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Define the relation on $\mathbb{Q}$ by $$[m,n]<[j,k]$$ if and only if $jn-mk$ belongs to $\mathbb{N}$, $j$ and $m$ belong to $\mathbb{Z}$, $n$ and $k$ belong to $\mathbb{N}$.

(a) Show that $<$ is well defined, that is if $(m,n)\sim (m',n')$ and $(j,k)\sim(j',k')$, then $jn-mk$ belongs to $\mathbb{N}$ if and only if $j'n'-m'k'$ belongs to $\mathbb{N}$. Here, $(m,n)\sim (j,k)$ means $mk=jn$.

(b) Show that $<$ is a total order relation on $\mathbb{Q}$.

I get stuck how to use the conditions: $mn'=m'n$ derived from $(m,n)\sim(m',n')$, $jk'=j'k$ derived from $(j,k)\sim(j',k')$ and $jn-mk$ is an natural number to show that $j'n'-m'k'$ is also an natural number in part (a). Thank you!

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What is your question about this exercise? – Trevor Wilson Feb 10 '13 at 0:44
Is the relation on $\mathbb Q$ or on $\mathbb{Q\times Q}$ or on $\mathbb{Z\times N}$? The first line indicates one thing, but the second another. – Asaf Karagila Feb 10 '13 at 2:17
@Asaf, I'm guessing that OP has forgotten to tell us that $[m,n]$ is notation for what the rest of us call $m/n$, so it really is a relation on the rationals. – Gerry Myerson Feb 10 '13 at 2:23
Elvis, I imagine you aren't just posting this collection of facts so we can all admire them; rather, you want help in establishing them. In which case, it will go better if you tell us how you came across these facts, why it is important to you to establish them, what you know about the terms used in them, how far you have progressed on your own, where you have gotten stuck, and so on. The more you tell us, the better we can help. – Gerry Myerson Feb 10 '13 at 2:26
Hi,[m,n]is an element of ZXZ, I get stuck how to use the conditions: mn'=m'n derived from(m,n)~(m',n'), jk'=j'k derived from (j,k)~(j',k') and jn-mk is an natural number to show that j'n'-m'k' is also an natural number in part (a). Thank you! – Elvis Feb 10 '13 at 19:33

What is it all about? How do you mean '$[m,n]\in\Bbb Q$' ? We are building the numbers out of almost nothing, first the natural numbers $1,2,3,..$ then -to ensure inverse for $+$- the integers, now this $[m,n]$ wants to represent the fraction $m/n$.

How to say with these formal pairs that $m/n < j/k$? This is equivalent to $mk<jn$, that is $jn-mk >0$. (I guess, $0\notin\Bbb N$ in your meaning.) Now, this form is acceptable, since division is not available yet, but the expression $jn-mk$ is already defined in $\Bbb Z$.

So, for the specific question: we are to prove (excluding division and fractions) basically that $m'/n'=m/n < j/k =j'/k'$ implies $m'/n'<j'/k'$. For this, first let's take only one step: $[m',n']=[m,n]<[j,k] \Rightarrow [m',n']<[j,k]$:

So we have $m'n=n'm$ and $mk<jn$. Then, since $n>0$ (denominator), the sign of $m$ and $m'$ is the same. Now assume that $m,m'>0$, and approaching $m'k$ for the proof: $$m'mk<jnm'=jn'm $$ Since $m>0$ is assumed, it follows that $m'k<jn'$. If $m,m'<0$ then the relation symbol will turn twice, and you can also check the case $m=m'=0$.

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Thank you, Beci! Your response gave me lots of hints. Just a little modification on your methods. In the case we assume m>0,m'>0,j>0 and j'>0, we can multiply both sides of the inequality of jn>mk by j'm', then we have j'm'jn>j'm'mk. Then we use the fact that mn'=m'n and jk'=j'k, and we have j'jmn'>jk'm'm, that's j'n'jm>m'k'jm. Since by assumption jm>0, we have j'n'>m'k'as we need. – Elvis Feb 11 '13 at 3:53

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