mock test question 5

[Kalyani Manoj]

A function is called onto if the range of the function is same as the codomain of the function.

 • f : R −→ N f(n) = ( 0 if n ∈ Q 1 otherwise Range(f) = {0, 1}, is not the same as Codomain(f) = N.

 • f : N −→ N f(n) = ( n if n is even n + 1 otherwise Range(f) = Set of even numbers, is not same as Codomain(f) = N.

 • f : N −→ N f(n) = ( n if n is odd 2n otherwise Range(f) = N \ {2}, is not the same as Codomain(f) = N.

 • f : N −→ N f(n) = ( n 2 if n is even n + 1 otherwise Range(f) = N, is same as Codomain(f) = N. Hence f is onto.

 • f : R −→ R f(n) = n Range(f) = R, is same as Codomain(f) = R. Hence f is onto  

can you please explain cuz i still dont understand this...how did they find out the range.. please help

[Kalyani Manoj]

why is   f : N −→ N f(n) = ( n/ 2 if n is even n + 1 )    surjective (isnt n is set of even the range???)

and   f : N −→ N f(n) = ( n 2 if n is even n + 1  not surjective???

 please help me



[Mathematics 1 Support 2]

Dear Kalyani,

Take numbers from 0 to 10 and try to solve using the conditions given for better clarity. 

Thanks and regards 

Vicky Kumar Sharma 

Maths-1 instructor

IITM Online Degree 

[D D]

but 0 is not there in the range but in co domain, 0 is also a natural number. it should not be surjective