Using the fact that congruence modulo m is an equivalence relation on Z and without reference to Theorems 3

x is related to y modulo m iff m divide x-y. 

In part e I have used the statement of c. 

And part f I have proved by contrapoeitive means if we want to prove A implies B we should prove not B implies not A. 

Step-by-step explanation

Ans CJ. x = y( modm) means m divides x-y m x -y = ) m / y - - y = 16 ( modm ) means x is inclass of y X = y ( modm ) means y is in class of X XE [ x ] of X E X implies x E y so xcy - O ye ly implies yEX so ycx - 3 and ( gives * = es If any to then there exists a element aczny implies Ex and also afy aEx implies x. = almodm) implies m/ x-a - aEy implies y= a ( modm) implies m/ y-a so mla-y - 1 and gives ma-y+ x-a m / x - y = ) x = y mod (m ) By using part ( c ) X = y FJ we have to prove by contrapositive ways So we have to prove that Not ( X # 1 ) implies not ( X ny = ) So toke * = y since XEX implies xe y then xexny so kny + p We have proved x=y implies any + $ By contrapositive If xny = p then x ty