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.
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Solution f is not one-one, as f(1) = f(2) = 1. But f is onto, as given any y ∈ N, y ≠ 1,we can choose x as y + 1 such that f(y + 1) = y + 1 – 1 = y. Also for 1 ∈ N, wehave f(1) = 1.
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