If $y,z$ are elements of an archimedean field $F$ and if $y<z$, then there is a rational element $r$ of $F$ such that $y<r<z$

Question

If $y,z$ are elements of an archimedean field $F$ and if $y<z$, then there is a rational element $r$ of $F$ such that $y<r<z$

The proof begins with saying that it is no loss of generality that we assume that $0<y<z$

I don't understand well why this the case. Please guide me .

I think there will be loss of generality because suppose : $z>0$ but still it's possible that $y<0$ i.e. $z \in P ~;~ y \notin P$ where $P$ is a positive class in $F$

Thank you for your help.

Answer 1

It's not because in an ordered field if you have arbitrary elements $a,b,c\in F$ then

$$a<b\iff a+c<b+c$$

Hence if $0<y$, for example, then you can just add $1-y$ to both sides to get

$$0<1<z+1-y$$

and you have $y'=1, z'=z+1-y$ so that your new question is only about positives, and is equivalent.

In particular, say that $r$ is rational between any $y,z\in F$ and say $-N<y$ is an integer less than $y$, then

$$0<y+N<r+N<z+N$$

so that

$$\begin{cases} y'=y+N\\ r'=r+N \\ z'=z+N \end{cases}$$

and $y<r<z\iff y'<r'<z'$, so that things work the same with positives as with any choice. Also, clearly $r'$ is rational.

Answer 2

It seems to me that by "without loss of generality," the author really means "since the other case is much easier:" if $z>0$, $y<0$, it's clear that $0$ is the desired rational element. If both are negative, multiply by $-1$.