Question

Complete the following: (With Details and example if needed)

Given set S = {-3, -2, 0, 1, 2, 4}

Relation R is a relation on S such that R = { (?, ?) | ? ∈ ?, ? ∈ ? ??? ?? ≥ 1}

1. Using roster method, what is R?

2. Show the digraph representation of R

3. Show the matrix representation of R

4. if the relation does not have the property, give an example of why:

- Is R reflexive?

- Is R symmetric?

-Is R antisymmetric?

- Is R transitive?

Answer 1

1). Relation R in roster method:-

R = { (-3,-3),(-2,-2),(1,1),(2,2),(4,4),(-3,-2),(-2,-3),(1,2),(2,1),(1,4),(4,1),(2,4),(4,2) }

2). Diagraph:-

4). Is R reflexive?

Any relation is reflexive if for every element in set S is related to each other.

Relation R is non-Reflexive relation because 0 which is element of S is not related to 0 .

that is 0 * 0 is not greater or equal to 1.so , (0,0) doesn't belong to R.

Is R symmetric?

If for every (x,y) R, (y,x) must also belongs to R then the relation R is symmetric relation.

Relation R is clearly symmetric relation.

proof :-

(x,y) R means that xy>=1

xy >=1 can be written as yx>=1

yx>=1 means (y,x)R

therefore every (x,y)R , (y,x) also belongs to R.

Hence, relation R is symmetric.

Is R antisymmetric?

Any relation R is antisymmetric if for every (x,y)R and (y,x)R then y = x.

Clearly , from roster form we can see that (1,4) and (4,1) both are in R but 1 is not equal to 4.

Therefore, R is not antisymmetric.

Is R transitive?

Any relation R is transitive if for every (x,y)R and (y,z)R then (x,z) must also belongs to R.

R is clearly Transitive relation.

Proof :-

(x,y) R means that xy>=1

(y,z) R means that yz>=1

now , either x , y and z are all negative or all positive .

suppose x, y and z are all negative then in that case xz>=1 so , (x,z)R.

suppose x,y and z are all positive then in that case xz>=1 so , (x,z)R

so ,in all cases if (x,y) R and (y,z) R then (x,z) also belongs to R.

Therefore , R is transitive relation.