# Naive Set Theory: Constructing R

## Dedekind Cuts

We come now to the construction of a very important set - the real numbers. The rationals contain a great many useful values, but like the integers and natural numbers before them, they still don't contain "everything." For example, the ratio of the hypotenuse of an isosceles right triangle to one of its adjacent sides is not a rational number. Thus, for the third time, we must construct a new number system that can handle everything its predecessor could, plus more.

### Completeness Part I

The question is, what exactly is more? The concepts of counting, debts, and fractions were intuitive enough to understand, but what new property are we after now? Based on the problem posed by the isosceles right triangle above, we might first guess that the reals are the rationals plus all of the square roots (or rational roots) of rational numbers. However, the ratio of a circle's circumference to its diameter dashes these hopes. Perhaps we could try to formalize the notion of geometric ratios in general? But then what about the number $e$?

It turns out this notion of "the rest of the numbers" was not easy to nail down. The ancient Greeks knew that the rationals didn't have "all" the numbers, but it took another two thousand years for mathematicians to figure out how to express the concept of "all." Today, we call this property completeness. Intuitively, a set is complete if it doesn't have any holes between its elements. Formally, a set is complete if it has the least upper bound property.

### The Real Numbers

The set of real numbers, $\mathbb{R}$, is the ordered field with the least upper bound property that contains $\mathbb{Q}$ as a subfield. This terse definition can have its jargon expanded into its constituent parts. A field is a set fulfilling a number of algebraic requirements regarding addition and multiplication; an ordered field is a field for which an order relation is defined and that relates ti its algebraic properties; a subfield is a subset of a field that is itself a field; and finally the least upper bound property means that every nonempty subset of $\mathbb{R}$ with an upper bound has a least upper bound in $\mathbb{R}$.

Formally, $\mathbb{R}$ is the set with the following properties:

• $\mathbb{R}$ is closed under the addition operation, $+ : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$, which has the following properties:
• Associativity of addition: $x + (y + z) = (x + y) + z$ for all $x, y, z \in \mathbb{R}$.
• Commutativity of addition: $x + y = y + x$ for all $x, y \in \mathbb{R}$.
• Additive identity: There exists an element $0 \in \mathbb{R}$ such that $x + 0 = x$ for all $x \in \mathbb{R}$
• Additive inverse: For each $x \in \mathbb{R}$ there exist an element $-x \in \mathbb{R}$ such that $x + (-x) = 0$.
• $\mathbb{R}$ is closed under the multiplication operation, $\cdot : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$, which has the following properties:
• Associativity of multiplication: $x \cdot (y \cdot z) = (x \cdot y) \cdot z$ for all $x, y, z \in \mathbb{R}$.
• Commutativity of multiplication: $x \cdot y = y \cdot x$ for all $x, y \mathbb{R}$.
• Multiplicative identity: There exists an element $1 \in \mathbb{R}$ such that $x \cdot 1 = x$ for all $x \in \mathbb{R}$
• Multiplicative inverse: For each $x \in \mathbb{R}$ where $x \neq 0$ there exists an element $x^{-1}$ such that $x \cdot x^{-1} = 1$.
• Distributivity: $x \cdot ( y + z ) = (x \cdot y) + (x \cdot z)$ for all $x, y, z \in \mathbb{R}$.
• There exists an order relation $<$ on $\mathbb{R}$ with the following properties:
• If $x < y$, then $x + z < y + z$ for all $x, y, z, \in \mathbb{R}$.
• If $x, y \in \mathbb{R}$ and $x > 0$, and $y > 0$, then $x \cdot y > 0$.
• Least Upper Bound Property (Completeness): Every nonempty subset of $\mathbb{R}$ with an upper bound has a least upper bound in $\mathbb{R}$.
• $\mathbb{Q} \subset \mathbb{R}$

The set of real numbers that are not rational are called the irrational numbers.

Style note: The multiplication of two numbers, $\alpha \cdot \beta$, can be shortened to $\alpha\beta$. Likewise, to reduce the number of parentheses on the page, multiplication takes precedence over addition, so $(\alpha \cdot \beta) + (\gamma \cdot \delta)$ can be unambiguously shortened to $\alpha\beta + \gamma\delta$.

### Completeness Part II

How exactly does the least upper bound property relate to completeness, though? These two concepts do not seem at first related. The trick lies in observing that we can conjure up a rational number that is arbitrarily close to an irrational number. We can approximate pi to 2 decimal places or over 10 trillion, but no matter how far we go, pi itself is always just a tiny bit larger. These approximations are all in the set of rational numbers less than pi. The least upper bound for this set is, of course, pi itself. If we can construct the a new set of numbers out of sets that are defined by their least upper bounds like this, we're guaranteed to have "all" the numbers. Since this new number system will be complete, we'll know we'll finally be "done," and we won't need to construct any additional number systems to fill in any more holes.

### Dedekind Cuts

The formalization of this kind of bounded set is called a Dedekind cut. A Dedekind cut, or simply a cut, is a subset $\alpha \subset \mathbb{Q}$ with the following three properties:

1. Nonempty and proper: $\alpha \neq \varnothing$ and $\alpha\neq \mathbb{Q}$

2. Closed downward: If $a \in \alpha$, $y \in \mathbb{Q}$, and $b \le a$, then $b \in \alpha$.

3. No largest member: If $a \in \alpha$, then there exists a $b \in \alpha$ such that $a < b$.

A Dedekind cut is similar to a right open ray in $\mathbb{Q}$, except its supremum is not specified. In fact, its supremum may not be rational (that's the point!). The following diagram shows a cut as though it were a ray: A Dedekind cut whose supremum is rational is given by $\alpha = \{ a \in \mathbb{Q} : a < 4 \}$, as visualized below: A Dedekind cut whose supremum is not rational is given by $\alpha = \{ a \in \mathbb{Q} : a^2 < 2 \}$, as visualized below: We can now construct the real numbers. The set of real numbers, denoted $\mathbb{R}$, is the set of all Dedekind cuts. Remember that this definition is a construction, whose main purpose is to show that, given only the basic axioms of set theory, a set with all of the qualities $\mathbb{R}$ actually exists. There are in fact other ways to construct $\mathbb{R}$, but all constructions have the same properties by definition and are thus indistinguishable from one another. The "real" or "working" definition of $\mathbb{R}$ is in terms of its actual properties, which were listed above.

The proofs below show that the construction of $\mathbb{R}$ as the set of all Dedekind cuts has all of these properties.

Style note: To aid in clarity in this section, we make a few stylistic choices. Real numbers are denoted with Greek letters (such as $\alpha$, $\beta$, and $\gamma$), and rational numbers are denoted with Latin letters (such as $a$, $b$, and $c$). Likewise, the real numbers zero and one are denoted as $0_{\mathbb{R}}$ and $1_{\mathbb{R}}$, while their rational counterparts remain unadorned as $0$ and $1$.

## Problems

1. Prove the following two statements, call them P1 and P2, which are useful for the subsequent proofs in this construction:

Let $\alpha$ be a cut.

• P1: If $x \in \alpha$, and $y \notin \alpha$, then $x < y$.

• P2: If $x \notin \alpha$, and $x \le y$, then $y \notin \alpha$.

• P1: Let $x \in \alpha$, and let $y \notin \alpha$. Then $x \neq y$. If $y < x$, then by property 2 of cuts, then $y \in \alpha$. But this is a contradiction. Thus by trichotomy on $\mathbb{Q}$, $x < y$.

• P2: Let $x \notin \alpha$, and let $x \le y$. Proof by contradiction: Assume $y \in \alpha$. Then by property 2 of cuts, $x \in \alpha$. But this is a contradiction. Therefore $y \notin \alpha$.

2. Define $\alpha <\beta$ to mean that $\alpha \subset \beta$. Show that this relation forms a linear ordering on $\mathbb{R}$.

1. Comparability: Let $\alpha, \beta \in \mathbb{R}$ such that $\alpha \neq \beta$. If $\alpha \subset \beta$, then $\alpha < \beta$. If $\alpha$ is not a proper subset of $\beta$, then there exists a $a \in \alpha$ such that $a \notin \beta$. By the first property of cuts proven above, this means that $b < a$ for all $b \in \beta$. By the second property of cuts, this means that $b \in \alpha$. Therefore $\beta \subset \alpha$, and thus $\beta < \alpha$.

2. Nonreflexivity: Proof by contradiction. Let $\alpha \in \mathbb{R}$, but assume $\alpha < \alpha$. Then $\alpha$ is a proper subset of itself. But this is a contradiction, no set is a proper subset of itself. Therefore therefore it is not possible that $\alpha < \alpha$.

3. Transitivity: Let $\alpha, \beta, \gamma \in \mathbb{R}$ such that $\alpha < \beta$ and $\beta < \gamma$. Thus $\alpha \subset \beta$ and $\beta \subset \gamma$. By transitivity of subsets, $\alpha \subset \gamma$, therefore $\alpha < \gamma$.

3. Prove that $\mathbb{R}$ has the least upper bound property.

Hint: Let $R$ be a nonempty subset of $\mathbb{R}$ such that $R$ is bounded from above. Let $\mathcal{R} = \bigcup R$ and show that $\mathcal{R} \in \mathbb{R}$ and that $\mathcal{R} = \text{sup}(R)$.

A set has the least upper bound property if every subset bounded from above has a least upper bound. Let $R \subset \mathbb{R}$ that is bounded from above by some $\beta \in \mathbb{R}$. Let $\mathcal{R} = \bigcup R$. That is, $\mathcal{R}$ is the union of all the real numbers in $R$, which in our construction is the union of all the cuts in $R$. Part one of this proof is to show that $\mathcal{R}$ is itself a real number, and part two is to show that it is the least upper bound of $R$.

Part 1: Prove that $\mathcal{R} \in \mathbb{R}$: In order to show that $\mathcal{R} \in \mathbb{R}$, we must show that it meets the three properties of cuts.

1. $R$ is nonempty, therefore there exists an $\alpha_0 \in R$. Because $\alpha$ is a cut, it is nonempty. Since $\alpha_0 \subset \mathcal{R}$, it follows that $\mathcal{R}$ is also nonempty. Next, note that because $\beta$ is an upper bound of $R$, $\alpha \leq \beta$ for all $\alpha \in R$. By definition of the order relation, this means that $\alpha \subseteq \beta$. Therefore $\mathcal{R} \subseteq \beta$. Thus $\mathcal{R} \neq \mathbb{Q}$.

2. Let $x \in \mathcal{R}$. Then $x \in \alpha_1$ for some $\alpha_1 \in R$. Pick $y \in \mathbb{Q}$ such that $y < x$. By property 2 of cuts, $y \in \alpha_1$. Therefore $y \in \mathcal{R}$.

3. Let $x \in \mathcal{R}$. Then $x \in \alpha_1$ for some $\alpha_1 \in R$. By property 3 of cuts, there exists a $y \in \alpha_1$ such that $x < y$. Since $y \in \alpha_1$ and $\alpha_1 \subset \mathcal{R}$, it follows that $y \in \mathcal{R}$.

Part 2: Prove that $\mathcal{R} = \text{sup}(R)$: First we show that $\mathcal{R}$ is an upper bound of $R$, and then we show that it is the least such bound.

1. Let $\alpha \in R$. Then $\alpha \subseteq \mathcal{R}$, which by definition of the order relation means that $\alpha \leq \mathcal{R}$. Therefore $\mathcal{R}$ is an upper bound for $R$.

2. Suppose $\gamma < \mathcal{R}$. By definition of the order relation, $\gamma \subset \mathcal{R}$. By definition of proper subset, there exists an element $x \in \mathcal{R}$ such that $x \notin \gamma$. Because $x \in \mathcal{R}$, there exists some $\alpha \in R$ such that $x \in \alpha$. By P1, $y < x$ for every $y \in \gamma$. Therefore $\gamma \subset \alpha$, by property 2 of cuts.  Therefore $\gamma < \alpha$, and so $\gamma$ is not an upper bound of $R$. Thus $\mathcal{R}$ is the least upper bound of $R$.

4. Define the addition function $+$ as

$$\alpha + \beta = \{ x + y : x \in \alpha, y \in \beta \},$$

where $\alpha, \beta \in \mathbb{R}$. Prove that $\mathbb{R}$ is closed under addition.

Clearly the domain of $+$ is $\mathbb{R} \times \mathbb{R}$. To check that the range is $\mathbb{R}$, we show that $\alpha + \beta$ is a cut. Let $\gamma = \alpha + \beta$.

1. First we show that $\gamma$ is neither empty nor all of $\mathbb{Q}$. Since $\alpha$ and $\beta$ are not empty, there is an $a \in \alpha$ and a $b \in \beta$. Then $a + b \in \alpha + \beta$, so $\gamma$ is nonempty. To show that $\gamma$ is not all of $\mathbb{Q}$, note that both $\alpha$ and $\beta$ are both cuts and therefore not all of $\mathbb{Q}$. Therefore there exist a elements $x \notin \alpha$ and $y \notin \beta$. By P1, $x$ greater than all elements in $\alpha$ and $y$ is greater than all elements in $\beta$. Therefore $x + y$ is not in $\gamma$, and thus $\gamma \neq \mathbb{Q}$.

2. Let $x \in \gamma$ and $y \in \mathbb{Q}$ such that $y < x$. By definition of addition, $x = m + n$ where $m \in \alpha$ and $n \in \beta$. Because $y < x$, it follows that $y - n < m$, so by property 2 of cuts, $y - n \in \alpha$. Thus $y = (y - n) + n$, so $y \in \gamma$.

3. Let $x \in \gamma$. Then $x = m + n$ for some $m \in \alpha$ and $n \in \beta$. Because $\alpha$ is a cut, there exists a $p \in \alpha$ such that $p > m$. Then $p + n \in \gamma$ and $p + n > x$.

5. Associativity of Addition: Show that $a + (b + c) = (a + b) + c$

Let $\alpha, \beta, \gamma \in \mathbb{R}$. Then

$\alpha + (\beta + \gamma) = \alpha + \{ y + z : y \in \beta, z \in \gamma \} \\ \alpha + (\beta + \gamma) = \{ x + (y + z) : x \in \alpha, y \in \beta, z \in \gamma \} \\ \alpha + (\beta + \gamma) = \{ (x + y) + z : x \in \alpha, y \in \beta, z \in \gamma \} \\ \alpha + (\beta + \gamma) = \{ (x + y) : x \in \alpha, y \in \beta \} + \gamma \\ \alpha + (\beta + \gamma) = (\alpha + \beta) + \gamma \\$

6. Commutativity of Addition: Show that $\alpha + \beta = \beta + \alpha$.

$\alpha + \beta = \{ x + y : x \in \alpha, y \in \beta \} \\ \alpha + \beta = \{ y + x : x \in \alpha, y \in \beta \} \\ \alpha + \beta = \beta + \alpha$

7. Additive Identity: Show that there exists an element $0_{\mathbb{R}} \in \mathbb{R}$ such that $\alpha + 0_{\mathbb{R}} = a$ for all $\alpha \in \mathbb{R}$.

Hint: Show that $0_{\mathbb{R}} = \{ x \in \mathbb{Q} : x < 0 \}$.

Let $0_{\mathbb{R}} = \{ x \in \mathbb{Q} : x < 0 \}$, and let $\alpha \in \mathbb{R}$ be any other cut. To show that $\alpha + 0_{\mathbb{R}} = \alpha$, we must show that $\alpha + 0_{\mathbb{R}} \subseteq \alpha$ and $\alpha \subseteq \alpha + 0_{\mathbb{R}}$. First, let $x \in 0_{\mathbb{R}}$ and $y \in \alpha$. Because $x$ is negative, $x + y < y$. Therefore, by property 2 of cuts, $x + y \in \alpha$, and so $\alpha + 0_{\mathbb{R}} \subseteq \alpha$. To show the opposite inclusion, pick $z \in \alpha$ such that $x < z$. Then $x - z < 0$, so $x - z \in 0_{\mathbb{R}}$. We can see that, $y = z + (y - z) \in \alpha + 0_{\mathbb{R}}$, so therefore $\alpha \subseteq \alpha + 0_{\mathbb{R}}$.

8. Additive Inverses: Show that for every element $\alpha \in \mathbb{R}$ there exists an element $-a \in \mathbb{R}$ such that $\alpha + (-\alpha) = 0$.

Hint: Let $-\alpha = \{ x \in \mathbb{Q} : \text{ there exists a } y \in \mathbb{Q} \text{ such that } -x - y \notin \alpha \}$.

Let $\alpha \in \mathbb{R}$, and let $-\alpha = \{ x \in \mathbb{Q} : \text{ there exists a } y \in \mathbb{Q} \text{ such that } -x - y \notin \alpha \}$. We must first show that $-\alpha$ is a cut and then show that $\alpha + (-\alpha) = 0$.

Part 1: To show that $-\alpha$ is a cut, we verify that it meets the three properties of cuts:

1. To show $-\alpha$ is not empty, let $x \notin \alpha$, and let $y = -x - 1$. Then $y \in -\alpha$, so $\alpha \neq \varnothing$. To show $-\alpha$ is not all of $\mathbb{Q}$, let $x \in \alpha$. Then $-x \notin -\alpha$. Therefore $-\alpha \neq \mathbb{Q}$.

2. Let $x \in -\alpha$ and $z > 0$ such that $-x -z \notin \alpha$. Let $y \in \mathbb{Q}$ such that $y < x$.  Then $-x < -y$, so $-x - z < -y - z$. By P2, $-y - z \notin \alpha$, therefore $y \in -\alpha$.

3. Let $x \in -\alpha$. Then there exists a $z > 0$ such that $-x - z \notin \alpha$. Let $p = z/2$. Clearly $p > 0$. Let $q = x + p$. Then $x < q$ and $-q - p = -x - z$. Therefore $-q - p \notin \alpha$, so $q \in -\alpha$.

Part 2: To show that $\alpha + (-\alpha) = 0_{\mathbb{R}}$, we must show that $\alpha + (-\alpha) \subseteq 0_{\mathbb{R}}$ and that $0_{\mathbb{R}} \subseteq \alpha + (-\alpha)$.

Let $x \in \alpha$, $y \in -\alpha$, and $z > 0$ such that $-y - z \notin \alpha$ By P1, $-y - z > x$. Therefore $x + y < -z < 0$, so $\alpha + (-\alpha) \subseteq 0$.

Conversely, let $p \in 0_{\mathbb{R}}$. Then $p < 0$. Let $q = -p/2$, and note that $q > 0$. By the Archimedian property on $\mathbb{Q}$, there exists an $n \in \mathbb{N}$ such that $n\cdot q \in \alpha$ but $(n + 1) \cdot q \notin \alpha$. Let $r = -(n + 2) \cdot q$. Then $-r - q \notin \alpha$, so $r \in -\alpha$. Therefore $p = n\cdot q + r \in \alpha + (-\alpha)$, and so $0 \subseteq \alpha + (-\alpha)$.

9. Additive Ordering Property: Show that if $\alpha < \beta$, then $\alpha + \gamma < \beta + \gamma$ for all $\alpha, \beta, \gamma \in \mathbb{R}$.

Let $x \in \alpha + \gamma$. Then $x < a + c$ for some $a \in \alpha$ and $c \in \gamma$. By the order definition, $\alpha \subset \beta$. Therefore there exists some $b \in \beta$ such that $b \notin \alpha$. By P2, $b > a$. Therefore $a + c < b + c$. Thus $x < b + c$, so $x \in \beta + \gamma$, and therefore $a + c < b + c$.

10. Show that $-(-\alpha) = \alpha$ for all $\alpha \in \mathbb{R}$.

$\alpha + (-\alpha) = 0_{\mathbb{R}} \\ -\alpha + (-(-\alpha)) = 0_{\mathbb{R}} \\ -\alpha + (-(-\alpha)) = \alpha + (-\alpha) \\ \alpha + ((-\alpha) + (-(-\alpha)) = \alpha + (\alpha + (-\alpha)) \\ (\alpha + (-\alpha)) + (-(-\alpha)) = \alpha + (\alpha +(-\alpha)) \\ 0_{\mathbb{R}} + (-(-\alpha)) = \alpha + 0_{\mathbb{R}} \\ -(-\alpha) = \alpha \\$

11. Show that $-(\alpha + \beta) = -\alpha - \beta$.

$(\alpha + \beta) + (-\alpha - \beta) = (\alpha - \alpha) + (\beta - \beta) \\ (\alpha + \beta) + (-\alpha - \beta) = 0_{\mathbb{R}} \\ (-(\alpha + \beta) + (\alpha + \beta)) + (-\alpha - \beta) = -(\alpha + \beta) \\ 0_{\mathbb{R}} + (-\alpha - \beta) = -(\alpha + \beta) \\ -\alpha - \beta = -(\alpha + \beta)$

12. The multiplication function for Dedekind cuts requires a preliminary multiplication function, $p : \mathbb{R}^+ \times \mathbb{R}^+ \rightarrow \mathbb{R}$, which is defined only for positive real numbers, denoted $\mathbb{R}^+$:

$$p(\alpha, \beta) = \{ z < x \cdot y : x \in \alpha, x > 0, y \in \beta, y > 0 \}$$

1. Explain why only the positive rationals in each cut are included.

2. Show that $p(\alpha, \beta)$ is a cut and therefore in $\mathbb{R}$.

1. Cuts are unbounded below, which means they all contain negative rationals of arbitrary magnitude. Since the product of two negative numbers is positive, setting $p$ less than that product would include arbitrarily large positive numbers. For example, $-5_{\mathbb{Q}} \in 1_{\mathbb{R}}$, and $-10_{\mathbb{Q}} \in 2_{\mathbb{R}}$, but $(-5_{\mathbb{Q}}) \cdot (-10_{\mathbb{Q}}) = 50$, and we know $1_{\mathbb{R}} \cdot 2_{\mathbb{R}}$ should equal $2_{\mathbb{R}}$, but $50_{\mathbb{Q}} \notin 2_{\mathbb{R}}$.

2. We show that $p(\alpha, \beta)$ meets all three criteria of a cut.

1. Because $\alpha$ and $\beta$ are cuts, they are nonempty. Because they are positive, there exist a positive $x \in \alpha$ and $y \in \beta$, and their product $x \cdot y$ is positive. The set of all $p < x \cdot y$ is clearly nonempty, so $p(\alpha, \beta)$ is therefore nonempty. Likewise, let $m \notin \alpha$ and $n \notin \beta$. Then by P1, $m$ is greater than all $x \in \alpha$ and $n$ is greater than all $y \in \beta$. Restricting $x$ and $y$ to only positive values, $m \cdot n$ is greater than all $x \cdot y$, and therefore is not in $p(\alpha, \beta)$. Therefore $p(\alpha, \beta)$ is not all of $\mathbb{Q}$.

2. Let $m \in p(\alpha, \beta)$, $n \in \mathbb{Q}$, and $n < m$. Then there exist some positive $x \in \alpha$ and positive $y \in \beta$ such that $m < x \cdot y$. By transitivity, $n < x \cdot y$, so $n \in p(\alpha, \beta)$.

3. Let $m \in p(\alpha, \beta)$. Then there exit some positive $x \in \alpha$ and positive $y \in \beta$ such that $m < x \cdot y$. By property 3 of cuts, there exist some positive $a \in \alpha$ and positive $b \in \beta$ such that $x < a$ and $y < b$. Then $m < xy < ab$ and $ab \in p(\alpha, \beta)$.

13. Define the multiplication function $\cdot$ as

$$\alpha \cdot \beta = \left\{ \begin{array}{ll} p(\alpha, \beta) & \text{if both } \alpha > 0_{\mathbb{R}} \text{ and } \beta > 0_{\mathbb{R}} \\ -p(-\alpha, \beta) & \text{if } \alpha < \text{ and } \beta > 0_{\mathbb{R}} \\ -p(\alpha, -\beta) & \text{if } \alpha > 0_{\mathbb{R}} \text{ and } \beta < 0_{\mathbb{R}} \\ p(-\alpha, -\beta) & \text{if both } \alpha < 0_{\mathbb{R}} \text{ and } \beta < 0_{\mathbb{R}} \\ 0_{\mathbb{R}} & \text{if either } \alpha = 0_{\mathbb{R}} \text{ or } \beta = 0_{\mathbb{R}} \end{array} \right.$$

where $p$ is the restricted multiplication function defined above.

Show that $\mathbb{R}$ is closed under multiplication.

To check that $\mathbb{R}$ is closed under multiplication, we must check that the domain of $\cdot$ is $\mathbb{R} \times \mathbb{R}$ and its range is $\mathbb{R}$. The definition handles all combinations of positive, negative, and zero values for both $\alpha$ and $\beta$, thus $\cdot$ is defined over $\mathbb{R} \times \mathbb{R}$. Likewise, its codomain is $\mathbb{R}^{+} \cup \mathbb{R}^{-} \cup \{0_{\mathbb{R}}\} = \mathbb{R}$. Thus $\mathbb{R}$ is closed under multiplication.

14. Associativity of Multiplication: Show that $\alpha \cdot (\beta \cdot \gamma) = (\alpha \cdot \beta) \cdot \gamma$.

Hint: Show that multiplication is associative for the case when $\alpha$, $\beta$, and $\gamma$ are all positive, then work out the cases for when each of them may not be positive. There are quite a few cases.

First, we prove that the restricted multiplication function $p$ is associative for when $\alpha$, $\beta$, and $\gamma$ are all positive:

$\alpha \cdot (\beta \cdot \gamma) = p(\alpha, p(\beta, \gamma)) \\ \alpha \cdot (\beta \cdot \gamma) = \{ a < x \cdot m : x \in \alpha, m \in p(\beta, \gamma), x > 0_{\mathbb{Q}}, m > 0_{\mathbb{Q}} \} \\ \alpha \cdot (\beta \cdot \gamma) = \{ a < x \cdot (y \cdot z) : x \in \alpha, y \in \beta, z \in \gamma, x > 0_{\mathbb{Q}}, y > 0_{\mathbb{Q}}, z > 0_{\mathbb{Q}} \} \\ \alpha \cdot (\beta \cdot \gamma) = \{ a < (x \cdot y) \cdot z : x \in \alpha, y \in \beta, z \in \gamma, x > 0_{\mathbb{Q}}, y > 0_{\mathbb{Q}}, z > 0_{\mathbb{Q}} \} \\ \alpha \cdot (\beta \cdot \gamma) = \{ a < n \cdot z : n \in p(\alpha, \beta), z \in \gamma, n > 0_{\mathbb{Q}}, y > 0_{\mathbb{Q}} \} \\ \alpha \cdot (\beta \cdot \gamma) = p(p(\alpha, \beta), \gamma)) \\ \alpha \cdot (\beta \cdot \gamma) = (\alpha \cdot \beta) \cdot \gamma \\$

From here, we exhaust all of the other cases where $\alpha$, $\beta$, and $\gamma$ are either positive or negative:

• If $\alpha$ and $\beta$ are positive and $\gamma$ is negative

$\alpha \cdot (\beta \cdot \gamma) = \alpha \cdot (-(\beta \cdot (-\gamma))) \\ \alpha \cdot (\beta \cdot \gamma) = -(\alpha \cdot (\beta \cdot (-\gamma))) \\ \alpha \cdot (\beta \cdot \gamma) = -((\alpha \cdot \beta) \cdot (-\gamma)) \\ \alpha \cdot (\beta \cdot \gamma) = (\alpha \cdot \beta) \cdot \gamma \\$

• If $\alpha$ is positive, $\beta$ is negative, and $\gamma$ is positive, then

$\alpha \cdot (\beta \cdot \gamma) = \alpha \cdot (-((-\beta) \cdot \gamma)) \\ \alpha \cdot (\beta \cdot \gamma) = -(\alpha \cdot ((-\beta) \cdot \gamma)) \\ \alpha \cdot (\beta \cdot \gamma) = -((\alpha \cdot (-\beta)) \cdot \gamma) \\ \alpha \cdot (\beta \cdot \gamma) = -(\alpha \cdot (-\beta)) \cdot \gamma \\ \alpha \cdot (\beta \cdot \gamma) = (\alpha \cdot \beta) \cdot \gamma \\$

• If $\alpha$ is positive and $\beta$ and $\gamma$ are both negative, then

$\alpha \cdot (\beta \cdot \gamma) = \alpha \cdot ((-\beta) \cdot (-\gamma)) \\ \alpha \cdot (\beta \cdot \gamma) = (\alpha \cdot (-\beta)) \cdot (-\gamma) \\ \alpha \cdot (\beta \cdot \gamma) = -(\alpha \cdot (-\beta)) \cdot \gamma) \\ \alpha \cdot (\beta \cdot \gamma) = (\alpha \cdot \beta) \cdot \gamma) \\$

• If $\alpha$ is negative and $\beta$ and $\gamma$ are both positive, then

$\alpha \cdot (\beta \cdot \gamma) = -((-\alpha) \cdot (\beta \cdot \gamma)) \\ \alpha \cdot (\beta \cdot \gamma) = -(((-\alpha) \cdot \beta) \cdot \gamma) \\ \alpha \cdot (\beta \cdot \gamma) = -((-\alpha) \cdot \beta) \cdot \gamma \\ \alpha \cdot (\beta \cdot \gamma) = (\alpha \cdot \beta) \cdot \gamma \\$

• If $\alpha$ is negative, $\beta$ is positive, and $\gamma$ is is negative, then

$\alpha \cdot (\beta \cdot \gamma) = -((-\alpha) \cdot (-(\beta \cdot (-\gamma)))) \\ \alpha \cdot (\beta \cdot \gamma) = (-\alpha) \cdot (\beta \cdot (-\gamma)) \\ \alpha \cdot (\beta \cdot \gamma) = ((-\alpha) \cdot \beta) \cdot (-\gamma) \\ \alpha \cdot (\beta \cdot \gamma) = -(\alpha \cdot \beta) \cdot (-\gamma) \\ \alpha \cdot (\beta \cdot \gamma) = (\alpha \cdot \beta) \cdot \gamma \\$

• If $\alpha$ and $\beta$ are both negative and $\gamma$ is is positive, then

$\alpha \cdot (\beta \cdot \gamma) = -((-\alpha) \cdot (-((-\beta) \cdot \gamma))) \\ \alpha \cdot (\beta \cdot \gamma) = (-\alpha) \cdot ((-\beta) \cdot \gamma)) \\ \alpha \cdot (\beta \cdot \gamma) = ((-\alpha) \cdot (-\beta)) \cdot \gamma) \\ \alpha \cdot (\beta \cdot \gamma) = (\alpha \cdot \beta) \cdot \gamma \\$

• If $\alpha$, $\beta$, and $\gamma$ are all negative, then

$\alpha \cdot (\beta \cdot \gamma) = \alpha \cdot ((-\beta) \cdot (-\gamma)) \\ \alpha \cdot (\beta \cdot \gamma) = -((-\alpha) \cdot ((-\beta) \cdot (-\gamma))) \\ \alpha \cdot (\beta \cdot \gamma) = -(((-\alpha) \cdot (-\beta)) \cdot (-\gamma)) \\ \alpha \cdot (\beta \cdot \gamma) = -((\alpha \cdot \beta) \cdot (-\gamma)) \\ \alpha \cdot (\beta \cdot \gamma) = (\alpha \cdot \beta) \cdot \gamma \\$

Finally, we note the three cases where $\alpha$, $\beta$, or $\gamma$ is zero:

• If $\alpha = 0_{\mathbb{R}}$, then $\alpha \cdot (\beta \cdot \gamma) = 0_{\mathbb{R}} \cdot (\beta \cdot \gamma) = 0_{\mathbb{R}}$

• If $\beta = 0_{\mathbb{R}}$, then $\alpha \cdot (\beta \cdot \gamma) = \alpha \cdot (0_{\mathbb{R}} \cdot \gamma) = \alpha \cdot 0_{\mathbb{R}} = 0_{\mathbb{R}}$

• If $\gamma = 0_{\mathbb{R}}$, then $\alpha \cdot (\beta \cdot \gamma) = \alpha \cdot (\beta \cdot 0_{\mathbb{R}}) = \alpha \cdot 0_{\mathbb{R}} = 0_{\mathbb{R}}$

15. Commutativity of Multiplication: Show that $\alpha \cdot \beta = \beta \cdot \alpha$.

Hint: Prove the case when $\alpha$ and $\beta$ are positive, then handle the non-positive cases.

First we show that the restricted multiplication function $p$ is commutative for when $\alpha$ and $\beta$ are both positive:

$\alpha \cdot \beta = p(\alpha, \beta) \\ \alpha \cdot \beta = \{ a < x \cdot y : x \in \alpha, x > 0_{\mathbb{Q}}, y \in \beta, y > 0_{\mathbb{Q}} \} \\ \alpha \cdot \beta = \{ a < y \cdot x : x \in \alpha, x > 0_{\mathbb{Q}}, y \in \beta, y > 0_{\mathbb{Q}} \} \\ \alpha \cdot \beta = p(\beta, \alpha) \\$

Next we cover the cases where $\alpha$ and $\beta$ may be negative:

• If $\alpha$ is positive and $\beta$ is negative, then

$\alpha \cdot \beta = -(\alpha \cdot (-\beta)) \\ \alpha \cdot \beta = -((-\beta) \cdot \alpha) \\ \alpha \cdot \beta = \beta \cdot \alpha \\$

• If $\alpha$ is negative and $\beta$ is positive, then

$\alpha \cdot \beta = -((-\alpha) \cdot \beta) \\ \alpha \cdot \beta = -(\beta \cdot (-\alpha)) \\ \alpha \cdot \beta = \beta \cdot \alpha \\$

• If $\alpha$ and $\beta$ are both negative, then

$\alpha \cdot \beta = (-\alpha) \cdot (-\beta) \\ \alpha \cdot \beta = (-\beta) \cdot (-\alpha) \\ \alpha \cdot \beta = \beta \cdot \alpha \\$

Finally, we note that if $\alpha = 0_\mathbb{R}$, then $\alpha \cdot \beta = 0_\mathbb{R} = \beta \cdot \alpha$ by definition. The same is true if $\beta = 0_{\mathbb{R}}$.

16. Multiplicative Identity: Show that there exists an element $1_{\mathbb{R}} \in \mathbb{R}$ such that $\alpha \cdot 1_{\mathbb{R}} = \alpha$ for all $\alpha \in \mathbb{R}$.

Let $1_{\mathbb{R}} = \{ x \in \mathbb{Q} : x < 1 \}$.

Part 1: We show that $1_{\mathbb{R}}$ is a cut.

1. Clearly $0 < 1$, so $0 \in 1_{\mathbb{R}}$, and therefore $1_{\mathbb{R}} \neq \varnothing$. Likewise, $2 > 1$, so $2 \notin 1_{\mathbb{R}}$, and therefore $1_{\mathbb{R}} \neq \mathbb{Q}$.

2. Let $x \in 1_\mathbb{Q}$. If $y < x$ for some $y \in \mathbb{Q}$, then $y < 1$ by transitivity, and therefore $y \in 1_{\mathbb{R}}$

3. Let $x \in 1_{\mathbb{R}}$. Then $x < 1$. Note that $x < \dfrac{x + 1}{2} < 1$, and therefore $\dfrac{x + 1}{2} \in 1_{\mathbb{R}}$.

Part 2: We show that $\alpha \cdot 1_{\mathbb{R}} = \alpha$ for all $\alpha \in \mathbb{R}$.

First we consider the case where $\alpha > 0_{\mathbb{R}}$. Let $x \in \alpha$ such that $x > 0$, and let $y \in 1_{\mathbb{R}}$ such that $y > 0$. First, we note that $x \cdot y < x$, therefore by property 3 of cuts, $p(\alpha, 1_{\mathbb{R}}) \subseteq \alpha$. Conversely, by property 3 of cuts, there exists a $z \in \alpha$ such that $x < z$. Then $0 < \frac{x}{z} < 1$, and so $\frac{x}{z} \in 1_{\mathbb{R}}$. We can see that $x = z \cdot \frac{x}{z} \in p(\alpha, 1_{\mathbb{R}})$, so $\alpha \subseteq p(\alpha, 1_{\mathbb{R}})$.

Next we consider the case where $\alpha < 0$. By definition of multiplication, $\alpha \cdot 1_{\mathbb{R}} = -p(-\alpha, 1_{\mathbb{R}}) = -(-\alpha) = \alpha$.

Finally, we note that $0_{\mathbb{R}} \cdot 1_{\mathbb{R}} = 0_{\mathbb{R}}$ by definition.

17. Let $\alpha \in \mathbb{R}$, and let $q \in \mathbb{Q}$ such that $q > 0$. Show that there exist and $a \in \alpha$ and $b \notin \alpha$ such that $b - a = q$.

Pick any $a_1 \in \alpha$, and pick $b_1$ such that $b_1$ is not a least upper bound of $\alpha$. Then by the Archimedean property on $\mathbb{Q}$, there exists an $n \in \mathbb{N}$ such that $nq > b_1 - a_1$. Then $a_1 + nq > b_1$, and so by P2, $a_1 + nq$ is an upper bound of $\alpha$. Let $m \in \mathbb{N}$ be the least natural number such that $a_1 + mq$ is an upper bound of $\alpha$. Then $a_1 + (m-1)q \in \alpha$. Let $a = a_1 + (m-1)q$, and let $b = a_1 + mq$. Then $c = b - a$.

18. Multiplicative Inverses: Show that for each nonzero $\alpha \in \mathbb{R}$, there exists an element $\alpha^{-1}$ such that $\alpha \cdot \alpha^{-1} = 1_{\mathbb{R}}$.

Positive Case: For $\alpha > 0_{\mathbb{R}}$, let $\alpha^{-1} = \{ x < \frac{1}{y} : y \notin \alpha, y > 0\}$.

Part 1: First we show that $\alpha^{-1}$ is a cut.

1. Because $\alpha \neq \varnothing$, there is an element $x \notin \alpha$. Then $\frac{1}{x} - 1 < \frac{1}{x}$, so $\frac{1}{x} \in \alpha^{-1}$, so $\alpha^{-1} \neq \varnothing$. Likewise, because $\alpha \neq \varnothing$, there exsits an $x \in \alpha$. Then $\frac{1}{x} > \frac{1}{y}$ for all $y \in \alpha$, so $\frac{1}{x} \notin \alpha^{-1}$. Thus $\alpha^{-1} \neq \mathbb{Q}$.

2. Let $x \in \alpha^{-1}$. Then there is some $y \notin \alpha$ such that $x < \frac{1}{y}$. If $z \in \mathbb{Q}$ such that $z < x$, then $z < \frac{1}{y}$, and therefore $z \in \alpha^{-1}$.

3. Let $x \in \alpha^{-1}$. Then there is some $y \notin \alpha$ such that $x < \frac{1}{y}$. Note that $x < \dfrac{x + \frac{1}{y}}{2} < \frac{1}{y}$, and therefore $\dfrac{x+\frac{1}{y}}{2} \in \alpha^{-1}$.

Part 2: Next we show that $\alpha \cdot \alpha^{-1} = 1_{\mathbb{R}}$.

By definition of multiplication on positive real numbers, we see that

$\alpha \cdot \alpha^{-1} = \{x < y \cdot \frac{1}{z} : y \in \alpha, y > 0, z \notin \alpha \}$.

We will show that $1_{\mathbb{R}} \subseteq \alpha \cdot \alpha^{-1}$ and that $\alpha \cdot \alpha^{-1} \subseteq 1_{\mathbb{R}}$.

First, let $x \in \alpha \cdot \alpha^{-1}$. Then $x < y \cdot \frac{1}{z}$, where $y \in \alpha$ and $z \notin \alpha$ and both $y$ and $z$ are positive. By P2, $y < z$. Therefore $y \cdot \frac{1}{z} < z \cdot \frac{1}{z} = 1$, so $x \in 1_{\mathbb{R}}$.

Conversely, let $z \in 1$ such that $0 < z < 1$, and let $a \in \alpha$ such that $\alpha > 0_{\mathbb{R}}$. Then $(1-z)a > 0$. By the previous lemma, there exists an $x \in \alpha$ and a $y \notin \alpha$ such that $y - x = (1-z)a$. Since $y \notin \alpha$, by P2 it follows that $a < y$, and thus $y - x < (1-z)y$. Rearrange this inequality to see that $zy < x$, and thus $y < \frac{x}{z}$. Therefore by P2 again, $\frac{x}{z} \notin \alpha$. Finally, we see that that $z = x \cdot \frac{z}{x} = x \cdot \dfrac{1}{\frac{x}{z}} \in p(\alpha, \alpha^{-1})$, and thus $1 \subseteq \alpha \cdot \alpha^{-1}$.

Negative Case: For $\alpha < 0_{\mathbb{R}}$, note that $-\alpha > 0_{\mathbb{R}}$, so there exists a $(-\alpha)^{-1}$ such that $-\alpha \cdot (-\alpha)^{-1} = 1_{\mathbb{R}}$. Thus

$\alpha \cdot -((-\alpha)^{-1}) = -(\alpha \cdot (-\alpha)^{-1}) \\ \alpha \cdot -((-\alpha)^{-1}) = -\alpha \cdot (-\alpha)^{-1} \\ \alpha \cdot -((-\alpha)^{-1}) = 1_{\mathbb{R}} \\ \alpha^{-1} \cdot (\alpha \cdot -((-\alpha)^{-1})) = \alpha^{-1} \cdot 1_{\mathbb{R}} \\ (\alpha^{-1} \cdot \alpha) \cdot -((-\alpha)^{-1}) = \alpha^{-1} \\ 1_{\mathbb{R}} \cdot -((-\alpha)^{-1}) = \alpha^{-1} \\ -((-\alpha)^{-1}) = \alpha^{-1} \\$.

19. Multiplicative Ordering Property: Show that if $0_{\mathbb{R}} < \alpha$ and $0_{\mathbb{R}} < \beta$, then $0_{\mathbb{R}} < \alpha \cdot \beta$.

By the definition of the order relation, we must show that $0_{\mathbb{R}} \subset \alpha \cdot \beta$.

Let $x \in 0_{\mathbb{R}}$. Then $x < 0$. Because $\alpha$ and $\beta$ are positive, there is a positive $a \in \alpha$ and a positive $b \in \beta$.  Therefore $0 < a \cdot b$, so by transitivity $x < a \cdot b$, and and therefore $0_{\mathbb{R}} \subset \alpha \cdot \beta$.

20. Distributive Property: Show that $\alpha (\beta + \gamma) = \alpha\beta + \alpha\gamma$.

Hint: First show that the distributive property holds when $\alpha$, $\beta$, and $\gamma$ are all positive. Then handle the cases when each of them may not be positive.

Note: There are a lot of cases. This is a tedious proof.

We start with the case where $\alpha$, $\beta$, and $\gamma$ are all positive and show that $p(\alpha, \beta + \gamma) = p(\alpha, \beta) + p(\alpha, \gamma)$.

1. $\alpha(\beta + \gamma) = p(\alpha, \beta + \gamma) \\ \alpha(\beta + \gamma) = \{ x < ay : a \in \alpha, a > 0, y \in (\beta + \gamma), y > 0\} \\ \alpha(\beta + \gamma) = \{ x < a (b + c) : a \in \alpha, a > 0, b \in \beta, b < 0, c \in \gamma, c > 0_{\mathbb{Q}}\} \\ \alpha(\beta + \gamma) = \{ x < ab + ac : a \in \alpha, a > 0, b \in \beta, b < 0, c \in \gamma, c > 0\} \\ \alpha(\beta + \gamma) = \{ x < y + z : y \in (\alpha + \beta), y > 0, z \in (\alpha + \gamma), z > 0\} \\ \alpha(\beta + \gamma) = p(\alpha, \beta) + p(\alpha, \gamma) \\ \alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$

From here we list the ten additional cases where $\alpha$, $\beta$, and $\gamma$ are variously positive, negative, or zero.

1. If $\alpha = 0_{\mathbb{R}}$, then

$\alpha(\beta + \gamma) = 0_{\mathbb{R}}(\beta + \gamma) \\ \alpha(\beta + \gamma) = 0_{\mathbb{R}} \\ \alpha(\beta + \gamma) = 0_{\mathbb{R}} + 0_{\mathbb{R}} \\ \alpha(\beta + \gamma) = 0_{\mathbb{R}}\beta + 0_{\mathbb{R}}\gamma \\ \alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma \\$

2. If $\beta = 0_{\mathbb{R}}$, then

$\alpha(\beta + \gamma) = \alpha (0_{\mathbb{R}} + \gamma) \\ \alpha(\beta + \gamma) = \alpha\gamma \\ \alpha(\beta + \gamma) = 0_{\mathbb{R}} + \alpha \gamma \\ \alpha(\beta + \gamma) = \alpha0_{\mathbb{R}} + \alpha\gamma \\ \alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma \\$

3. If $\gamma = 0_{\mathbb{R}}$, then $\alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$ holds by commutativity in case 2.

4. If $\alpha$ is negative and $\beta$ and $\gamma$ are positive, then

$\alpha(\beta + \gamma) = -(-\alpha(\beta + \gamma)) \\ \alpha(\beta + \gamma) = -((-\alpha)\beta (-\alpha)\gamma) \\ \alpha(\beta + \gamma) = -(-(\alpha\beta) -(\alpha\gamma)) \\ \alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma \\$

5. If $\alpha$ and $\beta$ are positive and $\gamma$ is negative, then there are three sub-cases:

• If $\beta + \gamma$ is positive, then

$\alpha\beta + \alpha\gamma = \alpha\gamma + \alpha((\beta + \gamma) - \gamma) \\ \alpha\beta + \alpha\gamma = \alpha\gamma + (\alpha(\beta + \gamma) - \alpha\gamma) \\ \alpha\beta + \alpha\gamma = \alpha(\gamma + \beta) \\$

• If $\beta + \gamma = 0_{\mathbb{R}}$, then $\gamma = -\beta$

$\alpha(\beta + \gamma) = \alpha0_{\mathbb{R}} \\ \alpha(\beta + \gamma) = 0_{\mathbb{R}} \\ \alpha(\beta + \gamma) = ((-\alpha)\beta) - ((-\alpha)\beta) \\ \alpha(\beta + \gamma) = \alpha(-\beta) + \alpha\beta \\ \alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma\\$

• If $\beta + \gamma$ is negative, then

$\alpha(\beta + \gamma) = -\alpha(-(\beta + \gamma)) \\ \alpha(\beta + \gamma) = -\alpha(-\beta - \gamma) \\ \alpha(\beta + \gamma) = (-(\alpha\gamma) -\alpha(-\beta - \gamma)) + \alpha\gamma \\ \alpha(\beta + \gamma) = -(\alpha\gamma + \alpha(-\beta - \gamma)) + \alpha\gamma \\ \alpha(\beta + \gamma) = -(\alpha(\gamma + (-\beta - \gamma)) + \alpha\gamma \\ \alpha(\beta + \gamma) = -(\alpha(-\beta)) + \alpha\gamma \\ \alpha(\beta + \gamma) = \alpha\beta) + \alpha\gamma \\$

6. If $\alpha$ is positive, and instead $\beta$ is negative and $\gamma$ is positive, $\alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$ by commutativity on the above case.

7. If $\alpha$ is positive and $\beta$ and $\gamma$ are negative, then $\beta + \gamma$ is negative, and thus

$\alpha(\beta + \gamma) = -\alpha(-(\beta + \gamma)) \\ \alpha(\beta + \gamma) = -\alpha(-\beta - \gamma) \\ \alpha(\beta + \gamma) = (-\alpha)(-\beta) + (-\alpha)(-\gamma)) \\ \alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma \\$

8. If $\alpha$ and $\beta$ are negative and $\gamma$ is positive, then there are three sub-cases:

• If $\beta + \gamma$ is positive, then

$\alpha(\beta + \gamma) = -((-\alpha)(\beta + \gamma)) \\ \alpha(\beta + \gamma) = -(((-\alpha)\beta) + (-\alpha)\gamma) \\ \alpha(\beta + \gamma) = -((-\alpha)\beta)) + (-(-\alpha)\gamma)) \\ \alpha(\beta + \gamma) = -(-(\alpha\beta)) + \alpha\gamma) \\ \alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma \\$

Note that the second step follows from case 6 above.

• If $\beta + \gamma = 0_{\mathbb{R}}$, then $\beta = -\gamma$ and thus

$\alpha(\beta + \gamma) = \alpha\cdot 0_{\mathbb{R}} \\ \alpha(\beta + \gamma) = 0_{\mathbb{R}} \\ \alpha(\beta + \gamma) = ((-\alpha)\gamma) + (-((-\alpha)\gamma)) \\ \alpha(\beta + \gamma) = (\alpha(-\gamma)) + (-(-(\alpha\gamma))) \\ \alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma \\$

• If $\beta + \gamma$ is negative, then

$\alpha(\beta + \gamma) = (-\alpha)(-(\beta + \gamma)) \\ \alpha(\beta + \gamma) = (-\alpha)(-(\beta + \gamma)) \\ \alpha(\beta + \gamma) = (-\alpha)(-\beta - \gamma) \\ \alpha(\beta + \gamma) = (-\alpha)(-\beta) + (-\alpha)(-\gamma) \\ \alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma \\$

9. If $\alpha$ is negative, and instead $\beta$ is positive and $\gamma$ is negative, $\alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$ by commutativity in the previous case.

10. Finally, if $\alpha$, $\beta$, and $\gamma$ are all negative, then

$\alpha (\beta + \gamma) = -\alpha(-\beta - \gamma) \\ \alpha (\beta + \gamma) = (-\alpha)(-\beta) + (-\alpha)(-\gamma) \\ \alpha (\beta + \gamma) = \alpha\beta + \alpha\gamma \\$

21. Embedding of $\mathbb{Q}$: Consider the function $E : \mathbb{Q} \rightarrow \mathbb{R}$, where $E(x) = \{ y : y < x \}$. Show that $E(x)$ is a cut and that the following properties hold:

• $E(x) + E(y) = E(x + y)$

• $E(x)E(y) = E(xy)$

• $E(x) < E(y) \text{ if and only if } x < y$

This embedding allows us to consider $\mathbb{Q}$ as a subset of $\mathbb{R}$, as $\mathbb{Q}$ and $E(\mathbb{Q})$ are indistinguishable from one another as far as arithmetic and ordering properties are concerned.

We show that $E(x)$ meets the three criteria of a cut:

1. $x-1 \in E(x)$, so $E(x) \neq \varnothing$. Likewise, $x+1 \notin E(x)$, so $E(x) \neq \mathbb{Q}$

2. Let $a \in E(x)$ and $b < a$. Then $b < x$, so $b \in E(x)$

3. Let $a \in E(x)$. Then $a < \frac{a+x}{2} < x$, so $\frac{a+x}{2} \in E(x)$.

Next, we show that $E(x)$ has the desired arithmetic and ordering properties:

• Let $p \in E(x) + E(y)$. Then $p = a + b$, where $a < x$ and $b < y$. Then $p < x + y$, and therefore $E(x + y)$. Conversely, let $p \in E(x + y)$. Then $p < x + y$. Let $2q = x + y - p$, and let $x' = x - q$ and $y' = y - q$. Then $x' \in E(x)$ and $y' \in E(y)$. Therefore $p = x' + y'$, so $p \in E(x) + E(y)$.

• There are 3 cases:

• Both $x$ and $y$ are positive: Let $p \in E(x)E(y)$. Then $p < ab$, where $a \in E(x)$, $a > 0$, $b \in E(y)$, $b > 0$. Since $a < x$ and $b < y$, $ab < xy$, and so $p < xy$. Therefore $p \in E(xy)$. Conversely, let $p \in E(xy)$. Then $p < xy$. Let $x' \in \mathbb{Q}$ such that $\frac{p}{y} < x' < x$. Then $x' \in E(x)$. Note that $p < x'y < xy$. Let $y' = \frac{p}{x'}$. Then $y' < y$, and so $y' \in E(y)$. Clearly $p = x'y'$, so $p \in E(x)E(y)$.

• One of $x$ and $y$ is positive (say $x$) and the other is negative (say $y$):

$E(x)E(y) = E(x)(-E(-y)) \\ E(x)E(y) = -(E(x)E(-y)) \\ E(x)E(y) = -E(x(-y)) \\ E(x)E(y) = -E(-(xy)) \\ E(x)E(y) = E(xy)$

• Both $x$ and $y$ are negative:

$E(x)E(y) = (-E(-x))(-E(-y)) \\ E(x)E(y) = E(-x)E(-y) \\ E(x)E(y) = E((-x)(-y)) \\ E(x)E(y) = E(xy)$

• Let $x < y$. Then $x \in E(y)$ and $x \notin E(x)$. Therefore $E(x) < E(y)$. Conversely, let $E(x) < E(y)$. Then there is a $p \in E(y)$ such that $p \notin E(x)$. Therefore $x \leq p < y$, and thus $x < y$.