Q. 25.0( 1 Vote )

# Let A be the set of all triangles in a plane. Show that the relationR = {(∆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.

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