Question

Test whether the following relation R₁, R₂, and R₃ are (a) reflexive (b) symmetric and (c) transitive:

(i) R₁ on Q₀ defined by (a, b) ∈ R₁ ⇔ a = 1/b.

(ii) R₂ on Z defined by (a, b) ∈ R₂ ⇔ |a – b| ≤ 5

(iii) R₃ on R defined by (a, b) ∈ R₃ ⇔ a² – 4ab + 3b² = 0.

Answer 1

(i) Given as R₁ on Q₀ defined by

(a, b) ∈ R₁ ⇔ a = 1/b.

Reflexivity:

Let a be an arbitrary element of R₁.

Then, a ∈ R₁

⇒ a ≠ 1/a for all a ∈ Q₀

Clearly, R₁ is not reflexive.

Symmetry:

Let (a, b) ∈ R₁

Then, (a, b) ∈ R₁

Therefore we can write ‘a’ as a = 1/b

⇒ b = 1/a ⇒ (b, a) ∈ R₁

So, R₁ is symmetric.

Transitivity: 

Here, (a, b) ∈ R₁ and (b, c) ∈ R₂

⇒ a = 1/b and b = 1/c

⇒ a = 1/ (1/c) = c

⇒ a ≠ 1/c ⇒ (a, c) ∉ R₁

So, R₁ is not transitive.

(ii) Given as R₂ on Z defined by

(a, b) ∈ R₂ ⇔ |a – b| ≤ 5

Now let us check whether R₂ is reflexive, symmetric and transitive.

Reflexivity:

Let a be an arbitrary element of R₂.

Then, a ∈ R₂

On applying the given condition we get,

⇒ |a − a| = 0 ≤ 5

So, R₁ is reflexive.

Symmetry:

Let (a, b) ∈ R₂

⇒ |a − b| ≤ 5 [Since, |a − b| = |b − a|]

⇒ |b − a| ≤ 5

⇒ (b, a) ∈ R₂

So, R₂ is symmetric.

Transitivity:

Let (1, 3) ∈ R₂ and (3, 7) ∈ R₂

⇒ |1 − 3| ≤ 5 and |3 − 7| ≤ 5 But |1 − 7| ≰ 5

⇒ (1, 7) ∉ R₂

Clearly, R₂ is not transitive.

(iii) Given as R₃ on R defined by

(a, b) ∈ R₃ ⇔ a² – 4ab + 3b² = 0.

Now let us check whether R₂ is reflexive, symmetric and transitive.

Reflexivity:

Let a be an arbitrary element of R₃.

Then, a ∈ R₃ ⇒ a² − 4a × a + 3a² = 0

Clearly, R₃ is reflexive Symmetry:

Let (a, b) ∈ R₃ ⇒ a² − 4ab + 3b² = 0

But b² − 4ba + 3a² ≠ 0 for all a, b ∈ R

So, R₃ is not symmetric.

Transitivity:

Let (1, 2) ∈ R₃ and (2, 3) ∈ R₃

⇒ 1 − 8 + 6 = 0 and 4 – 24 + 27 = 0

But 1 – 12 + 9 ≠ 0

Clearly, R₃ is not transitive.

Comments

The second relation R2 on Z defined by (a, b) ∈ R2 ⇔ |a – b| ≤ 5  I found in this relation that this relation is transitive.  As the three numbers had been taken by you (1,3) ,(3,7) according to it, it is not transitive but according  to me we can also chose any other  number instead of it such as (1,2),(2,3) and when I put that number as the condition is given (a, b) ∈ R2 ⇔ |a – b| ≤ 5 I found that the given relation is transitive and if it is still not a transitive relation then why?

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We can disprove any result by giving one counter example but can not prove a result by given an example in favour of the result. Here we find a counter example, so it is not a transitive relation.