A description/proof of that there are rational and irrational Dedekind cuts between 2 Dedekind cuts
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of rational Dedekind cut.
- The reader knows a definition of irrational Dedekind cut.
- The reader admits the proposition that the square root of any non-negative real number exists as a real number.
Target Context
- The reader will have a description and a proof of the proposition that for any 2 Dedekind cuts, there are a rational Dedekind cut and an irrational Dedekind cut between them.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any Dedekind cuts \(\tilde{r_1}, \tilde{r_4}\), such that \(\tilde{r_1} \subset \tilde{r_4}\), there are a rational Dedekind cut, \(\tilde{r_2}\), such that \(\tilde{r_1} \subset \tilde{r_2} \subset \tilde{r_4}\) and an irrational Dedekind cut, \(\tilde{r_3}\), such that \(\tilde{r_1} \subset \tilde{r_3} \subset \tilde{r_4}\).
2: Proof
Note that for any rational number, \(r\) denotes the rational number regarded as a rational number; \(\tilde{r}\) denotes the Dedekind cut of \(r\) regarded as a real number; any irrational number is always denoted as \(\tilde{r}\), because it is always as a Dedekind cut.
There is a rational number, \(q_2\), such that \(q_2 \notin \tilde{r_1}\) and \(q_2 \in \tilde{r_4}\). There is a rational number, \(q'_2\), such that \(q_2 \lt q'_2 \in \tilde{r_4}\), because \(\tilde{r_4}\) does not have any largest element, by the definition of Dedekind cut. \(\tilde{q'_2} = \{q \in \mathbb{Q}\vert q \lt q'_2\}\). \(\tilde{q'_2} \subset \tilde{r_4}\), because \(\tilde{r_4}\) contains all the rational numbers smaller than \(q'_2\), but \(q'_2\) is contained only in \(\tilde{r_4}\). \(\tilde{r_1} \subset \tilde{q'_2}\), because \(\tilde{q'_2}\) contains all the rational numbers smaller than \(\tilde{r_1}\), but \(q_2\) is contained only in \(\tilde{q'_2}\). \(\tilde{r_2}\) can be taken to be \(\tilde{q'_2}\).
As a fact that will be used repeatedly hereafter, for any real numbers, \(\tilde{r}, \tilde{r'}\), such that \(0 \leq \tilde{r} \lt \tilde{r'}\), there is a rational number, \(q\), such that \(\tilde{r} \lt \tilde{q}^2 \lt \tilde{r'}\), because \(\sqrt{\tilde{r}} \lt \sqrt{\tilde{r'}}\) (the existences of the square roots are presupposed) and there is a rational number, \(q\), such that \(\sqrt{\tilde{r}} \subset \tilde{q} \subset \sqrt{\tilde{r'}}\), by the previous paragraph, and \(\sqrt{\tilde{r}} \sqrt{\tilde{r}} \subset \tilde{q} \tilde{q} \subset \sqrt{\tilde{r'}} \sqrt{\tilde{r'}}\), so, \(\tilde{r} \lt \tilde{q}^2 \lt \tilde{r'}\).
Let us find an irrational Dedekind cut, \(\tilde{r_3}\), such that \(\tilde{r_1} \subset \tilde{r_3} \subset \tilde{r_2}\).
Let us suppose that \(0 \lt \tilde{r_2}\). If \(\tilde{r_1} \lt 0\), we will take \(\tilde{r'_1} = 0\), and otherwise, \(\tilde{r'_1} = \tilde{r_1}\), so, \(0 \leq \tilde{r'_1} \lt \tilde{r_2}\). There is a positive rational number, \(q_3\), such that \({\tilde{r'_1}}^2 \lt 2 {\tilde{q_3}}^2 \lt {\tilde{r_2}}^2\), because it is equivalent with \(\frac{{\tilde{r'_1}}^2}{2} \lt {\tilde{q_3}}^2 \lt \frac{{\tilde{r_2}}^2}{2}\). Then, define \(\tilde{r_3} := \{q \in \mathbb{Q}\vert (q \lt 0) \lor (0 \leq q \land q^2 \lt 2 {q_3}^2)\}\).
\(\tilde{r_3}\) is a Dedekind cut, because \(\tilde{r_3} \neq \emptyset\) and \(\tilde{r_3} \neq \mathbb{Q}\) and if \(q' \in \tilde{r_3}\), for any \(q'' \lt q'\), \(q'' \in \tilde{r_3}\), and \(\tilde{r_3}\) does not contain any largest element (for any \(q'\) such that \(q'^2 \lt 2 {q_3}^2\), there is a \(q''\) such that \(q'^2 \lt q''^2 \lt 2 {q_3}^2\)). The complement of \(\tilde{r_3}\) is \(\{q \in \mathbb{Q}\vert 0 \leq q \land 2 {q_3}^2 \leq q^2\}\), which does not have any smallest element as there is no \(q\) that satisfies \(2 {q_3}^2 = q^2\) (for any \(q'\) such that \(2 {q_3}^2 \lt q'^2\), there is a \(q''\) such that \(2 {q_3}^2 \lt q''^2 \lt q'^2\)), which means that \(\tilde{r_3}\) is an irrational number.
\(\tilde{r'_1} \subset \tilde{r_3} \subset \tilde{r_2}\), because while \(\tilde{r'_1} \subseteq \tilde{r_3} \subseteq \tilde{r_2}\) is obvious, there are \(q', q''\) such that \(\tilde{r'_1}^2 \lt \tilde{q'}^2 \lt \tilde{r_3}^2 \lt \tilde{q''}^2 \lt \tilde{r_2}^2\), and \(q' \notin \tilde{r'_1}\), \(q' \in \tilde{r_3}\), \(q'' \notin \tilde{r_3}\), and \(q'' \in \tilde{r_2}\).
Let us suppose that \(\tilde{r_2} \leq 0\). Then, \(\tilde{r_1} \lt \tilde{r_2} \leq 0\). There is a negative rational number, \(q_3\), such that \({\tilde{r_2}}^2 \lt 2 {\tilde{q_3}}^2 \lt {\tilde{r_1}}^2\). Then define \(\tilde{r_3} := \{q \in \mathbb{Q}\vert q \leq 0 \land 2 {q_3}^2 \lt q^2\}\).
\(\tilde{r_3}\) is a Dedekind cut, because \(\tilde{r_3} \neq \emptyset\) and \(\tilde{r_3} \neq \mathbb{Q}\) and if \(q' \in \tilde{r_3}\), for any \(q'' \lt q'\), \(q'' \in \tilde{r_3}\), and \(\tilde{r_3}\) does not contain any largest element (for any \(q'\) such that \(2 {q_3}^2 \lt q'^2\), there is a \(q''\) such that \(2 {q_3}^2 \lt q''^2 \lt q'^2\)). The complement of \(\tilde{r_3}\) is \(\{q \in \mathbb{Q}\vert (0 \lt q) \lor (q \leq 0 \land q^2 \leq 2 {q_3}^2)\}\), which does not have any smallest element as there is no \(q\) that satisfies \(q^2 = 2 {q_3}^2\) (for any \(q'\) such that \(q'^2 \lt 2 {q_3}^2\), there is a \(q''\) such that \(q'^2 \lt q''^2 \lt 2 {q_3}^2\)), which means that \(\tilde{r_3}\) is an irrational number.
\(\tilde{r_1} \subset \tilde{r_3} \subset \tilde{r_2}\), because while \(\tilde{r_1} \subseteq \tilde{r_3} \subseteq \tilde{r_2}\) is obvious, there are negative \(q', q''\) such that \(\tilde{r_2}^2 \lt \tilde{q'}^2 \lt \tilde{r_3}^2 \lt \tilde{q''}^2 \lt \tilde{r_1}^2\), and \(q'' \notin \tilde{r_1}\), \(q'' \in \tilde{r_3}\), \(q' \notin \tilde{r_3}\), and \(q' \in \tilde{r_2}\).
