- Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x}
- Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer}
(iii) R = {(x, y) : y is separable by x} in A = {1, 2, 3, 4, 5, 6} From above we have,
R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5), (6,
6)}
According to reflexive property: (x, x) ∈ R, then, at that point R is reflexive.
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) ∈ R . In this way, R is reflexive. According to symmetric property: (x, y) ∈ R and (y, x) ∈ R, then, at that point R is symmetric. (1, 2) ∈ R yet (2, 1) ∉ R. So R isn’t symmetric.
According to transitive property: If (x, y) ∈ R and (y, z) ∈ R, then, at that point (x, z) ∈ R. In this way R is transitive. Additionally (1, 4) ∈ R and (4, 4) ∈ R and (1, 4) ∈ R, So R is transitive.
In this way, R is reflexive and transitive however nor symmetric.
(iv) R = {(x, y) : x – y is an integer} in set Z, all things considered.
Presently, (x, x), say (1, 1) = x – y = 1 – 1 = 0 ∈ Z => R is reflexive. (x, y) ∈ R and (y, x) ∈ R, i.e.,
x – y and y – x are numbers => R is symmetric. (x, y) ∈ R and (y, z) ∈ R, then, at that point (x, z) ∈ R i.e.,
x – y and y – z and x – z are numbers.
(x, z) ∈ R => R is transitive
Hence, R is reflexive, symmetric and transitive.