Question

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 x Z.

(x,y) ∈ R means that (x+y)/2 is an integer

1.) State explicitly which ordered pairs are in Y x Z and which are in R.
2.) Is 1 R 0?
3.) Is 2 R 0?
4.) Is 2 R 1?
5.) What are the domain and co-domain of R?​

Answer 1

Answer:

R={(x,y):x−yis an integer}

Now, for every x∈Z,(x,x)∈R as x−x=0 is an integer.

∴R is reflexive.

Now, for every x,y∈Z if (x,y)∈R, then x−y is an integer.

⇒−(x−y) is also an integer.

⇒(y−x) is an integer.

∴(y,x)∈R

⇒R is symmetric.

Now,

Let (x,y) and (y,z)∈R, where x,y,z∈Z.

⇒(x−y) and (y−z) are integers.

⇒x−z=(x−y)+(y−z) is an integer.

∴(x,z)∈R

∴R is transitive.

Hence, R is reflexive, symmetric, and transitive.

Answer 2

Answer:

Step-by-step explanation:

Given Y = {0,1,2} and Z = {0,1}

R is a relation from Y to Z such that (x+y)/2 is an integer

We can find Y x Z = {(0,0), (0,1), (1,0) , (1,1), (2,0), (2,1)}

(1) For the given ordered pairs,

(0,0) = (0+0)/2 = 0/2 = 0 which is a integer. => (0,0) ∈ R

(0,1) = (0+1)/2 = 1/2 which is not an integer  => (0,1) ∈ Y x Z

(1,0) = (1+0)/2 = 1/2 which is not an integer  => (1,0) ∈ Y x Z

(1,1) = (1+1)/2 = 2/2 = 1 which is a integer. => (1,1) ∈ R

(2,0) = (2+0)/2 = 2/2 = 1 which is a integer. => (2,0) ∈ R

(2,1) = (2+1)/2 = 3/2 which is not an integer  => (2,1) ∈ Y x Z

(2) 1 R 0 = (1+0)/2 = 1/2 which is not an integer

Hence 1/2∉ R

(3) 2 R 0 = (2+0)/2 = 2/2 = 1 which is an integer

Hence 1∈ R

(4) 2 R 1 = (2+1)/2 = 3/2 which is not an integer

Hence 3/2∉ R

(5) As domain is the set of all possible inputs,

Domain of R = {0,1,2}

As codomain is the set of possible outputs,

Codomain of R = {0,1}