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Question
Check the injectivity and surjectivity of the following function:
f : Z → Z given by f(x) = x2
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Solution
f : Z → Z given by f(x) = x2
Here, Z = {0, ±1, ±2, ±3, ....}
Injectivity:
Suppose f(–1) = f(1) = 1
But –1 ≠ 1
∴ f is not a one-one function.
Surjective:
There are many elements in a codomain Z which have no pre-image in the domain Z.
For example, 2 ∈ Z is an element of a codomain, but for f(x) = x2 there is no x ∈ Z such that f(x) = 2, because `sqrt2` is not an integer.
∴ f is not onto.
Hence, f is neither injective nor surjective.
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