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Show that the function f : N → N given by f(1) = f (2) = 1 and f (x) = x – 1, for every x > 2, is onto but not one-one.


f : N → N is given by f (1) = f (2) = 1 and f (x) = x – 1.
Since    f (1) = f (2)
∴ f is not one-one Given any y ∈ N, y ≠ 1, we can choose x as y + 1 such that f(y +1) = y + 1 – 1 = y Also    f (1) = 1
∴ f is onto.

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