Show that the function `f: Z->Z` defined by `f(x)=x^2` for all `x in Z` , is a function but not bijective function.

Explanation

Let `f:Z↦Z` be defined as `f(x)=x^2,x∈Z`.

We know that the square of an integer is always a unique integer.

So, 'f'` is a function.

Now, since `f(−2)=f(2)=4`, ''f'' is not an injection.

There is no integer
`x∈Z:f(x)=−1`.

Thus ''f'' is neither one-one nor onto, ''f'' is not a bijection.