Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.

Class 12, NCERT Chapter 1, Exercise 1.2, Q8

f: A × B → B × A is defined as f(a, b) = (b, a).

∴ f is one-one.

Now, let (b, a) ∈ B × A be any element.

Then, there exists (a, b) ∈A × B such that f(a, b) = (b, a). [By definition of f]

∴ f is onto.

Hence, f is bijective.

Comments

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.

. Let f : R → R be defined as f(x) = 3x. Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto.

Class 12, NCERT Chapter 1, Exercise 1.2, Q12 f : R → R defined as f ( x ) = 3 x . Let x , y ∈ R such that f ( x ) = f ( y ). ⇒ 3 x = 3 y ⇒ x = y ∴ f is one-one. Also, for any real number ( y) in co-domain R , there exists in R such that . ∴ f is onto. Hence, function f is one-one and onto. The correct answer is A.

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