Test whether the following relations R1, R2 and R3 are (i) reflexive (ii) symmetric and (iii) transitive :
R2 on Z defined by (a, b) ϵ R2⇔ |a – b| ≤ 5
Here, R1, R2, R3, and R4 are the binary relations.
So, 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, using these results let us start determining given relations.
We have
R2 on Z defined by (a, b) ∈ R2⇔ |a – b| ≤ 5
Check for Reflexivity:
∀ a ∈ Z,
(a, a) ∈ R2 needs to be proved for reflexivity.
If (a, b) ∈ R2
Then, |a – b| ≤ 5 …(1)
So, for (a, a) ∈ R1
Replace b by a in equation (1), we get
|a – a| ≤ 5
⇒ 0 ≤ 5
⇒ (a, a) ∈ R2
So, ∀ a ∈ Z, then (a, a) ∈ R2
∴ R2 is reflexive.
Check for Symmetry:
∀ a, b ∈ Z
If (a, b) ∈ R2
We have, |a – b| ≤ 5 …(2)
Replace a by b & b by a in equation (2), we get
|b – a| ≤ 5
Since, the value is in mod, |b – a| = |a – b|
⇒ The statement |b – a| ≤ 5 is true.
⇒ (b, a) ∈ R2
So, if (a, b) ∈ R2, then (b, a) ∈ R2
∀ a, b ∈ Q0
∴ R1 is symmetric.
Check for Transitivity:
∀ a, b, c ∈ Z
If (a, b) ∈ R2 and (b, c) ∈ R2
⇒ |a – b| ≤ 5 and |b – c| ≤ 5
Since, inequalities cannot be added or subtract. We need to take example to check for,
|a – c| ≤ 5
Take values a = 18, b = 14 and c = 10
Check: |a – b| ≤ 5
⇒ |18 – 14| ≤ 5
⇒ |4| ≤ 5 is true.
Check: |b – c| ≤ 5
⇒ |14 – 10| ≤ 5
⇒ |4| ≤ 5
Check: |a – c| ≤ 5
⇒ |18 – 10| ≤ 5
⇒ |8| ≤ 5 is not true.
⇒ (a, c) ∉ R2
So, if (a, b) ∈ R2 and (b, c) ∈ R2, then (a, c) ∉ R1
∀ a, b, c ∈ Z
∴ R2 is not transitive.