I know this is a common proof. I'm following Rudin's proof and I'm following everything except for one step.

Suppose $x, y \in \Bbb R$ and $x < y$. Then there exists an $n \in \Bbb N$ such that $n(y-x) > 1$.

Again by the Archimedean property, there exist $m_{1}, m_{2} \in \Bbb N$ such that $m_{1} > nx$ and $m_{2} > -nx$, i.e.

$$

-m_{2} < nx < m_{1}

$$

From here, Rudin says there must be an $m \in \Bbb Z$ with $-m_{2} \le m \le m_{1}$ and that

$$

m-1 \le nx < m

$$

I'm confused about these two steps. If $-m_{2} < nx < m_{1}$, then isn't $-m_{2} < m_{1}$?

edit: to be clear, I follow everything up until the introduction of $m$.

## Best Answer

Yes, $-m_2<m_1$, but we knew this anyway: $m_1$ and $m_2$ are positive integers, so $-m_2$ is negative and $m_1$ is positive.

Rudin introduces both $m_1$ and $m_2$ in order to avoid having to split the argument into cases depending on whether $x>0$, $x=0$, or $x<0$. If you simply take $m_1$ to be the minimal such that $m_1>nx$, you’re in trouble if $x$ is negative: $m_1=1$, which doesn’t do what you want.

Once you have $-m_2<nx<m_1$, you can use the well-ordering principle to set

$$k_0=\min\{k\in\Bbb N:-m_2+k>nx\}\;;$$

$\{k\in\Bbb N:-m_2+k>nx\}$ is non-empty, because it contains $m_1-(-m_2)$ so the well-ordering principle ensures that $k_0$ exists. Now let $m=-m_2+k_0$, and you can easily check that $m-1\le nx<m$.