Skip to main content

Show that the relation R defined in the set A of all polygons as R = {(P1 , P2 ) : P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?



Class 12, NCERT Chapter 1,  Exercise 1.1, Q13


It is given that the relation R defined in the set A of all polygons as
R = {(P1, P2): P1 and P2 have same number of sides},
Then, R is reflexive since (P1, P2) ϵ R as the same polygon has the same number of sides with itself.
Let (P1, P2) ϵ R
 P1 and P2 have the same number of sides.
 P2 and P1 have the same number of sides.
 (P2, P1) ϵ R
Therefore, R is symmetric.

Now, let (P1, P2), (P2, P3) ϵ R
 P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of sides.
 P1 and P3 have the same number of sides.
 (P1, P3) ϵ R
Therefore, R is transitive.

Thus, R is an equivalence relation.

The elements in A related to the right-angled triangle (T) with sides 3, 4 and 5 are those polygons that have 3 sides.
Therefore, the set of all elements in A related to triangle T is the set of all triangles.

Comments

Popular posts from this blog

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. (i) f : R → R defined by f(x) = 3 – 4x (ii) f : R → R defined by f(x) = 1 + x^2

Class 12, NCERT Chapter 1,  Exercise 1.2, Q7 (i)  Here f(x) = 3 – 4x Then f(x 1 ) = 3 – 4x 1 and f (x 2 ) = 3 – 4x 2 Now f(x 1 ) = f (x 2 ) ⇒ 3 – 4x 1  = 3 – 4x 2 ⇒ – 4x 1  = –4x 2 ⇒ x 1  = x 2 ∴ f(x 1 ) = f(x 2 ) & x 1  = x 2 So, f is one-one function. Let f(x) = y ∈ R Then y=3 - 4x ⇒ x= 3-y/4 (ii) Let x 1  = 2 and x 2  = –2 ∈ R Here f(x) = 1 + x 2 Then f(x 1 ) = f (2) = 1 + (2)  2  = 5 and f (x 2 ) = f(–2) = 1 + (–2)  2  = 5 ∴ f (x 1 ) = f(x 2 ) but x 1  ≠ x 2 So, f is not one-one function. Let f(x) = – 2 ⇒R Then 1 + x 2  = –2 ⇒ x 2  = –3 = ± √–3 ∈R So, f is not onto function.

Let A be the set of all 50 students of Class X in a school. Let f : A → N be function defined by f(x) = roll number of the student x. Show that f is one-one but not onto.

Class 12, NCERT Chapter 1,  Example7 No two different students in the class can have the same roll number. Therefore, f must be one-one. We can assume without any loss of generality that roll numbers of students are from 1 to 50. This implies that 51,52,53... in N is not roll number of any student of the class, so that 51,52,53...  can not be an image of any element of X under f. Hence, f is not onto.

Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation.

Class 12, NCERT Chapter 1,  Exercise 1.1, Q7 Set  A  is the set of all books in the library of a college. Given, R = { x ,  y ):  x  and  y  have the same number of pages} Now, R is reflexive since ( x ,  x ) ∈ R as  x  and  x  has the same number of pages. Let ( x ,  y ) ∈ R  ⇒  x  and  y  have the same number of pages. ⇒  y  and  x  have the same number of pages. ⇒ ( y ,  x ) ∈ R ∴R is symmetric. Now, let ( x ,  y ) ∈R and ( y ,  z ) ∈ R. ⇒  x  and  y  and have the same number of pages and  y  and  z  have the same number of pages. ⇒  x  and  z  have the same number of pages. ⇒ ( x ,  z ) ∈ R ∴R is transitive. Hence, R is an equivalence relation.