Theorem
\(\mathbb{R}-\mathbb{Q}\) is dense in \(\mathbb{R}\). If \(a \in \mathbb{R}\), \(b \in \mathbb{R}\), and \(a < b\), then there exists an irrational number \(r \in \mathbb{R}-\mathbb{Q}\) such that
$$
a < r < b.
$$
Strategy
We want to use the density of \(\mathbb{Q}\) theorem that we proved here.
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
$$
\begin{align*}
\frac{s + 1}{n} &= \frac{s}{n} + \frac{1}{n} \\
&\leq a + \frac{1}{n} \quad \left(\text{By Theorem 1.15, }\frac{s}{n} \leq a\right) \\
&< a + (b - a) \quad \left(\text{We defined $n$ such that } n > \frac{1}{b - a}\right) \\
&= b. \ \blacksquare
\end{align*}
$$
References
- Lecture Notes by Professor Chun Kit Lai