Question

show that the relation "is congruent to",on the set of all triangles in a plane is an equivalence relation
plz answer me correctly ​

Answer 1

Answer:

Step-by-step explanation:

R : {(Δ1 , Δ2)} ; Δ1 ≈ Δ2 (Δ1 , Δ2 € a single plane)

(i) Reflexive : (a,a) € R

Δ1 ≈ Δ2 therefore, (Δ1 , Δ2) € R

=> R is reflexive

(ii) Symmetric : (a,b) € R ; (b,a) € R

Let (Δ1 , Δ2) € R

To prove : (Δ2 , Δ1) € R ; Δ2 ≈ Δ1

Δ1 ≈ Δ2 so therefore, Δ2 ≈ Δ1

Therefore, (Δ2 , Δ1) € R

=> R is symmetric

(iii) Transitive (a,b) € R (b,c) € R

(a,c) € R

Let (Δ1 , Δ2) € R , (Δ2 , Δ3) € R

(Δ3 , Δ1) € R

To prove : (Δ1 , Δ3) € R

Proof : Δ1 ≈ Δ2 ------- (1)

Δ2≈Δ3 --------(3)

From (1) and (2)

Δ1≈Δ2≈Δ3

Δ1≈Δ3

=> R is transitivity

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Answer 2

$\huge\bold\pink\bigstar{Answer}$:

R : {(Δ1 , Δ2)} ; Δ1 ≈ Δ2 (Δ1 , Δ2 € a single plane)

(i) Reflexive : (a,a) € R

Δ1 ≈ Δ2 therefore, (Δ1 , Δ2) € R

=> R is reflexive

(ii) Symmetric : (a,b) € R ; (b,a) € R

Let (Δ1 , Δ2) € R

To prove : (Δ2 , Δ1) € R ; Δ2 ≈ Δ1

Δ1 ≈ Δ2 so therefore, Δ2 ≈ Δ1

Therefore, (Δ2 , Δ1) € R

=> R is symmetric

(iii) Transitive (a,b) € R (b,c) € R

(a,c) € R

Let (Δ1 , Δ2) € R , (Δ2 , Δ3) € R

(Δ3 , Δ1) € R

To prove : (Δ1 , Δ3) € R

Proof : Δ1 ≈ Δ2 ------- (1)

Δ2≈Δ3 --------(3)

From (1) and (2)

Δ1≈Δ2≈Δ3

Δ1≈Δ3

=> R is transitivity