Question

prove that the relation R in the set of integers z defined by R=(x, y):x-y is an integer is an equivalent relation​

Answer 1

Note :

★ Reflexive relation :

A relation R on a set A is said to be reflexive if aRa for all a € A ,

ie. (a,a) € R for all a € A .

★ Symmetric relation :

A relation R on a set A is said to be symmetric if aRb => bRa ,

ie. (a,b) € R => (b,a) € R for all a,b € R .

★ Transitive relation :

A relation R on a set A is said to be transitive if aRb and bRc => aRc

ie. (a,b) € R and (b,c) € R => (a,c) € R .

★ Equivalence relation :

A relation R on a set A is said to be an equivalence if it is reflexive , symmetric as well as transitive .

Proof :

Here ,

The given relation R on the set Z (set of integers) is ; R = { (x,y) : x - y € Z } .

• Whether R is reflexive :

Let a € Z , then

→ a - a = 0 € Z { 0 is an integer }

→ a - a € Z

→ (a,a) € R

Since ,

(a,a) € R for all a € Z , hence R is reflexive .

• Whether R is symmetric :

Let (a,b) € R , then

=> a - b € Z

=> -(b - a) € Z

=> b - a € Z { negative of an integer is again an integer }

=> (b,a) € R

Since ,

(a,b) € R => (b,a) € R for all a,b € Z , hence R is symmetric .

• Whether R is transitive :

Let (a,b) € Z and (b,c) € Z , then

=> a - b € Z and b - c € Z

=> (a - b) + (b - c) € Z { sum of two integers is again an integer }

=> a - c € Z

=> (a,c) € R

Since ,

(a,b) € R and (b,c) € R => (a,c) € R for all a,b,c € Z , hence R is transitive .

• Whether R is equivalence :

Since , the relation R is reflexive , symmetric as well as transitive , hence R is equivalence .

Hence proved .