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.