Math Problem Statement

R1 is transitive?

Solution

To determine if the relation R1R_1 is transitive, we need to check if whenever (a,b)R1(a, b) \in R_1 and (b,c)R1(b, c) \in R_1, it implies that (a,c)R1(a, c) \in R_1.

Given:

  • AA is the set of even positive integers less than or equal to 20.
  • BB is the set of positive integers less than 20 which are divisible by 6.
  • R1={(a,b)aA,bB,a is a factor of b}R_1 = \{ (a, b) \mid a \in A, b \in B, a \text{ is a factor of } b \}.

First, identify the sets AA and BB:

  • A={2,4,6,8,10,12,14,16,18,20}A = \{ 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 \}
  • B={6,12,18}B = \{ 6, 12, 18 \}

The relation R1R_1 contains pairs (a,b)(a, b) such that aa is a factor of bb.

Now, let's list the pairs in R1R_1:

  • (2,6)(2, 6), (2,12)(2, 12), (2,18)(2, 18)
  • (6,6)(6, 6), (6,12)(6, 12), (6,18)(6, 18)
  • (12,12)(12, 12), (12,18)(12, 18)
  • (18,18)(18, 18)

For R1R_1 to be transitive, if (a,b)R1(a, b) \in R_1 and (b,c)R1(b, c) \in R_1, then (a,c)R1(a, c) \in R_1.

Check transitivity:

  1. If a=2a = 2:

    • (2,6)(2, 6) and (6,6)(6, 6) implies (2,6)(2, 6), which is true.
    • (2,6)(2, 6) and (6,12)(6, 12) implies (2,12)(2, 12), which is true.
    • (2,6)(2, 6) and (6,18)(6, 18) implies (2,18)(2, 18), which is true.
    • (2,12)(2, 12) and (12,12)(12, 12) implies (2,12)(2, 12), which is true.
    • (2,12)(2, 12) and (12,18)(12, 18) implies (2,18)(2, 18), which is true.
    • (2,18)(2, 18) and (18,18)(18, 18) implies (2,18)(2, 18), which is true.
  2. If a=6a = 6:

    • (6,6)(6, 6) and (6,6)(6, 6) implies (6,6)(6, 6), which is true.
    • (6,6)(6, 6) and (6,12)(6, 12) implies (6,12)(6, 12), which is true.
    • (6,6)(6, 6) and (6,18)(6, 18) implies (6,18)(6, 18), which is true.
    • (6,12)(6, 12) and (12,12)(12, 12) implies (6,12)(6, 12), which is true.
    • (6,12)(6, 12) and (12,18)(12, 18) implies (6,18)(6, 18), which is true.
    • (6,18)(6, 18) and (18,18)(18, 18) implies (6,18)(6, 18), which is true.
  3. If a=12a = 12:

    • (12,12)(12, 12) and (12,12)(12, 12) implies (12,12)(12, 12), which is true.
    • (12,12)(12, 12) and (12,18)(12, 18) implies (12,18)(12, 18), which is true.
    • (12,18)(12, 18) and (18,18)(18, 18) implies (12,18)(12, 18), which is true.
  4. If a=18a = 18:

    • (18,18)(18, 18) and (18,18)(18, 18) implies (18,18)(18, 18), which is true.

Since all possible combinations satisfy the transitivity condition, the relation R1R_1 is transitive.

Would you like any further details or have any questions?

Here are 5 questions you might ask:

  1. Is R1R_1 reflexive?
  2. Is R1R_1 symmetric?
  3. What are the properties of relation R2R_2?
  4. Can you explain the concept of reflexive, symmetric, and transitive relations?
  5. How do we determine if a relation is an equivalence relation?

Tip: When analyzing relations, always consider their reflexive, symmetric, and transitive properties to fully understand their behavior.

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Math Problem Analysis

Mathematical Concepts

Relations
Transitivity

Formulas

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Theorems

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Suitable Grade Level

Grades 11-12