Math Problem Statement

Didefinisikan himpunan A = {0,1,2,3} dan B = {-3, -1, 3, 6}. Buatlah relasi sebagai pasangan berurutan dengan aturan: a. R1 = {(a, b) | ab > 0; a ∈ A, b ∈ B} b. R2 = {(a, b) | a ≤ b; a ∈ A, b ∈ B}

Solution

Let's solve the problem presented.

Given:

  • A={0,1,2,3}A = \{0, 1, 2, 3\}
  • B={3,1,3,6}B = \{-3, -1, 3, 6\}

We are asked to form two relations R1R_1 and R2R_2 from set AA to set BB based on the specified conditions.

a. Relation R1={(a,b)ab>0;aA,bB}R_1 = \{(a, b) | ab > 0; a \in A, b \in B\}

This relation consists of pairs (a,b)(a, b) where the product abab is greater than zero. Let's evaluate each possible pair:

  • For a=0a = 0: 0×b=00 \times b = 0 for all bBb \in B. So, no pairs with a=0a = 0 will satisfy ab>0ab > 0.
  • For a=1a = 1:
    • 1×(3)=31 \times (-3) = -3 (not > 0)
    • 1×(1)=11 \times (-1) = -1 (not > 0)
    • 1×3=31 \times 3 = 3 (satisfies ab>0ab > 0)
    • 1×6=61 \times 6 = 6 (satisfies ab>0ab > 0)
    • Thus, pairs are (1,3)(1, 3) and (1,6)(1, 6).
  • For a=2a = 2:
    • 2×(3)=62 \times (-3) = -6 (not > 0)
    • 2×(1)=22 \times (-1) = -2 (not > 0)
    • 2×3=62 \times 3 = 6 (satisfies ab>0ab > 0)
    • 2×6=122 \times 6 = 12 (satisfies ab>0ab > 0)
    • Thus, pairs are (2,3)(2, 3) and (2,6)(2, 6).
  • For a=3a = 3:
    • 3×(3)=93 \times (-3) = -9 (not > 0)
    • 3×(1)=33 \times (-1) = -3 (not > 0)
    • 3×3=93 \times 3 = 9 (satisfies ab>0ab > 0)
    • 3×6=183 \times 6 = 18 (satisfies ab>0ab > 0)
    • Thus, pairs are (3,3)(3, 3) and (3,6)(3, 6).

So, the relation R1R_1 is: R1={(1,3),(1,6),(2,3),(2,6),(3,3),(3,6)}R_1 = \{(1, 3), (1, 6), (2, 3), (2, 6), (3, 3), (3, 6)\}

b. Relation R2={(a,b)ab;aA,bB}R_2 = \{(a, b) | a \leq b; a \in A, b \in B\}

This relation consists of pairs (a,b)(a, b) where aba \leq b.

  • For a=0a = 0:
    • 030 \leq -3 (false)
    • 010 \leq -1 (false)
    • 030 \leq 3 (true, so (0,3)(0, 3))
    • 060 \leq 6 (true, so (0,6)(0, 6))
  • For a=1a = 1:
    • 131 \leq -3 (false)
    • 111 \leq -1 (false)
    • 131 \leq 3 (true, so (1,3)(1, 3))
    • 161 \leq 6 (true, so (1,6)(1, 6))
  • For a=2a = 2:
    • 232 \leq -3 (false)
    • 212 \leq -1 (false)
    • 232 \leq 3 (true, so (2,3)(2, 3))
    • 262 \leq 6 (true, so (2,6)(2, 6))
  • For a=3a = 3:
    • 333 \leq -3 (false)
    • 313 \leq -1 (false)
    • 333 \leq 3 (true, so (3,3)(3, 3))
    • 363 \leq 6 (true, so (3,6)(3, 6))

So, the relation R2R_2 is: R2={(0,3),(0,6),(1,3),(1,6),(2,3),(2,6),(3,3),(3,6)}R_2 = \{(0, 3), (0, 6), (1, 3), (1, 6), (2, 3), (2, 6), (3, 3), (3, 6)\}

Would you like further details on how these pairs were derived?


Here are some related questions for further exploration:

  1. How would these relations change if sets AA and BB had different elements?
  2. Can you create a new relation using a different rule for pairs?
  3. What happens if we restrict the elements in BB to positive values only?
  4. How can we represent these relations visually using a Cartesian plane?
  5. What are some real-life examples where such relations could apply?

Tip: When forming relations between sets, always test each possible pair to see if it satisfies the given condition.

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Math Problem Analysis

Mathematical Concepts

Set Theory
Relations
Cartesian Product

Formulas

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Theorems

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Suitable Grade Level

Grades 10-11