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Question

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

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Solution

(i) f: NN is given by,

f(x) = x2

It is seen that for x, yN, f(x) = f(y) ⇒ x2 = y2x = 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.

(ii) f: ZZ 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.

(iii) f: RR 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.

(iv) f: NN given by,

f(x) = x3

It is seen that for x, yN, f(x) = f(y) ⇒ x3 = y3x = 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.

(v) f: Z Z is given by,

f(x) = x3

It is seen that for x, yZ, f(x) = f(y) ⇒ x3 = y3x = 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.


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