Is There a Surjective Function from Z+ to Z?

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Homework Help Overview

The discussion revolves around the existence of a surjective function from the set of positive integers (Z+) to the set of all integers (Z). Participants explore the implications of countability and the challenges posed by the inclusion of zero in Z.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster questions whether a function can be constructed to map Z+ onto Z, expressing concern about the subset relationship and the presence of zero in Z. Another participant presents a specific function they derived, which appears to map Z+ to Z, prompting further discussion about the function's complexity and the possibility of simpler alternatives.

Discussion Status

The conversation is active, with participants sharing ideas and questioning the nature of the function. While one participant expresses confidence in the existence of a surjective function, there is still exploration of alternative forms and simplifications.

Contextual Notes

The original poster's hesitation stems from the assumption that Z+ may not have enough elements to cover all integers in Z, particularly due to the inclusion of zero. The discussion is framed within the context of homework help, suggesting constraints on the types of solutions being sought.

phoenixy
Hi,

Does there exist a function f: Z+ --> Z which is onto?

I had been told there such funciton exists, since both Z+ and Z are countable infinite series. Thus there exists some transformation that could map Z+ to every single Z

However, I still can't shake off the idea that since Z+ is a subset of Z, there just aren't "enough" Z+ to cover every single Z, and the 0 in Z is giving me trouble as well


Thanks for any input
 
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After goofing around with pencil and paper,

If N is a positive integer, it seems like this does the trick:

f(N)= (N/2)(-1)^N + 1/4 + (1/4)(-1)^(N+1).

This gives:
f(1)=0
f(2)=1
f(3)=-1
f(4)=2
f(5)=-2
f(6)=3
f(7)=-3

and so on. Is that the sort of function that you are talking about?
 
Oh wow, that looks like it.

Now I'm a firm believer of countable infinity. :smile:


Your equation will do, thanks!

I'm wondering if there is any easier function. This question isn't suppose to be a tough one.
 
I'll bet there is one that looks less messy, given that I just kludged that one up by trial & error.
 

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