Show that f : N → N, given by f(x)={ x+1, if x is odd, and x-1, if x is even} is both one-one and onto.

Class 12, NCERT Chapter 1, Example12

Suppose f(x₁ ) = f(x₂ ).
Note that if x₁ is odd and x₂ is even, then we will have
x₁ + 1 = x₂ – 1, i.e., x₂ – x₁ = 2, which is impossible.
Similarly, the possibility of x₁ being even and x₂ being odd can also be ruled out, using a similar argument.
Therefore, both x₁ and x₂ must be either odd or even.
Suppose both x₁ and x₂ are odd.
Then f(x₁ ) = f(x₂ )
⇒ x₁ + 1 = x₂ + 1
⇒ x₁ = x₂
Similarly, if both x₁ and x₂ are even, then also
f(x₁ ) = f(x₂ )
⇒ x₁ – 1 = x₂ – 1
⇒ x₁ = x₂ .
Thus, f is one-one.
Also, any odd number 2r + 1 in the co-domain N is the image of 2r + 2 in the domain N and
any even number 2r in the co-domain N is the image of 2r – 1 in the domain N.

Thus, f is onto.

Comments

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.

. 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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