Theorem
\(Q\) is dense in \(R\). If \(a \in \mathbb{R}\), \(b \in \mathbb{R}\), and \(a < b\), then there exists a rational number \(r \in \mathbf{Q}\) such that $$ a < r < b. $$

Proof (1, Lecture/Textbook)

Strategy

We want to find a rational number \(\frac{m}{n}\) such that \(a < \frac{m}{n} < b\). Suppose that \(0 < a < b\). Now construct the set

$$ \begin{align*} E = \{ k \in \mathbb{N} \mid \frac{k}{n} \leq a\} \end{align*} $$

So we’re walking in steps of size \(\frac{1}{n}\). Note that \(S \subset \mathbb{N}\). This set is bounded by \(na\). If we let \(k_0 = \sup(E)\), then we know that \(k_0 + 1\) is not in \(E\) and we know that \(\frac{k_0+1}{n} > a\). So we have a potential fraction that is strictly greater than \(a\). We still want to make this fraction be strictly less than \(b\) just like the picture above.

How can we do it? We can derive what \(n\) needs to be since \(n\) is fixed. We specifically want that

$$ \begin{align*} b &> \frac{k_0 + 1}{n} \\ b &> \frac{k_0}{n} + \frac{1}{n}\\ b &> \frac{k_0}{n} + \frac{1}{n} \\ b &> a + \frac{1}{n} \quad (\text{since }\frac{k_0}{n} \leq a) \\ \end{align*} $$

But this means that we want \(n\) to be

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

So the construction of the set ensures that the element that comes right after the supremum is greater than \(a\). But this relies on the set not being empty. To ensure this, we also want to make sure that \(n \geq \frac{1}{a}\). The second condition that we just derived ensures that this element is strictly less than \(b\). Hence, we have found a rational number in between \(a\) and \(b\).

Formal Proof

By the Archimedean principle, choose \(n\) such that

$$ \begin{align*} n &> \frac{1}{b - a} \quad \text{ and } \quad n > \frac{1}{a} \end{align*} $$

From this, we get that

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

Now, consider the set

$$ \begin{align*} E = \{ k \in \mathbb{N} \mid \frac{k}{n} \leq a\} \end{align*} $$

Observe that \(1 \in E\). Therefore, \(E\) is non-empty. Moreover, \(E\) is bounded by \(na\). By the completeness axiom, there exists a supremum of \(E\). Let \(s = \sup(E)\). By Theorem 1.15, \(s = \sup(E) \in E\). Now, since \(s = \sup(E)\), then we know that \(s+1 \not\in E\). By the definition of \(E\), this implies that

$$ \begin{align*} \frac{s + 1}{n} > a \end{align*} $$

So this establishes that the fraction is strictly greater than \(a\). Next, we want to show that it is less than \(b\). To see this observe that

$$ \begin{align*} \frac{s + 1}{n} &= \frac{s}{n} + \frac{1}{n} \\ &\leq a + \frac{1}{n} \quad \left(\text{By Theorem 1.15, }\frac{s}{n} \leq a\right) \\ &< a + (b - a) \quad \left(\text{We defined $n$ such that } n > \frac{1}{b - a}\right) \\ &= b. \ \blacksquare \end{align*} $$

Proof (2, Dr. Peyam)

Suppose that \(a < b\) and assume that \(0 < a < b\). Since \(a < b\), then we know that \(b - a > 0\). Since \(b - a\) is a positive number, then by the Archimedean Principle (with \(x = b-a\) and \(y = 1\)), we can always find \(n \in \mathbb{N}\) such that

$$ \begin{align*} n(b - a) > 1 \end{align*} $$

which implies that

$$ \begin{align*} b - a > \frac{1}{n} \end{align*} $$
Side Note: The idea here is to start from \(0\) and then walk in small steps of size \(\frac{1}{n}\) until we reach the last step that is less than \(b\) with the additional property that if we added another step, then we get out of this interval (in this case \(\frac{m+1}{n}\)). For this to work, we want the step to be small enough not to skip over the entire interval. Swe need to have \(\frac{1}{n} < b-a\). We formalize this idea next.

So let

$$ \begin{align*} S = \{ \frac{m}{n} \mid m = 0,1,2,\ldots, \frac{m}{n} < b\} \end{align*} $$

We know that \(S\) is non-empty since \(0 \in S\) because \(b > 0\). Moreover, \(S\) is bounded above by \(b\). Therefore, by the completeness axiom, \(S\) has a least upper bound. So let \(r = \sup(S)\). We claim that \(a < r < b\). Observe that \(S\) is finite. This means that \(r = \sup(S) = \max(S)\). But since it’s the maximum element, then \(r \in S\). So \(r\) is rational. By definition, we know that \(r < b\). So the only thing left to show is that \(r \geq a\).

Suppose for the sake of contradiction that \(r = \frac{m}{n} \leq a\). We know that

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

Since \(r = \frac{m}{n} \leq a\), then

$$ \begin{align*} b &\geq \frac{m}{n} + \frac{1}{n} \\ b &\geq \frac{m+1}{n} \end{align*} $$

But this implies that \(\frac{m+1}{n} \in S\). This is a contradiction since \(\sup(S) = \frac{m}{n}\). Then we must have that \(r = \frac{m}{n} < a\). \(\ \blacksquare\)

References