Show that the function f - N - N given by f 1- f 2-1 and f x - x -1- for every x - 3- is onto but not one-one-

F(1) = f(2) = 1. Therefore f is not one one. Given any y ϵ N, y ≠ 1, we can choose x as y + 1 such that f(y + 1) = y + 1 – 1 = y. Also for 1 ϵ N, we have f (1) ...

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Show that the function f: N $\to$ N given by f(1) = f(2) = 1 and f(x) = x – 1, for every x $\geq$ 3, is onto but not one-one.

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Solution

f(1) = f(2) = 1. Therefore f is not one-one.

Given any y $ϵ$ N, y $\neq$ 1, we can choose x as y + 1 such that f(y + 1) = y + 1 – 1 = y.

Also for 1 $ϵ$ N, we have f (1) = 1.

Therefore f is onto.

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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 ≥ 3, is onto but not one-one.

f(1) = f(2) = 1. Therefore f is not one-one. Given any y ϵ N, y ≠ 1, we can choose x as y + 1 such that f(y + 1) = y + 1 – 1 = y. Also for 1 ϵ N, we have f (1) = 1. Therefore f is onto.

Show that the function f: N $\to$ N given by f(1) = f(2) = 1 and f(x) = x – 1, for every x $\geq$ 3, is onto but not one-one.

f(1) = f(2) = 1. Therefore f is not one-one.

Given any y $ϵ$ N, y $\neq$ 1, we can choose x as y + 1 such that f(y + 1) = y + 1 – 1 = y.

Also for 1 $ϵ$ N, we have f (1) = 1.

Therefore f is onto.