Question

1. A relation R in the set Z of all integers defined as :
R = {(x, y) : x - y is an integer),
then Ris :
(B) Transitive
(A) Reflexive
(D)
None of these
(C) Symmetric​

Answer 1

R={(x,y):x−yis an integer}

R={(x,y):x−yis an integer}Now, for every x∈Z,(x,x)∈R as x−x=0 is an integer.

R={(x,y):x−yis an integer}Now, for every x∈Z,(x,x)∈R as x−x=0 is an integer.∴R is reflexive.

R={(x,y):x−yis an integer}Now, for every x∈Z,(x,x)∈R as x−x=0 is an integer.∴R is reflexive.Now, for every x,y∈Z if (x,y)∈R, then x−y is an integer.

R={(x,y):x−yis an integer}Now, for every x∈Z,(x,x)∈R as x−x=0 is an integer.∴R is reflexive.Now, for every x,y∈Z if (x,y)∈R, then x−y is an integer.⇒−(x−y) is also an integer.

R={(x,y):x−yis an integer}Now, for every x∈Z,(x,x)∈R as x−x=0 is an integer.∴R is reflexive.Now, for every x,y∈Z if (x,y)∈R, then x−y is an integer.⇒−(x−y) is also an integer.⇒(y−x) is an integer.

R={(x,y):x−yis an integer}Now, for every x∈Z,(x,x)∈R as x−x=0 is an integer.∴R is reflexive.Now, for every x,y∈Z if (x,y)∈R, then x−y is an integer.⇒−(x−y) is also an integer.⇒(y−x) is an integer.∴(y,x)∈R

R={(x,y):x−yis an integer}Now, for every x∈Z,(x,x)∈R as x−x=0 is an integer.∴R is reflexive.Now, for every x,y∈Z if (x,y)∈R, then x−y is an integer.⇒−(x−y) is also an integer.⇒(y−x) is an integer.∴(y,x)∈R⇒R is symmetric.

Now,

Now,Let (x,y) and (y,z)∈R, where x,y,z∈Z.

Now,Let (x,y) and (y,z)∈R, where x,y,z∈Z.⇒(x−y) and (y−z) are integers.

Now,Let (x,y) and (y,z)∈R, where x,y,z∈Z.⇒(x−y) and (y−z) are integers.⇒x−z=(x−y)+(y−z) is an integer.

Now,Let (x,y) and (y,z)∈R, where x,y,z∈Z.⇒(x−y) and (y−z) are integers.⇒x−z=(x−y)+(y−z) is an integer.∴(x,z)∈R

Now,Let (x,y) and (y,z)∈R, where x,y,z∈Z.⇒(x−y) and (y−z) are integers.⇒x−z=(x−y)+(y−z) is an integer.∴(x,z)∈R∴R is transitive.

Now,Let (x,y) and (y,z)∈R, where x,y,z∈Z.⇒(x−y) and (y−z) are integers.⇒x−z=(x−y)+(y−z) is an integer.∴(x,z)∈R∴R is transitive.Hence, R is reflexive, symmetric, and transitive.