L.et 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 student x. Show that f is one-one but not onto.
A is the set of all 50 students of class X in a school.
No two different students of the class can have same roll number.
Therefore, f must be one-one.
We can assume without any loss of generality that roll numbers of students arc from 1 to 50. This implies that 51 in N is not roll number of any student of the class. so that 51 can not be image of any element of X under f. Hence, f is not onto.
f : N → N is given by f (x) = 2x Let ,x1, x2 ∈ N such that f (x1) = f (x2)
∴ 2 x1 = 2 x2 ⇒ x1 = x2 ∴ f is one-one.
f is not onto as for 1 ∈ N, there does not exist any x in N such that f (x) = 2 x = 1.
Check the injectivity and surjectivity of the following functions :
(i) f : N → N given by f (x) = x2
(ii) f : Z → Z given by f (x) = x2
(iii) f : R → R given by f (x) = x2 (iv) f : N → N given by f (x) = x3
(v) f : Z → Z given by f (x) = x3
(i) f : N → N is given by f (x) = x2
Let x1 x2 ∈ N be such that f(x1) = f(x2)
∴ x12 = x 22 ⇒ x2 2 x1 2 = 0
⇒ (x2 –x1) (x2 + x1) = 0
⇒ x2 – x1 = 0 [x1 + x2 ≠ 0 as x1, x2 ∈ N]
⇒ x2 = x1 ⇒ x1 = x2
∴ f is one-one, i.e.. f is injective.
Since range of f = { 12 , 22, 32............}
= {1.4.9......} ≠ N.
∴ f is not surjective.
(ii) f : Z → Z is given by f (x) = x2
Let x1, x2 ∈ Z be such that f(x1)= f (x2)
∴ x22 =x22 ⇒ x22 = 0
⇒ (x2 – x1) (x2 + x1) = 0
⇒ x2 = x1 or x2 = – x1
∴ f (x1) = f(–x1) ∀ x1 ∈ Z
∴ f is not one-one, i.e. f is not injective.
Also range of f = { 02, 12, 22,.....}
= {0, 1,4, 9,.........}
≠ Z
∴ f is not onto i.e.. f is not surjective (iii) f : R → R is given by f (x1) = x2 Let x1, x2 ∈ R be such that f (x1) = f (x2)
⇒ x12 = x22 ⇒ (x2 – x2) (x2 + x1) = 0
⇒ x2 = x1 or x2 = – x1
⇒ f(x1) = f (–x1) ∀ x1 ∈ R
∴ f is not one-one, i.e., f is not injective.
As range of f does not contain any negative real, therefore, range of ≠ R.
Hence. f is not onto, i.e., f is not surjective.
(iv) f : N → N is given by f (x) = x3 Let x1 ,. x2 ∈ N be such that f (x1) = f(x2)
⇒ x13 = x23 ⇒ x1 = x2
∴ f is one-one, i.e., injective.
Also range of f = {13, 23, 33,.........}
= {1,8,27,.....}
≠ N
∴ f is not onto, i.e.,f is not surjective.
(v) f : Z → Z is given by f (x) = x3 Let x1, x2 ∈ Z be such that f (x1) = f (x2)
⇒ x13 = x23 ⇒ x1 = x2
Also range of f = {03 ± 13, ± 23, ± 33,....}
= {0, ± 1, ± 8, ± 27.............}
≠ Z ∴ f is not onto, i.e., f is not surfective.
f : N → N is given by f (x) = 5 x
Let x1, x2 ∈ N such that f (x1) = f (x2)
∴ 5 x1 = 5 x2 ⇒ x1 = x2 ∴ f is one-one i.e. injective.
f is not onto i.e. surjective as for 1 ∈ N, there docs not exist any in N such that f (x) = 5 x = 1
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