Show that the relation “≥” on the set R of all real numbers is reflexive and transitive but not symmetric.
We have
The relation “≥” on the set R of all real numbers.
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.
So, let the relation having “≥” be P.
We can write
P = {(a, b): a ≥ b, a, b ∈ R}
Check for Reflexivity:
∀ a ∈ R
If (a, a) ∈ P
⇒ a ≥ a, which is true.
Since, every real number is equal to itself.
So, ∀ a ∈ R, then (a, a) ∈ P.
⇒ P is reflexive.
∴ P is reflexive.
Check for Symmetry:
∀ a, b ∈ R
If (a, b) ∈ P
⇒ a ≥ b
Now, replace a by b and b by a. We get
b ≥ a, might or might not be true.
Let us check:
Take a = 7 and b = 5.
a ≥ b
⇒ 7 ≥ 5, holds.
b ≥ a
⇒ 5 ≥ 7, is not true as 5 < 7.
⇒ b ≥ a, is not true.
⇒ (b, a) ∉ P
So, if (a, b) ∈ P, then (b, a) ∉ P
∀ a, b ∈ R
⇒ P is not symmetric.
∴ P is not symmetric.
Check for Transitivity:
∀ a, b, c ∈ R
If (a, b) ∈ P and (b, c) ∈ P
⇒ a ≥ b and b ≥ c
⇒ a ≥ b ≥ c
⇒ a ≥ c
⇒ (a, c) ∈ P
So, if (a, b) ∈ P and (b, c) ∈ P, then (a, c) ∈ P
∀ a, b, c ∈ R
⇒ P is transitive.
∴ P is transitive.
Thus, shown that the relation “≥” on the set R of all the real numbers are reflexive and transitive but not symmetric.