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

Solution 1:

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 $m_2>-nx$, $-m_2<nx$. So we know that $-m_2<m_1$.

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 $m-1\leq nx< m$.

Solution 2:

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).

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