The function f(n)=1−n2,n∈Z is not one-to-one:
f(−1)=1−(−1)2=0
f(1)=1−(1)2=0
Hence f(−1)=0=f(1), but (−1=1)
The function f(n)=1−n2,n∈Z is not onto:
there is no integer n with f(n)=1−n2=2, for instance.
If n∈Z, then −n∈Z and
f(−n)=1−(−n)2=1−n2=f(n),n∈Z The function f(n)=1−n2,n∈Z is even.
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