In an Archimedian Field $F$, there is a positive rational element $r$ such that $r < z$ for any $ z>0$ in $F$ . Is this statement true?
Attempt: An archimedian field $F$ is an ordered field in which $\forall z \in F,~~ \exists ~n\in N$ such that $n-z>0 ~~i.e.~ n-z \in P$ where $P$ is a positive class in $F$.
Since, $z>0 \implies z \in P$. Hence, by a known result, we have a natural number $n$ such that $0< \dfrac {1}{n} <z$.
Hence, clearly the positive rational number $r <z$ is $\dfrac {1}{n}$ here.
Then, why does Bartle say that this statement is in general false?
Edit: Let $F$ be an archimedian field, then , if $z>0~~\exists n\in \mathbb N$ such that $0< \dfrac {1}{n} <z$
Proof: if $z>0$, then $1/z >0$. Hence, there exists a natural number $n$ such that $n >\dfrac {1}{z} \implies n- \dfrac {1}{z} >0 \implies \dfrac {nz-1}{z} >0$
$\implies z (\dfrac {nz-1}{z}) >0$
$ \implies nz-1>0$
$ \implies n^{-1} (nz-1) >0$ (As $n \in P)$
$\implies (z-\dfrac {1}{n})>0$ or $z>\dfrac {1}{n}>0$
Thank you for your help.
The proof looks good. You might just use that if $a>b$ and $c>0$ then $ac>bc$, rather than doing all that algebra. (If you don't have this result, it is easy to prove as a lemma.) Then you'd have:
$$n>\frac{1}{z}\implies nz>1\implies z>\frac{1}{n}$$
Where we first use $c=z>0$ then use $c=\frac{1}{n}>0$.
Without further context from the book, it is hard to know what Bartle means, but perhaps he means that the statement:
$$\forall z>0\,\exists r\in\mathbb Q\,( 0<r<z)$$
is true, but:
$$\exists r\in\mathbb Q\,\forall z>0\,( 0<r<z)$$
is false.