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) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Equal Sets

Show that the...

Question

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.

Open in App

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.

Suggest Corrections

Similar questions

Q. Show that the function

f : N \to N

, given by

f (1) = f (2) = 1

and

f (x) = x - 1

for every

x > 2

, is onto but not one-one.

Q. Prove that the function f : N → N, defined by f(x) = x² + x + 1, is one-one but not onto.

Q. Show that function

f : N \to N

given by

f (x) = 3 x

is one-one but not onto.

Q. Show that the function f : N → N defined by f(x) = x² + x + 1 is one-one but not onto. Find the inverse of f : N → S, where S is range of f.

Show that

f : N \to N

, given by

f (x) = {\begin{matrix} x + 1, if x is odd x - 1, if x is even \end{matrix}

is both one-one and onto.

Join BYJU'S Learning Program

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.