2Q. No. 18 Let R be a relation on Q, defined by R={(a,b)/a,b EQ and a-be Z} .Show that R isan equivalence relation. About the author Hadley
Answer: R is equivalence relation. Step-by-step explanation: Given : R = {(a,b) : a, b € Q and a- b € Z}. 1. Let a € Q . Then , a – a = 0 € Z. .;. (a,a) € R for all a € Q. So, R is reflexive. 2. (a,b) € R => (a-b) € Z, (a -b ) is an integer => – (a-b) is an integer => (b-a) is an integer => (a,b) € R. .;. R is symmetric. 3. (a,b) € R and (b,a) € R = (a- b) is an integer and (b-c) is an integer = {(a-b) + (b-c)} is an integer = (a-c) is an integer = (a,c) € R. Thus (a,b) € R and (b,c) € R = (a,c) € R. .;. R is transitive. Reply
Answer:
R is equivalence relation.
Step-by-step explanation:
Given : R = {(a,b) : a, b € Q and a- b € Z}.
1. Let a € Q . Then , a – a = 0 € Z.
.;. (a,a) € R for all a € Q.
So, R is reflexive.
2. (a,b) € R => (a-b) € Z, (a -b ) is an integer
=> – (a-b) is an integer
=> (b-a) is an integer
=> (a,b) € R.
.;. R is symmetric.
3. (a,b) € R and (b,a) € R
= (a- b) is an integer and (b-c) is an integer
= {(a-b) + (b-c)} is an integer
= (a-c) is an integer
= (a,c) € R.
Thus (a,b) € R and (b,c) € R = (a,c) € R.
.;. R is transitive.