Each of the following defines a relation on N :
x + y = 10, 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 = 10, 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 = 10} on N
Check for Reflexivity:
∀ x ∈ N
We should have, (x, x) ∈ R
⇒ x + x = 10, which is not true everytime.
Take x = 4.
x + x = 10
⇒ 4 + 4 = 10
⇒ 8 = 10, which is not true.
That is 8 ≠ 10.
So, ∀ x ∈ N, then (x, x) ∉ R
⇒ R is not reflexive.
Check for Symmetry:
∀ x, y ∈ N
If (x, y) ∈ R
⇒ x + y = 10
Now, replace x by y and y by x. We get
y + x = 10, which is as same as x + y = 10.
⇒ y + x = 10
⇒ (y, x) ∈ R
So, if (x, y) ∈ R, and then (y, x) ∈ R ∀ x, y ∈ N
⇒ R is symmetric.
Check for Transitivity:
∀ x, y, z ∈ N
If (x, y) ∈ R and (y, z) ∈ R
⇒ x + y = 10 and y + z = 10
⇒ x + z = 10, may or may not be true.
Let us take x = 6, y = 4 and z = 6
x + y = 10
⇒ 6 + 4 = 10
⇒ 10 = 10, which is true.
y + z = 10
⇒ 4 + 6 = 10
⇒ 10 = 10, which is true.
x + z = 10
⇒ 6 + 6 = 10
⇒ 12 = 10, which is not true
That is, 12 ≠ 10
⇒ x + z ≠ 10
⇒ (x, z) ∉ R
So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∉ R
∀ x, y, z ∈ N
⇒ R is not transitive.
Hence, the relation is symmetric but neither reflexive nor transitive.