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\noindent
{\bf Math 3220}
\hfill {\Large \bf Rationals and Irrationals are Densely Packed} \hfill 
{\bf Intro. to Real Analysis}

\vspace{0.2in}

\THM Let $a,b \in \mathbb{R}$ such that $a<b$. There exists both a rational number $r$ and an irrational number $s$ such that $a<r<b$ and $a<s<b$. Moreover, there are infinitely many rational and irrational numbers between $a$ and $b$.

\bproof  First, we will prove the existence of our rational number $r$. To do this we consider three cases: $a \geq 0$, $a<0 \leq b$, and finally $a<b<0$.

\begin{itemize}
\item Suppose $a \geq 0$.  We have that $a<b$ so that $0<b-a$. By the squeezing in lemma, there exists some positive integer $n$ (i.e., $n \in \mathbb{N}$) such that $0 < \dfrac{1}{n} < b-a$. 

Let $M = \{ k \in \mathbb{N} \;|\; k/n > a \}$.  If $a<1/n$, then $1 \in M$. Otherwise, $0<1/n<a$ ({\it Note:} here we have used our assumption that $a \geq 0$) and the Archimedean principle states that there is some positive integer $n'$ such that $n'/n > a$ so that $n' \in M$. Either way, we have that $M$ is not empty. Now $M$ is a non-empty set of positive integers. The Well-Ordering Principle states that $M$ must have a least element. Let $m=\mathrm{min}(M)$ be this least element of $M$.

Therefore, $\dfrac{m}{n} > a$ but $\dfrac{m-1}{n} \leq a$. Clearing denominators, we have $m-1 \leq an$ so that $m \leq an+1$. Therefore, $\dfrac{m}{n} \leq a + \dfrac{1}{n} < a+(b-a)=b$ (since $1/n<b-a$). Therefore, we have established that $a < r < b$ where $r=\dfrac{m}{n} \in \mathbb{Q}$.   

\item Now consider that case when $a<0\leq b$. We have $0<-a$ so by the squeezing in lemma there is a positive integer $n$ such that $0<-\dfrac{1}{n}<a$. Therefore, $a<\dfrac{1}{n}<0\leq b$. Thus $r=-\dfrac{1}{n}$ is our desired rational number.

\item Finally, consider when $a<b<0$. Then $0<-b<-a$. So by our first case we have some $r' \in \mathbb{Q}$ such that $-b< r' <-a$. Therefore, $a<r<b$ where $r=-r' \in \mathbb{Q}$. 
\end{itemize}

Once we can squeeze in rational numbers, irrationals follow easily. First, suppose that we have two rational numbers: $\dfrac{u}{v}$ and $\dfrac{w}{x}$ (where $u,v,w,$ and $x$ are all integers and of course $v \not=0$ and $x \not=0$). Assume that $\dfrac{u}{v} < \dfrac{w}{x}$. Thus we have $\dfrac{ux}{vx} = \dfrac{u}{v} <  \dfrac{w}{x} = \dfrac{vw}{vx}$ so that $ux<vw$ (because these are integers we have: $ux<ux+1\leq vw$).


Notice that $1<\sqrt{2}<2$ (since squaring gives $1<2<4$). So $0<\sqrt{2}-1<1$.  Thus $ux < ux + (\sqrt{2}-1) < ux+1 \leq vw$. Therefore, $\dfrac{u}{v} = \dfrac{ux}{vx} < \dfrac{ux-1+\sqrt{2}}{vx} < \dfrac{ux+1}{vx} \leq \dfrac{vw}{vx} = \dfrac{w}{x}$. We observe that $p = \dfrac{ux-1+\sqrt{2}}{vx}$ is irrational.

Therefore, between any two rational numbers there is an irrational number. Now consider $a<b$. We can squeeze in a rational number $r_1$ such that $a<r_1<b$. Now squeeze
a rational number in between $r_1$ and $b$ and get $a<r_1<r_2<b$. Now squeeze an irrational number between those rational numbers and get $a<r_1<p<r_2<b$. Thus an irrational number, $p$, can be squeezed between any two real numbers.

We can continue such a process squeezing in rational and irrational numbers and get and endless list: $a<r_1<p_1<r_2<p_2<\cdots<b$. Thus there are an infinite number of irrational and rational numbers between any two real numbers.
\eproof



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