# A simple question from Rudin's principles of mathematical analysis

1. Jul 25, 2011

It's NOT a homework. I'm self-studying Rudin's Analysis and I came to this part that I can't follow the argument after a certain point:

statement: if x,y $\in$ R, and x<y, then there exists a p $\in$ Q such that x<p<y.

Proof: since x<y, we have y-x>0, and the Archemedean property furnishes a positive integer n such that:
n(y-x)>1
Apply the Archemedes' property again to obtain positive integers m1 and m2 such that m1>nx, m2>-nx. Then:
-m2<nx<m1.

[Here is where I get confused].
Hence there is an integer m (with -m2<=m<=m1) such that: *
m-1 <= nx < m. *
If we combine these inequalities, we obtain
nx < m <= 1 + nx < ny. *

Since n>0, it follows that
x < m/n <y.

I don't understand the places that I've marked with *. I fail to follow the argument after the place that I've mentioned it. Can anyone clarify up these statements for me please?

2. Jul 25, 2011

### micromass

Denote

$$m=min\{k~\vert~nx<k\}$$

Then m will satisfy $m-1\leq nx<m$ (try to prove this). There is only one question: does the minimum exist? That is, does the set $\{k~\vert~nx<k\}$ nonempty and does it has a lower bound.

What exactly don't you understand here?
$nx<m$ by the choice of m
$m\leq 1+nx$ since $m-1\leq nx$
$1+nx<ny$ since $n(y-x)>1$.

3. Jul 25, 2011

well, since x is a positive real number and n is a natural number then nx>0. k>nx by definition, therefore k is a positive integer. to prove that the set is nonempty I should show that no matter how large nx is, there is always a k such that nx<k. well, suppose that It wasn't true. therefore for any positive integer k we had k<=nx which leads to a false statement that the set of positive integers has an upper bound. so there exists a positive integer k such that nx<k which means the set is non-empty. Now we know that the set $$A = \{k~\vert~nx<k\}$$ is non-empty and its members are positive integers. the well-ordering principle now guarantees the existence of the smallest k in the set which is the m we want. so the existence part of the theorem is now completed.
Now to prove that $m-1\leq nx<m$ one need to realize that if [itext]m-1\leq nx[/itext] wasn't true, then we had nx<m-1. but m-1<m and m was the smallest positive number s.t nx<k, which is a contradiction. that proves the inequality part.

Now I understand it completely.

Are we done or there are still holes in the argument that we need to fill?

4. Jul 25, 2011

### micromass

OK, a few remarks: we don't know if x is positive or not. Second, it was easier if you would just use m1 and m2 in your argument. The existence of m
2
provides a lower bound, and the existence of m1 says that the set is nonempty.

It's alright, I think.

5. Jul 25, 2011