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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.

MathMan
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  • What is the "known result?" This is the key of the proof, and you've just elided it. It is where you are using the Archimedian property.... – Thomas Andrews Aug 28 '14 at 13:38
  • Ohkay, I will write that down.. – MathMan Aug 28 '14 at 13:38
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    Also, your statement is a bit confused. You want to prove: $\forall z>0, \exists r>0,\left( (r\text{ rational }) \land (0<r<z)\right)$. But, because you put the $\forall$ after the statement, it could be read as $\exists r>0,\forall z>0 ...$ which isn't true. $r$ depends on $z$... – Thomas Andrews Aug 28 '14 at 13:41
  • Also, who is Bartle, and what statement does he say is false in general? Is this a statement from Bartle's Introduction to Real Analysis? – Thomas Andrews Aug 28 '14 at 13:43
  • Presumably Bartle means there is no positive infinitesimal. Check the placement of the universal quantifier. – Bill Dubuque Aug 28 '14 at 13:45
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    @VHP The step should be: $z\left(\frac{nz-1}{z}\right)>0$, not $z^{-1}\left(\frac{nz-1}{z}\right)>0$, which doesn't cancel. – Thomas Andrews Aug 28 '14 at 13:48
  • Okay, the proof looks complete. Unless you give us more text from Bartle, it isn't clear how to answer your last question. Exact phrasing matters here... – Thomas Andrews Aug 28 '14 at 13:50
  • @ThomasAndrews Made the changes. It's called the elements of real analysis by Robert G Bartle. – MathMan Aug 28 '14 at 13:50
  • @BillDubuque I do understand if the book actually means there is no positive infinitesimal – MathMan Aug 28 '14 at 13:51
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    You might find this answer of interest for some intuition on related matters. – Bill Dubuque Aug 28 '14 at 16:23

1 Answers1

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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.

Thomas Andrews
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  • The book says - There is a positive rational element $r$ such that $r<z$ for any positive $z$ in $F$.

    The reader should convince himself that this statement is false

     – MathMan Aug 28 '14 at 13:56
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    Yeah, it depends on how you parse that sentence. It could be read: "(There is a positive rational element $r$ such that $r<z$) for any positive $z$ in $F$." That's the first sentence above. It could also mean: "There is a positive rational element $r$ such that ($r<z$ for any positive $z$ in $F$.)" The first reading is what you proved above, and it is true, the second reading is the second sentence in my answer, and is false. – Thomas Andrews Aug 28 '14 at 14:01
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    That's the problem with putting the universal quantifier at the end like that in the written sentence - it is ambiguous. – Thomas Andrews Aug 28 '14 at 14:02