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

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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.

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.

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