Problem:

Define $\displaystyle h: Z \rightarrow Z$ by the rule

h(n) = 4n-1 for all $\displaystyle n \in Z$

Is h onto?

Solution:

I know that h is not onto because if we suppose $\displaystyle 4n-1 = m \in Z$ then we have to search an $\displaystyle n \in Z$ such that h(n) = m, and for $\displaystyle m=0 \in Z$ there doesn't exist any integer n.

It is all right.As A Counter Example

h(n) = 0

4n-1 = 0

4n = 1

n = 1/4 $\displaystyle \notin$ Z

Role of Rule in Proving a Function? Help!

But, what is the purpose of the rule h(n) = 4n-1 ? because if we stuck on this rule then 4n-1 will never be equal to 0.

For Example:

Any help??According to h(n) = 4n-1 For All $\displaystyle n \in Z$

n = -1 Then 4(-1) - 1

n = 0 Then 4(0) - 1

n = 1 Then 4(1) - 1

n = 2 Then 4(2) - 1

...So on, and in this case 4n-1 will always be an integer thus h will become onto.