is an integer aRa Symmetric let a,b and aRb This means that a b is an

# Partial problem set 10 - 1 2 3 4 5 6 A A A A A A Reflexive...

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1. A 2. A 3. A 4. A 5. A 6. A) Reflexive: let a ∈ ℝ. a - a = 0. Since 0 is an integer, aRa. Symmetric: let a,b ∈ ℝ and aRb. This means that a - b is an integer. since b - a is equal to -1 times (a - b), b - a must be an integer as well, so bRa and aRb. Transitive: let a,b,c ∈ ℝ. This means that a - b is an integer and b - c is an integer. The sum of those two expressions is a - c, which must also be an integer, so aRc. Since R is reflexive, symmetric, and transitive, R is an equivalence relation on ℝ. B) All integers are in [1]. C) All odd multiples of ½ are in [1/2]. 7. A) 3 equivalence relations on {1,2,3} 1. All 3 in the same equivalence class 2. 1, 2, and 3 by themselves in their own classes 3. Two of the elements together in one class and the other by itself B) 5 equivalence relations on {1,2,3,4} 1. All 4 elements in the same equivalence class 2. Each element in its own class 3. Three elements in the same class, one on its own 4. One pair in one class, one pair in another 5. One pair in one class, and the other two in their own separate classes.

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• Fall '08
• Knight,J
• Binary relation, Transitive relation, Symmetric relation, Total order, aRb, total order relation

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