1:1 and onto proof of Z+ and Q+

[takk]

Homework Statement

After exploring this problem a bit out of curiosity I decided to attempt to write a full fledged proof of it for my final exam.
I will try my best to explain what I am doing. i am pretty frustrated with it but I've put so much time into it I don't want to give up and do something else now.

For this theorem Equality in Q+: \frac{a}{b}=\frac{c}{d} \Leftrightarrow a=c, and b=d

now I'm using the grid system, i.e.
1/1 1/2 1/3 1/4 1/5 1/6..
2/1 2/2 2/3 2/4 2/5 2/6..
3/1 3/2 3/3 3/4 3/5 3/6..
.
.
but adjusting it to represent the diagonals and defining it as follows
1/1 \Rightarrow p+q=2 (1 element)
1/2 2/1 \Rightarrow p+q=3 (2 elements)
1/3 2/2 3/1 \Rightarrow p+q=4 (3 elements)
etc
as you can see each list represent one diagonal of the grid

next I went to the integers by observing the following number line and looking at the sums that represent the integers

(-----|](--|-----|](---|-----|-----|](---|-----|-----|-----|]-----|-----|
0 1 2 3 4 5 6 7 8 9 10 11 12

Representing each section with a summation we get
(0,1]
\sum^{1}_{k=1}k
(1,3]
\sum^{2}_{k=1}k
(3,6]
\sum^{3}_{k=1}k
(6,10]
\sum^{4}_{k=1}k

now I'm trying to show f:Z^{+}\rightarrow Q^{+}
f(1) = 1/1

let n\geq2\inZ^{+}
let S_{n}={s\inZ^{+}\leq\sum^{j}_{k=1}k}

then consider 2n \sum^{2n}_{k=1}k= 1+2+...+n+n+1+...+2n
thus n\leq\sum^{2n}_{k=1}k
therefore 2n\inS_{n} therefore S_{n}\neq\oslash
Then by the well ordering principle \exists m\inS_{n} \exists m\leqj \forallj\inS_{n}

i.e exists a "least" element
Call the least element M

therefore
\sum^{m-1}_{k=1}k<n\leq\sum^{m}_{k=1}k
or in other words m\in (\sum^{m-1}_{k=1}k, \sum^{m}_{k=1}k]

so let p= n-\sum^{m-1}_{k=1}k
1\leqp\leqm

then choose a q such that p+q = m+1
q= m+1-p

now f(n) = p/q
That ends the mapping

Now show f:Q^{+}\rightarrowZ^{+} is 1:1 and onto.
This is where I can't finish the proof but this is what I have of an attempt on both

One to one:
x_{1}\neqx_{2} \Rightarrow f(x_{1})\neqf(x_{2})
or
f(x_{1})=f(x_{2})\Rightarrow x_{1}=x_{2}

suppose f(n_{1})=f(n_{2})

then \frac{p_{1}}{q_{1}}=\frac{p_{2}}{q_{2}}
I have no idea how to get from here to n_{1}=n_{2}

Onto:
consider p+q
the diagonals the p/q is in has the property that the # of elements in that diagonal is p+q-1 (referring way back to the beginning)
or in other words (\sum^{p+q-2}_{k=1}k, \sum^{p+q-1}_{k=1}k]
then let n= p+\sum^{p+q-2}_{k=1}k

and then I get hung up again

 

[mfb]

You seem to prove some unnecessary or trivial stuff while other parts are wrong.

For this theorem Equality in Q+: \frac{a}{b}=\frac{c}{d} \Leftrightarrow a=c, and b=d

So $\frac{1}{2}=\frac{2}{4} \Leftrightarrow$ 1=2 and 2=4? That is wrong.

Representing each section with a summation

What does that mean?

let S_{n}={s\inZ^{+}\leq\sum^{j}_{k=1}k}

I guess the sum is supposed to run up to n instead of j?
Okay, so as an example:
S₃ = {1,2,3,4,5,6}
It is obvious that S_n is not empty (as an example, it always contains 1) and that it has a smallest element (it always contains 1 and 1 is the smallest number in Z⁺).

\sum^{m-1}_{k=1}k<n\leq\sum^{m}_{k=1}k

Why should $n \leq \sum^{1}_{k=1} k = 1$ be true?

This plus the wrong way to start make it impossible to finish your mapping or show that it has the desired properties.