Q. 25.0( 1 Vote )

# 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.

Answer :

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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Fill in the blanks in each of the

Let the relation R be defined in N by aRb if 2a + 3b = 30. Then R = ______.

Mathematics - ExemplarFill in the blanks in each of the

Let the relation R be defined on the set

A = {1, 2, 3, 4, 5} by R = {(a, b) : |a^{2} – b^{2}| < 8}. Then R is given by _______.

State True or False for the statements

Every relation which is symmetric and transitive is also reflexive.

Mathematics - ExemplarState True or False for the statements

Let R = {(3, 1), (1, 3), (3, 3)} be a relation defined on the set A = {1, 2, 3}. Then R is symmetric, transitive but not reflexive.

Mathematics - Exemplar