Each of the following defines a relation on N :
x > y, x, y ∈ N
Determine which of the above relations are reflexive, symmetric and transitive.
Recall that for any binary relation R on set A. We have,
R is reflexive if for all x ∈ A, xRx.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
We have
x > y, x, y ∈ N
This relation is defined on N (set of Natural Numbers)
The relation can also be defined as
R = {(x, y): x > y} on N
Check for Reflexivity:
∀ x ∈ N
We should have, (x, x) ∈ R
⇒ x > x, which is not true.
1 can’t be greater than 1.
2 can’t be greater than 2.
16 can’t be greater than 16.
Similarly, x can’t be greater than x.
So, ∀ x ∈ N, then (x, x) ∉ R
⇒ R is not reflexive.
Check for Symmetry:
∀ x, y ∈ N
If (x, y) ∈ R
⇒ x > y
Now, replace x by y and y by x. We get
y > x, which may or not be true.
Let us take x = 5 and y = 2.
x > y
⇒ 5 > 2, which is true.
y > x
⇒ 2 > 5, which is not true.
⇒ y > x, is not true as x > y
⇒ (y, x) ∉ R
So, if (x, y) ∈ R, but (y, x) ∉ R ∀ x, y ∈ N
⇒ R is not symmetric.
Check for Transitivity:
∀ x, y, z ∈ N
If (x, y) ∈ R and (y, z) ∈ R
⇒ x > y and y > z
⇒ x > y > z
⇒ x > z
⇒ (x, z) ∈ R
So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∈ R
∀ x, y, z ∈ N
⇒ R is transitive.
Hence, the relation is transitive but neither reflexive nor symmetric.