Question

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

Solution

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

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