# Role of Rule in Prooving a Function? Help!

• Mar 30th 2010, 11:17 AM
mrsenim
Role of Rule in Proving a Function? Help!
Problem:
Define $h: Z \rightarrow Z$ by the rule
h(n) = 4n-1 for all $n \in Z$
Is h onto?

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

Quote:

As A Counter Example
h(n) = 0
4n-1 = 0
4n = 1
n = 1/4 $\notin$ Z
It is all right.

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:

Quote:

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

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.
Any help??
• Mar 30th 2010, 01:48 PM
emakarov
Not sure what your question is.

Quote:

I know that h is not onto because if we suppose http://www.mathhelpforum.com/math-he...b94db2ea-1.gif then we have to search an http://www.mathhelpforum.com/math-he...cf2a7dbb-1.gif such that h(n) = m, and for http://www.mathhelpforum.com/math-he...83a1995d-1.gif there doesn't exist any integer n.
Correct.

Quote:

But, what is the purpose of the rule h(n) = 4n-1 ?
You need the definition of a function to prove that it is onto, don't you? You can't prove that some unknown function is a surjection because not all functions are.

Quote:

So on, and in this case 4n-1 will always be an integer thus h will become onto.
How does it follow that h is onto if its values are always integers? For h to be onto, its values must be all integers.