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!
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 $m=m'=0$.