# Clarification for proof of $\mathbb{Q}$ being dense in $\mathbb{R}$ (Rudin's PMA)

Theorem 1.20(b) on page 9 of Rudin's "Principles of Mathematical Analysis," 3rd edition. For those without the text handy:

1.20 Theorem

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

(b) If $x \in \mathbb{R}$, $y \in \mathbb{R}$, and $x < y$, then there exists a $p \in \mathbb{Q}$ such that $x < p < y$.

Also a picture to the page in question: http://i.imgur.com/bufiYkE.png

We are given that $x < y$, thus $y - x > 0$ is obvious to me. I quickly lose Rudin after this step. I understand that the archimedean property $(a)$ is being used for the next line, where it says $n(y - x) > 1$, however I have no clue where the number "$1$" came from.

Furthermore, he says to apply $(a)$ again, but I have no idea what it means to "apply" a theorem arbitrarily. He doesn't say what to apply it to, and if he meant to apply it on $n(y - x) > 1$, then I am even more confused with the following step. He "applies" $(a)$ to obtain positive integers $m_1$ and $m_2$ such that $m_1 > nx$ and $m_2 > -nx$. As far as I understand, the archimedean property says that for $x > 0$, there is a positive integer $n$ such that $nx > y$. I don't understand how $m_1 > nx$ and $m_2 > -nx$ follow this property. In $m_1 > nx$, the equality sign is reversed from the archimedean property definition, and for $m_2 > -nx$, there is a negative sign.

And then the next line, he says "hence" there is an integer $m$ (with $-m_2 \leq m \leq m_1$) such that $m - 1 \leq nx \leq m$. I don't understand how you can deduce from the previous lines ($m_1 > nx$ and $m_2 > -nx$) to arrive at this one. Where are all these $m$'s coming from? The only connection I see is that number $1$ from $n(y - x) > 1$ from the first step. I have no clue where these $m$'s appeared out of nowhere.

• For your first question, as $y - x > 0$, there is $n \in \Bbb{N}$ such that $y - x > \frac{1}{n}$. – Michael Albanese Jul 13 '14 at 1:40
• Take $x=1>0$ to prove that if $y$ is any real then there is an integer $n>y$. Then there is $m_1$ and $m_2$... – Thomas Andrews Jul 13 '14 at 1:44
• You ask where the number $1$ came from. Think about what the theorem wants to establish: That whenever $x<y$, there is a rational in between. If this is true, then in particular it will be true when $x=0$, so we need to verify that whenever $y>0$, there is a rational $r$ with $0<r<y$. But if $r$ is rational, we can write it in the form $k/n$ for some integers $k,n$, with $n>0$. Now, if $0<k/n<y$, then certainly $k>0$, so $k\ge 1$, so we also have $0<1/n\le k/n<y$, and it would be enough to prove that if $y>0$ then there is an $n$ such that $0<1/n<y$. (Cont.) – Andrés E. Caicedo Jul 13 '14 at 2:29
• This is the same as proving that if $y>0$, then there is an $n$ with $1<ny$. This is precisely what Rudin begins with, only that he begins not with a positive $y$, but with reals $x<y$, so that now it is $y-x$ that is positive, and therefore what we need is an $n$ with $n(y-x)>1$. It turns out that this particular case is actually useful to prove the general result (that between any two numbers there is a rational), and so he proves this first, and proceeds from there. – Andrés E. Caicedo Jul 13 '14 at 2:31
• I suggest to read Robert Bartles proof of this result. I get the feeling that Rudin is proving several important results when proving (b). In Bartle and Sherberts "Introduction to Real Analysis" he first proves a couple of corollaries that will be used to prove the density of $\mathbb Q$ and it reads nicely. – taue2pi Jul 13 '14 at 4:54

Your confusion seems to arise because the Archimedes principle is stated in terms of $$x,y$$, and you have different $$x,y$$ in (b). Restate the Archimedean principle as:

(a) If $$u,v$$ are real numbers, with $$u>0$$ then there is a positive integer $$k$$ such that: $$ku>v$$.

(All I've done is change the variables, I hope.)

Now, $$1$$ is a real number, $$y-x$$ is a real number, and you've proven that $$y-x>0$$. So we know from (a) that if $$u:=y-x$$ and $$v:=1$$ that there is a positive integer, which we will call $$n$$, such that $$(y-x)n>1$$.

Similarly, since we know that $$nx$$ is a real number, and we know that $$1$$ is a real number and $$1>0$$, that from (a), setting $$u:=1$$ and $$v:=nx$$, that there is a positive integer we'll call $$m_1$$ such that $$m_1\cdot 1 > nx$$.

Finally, set $$u:=1>0$$ and $$v:=-nx$$ to show that there must be an $$m_2$$ so that $$m_2\cdot 1>-nx$$.

The last step is subtler, and doesn't use (a). Since $$m_2>-nx$$, $$-m_2. So we know that $$-m_2.

Now, you need a property of the integers: If a non-empty set of integers has a lower bound, then it has a least element.

Take the set $$S=\{m\in\mathbb Z: m> nx\}$$. We know that $$m_1\in S$$, so $$S$$ is non-empty, and we know that $$-m_2$$ is a lower bound for $$S$$. So there is a least element $$m\in S$$. Then $$m-1\notin S$$, and therefore $$m> nx$$ and $$m-1\leq nx$$. So $$m-1\leq nx< m$$.

• Michael Albanese's explanation of the first part ($n(y-x) > 1$) was more intuitive. I don't really understand yours, so I'll just leave it at that. For the next lines, what does "$u = 1 > 0$" mean? Is that the same thing as $u = 1$ since $1 > 0$ is just a true equality? If so, how do you know to set $u = 1$? And how do you know to set $v = nx$ and $v = -nx$? Since we changed our variables to $u$, $k$, and $v$, what is $n$ then? Is it just from the first AP? And then we're using AP again on the first AP? – mr eyeglasses Jul 13 '14 at 2:05
• But his "explanation" did not Archimedes property as you've stated it, so you can't use that as a step in your proof. The point is, AP applies to any pair of real numbers with the first positive, no matter what we call them. – Thomas Andrews Jul 13 '14 at 2:08
• @nablablah Can you state the version of the Archimedian property as given in Rudin? Is it just (a)? – Thomas Andrews Jul 13 '14 at 2:25
• Okay, so property (a) is what the book is calling the Archimedean property. Cool, that was my assumption, but I realized I might have been mistaken. – Thomas Andrews Jul 13 '14 at 2:34
• I believe you require "we know $-m_2$ is a lower bound for $S$." – jII May 25 '16 at 23:34

In Bartle and Sherberts "Introduction to Real Analysis" this result is proved with preliminary corollaries (you'll note that are in some way results that Rudin also uses).

And using these results it's proven the following:

Hopefully with this approach now you can understand Rudins proof.

• How come in corollary 2.4.6, the inequalities are both '$\leq$' ($n_y - 1 \leq y \leq n_y$), but in the proof, we have $m - 1 \leq nx < m$ (one '$\leq$' and one '$<$')? – mr eyeglasses Jul 13 '14 at 13:16
• Your correct. In the proof of corollary 2.4.6, what is actually proven is that if $y>0$, then there exists $n_y\in\mathbb N$ such that $n_y-1\leq y\lt n_y$, hence $n_y-1\leq y\leq n_y$ – taue2pi Jul 13 '14 at 23:12
• The proof is as follows. The Archimedean Property ensures that the subset $E_y:=\{m\in\mathbb N :y\lt m\}$ of $\mathbb N$ is not empty. By the Well-Ordering Porperty, $E_y$ has a least element wich will be denoted by $n_y$. Then $n_y-1$ does not belong to $E_y$, and hence we have $n_y-1\leq y\lt n_y$. this last inequality implies the inequality seen in the corollary. – taue2pi Jul 13 '14 at 23:18