Math Problem Statement

Let Y = {0, 1, 2} and Z = {0, 1}, and define a relation R from A to B as follows: Given any (x, y) ∈ Y × Z, (x, y) ∈ R means that (x + y)/2 is an integer. a. State explicitly which ordered pairs are in Y × Z and which are in R. b. Is 1 R 0? Is 2 R 0? Is 2 R 1? c. What are the domain and co-domain of R?

Solution

The given question defines a relation RR from set AA to set BB, where the sets are:

  • Y={0,1,2}Y = \{ 0, 1, 2 \}
  • Z={0,1}Z = \{ 0, 1 \}

For any ordered pair (x,y)(x, y) where xYx \in Y and yZy \in Z, the relation (x,y)R(x, y) \in R is defined as: x+y2 is an integer.\frac{x + y}{2} \text{ is an integer}.

Let me break down the questions one by one:

a. State explicitly which ordered pairs are in Y×ZY \times Z and which are in RR.

First, let's identify all ordered pairs in Y×ZY \times Z (the Cartesian product): Y×Z={(0,0),(0,1),(1,0),(1,1),(2,0),(2,1)}Y \times Z = \{(0,0), (0,1), (1,0), (1,1), (2,0), (2,1)\}

Now, apply the condition x+y2\frac{x + y}{2} to see which pairs belong to RR:

  1. For (0,0)(0, 0): 0+02=0\frac{0 + 0}{2} = 0, which is an integer, so (0,0)R(0, 0) \in R.
  2. For (0,1)(0, 1): 0+12=0.5\frac{0 + 1}{2} = 0.5, which is not an integer, so (0,1)R(0, 1) \notin R.
  3. For (1,0)(1, 0): 1+02=0.5\frac{1 + 0}{2} = 0.5, which is not an integer, so (1,0)R(1, 0) \notin R.
  4. For (1,1)(1, 1): 1+12=1\frac{1 + 1}{2} = 1, which is an integer, so (1,1)R(1, 1) \in R.
  5. For (2,0)(2, 0): 2+02=1\frac{2 + 0}{2} = 1, which is an integer, so (2,0)R(2, 0) \in R.
  6. For (2,1)(2, 1): 2+12=1.5\frac{2 + 1}{2} = 1.5, which is not an integer, so (2,1)R(2, 1) \notin R.

Thus, the pairs that are in RR are: R={(0,0),(1,1),(2,0)}R = \{ (0, 0), (1, 1), (2, 0) \}

b. Is 1R01 \, R \, 0? Is 2R02 \, R \, 0? Is 2R12 \, R \, 1?

To determine this, we check if the pairs (1,0)(1, 0), (2,0)(2, 0), and (2,1)(2, 1) belong to RR based on the condition x+y2\frac{x + y}{2}:

  • 1R01 \, R \, 0? For (1,0)(1, 0): 1+02=0.5\frac{1 + 0}{2} = 0.5, so No.
  • 2R02 \, R \, 0? For (2,0)(2, 0): 2+02=1\frac{2 + 0}{2} = 1, so Yes.
  • 2R12 \, R \, 1? For (2,1)(2, 1): 2+12=1.5\frac{2 + 1}{2} = 1.5, so No.

c. What are the domain and co-domain of RR?

  • Domain: The domain of RR is the set of xx values from the ordered pairs in RR. The pairs in RR are (0,0)(0, 0), (1,1)(1, 1), and (2,0)(2, 0). So, the domain is: Domain of R={0,1,2}\text{Domain of } R = \{ 0, 1, 2 \}

  • Co-domain: The co-domain of RR is the set of yy values from the ordered pairs in RR. The pairs in RR are (0,0)(0, 0), (1,1)(1, 1), and (2,0)(2, 0). So, the co-domain is: Co-domain of R={0,1}\text{Co-domain of } R = \{ 0, 1 \}


Would you like further details or have any questions? Here are 5 related questions for further practice:

  1. Can a pair like (3,1)(3, 1) be considered for RR?
  2. What is the range of the relation RR?
  3. How would the relation change if Z={0,1,2}Z = \{0, 1, 2\}?
  4. What is the inverse of the relation RR?
  5. What happens to the relation if we change the condition to xy2\frac{x - y}{2}?

Tip: When dealing with relations, always verify the condition of the relation for each pair systematically to ensure accuracy.

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Math Problem Analysis

Mathematical Concepts

Relations
Cartesian Product
Division
Integers

Formulas

(x + y)/2

Theorems

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Suitable Grade Level

Grades 9-11