MHB Injection, Surjection, Bijection

[KOO]

Can anyone explain to me how to do these types of questions? I have the answers but I don't understand it.

The function f: N -> N, f(n) = n+1 is
(a) Surjection but not an injection
(B) Injection but not a surjection
(c) A Bijection
(d) Neither surjection not injection

The answer is B: Injection but not a surjection

How to do these types of questions?

 

[I like Serena]

Can anyone explain to me how to do these types of questions? I have the answers but I don't understand it.

The function f: N -> N, f(n) = n+1 is
(a) Surjection but not an injection
(B) Injection but not a surjection
(c) A Bijection
(d) Neither surjection not injection

The answer is B: Injection but not a surjection

How to do these types of questions?

We start with the definitions.
A function is surjective if each element in the codomain has at least 1 original.
A function is injective if each element in the codomain has at most 1 original.
A function is bijective if it is both surjective and injective.

If we pick some element in the codomain N, how many originals does it have?

Btw, the answer is not B. Where did you get that from?

 

[Plato2]

Can anyone explain to me how to do these types of questions? I have the answers but I don't understand it.
The function f: N -> N, f(n) = n+1 is
(a) Surjection but not an injection
(B) Injection but not a surjection
(c) A Bijection
(d) Neither surjection not injection
The answer is B: Injection but not a surjection
How to do these types of questions?

It is easy if one learns the definitions.

Any function is a set of ordered pairs,
This function is $$f=\{(n,n+1) :n\in\mathbb{N}\}$$.

A function is injective if no two pairs have the same second terms.

A function is surjective if every term if the finial set is the second term of some pair.