Determine whether f is a function from Z to R if (a) f(n) = ±n (b) f(n) = √ n2 + 1 (c) f(n) = 1 n2−4

(a). We have to check well defineness of f to be a function in domain and range set..

Now, for $n_1.n_2\in \mathbb{Z}$ such that $n_1=n_2$ , thus

Case-I: if f(n)=n then $f(n_1)=n_1=n_2=f(n_2)$

Case-II: if f(n)=-n, then $f(n_1)=-n_1=-n_2=f(n_2)$

Thus, both cases are well defined. Neveretheless, the function f should assign each n to a single value instead of two different values. Thus $f(n)=\pm n$ is not a function.

(b) As, $f(n)=\sqrt{n^2+1}$ , Now consider $n_1=n_2\in \mathbb{Z}$

Thus, $n_1^2=n_2^2\iff n_1^2+1=n_2^2+1\iff\sqrt{ n_1^2+1}=\sqrt{n_2^2+1}\iff f(n_1)=f(n_2)$ $f(n_1)=\sqrt{n_1^2+1}$ is well defined.

Hence, f is a function.

(c).Given $f(n)=n^2-4$ , thus for $n_1=n_2\in \mathbb{Z}\implies f(n_1)=n_1^2-4=n_2^2-4=f(n_2)$

Thus, f is well defined and hence a function.