Question

If f is a relation on Z defined by xfy > x divides y, then show that fis reflexive and transitive
relation on z​

Answer 1

Answer:

Here, x R y if x divides y, therefore,

(i) R={(2,2),(2,4),(2,6),(2,8),(3,3),(3,6),(3,9),(4,4),(4,8),(5,5),(6,6),(7,7),(8,8),(9,9)}

(ii) Domain of R ={2,3,4,5,6,7,8,9}=A

(iii) Range of R ={2,3,4,5,6,7,8,9}=A

(iv) R

−1

=(y,x):(x,y) in R=(2,2),(4,2),(6,2),(8,2),(3,3),(6,3),(9,3),(4,4),(4,8),(5,5),(6,6),(7,7),(8,8),(9,9)

Infact R

−1

is {(y,x):x,y∈A,y is divisible by x}

(a) As (2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8) and (9,9) belong to R, therefore, R reflexive.

(b) Here, R is not symmetric. We may observe the (2,4)∈R but (4,2)∈

/

R. Infact, 'x' divides 'y' does not imply 'y divides x' when x



=y.

(c) As 'x' divides y' and 'y divides z' imply 'x divides z'. the therefore, the relation R is transitive.

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