Functions-19

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Show that the function f: Z → Z defined by f(x) = x2 + x for all x ∈ Z is a many-one function.

Solution: Let x, y ∈ Z. Then,

f(x) = f(y) => x2 + x = y2 + y

=>(x2 - y2) + (x - y) = 0

=>(x - y)(x + y + 1) = 0

=>(x - y)(x + y + 1) = 0

=>x = y or y = -x -1

Since, f(x) = f(y) does not provide the unique solution x = y but it also provides y = -x -1, this means that x ≠ y but f(x) = f(y)

when y = -x – 1. For example, if we put x = 1 in y = -x – 1, we obtain y = -2. This shows that 1 and -2 have the same image under f.

Hence, f is a many-one function.

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