Let A be the set of all triangles in a plane. Show that the relation
R = {(∆1, ∆2) : ∆1 ~ ∆2} is an equivalence relation on A.
Let R = {(∆1, ∆2) : ∆1 ~ ∆2} be a relation defined on A.
Now,
R is Reflexive if (Δ, Δ) ∈ R ∀ Δ ∈ A
We observe that for each Δ ∈ A we have,
Δ ~ Δ since, every triangle is similar to itself.
⇒ (Δ, Δ) ∈ R ∀ Δ ∈ A
⇒ R is reflexive.
R is Symmetric if (∆1, ∆2) ∈ R ⇒ (∆2, ∆1) ∈ R ∀ ∆1, ∆2 ∈ A
Let (∆1, ∆2) ∈ R ∀ ∆1, ∆2 ∈ A
⇒ ∆1 ~ ∆2
⇒ ∆2 ~ ∆1
⇒ (∆2, ∆1) ∈ R
⇒ R is symmetric
R is Transitive if (∆1, ∆2) ∈ R and (∆2, ∆3)∈ R ⇒ (∆1, ∆3) ∈ R ∀ ∆1, ∆2, ∆3 ∈ A
Let (∆1, ∆2) ∈ R and ((∆2, ∆3) ∈ R ∀ ∆1, ∆2, ∆3 ∈ A
⇒ ∆1 ~ ∆2 and ∆2 ~ ∆3
⇒ ∆1 ~ ∆3
⇒ (∆1, ∆3) ∈ R
⇒ R is transitive.
Since R is reflexive, symmetric and transitive, it is an equivalence relation on A.