I'm looking at a proof that $\mathbb{Q}$ is dense in $\mathbb{R}$, using only the Archimedean Property of $\mathbb{R}$ and basic properties of ordered fields.

One step asserts that for any $n \in \mathbb{N}$, $x \in \mathbb{R}$, there is an integer $m$ such that $m - 1 \leq nx < m$. Why is this true? (Ideally, this fact can be shown using only the Archimedean property of $\mathbb{R}$ and basic properties of ordered fields...)

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    $\begingroup$ What's your version of the archimedean property? $\endgroup$ – Dylan Moreland Jan 30 '12 at 2:58
  • $\begingroup$ $\forall x, y \in \mathbb{R}$ with $x > 0$, $\exists n \in \mathbb{N}$ such that $nx > y$. $\endgroup$ – jamaicanworm Jan 30 '12 at 3:03

You mention in comments that your version of the Archimedean property is

For all $x,y\in\mathbb{R}$ with $x\gt 0$, there exists $n\in\mathbb{N}$ such that $nx\gt y$.

Let $z\in\mathbb{R}$. Assume first that $z\gt 0$. Now, using the Archimedean property with $y=z$ and $x=1$, it follows that there exists $n\in\mathbb{N}$ such that $n\gt z$. Thus, the set $$\{ n\in\mathbb{N}\mid n\gt z\}$$ is nonempty. By the well-ordering principle, there is a least natural number $n_0$ such that $n_0\gt z$. Then $n_0\gt z$; if $n_0=1$, then $0\lt z\lt 1$ and $m=1$ works. If $n_0\gt 1$, then $n_0-1\in\mathbb{N}$, and minimality of $n_0$ means that $n_0-1\leq z$, Thus, $n_0-1\leq z\lt n_0$ and $m=n_0$ works again.

If $z=0$, take $m=1$.

If $z\lt 0$, then let $w=-z$. Then there exists, by the previous case, a nonnegative integer $k$ such that $k-1\leq w \lt k$. Therefore, $-k\lt z\leq 1-k$. If $z\lt 1-k$, then $m=1-k$ does the trick. If $z=1-k$, then $1-k\leq z\lt 2-k$, so $m=2-k$ works.

  • $\begingroup$ What is your definition of "the well-ordering principle," and how do you define the "minimality of $n_0$"? $\endgroup$ – jamaicanworm Jan 30 '12 at 3:16
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    $\begingroup$ @jamaicanworm: The "well-ordering principle" states that every nonempty subset of the natural numbers has a smallest element. "Minimality of $n_0$" means that since $n_0$ is the smallest natural number with the given property, then any strictly smaller natural number does not have the property. In particular, $n_0-1$ is not in the set, which means that $n_0-1$ does not satisfy $n_0-1\gt z$, which means that $n_0-1\leq z$ must hold. $\endgroup$ – Arturo Magidin Jan 30 '12 at 3:17
  • $\begingroup$ Thanks! I appreciate your detailed explanation. I'm looking at Rudin, and he writes the following: "Apply [the Archimedean Property] again, to obtain positive integers $m_1$ and $m_2$ such that $m_1 > nx$, and $m_2 > -nx$. Then $$-m_2 < nx < m_1. (I)$$ Hence there is an integer $m$ (with $-m_2 < m < m_1$) such that $$m-1 \leq nx < m. (II)$$" Do you know how he gets from inequality $(I)$ to inequality $(II)$? Is he just skipping the proof that you wrote, or is there a way to do it using $(I)$? $\endgroup$ – jamaicanworm Jan 30 '12 at 3:26
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    $\begingroup$ @jamaicanworm: He is doing this so he doesn't have to separate into positive and negative $x$. Once you have $m_1$, you can find the smallest positive integer $m_1$ with the desired property; then find the smallest positive integer $m_2$ with the desired property; from there, figuring out $m$ is easy. $\endgroup$ – Arturo Magidin Jan 30 '12 at 3:34
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    $\begingroup$ @jamaicanworm: $m_1$ and $m_2$ are both positive. If $m_2=1$, then determine whether $x\lt 0$ (in which case you can take $m=0$), or $x\geq 0$ (in which case $m=m_1$); otherwise, (that is, if $m_2\gt 1$) then either $m=1-m_2$ or $2-m_2$. $\endgroup$ – Arturo Magidin Jan 30 '12 at 4:01

Assume first that $x>0$, so that $nx>0$. By the Archimedean property there is a $k\in\mathbb{N}$ such that $k>nx$; let $m$ be the least such $k$. Clearly $m-1\le nx<m$. If $x=0$, just take $m=1$. Finally, if $x<0$, then $-nx>0$, so by the first part of the argument there is an integer $k$ such that $k-1\le -nx<k$, and hence $-k<nx\le 1-k$. If $nx\ne 1-k$, you’re done: just take $m=1-k$. If $nx=1-k$, take $m=2-k$.


If $\mathbb{Q}$ is not dense in $\mathbb{R}$, then there are two members $x,y\in\mathbb{R}$ such that no member of $\mathbb{Q}$ is between them. I claim that the distance $\varepsilon=|x-y|$ between $x$ and $y$ is an infinitesimal. By the Archimedean property, this implies $\varepsilon=0$.

If $$ \underbrace{\varepsilon+\cdots+\varepsilon}_{n\text{ terms}} > 1 $$ then $\varepsilon>1/n$, so some rational number of the form $k/n$ is between $x$ and $y$.

  • $\begingroup$ This wasn't really my question... but thanks for the alternate perspective! $\endgroup$ – jamaicanworm Jan 30 '12 at 16:43

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