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Question 2

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) = x

Answer

(i) fN → N is given by,

f(x) = x2

It is seen that for xy ∈Nf(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.

 

(ii) fZ → 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.

 

(iii) fR → 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.

 

(iv) fN → N given by,

f(x) = x3

It is seen that for xy ∈Nf(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.

 

(v) fZ → Z is given by,

f(x) = x3

It is seen that for xy ∈ Zf(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.

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