[Density of \(\mathbf{Q}\) in \(\mathbf{R}\)] For every two real numbers \(a\) and \(b\) with \(a < b\), there exists a rational number \(r\) satisfying \(a < r < b\).



For the definitions of an upper bound and the least upper bound of a set, see this. For the Archimedean Principle, see this

Proof:

Let \(a\) and \(b\) be real numbers with with \(a < b\). We will produce \(q = \frac{m}{n} \in \mathbf{Q}\) where \(m \in \mathbf{Z}\) and \(n \in \mathbf{N}\) such that

$$ \begin{align*} a < \frac{m}{n} < b. \end{align*} $$

We want to pick \(n\) such that it is large enough where a step of size \(1/n\) won’t cross over the entire interval of \([a,b]\). (side note: think of the numbers 0.5 and 1. You don’t want \(n\) to be 1. since any fraction \(m/1\) is never going to be inside the interval). To pick such \(n\), we can use the Archimedean Principle to pick an \(n \in \mathbf{N}\) such that \(n > \frac{1}{b - a}\). Re-arrange the terms to get

$$ \begin{align*} \frac{1}{n} &< b - a. \end{align*} $$

So now all we need is to find \(m\) such that \(a < \frac{m}{n} < b\). Multiply the inequality by \(n\) to get

$$ \begin{align*} an < m < bn. \end{align*} $$

Clearly \(m\) needs to be an integer greater than \(an\) and less than \(bn\). The idea in the book’s proof was to pick \(m\) to be the smallest integer greater than \(na\). In other words,

$$ \begin{align*} m-1 \leq na < m. \end{align*} $$

So now we need to prove that this choice of \(m\) will satisfy the inequality \(an < m < bn\). Clearly, \(na < m\) satisfies the right inequality and we’re done with part. Next, we want to see how \(m - 1 \leq na\) will satisfy \(m < bn\). To see this, we will re-write the inequality \(\frac{1}{n} > b - a\) to isolate \(a\)

$$ \begin{align*} \frac{1}{n} > b - a \\ a > b - \frac{1}{n}. \\ \end{align*} $$

Now we can plug this in

$$ \begin{align*} m-1 &\leq na \\ m &\leq na - 1 \\ m &< n\big(b - \frac{1}{n}\big) - 1 \\ m &< nb - n/n - 1 \\ m &< nb. \\ \end{align*} $$

From this we see that this choice of \(m\) indeed satisfies \(an < m < bn\) as required. \(\blacksquare\)

Notes: So in a sense \(Q\) must be dense since we have a rational number between no matter what two real numbers are given.

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