Check the injectivity 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
Please explain me in terms that why the function (i) is injective but (ii) and (iii) are not injective though all functions has f(X)=x²
Answers
Answer:
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Step-by-step explanation:
ANSWER
(i)
f(x)=x
2
It is seen that for x,y∈N,
f(x)=f(y)⇒x
2
=y
2
⇒x=y
∴f is injective
Now, 2∈N, But, there does not exist any x
in N such that f(x)=x
2
=2.
∴f is not surjective.
Hence, function f is injective but not surjective.
(ii)
f(x)=x
2
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)=x
2
=−2
∴f is not surjective.
Hence, function f is neither injective nor surjective.
(iii)
f(x)=x
2
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)=x
2
=−2
∴f is not surjective.
Hence, function f is neither injective nor surjective.
(iv)
f(x)=x
3
It is seen that for x,y∈N,f(x)=f(y)⇒x
3
=y
3
⇒x=y
∴f is injective.
Now, 2∈N. But, there does not exist any element
x in domain N such that f(x)=x
3
=2
∴f is not surjective.
Hence, function f is injective but not surjective.
(v)
f(x)=x
3
It is seen that for x,y∈Z,f(x)=f(y)⇒x
3
=y
3
⇒x=y
∴f is injective.
Now, 2∈Z. But, there does not exist any element
x in domain Z such that f(x)=x
3
=2
∴f is not surjective.
Hence, function f is injective but not surjective.