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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

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(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.


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