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Arturo Magidin
Arturo Magidin
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Interestingly, all answers so far give Cauchy sequences, so let me give the Dedekind cut construction.

The development assumes that the rational numbers are known and constructed, as are their operations and basic properties.

Definition. A cut is a partition $(A_1,A_2)$ of $\mathbb{Q}$, satisfying the following properties:

  1. $A_1\neq\varnothing$ and $A_2\neq\varnothing$.
  2. $A_1\cup A_2=\mathbb{Q}$.
  3. If $a\in A_1$ and $b\in A_2$, then $a\lt b$.
  4. If there exists a rational $c\in\mathbb{Q}$ such that for every $a\in A_1$ and every $b\in A_2$, $a\leq c$ and $c\leq b$, then $c\in A_1$.

In particular, if $a\in A_1$ and $c\lt a$, then $c\in A_1$ (since we must have $c\in A_1\cup A_2$). Point 4 is somewhat arbitrary: you could also require this element to be in $A_2$. Or you could, as Dedekind does, simply ignore it and eventually identify some cuts together. Also, point 2 is somewhat superfluous, as it is included in the notion of "partition".

Let us call $A_1$ the "left set" of the cut, and $A_2$ the "right set".

Now, every rational number $r$ determines a cut, defined by $(A^r_1,A^r_2)$, with $A^r_1=\{q\in\mathbb{Q}\mid q\leq r\}$, and $A^r_2 = \{q\in\mathbb{Q}\mid q\gt r\}$. However, not every cut is determined by a rational: for example, the cut $(A_1,A_2)$ given by $$\begin{align*} A_1 &= \{q\in\mathbb{Q}\mid q\lt 0\} \cup \{q\in\mathbb{Q}\mid q\geq 0\text{ and }q^2\leq 2\}\\ A_2 &= \{q\in\mathbb{Q}\mid g\geq 0\text{ and }q^2\gt 2\} \end{align*}$$ can be shown not to be of the form $(A_1^r,A_2^r)$ for any rational number $r$.

We define the set of real numbers $\mathbb{R}$ as the set of all cuts.

We define the order in $\mathbb{R}$ by letting $(A_1,A_2)\leq (B_1,B_2)$ if and only if $A_1\subseteq B_1$.

We define addition of cuts as follows: given $(A_1,A_2)$ and $(B_1,B_2)$, we define the cut $(A_1,A_2)\oplus (B_1,B_2) = (C_1,C_2)$, where $C_1$ consists of all rational numbers $q$ for which there exists $a\in A_1$ and $b\in B_1$ with $q\leq a+b$; and letting $C_2=\mathbb{Q}\setminus C_1$.

It is not hard to verify that $-(A_1,A_2) = (-A_2,-A_1)$, where $-X = \{-q\mid q\in X\}$; except that you need to be a bit careful given my definition: to make sure it satisfies point 4, if there is a rational $c$ in $A_1$ Such that $q\leq c$ for all $q\in A_1$, then you should move $-c$ out of $-A_1$ and put it in $-A_2$.

Now, multiplication is a bit more complicated. First we let $\mathbf{0}=(0_L,0_R)$, where $0_L=\{q\in\mathbb{Q}\mid q\leq 0\}$, and $0_R=\{q\in\mathbb{Q}\mid q>0\}$.

Then, given two cuts $(A_1,A_2)$, $(B_1,B_2)$, with both greater than or equal to $(0_L,0_R)$, we define their product $(A_1,A_2)\otimes(B_1,B_2) = (C_1,C_2)$ by letting $$C_1 = \{q\in\mathbb{Q}\mid \exists a\in A_1, b\in B_1\text{ such that }a\geq 0,b\geq 0,\text{ and }q\leq ab\}$$ and then lettig $C_2 = \mathbb{Q}\setminus C_1$.

If $(A_1,A_2)\geq (0_L,0_R)$ and $(B_1,B_2)\lt (0_L,0_R)$, we define the product $(A_1,A_2)\otimes(B_1,B_2)$ to be $-\bigl( (A_1,A_2)\otimes(-B_2,-B_1)\bigr)$.

And if $(A_1,A_2)\lt (0_L,0_R)$, and $(B_1,B_2)\lt(0_L,0_R)$, we define $$(A_1,A_2)\otimes(B_1,B_2) = (-A_2,-A_1)\otimes(-B_2,-B_1).$$

One can then show that the map $\mathbb{Q}\to\mathbb{R}$ given by $r\mapsto (A_1^r,A_2^r)$ is one-to-one, and satisfies $$\begin{align*} (A_1^{r+s},A_2^{r+s}) &= (A_1^r,A_2^r) \oplus (A_1^s,A_2^s)\\ (A_1^{rs},A_2^{rs}) &= (A_1^r,A_2^r)\otimes (A_1^s,A_2^s)\\ r\leq s&\iff (A_1^r,A_2^r)\leq (A_1^s,A_2^s) \end{align*}$$ so that $\mathbb{R}$ contains a "copy" of $\mathbb{Q}$ inside of it. Finally, one can show completeness by showing that if $\{(A^{(i)}_1,A^{(i)}_2)\}_{i\in I}$ is a nonempty family of real numbers that is bounded above (so there exists a real number $(B_1,B_2)$ such that $A^{(i)}_1\subseteq B_1$ for all $i$), then the real number $(A_1,A_2)$ with $A_1 = \cup_{i\in I}A^{(i)}_1$, $A_2=\mathbb{Q}-A_1$ is the supremum of the set.

Interestingly, all answers so far give Cauchy sequences, so let me give the Dedekind cut construction.

The development assumes that the rational numbers are known and constructed, as are their operations and basic properties.

Definition. A cut is a partition $(A_1,A_2)$ of $\mathbb{Q}$, satisfying the following properties:

  1. $A_1\neq\varnothing$ and $A_2\neq\varnothing$.
  2. $A_1\cup A_2=\mathbb{Q}$.
  3. If $a\in A_1$ and $b\in A_2$, then $a\lt b$.
  4. If there exists a rational $c\in\mathbb{Q}$ such that for every $a\in A_1$ and every $b\in A_2$, $a\leq c$ and $c\leq b$, then $c\in A_1$.

In particular, if $a\in A_1$ and $c\lt a$, then $c\in A_1$ (since we must have $c\in A_1\cup A_2$). Point 4 is somewhat arbitrary: you could also require this element to be in $A_2$. Or you could, as Dedekind does, simply ignore it and eventually identify some cuts together. Also, point 2 is somewhat superfluous, as it is included in the notion of "partition".

Let us call $A_1$ the "left set" of the cut, and $A_2$ the "right set".

Now, every rational number $r$ determines a cut, defined by $(A^r_1,A^r_2)$, with $A^r_1=\{q\in\mathbb{Q}\mid q\leq r\}$, and $A^r_2 = \{q\in\mathbb{Q}\mid q\gt r\}$. However, not every cut is determined by a rational: for example, the cut $(A_1,A_2)$ given by $$\begin{align*} A_1 &= \{q\in\mathbb{Q}\mid q\lt 0\} \cup \{q\in\mathbb{Q}\mid q\geq 0\text{ and }q^2\leq 2\}\\ A_2 &= \{q\in\mathbb{Q}\mid g\geq 0\text{ and }q^2\gt 2\} \end{align*}$$ can be shown not to be of the form $(A_1^r,A_2^r)$ for any rational number $r$.

We define the set of real numbers $\mathbb{R}$ as the set of all cuts.

We define the order in $\mathbb{R}$ by letting $(A_1,A_2)\leq (B_1,B_2)$ if and only if $A_1\subseteq B_1$.

We define addition of cuts as follows: given $(A_1,A_2)$ and $(B_1,B_2)$, we define the cut $(A_1,A_2)\oplus (B_1,B_2) = (C_1,C_2)$, where $C_1$ consists of all rational numbers $q$ for which there exists $a\in A_1$ and $b\in B_1$ with $q\leq a+b$; and letting $C_2=\mathbb{Q}\setminus C_1$.

It is not hard to verify that $-(A_1,A_2) = (-A_2,-A_1)$, where $-X = \{-q\mid q\in X\}$.

Now, multiplication is a bit more complicated. First we let $\mathbf{0}=(0_L,0_R)$, where $0_L=\{q\in\mathbb{Q}\mid q\leq 0\}$, and $0_R=\{q\in\mathbb{Q}\mid q>0\}$.

Then, given two cuts $(A_1,A_2)$, $(B_1,B_2)$, with both greater than or equal to $(0_L,0_R)$, we define their product $(A_1,A_2)\otimes(B_1,B_2) = (C_1,C_2)$ by letting $$C_1 = \{q\in\mathbb{Q}\mid \exists a\in A_1, b\in B_1\text{ such that }a\geq 0,b\geq 0,\text{ and }q\leq ab\}$$ and then lettig $C_2 = \mathbb{Q}\setminus C_1$.

If $(A_1,A_2)\geq (0_L,0_R)$ and $(B_1,B_2)\lt (0_L,0_R)$, we define the product $(A_1,A_2)\otimes(B_1,B_2)$ to be $-\bigl( (A_1,A_2)\otimes(-B_2,-B_1)\bigr)$.

And if $(A_1,A_2)\lt (0_L,0_R)$, and $(B_1,B_2)\lt(0_L,0_R)$, we define $$(A_1,A_2)\otimes(B_1,B_2) = (-A_2,-A_1)\otimes(-B_2,-B_1).$$

One can then show that the map $\mathbb{Q}\to\mathbb{R}$ given by $r\mapsto (A_1^r,A_2^r)$ is one-to-one, and satisfies $$\begin{align*} (A_1^{r+s},A_2^{r+s}) &= (A_1^r,A_2^r) \oplus (A_1^s,A_2^s)\\ (A_1^{rs},A_2^{rs}) &= (A_1^r,A_2^r)\otimes (A_1^s,A_2^s)\\ r\leq s&\iff (A_1^r,A_2^r)\leq (A_1^s,A_2^s) \end{align*}$$ so that $\mathbb{R}$ contains a "copy" of $\mathbb{Q}$ inside of it. Finally, one can show completeness by showing that if $\{(A^{(i)}_1,A^{(i)}_2)\}_{i\in I}$ is a nonempty family of real numbers that is bounded above (so there exists a real number $(B_1,B_2)$ such that $A^{(i)}_1\subseteq B_1$ for all $i$), then the real number $(A_1,A_2)$ with $A_1 = \cup_{i\in I}A^{(i)}_1$, $A_2=\mathbb{Q}-A_1$ is the supremum of the set.

Interestingly, all answers so far give Cauchy sequences, so let me give the Dedekind cut construction.

The development assumes that the rational numbers are known and constructed, as are their operations and basic properties.

Definition. A cut is a partition $(A_1,A_2)$ of $\mathbb{Q}$, satisfying the following properties:

  1. $A_1\neq\varnothing$ and $A_2\neq\varnothing$.
  2. $A_1\cup A_2=\mathbb{Q}$.
  3. If $a\in A_1$ and $b\in A_2$, then $a\lt b$.
  4. If there exists a rational $c\in\mathbb{Q}$ such that for every $a\in A_1$ and every $b\in A_2$, $a\leq c$ and $c\leq b$, then $c\in A_1$.

In particular, if $a\in A_1$ and $c\lt a$, then $c\in A_1$ (since we must have $c\in A_1\cup A_2$). Point 4 is somewhat arbitrary: you could also require this element to be in $A_2$. Or you could, as Dedekind does, simply ignore it and eventually identify some cuts together. Also, point 2 is somewhat superfluous, as it is included in the notion of "partition".

Let us call $A_1$ the "left set" of the cut, and $A_2$ the "right set".

Now, every rational number $r$ determines a cut, defined by $(A^r_1,A^r_2)$, with $A^r_1=\{q\in\mathbb{Q}\mid q\leq r\}$, and $A^r_2 = \{q\in\mathbb{Q}\mid q\gt r\}$. However, not every cut is determined by a rational: for example, the cut $(A_1,A_2)$ given by $$\begin{align*} A_1 &= \{q\in\mathbb{Q}\mid q\lt 0\} \cup \{q\in\mathbb{Q}\mid q\geq 0\text{ and }q^2\leq 2\}\\ A_2 &= \{q\in\mathbb{Q}\mid g\geq 0\text{ and }q^2\gt 2\} \end{align*}$$ can be shown not to be of the form $(A_1^r,A_2^r)$ for any rational number $r$.

We define the set of real numbers $\mathbb{R}$ as the set of all cuts.

We define the order in $\mathbb{R}$ by letting $(A_1,A_2)\leq (B_1,B_2)$ if and only if $A_1\subseteq B_1$.

We define addition of cuts as follows: given $(A_1,A_2)$ and $(B_1,B_2)$, we define the cut $(A_1,A_2)\oplus (B_1,B_2) = (C_1,C_2)$, where $C_1$ consists of all rational numbers $q$ for which there exists $a\in A_1$ and $b\in B_1$ with $q\leq a+b$; and letting $C_2=\mathbb{Q}\setminus C_1$.

It is not hard to verify that $-(A_1,A_2) = (-A_2,-A_1)$, where $-X = \{-q\mid q\in X\}$; except that you need to be a bit careful given my definition: to make sure it satisfies point 4, if there is a rational $c$ in $A_1$ Such that $q\leq c$ for all $q\in A_1$, then you should move $-c$ out of $-A_1$ and put it in $-A_2$.

Now, multiplication is a bit more complicated. First we let $\mathbf{0}=(0_L,0_R)$, where $0_L=\{q\in\mathbb{Q}\mid q\leq 0\}$, and $0_R=\{q\in\mathbb{Q}\mid q>0\}$.

Then, given two cuts $(A_1,A_2)$, $(B_1,B_2)$, with both greater than or equal to $(0_L,0_R)$, we define their product $(A_1,A_2)\otimes(B_1,B_2) = (C_1,C_2)$ by letting $$C_1 = \{q\in\mathbb{Q}\mid \exists a\in A_1, b\in B_1\text{ such that }a\geq 0,b\geq 0,\text{ and }q\leq ab\}$$ and then lettig $C_2 = \mathbb{Q}\setminus C_1$.

If $(A_1,A_2)\geq (0_L,0_R)$ and $(B_1,B_2)\lt (0_L,0_R)$, we define the product $(A_1,A_2)\otimes(B_1,B_2)$ to be $-\bigl( (A_1,A_2)\otimes(-B_2,-B_1)\bigr)$.

And if $(A_1,A_2)\lt (0_L,0_R)$, and $(B_1,B_2)\lt(0_L,0_R)$, we define $$(A_1,A_2)\otimes(B_1,B_2) = (-A_2,-A_1)\otimes(-B_2,-B_1).$$

One can then show that the map $\mathbb{Q}\to\mathbb{R}$ given by $r\mapsto (A_1^r,A_2^r)$ is one-to-one, and satisfies $$\begin{align*} (A_1^{r+s},A_2^{r+s}) &= (A_1^r,A_2^r) \oplus (A_1^s,A_2^s)\\ (A_1^{rs},A_2^{rs}) &= (A_1^r,A_2^r)\otimes (A_1^s,A_2^s)\\ r\leq s&\iff (A_1^r,A_2^r)\leq (A_1^s,A_2^s) \end{align*}$$ so that $\mathbb{R}$ contains a "copy" of $\mathbb{Q}$ inside of it. Finally, one can show completeness by showing that if $\{(A^{(i)}_1,A^{(i)}_2)\}_{i\in I}$ is a nonempty family of real numbers that is bounded above (so there exists a real number $(B_1,B_2)$ such that $A^{(i)}_1\subseteq B_1$ for all $i$), then the real number $(A_1,A_2)$ with $A_1 = \cup_{i\in I}A^{(i)}_1$, $A_2=\mathbb{Q}-A_1$ is the supremum of the set.

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Arturo Magidin
Arturo Magidin
  • 421.8k
  • 60
  • 874
  • 1.2k

Interestingly, all answers so far give Cauchy sequences, so let me give the Dedekind cut construction.

The development assumes that the rational numbers are known and constructed, as are their operations and basic properties.

Definition. A cut is a partition $(A_1,A_2)$ of $\mathbb{Q}$, satisfying the following properties:

  1. $A_1\neq\varnothing$ and $A_2\neq\varnothing$.
  2. $A_1\cup A_2=\mathbb{Q}$.
  3. If $a\in A_1$ and $b\in A_2$, then $a\lt b$.
  4. If there exists a rational $c\in\mathbb{Q}$ such that for every $a\in A_1$ and every $b\in A_2$, $a\leq c$ and $c\leq b$, then $c\in A_1$.

In particular, if $a\in A_1$ and $c\lt a$, then $c\in A_1$ (since we must have $c\in A_1\cup A_2$). Point 4 is somewhat arbitrary: you could also require this element to be in $A_2$. Or you could, as Dedekind does, simply ignore it and eventually identify some cuts together. Also, point 2 is somewhat superfluous, as it is included in the notion of "partition".

Let us call $A_1$ the "left set" of the cut, and $A_2$ the "right set".

Now, every rational number $r$ determines a cut, defined by $(A^r_1,A^r_2)$, with $A^r_1=\{q\in\mathbb{Q}\mid q\leq r\}$, and $A^r_2 = \{q\in\mathbb{Q}\mid q\gt r\}$. However, not every cut is determined by a rational: for example, the cut $(A_1,A_2)$ given by $$\begin{align*} A_1 &= \{q\in\mathbb{Q}\mid q\lt 0\} \cup \{q\in\mathbb{Q}\mid q\geq 0\text{ and }q^2\leq 2\}\\ A_2 &= \{q\in\mathbb{Q}\mid g\geq 0\text{ and }q^2\gt 2\} \end{align*}$$ can be shown not to be of the form $(A_1^r,A_2^r)$ for any rational number $r$.

We define the set of real numbers $\mathbb{R}$ as the set of all cuts.

We define the order in $\mathbb{R}$ by letting $(A_1,A_2)\leq (B_1,B_2)$ if and only if $A_1\subseteq B_1$.

We define addition of cuts as follows: given $(A_1,A_2)$ and $(B_1,B_2)$, we define the cut $(A_1,A_2)\oplus (B_1,B_2) = (C_1,C_2)$, where $C_1$ consists of all rational numbers $q$ for which there exists $a\in A_1$ and $b\in B_1$ with $q\leq a+b$; and letting $C_2=\mathbb{Q}\setminus C_1$.

It is not hard to verify that $-(A_1,A_2) = (-A_2,-A_1)$, where $-X = \{-q\mid q\in X\}$.

Now, multiplication is a bit more complicated. First we let $\mathbf{0}=(0_L,0_R)$, where $0_L=\{q\in\mathbb{Q}\mid q\leq 0\}$, and $0_R=\{q\in\mathbb{Q}\mid q>0\}$.

Then, given two cuts $(A_1,A_2)$, $(B_1,B_2)$, with both greater than or equal to $(0_L,0_R)$, we define their product $(A_1,A_2)\otimes(B_1,B_2) = (C_1,C_2)$ by letting $$C_1 = \{q\in\mathbb{Q}\mid \exists a\in A_1, b\in B_1\text{ such that }a\geq 0,b\geq 0,\text{ and }q\leq ab\}$$ and then lettig $C_2 = \mathbb{Q}\setminus C_1$.

If $(A_1,A_2)\geq (0_L,0_R)$ and $(B_1,B_2)\lt (0_L,0_R)$, we define the product $(A_1,A_2)\otimes(B_1,B_2)$ to be $-\bigl( (A_1,A_2)\otimes(-B_2,-B_1)\bigr)$.

And if $(A_1,A_2)\lt (0_L,0_R)$, and $(B_1,B_2)\lt(0_L,0_R)$, we define $$(A_1,A_2)\otimes(B_1,B_2) = (-A_2,-A_1)\otimes(-B_2,-B_1).$$

One can then show that the map $\mathbb{Q}\to\mathbb{R}$ given by $r\mapsto (A_1^r,A_2^r)$ is one-to-one, and satisfies $$\begin{align*} (A_1^{r+s},A_2^{r+s}) &= (A_1^r,A_2^r) \oplus (A_1^s,A_2^s)\\ (A_1^{rs},A_2^{rs}) &= (A_1^r,A_2^r)\otimes (A_1^s,A_2^s)\\ r\leq s&\iff (A_1^r,A_2^r)\leq (A_1^s,A_2^s) \end{align*}$$ so that $\mathbb{R}$ contains a "copy" of $\mathbb{Q}$ inside of it. Finally, one can show completeness by showing that if $\{(A^{(i)}_1,A^{(i)}_2)\}_{i\in I}$ is a nonempty family of real numbers that is bounded above (so there exists a real number $(B_1,B_2)$ such that $A^{(i)}_1\subseteq B_1$ for all $i$), then the real number $(A_1,A_2)$ with $A_1 = \cup_{i\in I}A^{(i)}_1$, $A_2=\mathbb{Q}-A_1$ is the supremum of the set.