There is a problem that calls for the mapping the set of all the positive integers to the set of all positive rational numbers in order to prove they have the same cardinality. I know more the common answer of mapping it this way:
Map $\mathbb{Q}$ like a table of ℤ x ℤ:
\begin{array}{rl} & &1 &2 &3 &4 &5 \\ \\ 1 & &1/1 &2/1 &3/1 &4/1 &5/1 \\ 2 & &1/2 &2/2 &3/2 &4/2 &5/2 \\ \end{array}
Then you show the cardinality of that as in {1/1,2/1,3/1,4/1,/5/1,1/2/3/2 ... } is the same as the carnality of {1,2,3,4,5,6}
However, my first idea was much different. Now, I have not yet taken very high level courses, this is my intro to set theory. But why couldn't I do something like this:
The first element in the set $\mathbb{Q}$ is:
ε, where ε is an infinitesimal approaching 0. the second element of $\mathbb{Q}$ is ε, where ε is an infinitesimal approaching the first member, and so forth.
so
$$S(n) = \sum_{i=1}^{n} ε_i$$
where n ∈ $\mathbb{Z}$
This was the essence of my idea. I am sure it is probably way of base. But could someone explain why?
EDIT:
I think I understand where I was going wrong. Defining the infinitesimal here is non standard, and I can't think of it like a limit. The sum I use is also not correct.
$$S = \{\varepsilon_i : i \in \mathbb Z\}$$
is a far better way to represent what I am trying to say.
Then my question boils down to asking if $$S = \{\varepsilon_i : i \in \mathbb Z\}$$ is equal to the set of the positive rational numbers. In which case, no. Because the infinitesimal cant be used in this way.
@graydad mentioned another cool / non-standard way to map $\mathbb{Z}$ onto $\mathbb{Q}$
Let $f:\Bbb{Q}^+ \to \left\{2^n3^m :n,m \in \Bbb{Z}^+\right\}$ be defined by $f(n/m) = 2^n3^m$ where $n/m$ is in lowest reduced form. Then by the F.T. of Arithmetic we know $2^n3^m \neq 2^l3^p$ whenever $n/m \neq l/p$. You can show $f$ is a bijection and that the set $\left\{2^n3^m :n,m \in \Bbb{Z}^+\right\}$ has the same cardinality as $\Bbb{Z}$.
That is pretty interesting!
Interesting idea, but there are several major problems here. For one, the approach lacks rigour: what exactly do you mean for $\varepsilon$ to be an "infinitesimal approaching zero"? What is this thing? Does it have a fixed value, or is it moving around?
Just as importantly, if $\varepsilon$ is "infinitesimal" or "approaching zero", then it certainly doesn't sound like an element of $\mathbb Q$.
Third, what is $S(n)$? You are defining it to be the sum of the $\varepsilon_n$, but perhaps you want it to be a set of these objects: $$S = \{\varepsilon_i : i \in \mathbb Z\}$$
Also, even if this method with infinitesimals were possible, you haven't proved in any way that it leads to a countable set. How do you know you won't need uncountably many of these infinitesimals?