Show the relation R in the set

R equals open curly brackets open parentheses a comma b close parentheses : open vertical bar a minus b close vertical bar space i s space a space m u l t i p l e space o f space 3 close curly brackets

is an equivalence relation.

Asked by Topperlearning User | 25th Oct, 2016, 07:53: AM

Expert Answer:

Set A = {0,1, 2, 3, 4, 5, ....,10} and

or R = {(0, 3), (3, 0), (0, 6), (6, 0), (0, 9), (9, 0), (1, 4), (4, 1),(2, 5), (5, 2), (3, 6), (6, 3), (3, 9),(9, 3),(4, 7), (7, 4),(4, 10), (10, 4), (1, 7), (7, 1),(1, 10), (10, 1), (2, 8), (8, 2),(5, 8), (8, 5),(6, 9), (9, 6),(7, 10), (10, 7), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9), (10, 10)}
 
(i) a - a = 0 = 3k, where k = 0, . So, R is reflexive.
(ii) Here,
So, R is symmetric.
 
(iii) Here, open parentheses 0 comma 3 close parentheses element of R space a n d space open parentheses 3 comma 6 close parentheses element of R comma space a l s o space open parentheses 0 comma 6 close parentheses element of R 
So, the relation R is also transitive.
 
Since, relation R is reflexive, symmetric and transitive, therefore the R is an equivalence relation.

Answered by  | 25th Oct, 2016, 09:53: AM