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Question

Check the injectivity and surjectivity of the following functions:
(i) f:ZZ given by f(x)=x2


Solution

f:RR is given by f(x)=x2. It is seen that f(-1)=f(1)=1 but 11,
Therefore, f is not injective.
Now, 2R. But there does not exist any element xZ such that f(x)=x2=2
Therefore, f is not surjective.
Hence, function f is neither injective nor surjective.

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