2013,11,prove the relation

Asked by alanpeter9611 | 26th Feb, 2019, 02:23: AM

Expert Answer:

we space have space open vertical bar straight a minus straight b close vertical bar space divisible space by space 2

straight a equals straight b
open parentheses straight a comma straight a close parentheses element of straight R
open vertical bar straight a minus straight a close vertical bar equals 0 space divisible space by space 2
Reflexive

open parentheses straight a comma straight b close parentheses element of straight R
open vertical bar straight a minus straight b close vertical bar equals space divisible space by space 2
open parentheses straight b comma straight a close parentheses element of straight R
open vertical bar straight b minus straight a close vertical bar equals space divisible space by space 2
also
open vertical bar straight a minus straight b close vertical bar equals open vertical bar straight b minus straight a close vertical bar.... Symmetric

open parentheses straight a comma straight b close parentheses element of straight R
open vertical bar straight a minus straight b close vertical bar equals space divisible space by space 2
open parentheses straight b comma straight c close parentheses element of straight R
open vertical bar straight b minus straight c close vertical bar equals space divisible space by space 2
now
open vertical bar straight a minus straight c close vertical bar equals open vertical bar straight a minus straight b plus straight b minus straight c close vertical bar... addition space of space two space numbers space divisible space by space 2 space is space divisible space by space 2..... transitive
so space Equivalence space relation

for space 6 space element space rightwards arrow left curly bracket 8 right curly bracket

Answered by Arun | 26th Feb, 2019, 11:11: AM