# Proof of the Archimedean Property

1. Dec 22, 2014

### Bashyboy

I am reading Rudin's proof of this property, but I find one assertion he makes quite disagreeable to my understanding; I am hoping that someone could expound on this assertion. Here is the statement and proof of the archimedean property:

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

(a) Let A be the set of all nx. If (a) were false, then y would be an upper bound of A. But then A has a least upper bound in R...

I understand the method of proof he is employing, a proof by contradiction, which leads to y being larger than every element A (hence, it is an upper bound). However, I do not understand how this immediately implies that A must have a least upper bound. Which theorem is he invoking to come to such a conclusion?

EDIT: Also, what is the quantification on x and y?

2. Dec 22, 2014

### ShayanJ

I think its like this: If there is no n, such that $nx > y$, then y is greater than all the elements of A. But because the elements of A can be arbitrarily large, this means that $y=\infty$. But this is a contradiction because we chose y in the beginning and we could have chosen a finite real number.
And about your last question, the theorem can be written like below:
$\forall x,y \in \mathbb R \ \ \ (x>0 \rightarrow (\exists n\in \mathbb N \ \ \ (nx>y)))$

3. Dec 23, 2014

### Stephen Tashi

My guess is "the completeness of the real number system". http://en.wikipedia.org/wiki/Completeness_of_the_real_numbers

4. Dec 23, 2014

### Bashyboy

So, would the idea of $A$ having a least upper bound even be necessary for the proof?

5. Dec 23, 2014

### ShayanJ

Actually it isn't clear to me. That's why I proposed my own version of it. Maybe if you give the proof completely, I can tell!

EDIT:Well, I didn't use a least upper bound for A, so it doesn't seem necessary!

6. Dec 23, 2014

### jbriggs444

If A is the set of all real numbers of the form nx for some natural n and if y is an upper bound on A then by the least upper bound property, A must have a least upper bound. Call this bound B.

B is a real number. B minus x is a real number less than B. Since B is a least upper bound on A, B-x cannot be an upper bound on A. So there must be at least one member of A between B-x and B. That is, there is a real number of the form nx between B-x and B. Pick any such number and add x to it.

This new number is also of the form nx for some n. Thus it is a member of A and is greater than B. Contradiction.

Edit: Note that the proof offered by Shyan invokes infinity without definition and is, thus, invalid.

7. Dec 23, 2014

### ShayanJ

Actually I thought I have the idea right and people can fix the probable handwavy things in it. You mean it can't be fixed?

8. Dec 23, 2014

### DarthMatter

Normally you do not know that to each real number there is a bigger natural number when you prove this theorem.

9. Dec 23, 2014

### jbriggs444

Pulling that proof back up and looking at it...

$\infty$ is not a real number. So y cannot be equal to $\infty$. Further, there is no predicate "is finite" defined for the real numbers.

Edit: I missed a flaw. We didn't pick y in the beginning. The other team did. The assertion is that the relationship holds for all y. Even if there is such a thing as an infinite y, there still has to be an n large enough so that nx > y.

Last edited: Dec 23, 2014
10. Dec 24, 2014

### DarthMatter

Oops. Silly me. I thought we were on the same team. :D