Skip to main content

Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1 , L2 ) : L1 is perpendicular to L2 }. Show that R is symmetric but neither reflexive nor transitive.

RELATIONS AND FUNCTIONS


Class 12, NCERT Chapter 1,  Example 3

Solution

Figure

R is not reflexive, as a line L1 can not be perpendicular to itself, i.e., (L1 , L1 ) ∉ R. R is symmetric as (L1 , L2 ) ∈ R ⇒ L1 is perpendicular to L2 ⇒ L2 is perpendicular to L1 ⇒ (L2 , L1 ) ∈ R. R is not transitive. Indeed, if L1 is perpendicular to L2 and L2 is perpendicular to L3 , then L1 can never be perpendicular to L3 . In fact, L1 is parallel to L3 , i.e., (L1 , L2 ) ∈ R, (L2 , L3 ) ∈ R but (L1 , L3 ) ∉ R.

Example1     
Example2
Example3←you are here

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.