362: There Are Rational and Irrational Dedekind Cuts Between 2 Dedekind Cuts

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A description/proof of that there are rational and irrational Dedekind cuts between 2 Dedekind cuts

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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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.

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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.

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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.

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Main Body

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1: Description

For any Dedekind cuts $\widetilde{r_1},\widetilde{r_4}$, such that $\widetilde{r_1}\subset \widetilde{r_4}$, there are a rational Dedekind cut, $\widetilde{r_2}$, such that $\widetilde{r_1}\subset \widetilde{r_2}\subset \widetilde{r_4}$ and an irrational Dedekind cut, $\widetilde{r_3}$, such that $\widetilde{r_1}\subset \widetilde{r_3}\subset \widetilde{r_4}$.

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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 $q_2\notin \widetilde{r_1}$ and $q_2\in \widetilde{r_4}$. There is a rational number, $q_2^{\prime }$, such that $q_2<q_2^{\prime }\in \widetilde{r_4}$, because $\widetilde{r_4}$ does not have any largest element, by the definition of Dedekind cut. $\widetilde{q_2^{\prime }}=\{q\in \mathbb{Q}|q<q_2^{\prime }\}$. $\widetilde{q_2^{\prime }}\subset \widetilde{r_4}$, because $\widetilde{r_4}$ contains all the rational numbers smaller than $q_2^{\prime }$, but $q_2^{\prime }$ is contained only in $\widetilde{r_4}$. $\widetilde{r_1}\subset \widetilde{q_2^{\prime }}$, because $\widetilde{q_2^{\prime }}$ contains all the rational numbers smaller than $\widetilde{r_1}$, but $q_2$ is contained only in $\widetilde{q_2^{\prime }}$. $\widetilde{r_2}$ can be taken to be $\widetilde{q_2^{\prime }}$.

As a fact that will be used repeatedly hereafter, for any real numbers, $\tilde{r},\widetilde{r^{\prime }}$, such that $0\le \tilde{r}<\widetilde{r^{\prime }}$, there is a rational number, $q$, such that $\tilde{r}<{\tilde{q}}^2<\widetilde{r^{\prime }}$, because $\sqrt{\tilde{r}}<\sqrt{\widetilde{r^{\prime }}}$ (the existences of the square roots are presupposed) and there is a rational number, $q$, such that $\sqrt{\tilde{r}}\subset \tilde{q}\subset \sqrt{\widetilde{r^{\prime }}}$, by the previous paragraph, and $\sqrt{\tilde{r}}\sqrt{\tilde{r}}\subset \tilde{q}\tilde{q}\subset \sqrt{\widetilde{r^{\prime }}}\sqrt{\widetilde{r^{\prime }}}$, so, $\tilde{r}<{\tilde{q}}^2<\widetilde{r^{\prime }}$.

Let us find an irrational Dedekind cut, $\widetilde{r_3}$, such that $\widetilde{r_1}\subset \widetilde{r_3}\subset \widetilde{r_2}$.

Let us suppose that $0<\widetilde{r_2}$. If $\widetilde{r_1}<0$, we will take $\widetilde{r_1^{\prime }}=0$, and otherwise, $\widetilde{r_1^{\prime }}=\widetilde{r_1}$, so, $0\le \widetilde{r_1^{\prime }}<\widetilde{r_2}$. There is a positive rational number, $q_3$, such that ${\widetilde{r_1^{\prime }}}^2<2{\widetilde{q_3}}^2<{\widetilde{r_2}}^2$, because it is equivalent with $\frac{{\widetilde{r_1^{\prime }}}^2}{2}<{\widetilde{q_3}}^2<\frac{{\widetilde{r_2}}^2}{2}$. Then, define $\widetilde{r_3}:=\{q\in \mathbb{Q}|(q<0)\vee (0\le q\wedge q^2<2{q_3}^2)\}$.

$\widetilde{r_3}$ is a Dedekind cut, because $\widetilde{r_3}\ne \mathrm{\varnothing }$ and $\widetilde{r_3}\ne \mathbb{Q}$ and if $q^{\prime }\in \widetilde{r_3}$, for any $q^{″}<q^{\prime }$, $q^{″}\in \widetilde{r_3}$, and $\widetilde{r_3}$ does not contain any largest element (for any $q^{\prime }$ such that $q^{\prime 2}<2{q_3}^2$, there is a $q^{″}$ such that $q^{\prime 2}<q^{″2}<2{q_3}^2$). The complement of $\widetilde{r_3}$ is $\{q\in \mathbb{Q}|0\le q\wedge 2{q_3}^2\le q^2\}$, which does not have any smallest element as there is no $q$ that satisfies $2{q_3}^2=q^2$ (for any $q^{\prime }$ such that $2{q_3}^2<q^{\prime 2}$, there is a $q^{″}$ such that $2{q_3}^2<q^{″2}<q^{\prime 2}$), which means that $\widetilde{r_3}$ is an irrational number.

$\widetilde{r_1^{\prime }}\subset \widetilde{r_3}\subset \widetilde{r_2}$, because while $\widetilde{r_1^{\prime }}\subseteq \widetilde{r_3}\subseteq \widetilde{r_2}$ is obvious, there are $q^{\prime },q^{″}$ such that ${\widetilde{r_1^{\prime }}}^2<{\widetilde{q^{\prime }}}^2<{\widetilde{r_3}}^2<{\widetilde{q^{″}}}^2<{\widetilde{r_2}}^2$, and $q^{\prime }\notin \widetilde{r_1^{\prime }}$, $q^{\prime }\in \widetilde{r_3}$, $q^{″}\notin \widetilde{r_3}$, and $q^{″}\in \widetilde{r_2}$.

Let us suppose that $\widetilde{r_2}\le 0$. Then, $\widetilde{r_1}<\widetilde{r_2}\le 0$. There is a negative rational number, $q_3$, such that ${\widetilde{r_2}}^2<2{\widetilde{q_3}}^2<{\widetilde{r_1}}^2$. Then define $\widetilde{r_3}:=\{q\in \mathbb{Q}|q\le 0\wedge 2{q_3}^2<q^2\}$.

$\widetilde{r_3}$ is a Dedekind cut, because $\widetilde{r_3}\ne \mathrm{\varnothing }$ and $\widetilde{r_3}\ne \mathbb{Q}$ and if $q^{\prime }\in \widetilde{r_3}$, for any $q^{″}<q^{\prime }$, $q^{″}\in \widetilde{r_3}$, and $\widetilde{r_3}$ does not contain any largest element (for any $q^{\prime }$ such that $2{q_3}^2<q^{\prime 2}$, there is a $q^{″}$ such that $2{q_3}^2<q^{″2}<q^{\prime 2}$). The complement of $\widetilde{r_3}$ is $\{q\in \mathbb{Q}|(0<q)\vee (q\le 0\wedge q^2\le 2{q_3}^2)\}$, which does not have any smallest element as there is no $q$ that satisfies $q^2=2{q_3}^2$ (for any $q^{\prime }$ such that $q^{\prime 2}<2{q_3}^2$, there is a $q^{″}$ such that $q^{\prime 2}<q^{″2}<2{q_3}^2$), which means that $\widetilde{r_3}$ is an irrational number.

$\widetilde{r_1}\subset \widetilde{r_3}\subset \widetilde{r_2}$, because while $\widetilde{r_1}\subseteq \widetilde{r_3}\subseteq \widetilde{r_2}$ is obvious, there are negative $q^{\prime },q^{″}$ such that ${\widetilde{r_2}}^2<{\widetilde{q^{\prime }}}^2<{\widetilde{r_3}}^2<{\widetilde{q^{″}}}^2<{\widetilde{r_1}}^2$, and $q^{″}\notin \widetilde{r_1}$, $q^{″}\in \widetilde{r_3}$, $q^{\prime }\notin \widetilde{r_3}$, and $q^{\prime }\in \widetilde{r_2}$.

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References

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