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Can someone please explain help which proposition or axiom the following step comes from. I will highlight it in the proof.

$1.20\ (a)$ If $x \in R,\ y\in R$, and $x > 0$, then there is a positive integer $n$ such that $nx > y$.

Pf

Let $A = \{nx\ |\ n \in Z^{+}\}$. Suppose on the contrary that there exists no such element that is greater than $y$. Then $y$ is an upper bound and so there must exists an $\alpha = sup\ A$. Then since $x > 0$, $\alpha - x < \alpha$, and $\alpha - x$ is not an upper bound of $A$. Hence $$\alpha - x < mx$$ for some positive integer $m$. But then $$\alpha < (m+1)x \in A.$$

It seems obvious that we can say that, but I don't see yet exactly which axiom or propositions allows us to make the last statements. Can someone please help clear this up?

Thanks.

zzz2991
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    There should be an axiom that every bounded set of real numbers has a least upper bound, so that $\alpha = \sup A$ exists. Is that the step you mean? – Ben Grossmann May 17 '14 at 16:48
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    Not exactly, I see that. It is the step from $\alpha - x < mx$ to $\alpha < (m+1)x$ Which seems obvious. But i'm trying to stay in flavor of rigour and seeing what allows us to make such a statement – zzz2991 May 17 '14 at 16:51
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    Isn't there an axiom that says that if $a<b$, then $a+x<b+x$? – MJD May 17 '14 at 16:59
  • Oh, actually I think there is. It's in the definition of an ordered field. – zzz2991 May 17 '14 at 17:06

1 Answers1

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Pf

Let $A = \{nx\ |\ n \in Z^{+}\}$. Suppose on the contrary that there exists no such element that is greater than $y$. Then $y$ is an upper bound and

so there must exists an $\alpha = sup\ A$.

This follows from the least-upper-bound property and Theorem(1.19 in the book) which proves the existence of and ordered field $\mathbb{R}$ which has the least-upper-bound-property.

Then since $x > 0$, $\alpha - x < \alpha$, and $\alpha - x$ is not an upper bound of $A$.

Hence $$\alpha - x < mx$$ for some positive integer $m$.

Definition Suppose $S$ is an ordered set, $E \subset S$ and E is bounded above. Suppose there exists an $\alpha\in S$ with the following properties;

  1. $\alpha$ is an upper bound of $E$.
  2. If $\gamma < \alpha $ then $\gamma$ is not and upper bound of $E$

Then $\alpha$ is called as least upper bound of $E $ or simply supremum.

So that "hence" part comes from the 2nd condition for the definition of supremum.

(continuation of proof)

But then $$\alpha < (m+1)x \in A.$$

See that this statement contradicts the 1st condition of the definition of supremum.

How $\alpha-x<mx \implies \alpha <(m+1)x$?

Definition An ordered field is a field $F$ which is also an ordered set, such that

  1. $x+y<x+z$ if $x,y,z\in F$ and $y<z$
  2. $xy>0$ if $x\in F$, $y\in F$, $x>0$, and $y>0$

It follows from the 1st part of the above definition(1.17 in Baby Rudin)

hrkrshnn
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  • I'm more confused about how we got the last step $\alpha < (m+1)x$ from $\alpha -x $. I understand the use of the definition. – zzz2991 May 17 '14 at 17:01
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    I quote MJD's comment "Isn't there an axiom that says that if $a<b$, then $a+x<b+x$" – hrkrshnn May 17 '14 at 17:03
  • Yes, that's right I didn't see that. I guess I should read more carefully. But is that really just an axiom, there is no way to prove the relation? – zzz2991 May 17 '14 at 17:13
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    @DavidJhoo, It is a definition. An ordered field is a field $F$ which is also an ordered set, such that

    (1.) $x+y<x+z$ if $x,y,z\in F$ and $y<z$ (2.) $xy>0$ if $x\in F$, $y\in F$, $x>0$, and $y>0$

    It follows from the 1st part of the above definition(Def 1.17 in Baby Rudin)

     – hrkrshnn May 17 '14 at 17:22
  • Ok Ok, I see :). Thanks – zzz2991 May 17 '14 at 17:28
  • @DavidJhoo If you are satisfied with the answer you can click the tick mark to mark this as an accepted answer. – hrkrshnn May 18 '14 at 06:01