Question

Check the injectivity and surjectivity of the following functions:

1. f :**** N** ** $\to$ N given by f(x) = x2
2. f : Z** ** $\to$** ** Z given by f(x) = x2
3. f : R** ** $\to$** ** R**** given by f(x) = x2
4. f : N** ** $\to$** ** N given by f(x) = x3
5. f : Z $\to$ Z given by __ f(x) = x3

Answer 1

(i) f : N $\to$ N is given by,
f(x) = x2
It is seen that for x, y ∈ N, f(x) = f(y) ⇒ x2 = y2 ⇒ x = y.
∴ f is injective.
Now, 2 ∈ N. But, there does not exist any x in N**** such that f(x) = x2 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.

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(ii) f : Z** ** $\to$** ** Z is given by,
f(x) = x2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ f is not injective.
Now,−2 ∈ Z. But, there does not exist any element x ∈ Z**** such that f(x) = x2 = −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.

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(iii) f :**** R** ** $\to$** ** R**** is given by,
f(x) = x2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ f is not injective.
Now,−2 ∈ R. But, there does not exist any element x ∈ R**** such that f(x) = x2 = −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.

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(iv) f : N** ** $\to$** ** N given by,
f(x) = x3
It is seen that for x, y ∈ N, f(x) = f(y) ⇒ x3 = y3 ⇒ x = y.
∴f is injective.
Now, 2 ∈ N. But, there does not exist any element x in domain N**** such that f(x) = x3 = 2.
∴ f is not surjective
Hence, function f is injective but not surjective.

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(v) f : Z** ** $$$\to$** ** Z is given by,
f(x) = x3
It is seen that for x, y ∈ Z, f(x) =f(y) ⇒ x3 = y3 ⇒ x = y.
∴ f is injective.
Now, 2 ∈ Z. But, there does not exist any element x in domain Z**** such that f(x) = x3 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.