Yes. The two distinct pieces of the domain, the non-positive reals and the natural numbers, are mutually exclusive sets. Thus, having different function rules for each does not prevent R from being a function. If you wanted to have a visual representation to which you could apply the "vertical line test," the graph of R looks like the line y = -x for x ε (-∞ , 0] , and a line of points along y = x + 1 for n ∈ N.
Mike L.
asked 03/27/21Let R = {(n, −n) | n ∈ Z and n ≤ 0} ∪ {(n, n + 1) | n ∈ N}. Is R a function? Justify your answer.
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Mike L.
well apparently, No R is not a function. Because (0, 0) ∈ R and (0, 1) ∈ R.03/27/21