# Naive Set Theory: Number Systems

## Rational Numbers

With dreams of mathematics in our heads, we constructed the natural numbers. Alas, we found them wanting, as they are not closed under subtraction, and they only have the cancellation law to "undo" addition. To overcome this flaw, we constructed the integers, which are successfully closed under subtraction. However, they inherit another flaw of the natural numbers - the only way to "undo" multiplication is via the cancellation law, and in general neither the natural numbers nor the integers are closed under division. Though we won't bother to propose an initial division operation on $\mathbb{Z}$ as we did for subtraction on $\mathbb{N}$, it is easy to see that $\mathbb{Z}$ would not be closed under it, as what integer should we expect to be equal $1 / 2$?

Just as the integers were constructed with the ordered pair $(a, b)$ meant to signify $a - b$, with the corresponding equivalence relation managing everything in terms of the closed addition operation, the rationals are constructed with the ordered pair $(a, b)$ meant to signify $\dfrac{a}{b}$, with the corresponding equivalence relation managing everything in terms of the closed multiplication operation. However, there is one caveat. Division by zero is left undefined, as any other choice would lead to no good.

The rational numbers, denoted $\mathbb{Q}$, are formally defined as

$$\mathbb{Q} = \mathbb{Z} \times (\mathbb{Z} - {0} ) \; / \sim$$

where

$$(a, b) \sim (c, d) \text{ iff } a \cdot d = b \cdot c.$$

The fractional notation of $\dfrac{a}{b}$ is always preferred to the equivalence class notation of $[(a,b)]$. The top number $a$ is called the numerator, while the bottom number $b$ is called the denominator. Likewise, fractions are typically only used to represent rational numbers that cannot be represented by existing symbols. In general, rational numbers in the equivalence class $[(a,1)]$ are simply written as $a$. For example, here are the first few integers around $0$ in $\mathbb{Q}$:

$-2 = \dfrac{-2}{1} = [(-2,1)] \\ -1 = \dfrac{-1}{1} = [(-1,1)] \\ \phantom{-}0 = \dfrac{0}{1} = [(0,1)] \\ \phantom{-}1 = \dfrac{1}{1} = [(1,1)] \\ \phantom{-}2 = \dfrac{2}{1} = [(2,1)] \\$

(We could distinguish rational numbers from integers by subscripting them with either $\mathbb{Q}$ or $\mathbb{Z}$, accordingly, but this notation will be reserved for when the distinction is not already clear from the context.)

The exercises below cover the arithmetic and ordering properties of the rational numbers, as well as a few notational conventions. Further investigation of the rational numbers is left to other fields such as real analysis.

## Problems

1. Define the addition operation on $\mathbb{Q}$ as

$$+ : \mathbb{Q} \times \mathbb{Q} \rightarrow \mathbb{Q}$$

where

$$\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ a \cdot d + c \cdot b }{ b \cdot d }$$

for all $a,b,c,d \in \mathbb{Z}$.

Simplify the following sums:

1. $\dfrac{3}{4} + \dfrac{1}{2}$

2. $\dfrac{1}{2} + \dfrac{2}{3}$

3. $\dfrac{2}{3} + \dfrac{-5}{7}$

4. $\dfrac{-3}{-2} + \dfrac{-1}{-1}$

5. $\dfrac{1}{100} + \dfrac{-1}{101}$

1. $\dfrac{3}{4} + \dfrac{1}{2} = \dfrac{3 \cdot 2 + 1 \cdot 4}{2 \cdot 4} = \dfrac{6 + 4}{8} = \dfrac{10}{8} = \dfrac{5}{4}$

2. $\dfrac{1}{2} + \dfrac{2}{3} = \dfrac{1 \cdot 3 + 2 \cdot 2}{2 \cdot 3} = \dfrac{3 + 4}{6} = \dfrac{7}{6}$

3. $\dfrac{2}{3} + \dfrac{-5}{7} = \dfrac{2 \cdot 7 + (-5 \cdot 3)}{3 \cdot 7} = \dfrac{14 - 15}{21} = \dfrac{-1}{21}$

4. $\dfrac{-3}{-2} + \dfrac{-1}{-1} = \dfrac{ -3 \cdot -1 + (-1 \cdot -2)}{-2 \cdot -1} = \dfrac{3 + 2}{2} = \dfrac{5}{2}$

5. $\dfrac{1}{100} + \dfrac{-1}{101} = \dfrac{1 \cdot 101 + (-1 \cdot 100)}{100 \cdot 101} = \dfrac{101 - 100}{10100} = \dfrac{1}{10100}$

2. Additive Identity: Show that $q + 0 = q$ for all $q \in \mathbb{Q}$.

Let $q = \dfrac{x}{y}$, where $x, y \in \mathbb{Z}$. Then

$q + 0 = \dfrac{x}{y} + \dfrac{0}{1}\\ q + 0 = \dfrac{x \cdot 1 + y \cdot 0}{y \cdot 1} \\ q + 0 = \dfrac{x + 0}{y} \\ q + 0 = \dfrac{x}{y} \\ q + 0 = q \\$

3. Show that $\dfrac{x}{x} = 1$ for all $x \neq 0$.

Recall that two rational numbers $\dfrac{x}{y}$ and $\dfrac{u}{v}$ are equivalent if $x \cdot v = u \cdot y$. Rewriting $1$ as $\dfrac{1}{1}$, we see indeed that $x \cdot 1 = 1 \cdot x$.

4. Commutativity of Addition: Show that $a + b = b + a$ for all $a, b \in \mathbb{Q}$.

Let $a = \dfrac{x}{y}$ and let $b = \dfrac{u}{v}$, where $x,y,u,v \in \mathbb{Z}$. Then

$a + b = \dfrac{x}{y} + \dfrac{u}{v} \\ a + b = \dfrac{x \cdot v + u \cdot y}{y \cdot v} \\ a + b = \dfrac{v \cdot x + y \cdot u}{v \cdot y} \\ a + b = \dfrac{u}{v} + \dfrac{x}{y} \\ a + b = b + a \\$

5. Associativity of Addition: Show that $a + (b + c) = (a + b) + c$ for all $a, b, c \in \mathbb{Q}$.

Let $a = \dfrac{x}{y}$, $b = \dfrac{u}{v}$, and $c = \dfrac{p}{q}$, where $x,y,u,v,p,q \in \mathbb{Z}$. Then

$a + (b + c) = \dfrac{x}{y} + \left( \dfrac{u}{v} + \dfrac{p}{q} \right) \\ a + (b + c) = \dfrac{x}{y} + \dfrac{u \cdot q + p \cdot v}{v \cdot q} \\ a + (b + c) = \dfrac{ x \cdot (v \cdot q) + (u \cdot q + p \cdot v) \cdot y }{y \cdot (v \cdot q)} \\ a + (b + c) = \dfrac{ (x \cdot v) \cdot q + (u \cdot q) \cdot y + (p \cdot v) \cdot y }{y \cdot (v \cdot q)} \\ a + (b + c) = \dfrac{ (x \cdot v) \cdot q + u \cdot (q \cdot y) + (p \cdot v) \cdot y }{y \cdot (v \cdot q)} \\ a + (b + c) = \dfrac{ (x \cdot v) \cdot q + u \cdot (y \cdot q) + (p \cdot v) \cdot y }{y \cdot (v \cdot q)} \\ a + (b + c) = \dfrac{ (x \cdot v) \cdot q + (u \cdot y) \cdot q + (p \cdot v) \cdot y }{y \cdot (v \cdot q)} \\ a + (b + c) = \dfrac{ (x \cdot v + u \cdot y) \cdot q + (p \cdot v) \cdot y }{y \cdot (v \cdot q)} \\ a + (b + c) = \dfrac{ (x \cdot v + u \cdot y) \cdot q + p \cdot (v \cdot y) }{y \cdot (v \cdot q)} \\ a + (b + c) = \dfrac{ (x \cdot v + u \cdot y) \cdot q + p \cdot (y \cdot v) }{y \cdot (v \cdot q)} \\ a + (b + c) = \dfrac{ (x \cdot v + u \cdot y) \cdot q + p \cdot (y \cdot v) }{(y \cdot v) \cdot q} \\ a + (b + c) = \dfrac{x \cdot v + u \cdot y}{y \cdot v} + \dfrac{p}{q} \\ a + (b + c) = \left( \dfrac{x}{y} + \dfrac{u}{v} \right) + \dfrac{p}{q} \\ a + (b + c) = (a + b) + c \\$

6. Additive Inverse: Show that for every $\dfrac{a}{b} \in \mathbb{Q}$, there exists an element $\dfrac{-a}{b} \in \mathbb{Q}$ such that $\dfrac{a}{b} + \dfrac{-a}{b} = 0$.

First we confirm that $\dfrac{-a}{b} \in \mathbb{Q}$. If $\dfrac{a}{b} \in \mathbb{Q}$, then $a, b \in \mathbb{Z}$. Thus there exists an additive inverse for $a$, namely $-a$. Therefore $\dfrac{-a}{b} \in \mathbb{Q}$.

Next we confirm that $\dfrac{a}{b} + \dfrac{-a}{b} = 0$.

$\dfrac{a}{b} + \dfrac{-a}{b} = \dfrac{a \cdot b + (-a) \cdot b}{b \cdot b} \\ \dfrac{a}{b} + \dfrac{-a}{b} = \dfrac{a \cdot b - a \cdot b}{b \cdot b} \\ \dfrac{a}{b} + \dfrac{-a}{b} = \dfrac{0}{b \cdot b} \\ \dfrac{a}{b} + \dfrac{-a}{b} = 0 \\$

7. Define the multiplication operation on $\mathbb{Q}$ as

$$\cdot : \mathbb{Q} \times \mathbb{Q} \rightarrow \mathbb{Q}$$

where

$$\dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{a \cdot c}{b \cdot d}$$

Simplify the following products:

1. $2 \cdot 99$

2. $\dfrac{7}{2} \cdot \dfrac{2}{7}$

3. $\dfrac{4}{2} \cdot \dfrac{3}{9}$

4. $\dfrac{2}{3} \cdot \dfrac{5}{7}$

1. $2 \cdot 99 = 198$

2. $\dfrac{7}{2} \cdot \dfrac{2}{7} = \dfrac{7 \cdot 2}{2 \cdot 7} = \dfrac{14}{14} = 1$

3. $\dfrac{4}{2} \cdot \dfrac{3}{9} = \dfrac{4 \cdot 3}{2 \cdot 9} = \dfrac{12}{18} = \dfrac{2}{3}$

4. $\dfrac{2}{3} \cdot \dfrac{5}{7} = \dfrac{2 \cdot 5}{3 \cdot 7} = \dfrac{10}{21}$

8. Multiplicative Identity: Show that $1 \cdot q = q$ for all $q \in \mathbb{Q}$.

Let $q = \dfrac{x}{y}$. Then

$1 \cdot q = \dfrac{1}{1} \cdot \dfrac{x}{y} \\ 1 \cdot q = \dfrac{1 \cdot x}{1 \cdot y} \\ 1 \cdot q = \dfrac{x}{y} \\ 1 \cdot q = q \\$

9. Commutativity of Multiplication: Show that $a \cdot b = b \cdot a$ for all $a,b \in \mathbb{Q}$.

Let $a = \dfrac{x}{y}$ and $b = \dfrac{u}{v}$, where $x,y,u,v \in \mathbb{Z}$. Then

$a \cdot b = \dfrac{x}{y} \cdot \dfrac{u}{v} \\ a \cdot b = \dfrac{x \cdot u}{y \cdot v} \\ a \cdot b = \dfrac{u \cdot x}{v \cdot y} \\ a \cdot b = \dfrac{u}{v} \cdot \dfrac{x}{y} \\ a \cdot b = b \cdot a \\$

10. Associativity of Multiplication: Show that $a \cdot (b \cdot c) = (a \cdot b) \cdot c$ for all $a,b,c \in \mathbb{Q}$.

Let $a = \dfrac{x}{y}$, $b = \dfrac{u}{v}$, and $c = \dfrac{p}{q}$, where $x,y,u,v,p,q \in \mathbb{Z}$. Then

$a \cdot (b \cdot c) = \dfrac{x}{y} \cdot \left( \dfrac{u}{v} \cdot \dfrac{p}{q} \right) \\ a \cdot (b \cdot c) = \dfrac{x}{y} \cdot \dfrac{u \cdot p}{v \cdot q} \\ a \cdot (b \cdot c) = \dfrac{x \cdot (u \cdot p)}{y \cdot (v \cdot q)} \\ a \cdot (b \cdot c) = \dfrac{(x \cdot u) \cdot p}{(y \cdot v) \cdot q} \\ a \cdot (b \cdot c) = \dfrac{x \cdot u}{y \cdot v} \cdot \dfrac{p}{q} \\ a \cdot (b \cdot c) = \left(\dfrac{x}{y} \cdot \dfrac{u}{v} \right) \cdot \dfrac{p}{q} \\ a \cdot (b \cdot c) = (a \cdot b) \cdot c \\$

11. Show that $\dfrac{a}{c} + \dfrac{b}{c} = \dfrac{a + b}{c}$.

$\dfrac{a}{c} + \dfrac{b}{c} = \dfrac{a \cdot c + b \cdot c }{c \cdot c} \\ \dfrac{a}{c} + \dfrac{b}{c} = \dfrac{c \cdot a + c \cdot b }{c \cdot c} \\ \dfrac{a}{c} + \dfrac{b}{c} = \dfrac{c \cdot (a + b) }{c \cdot c} \\ \dfrac{a}{c} + \dfrac{b}{c} = \dfrac{c}{c} \cdot \dfrac{a + b}{c} \\ \dfrac{a}{c} + \dfrac{b}{c} = 1 \cdot \dfrac{a + b}{c} \\ \dfrac{a}{c} + \dfrac{b}{c} = \dfrac{a + b}{c} \\$

12. Multiplicative Inverse: Show that for all nonzero $\dfrac{a}{b} \in \mathbb{Q}$, there exists a number $\dfrac{b}{a}$ such that $\dfrac{a}{b} \cdot \dfrac{b}{a} = 1$.

First we check that $\dfrac{b}{a}$ exists. If $\dfrac{a}{b} \in \mathbb{Q}$, then $a, b \in \mathbb{Z}$, and if $\dfrac{a}{b} \neq 0$, then $a \neq 0$. Therefore the number $\dfrac{b}{a} \in \mathbb{Q}$.

Next we check that the product equals $1$:

$\dfrac{a}{b} \cdot \dfrac{b}{a} = \dfrac{a \cdot b}{b \cdot a} \\ \dfrac{a}{b} \cdot \dfrac{b}{a} = \dfrac{a \cdot b}{a \cdot b} \\ \dfrac{a}{b} \cdot \dfrac{b}{a} = 1$

13. Distributive Law: Show that $a \cdot (b + c) = a \cdot b + a \cdot c$ for all $a,b,c \in \mathbb{Q}$.

Rather than convert the lefthand side of the equality into the righthand side, we will convert the righthand side into the lefthand side. Let $a = \dfrac{x}{y}$, $b = \dfrac{u}{v}$, and $c = \dfrac{p}{q}$, where $x,y,u,v,p,q \in \mathbb{Z}$. Then

$a \cdot b + a \cdot c = \dfrac{x}{y} \cdot \dfrac{u}{v} + \dfrac{x}{y} \cdot \dfrac{p}{q} \\ a \cdot b + a \cdot c = \dfrac{x \cdot u}{y \cdot v} + \dfrac{x \cdot p}{y \cdot q} \\ a \cdot b + a \cdot c = \dfrac{(x \cdot u) \cdot (y \cdot q) + (x \cdot p) \cdot (y \cdot v)}{(y \cdot v) \cdot (y \cdot q)} \\ a \cdot b + a \cdot c = \dfrac{(y \cdot q) \cdot (x \cdot u) + (y \cdot v) \cdot (x \cdot p)}{(y \cdot v) \cdot (y \cdot q)} \\ a \cdot b + a \cdot c = \dfrac{y \cdot (q \cdot (x \cdot u)) + y \cdot (v \cdot (x \cdot p))}{y \cdot (v \cdot (y \cdot q))} \\ a \cdot b + a \cdot c = \dfrac{y \cdot ( q \cdot (x \cdot u) + v \cdot (x \cdot p))}{y \cdot (v \cdot (y \cdot q))} \\ a \cdot b + a \cdot c = \dfrac{y}{y} \cdot \dfrac{ q \cdot (x \cdot u) + v \cdot (x \cdot p)}{v \cdot (y \cdot q)} \\ a \cdot b + a \cdot c = 1 \cdot \dfrac{ q \cdot (x \cdot u) + v \cdot (x \cdot p)}{v \cdot (y \cdot q)} \\ a \cdot b + a \cdot c = \dfrac{ q \cdot (x \cdot u) + v \cdot (x \cdot p)}{v \cdot (y \cdot q)} \\ a \cdot b + a \cdot c = \dfrac{ (x \cdot u) \cdot q + (x \cdot p) \cdot v}{(y \cdot q) \cdot v} \\ a \cdot b + a \cdot c = \dfrac{ x \cdot (u \cdot q) + x \cdot (p \cdot v)}{y \cdot (q \cdot v)} \\ a \cdot b + a \cdot c = \dfrac{ x \cdot ((u \cdot q) + (p \cdot v))}{y \cdot (q \cdot v)} \\ a \cdot b + a \cdot c = \dfrac{x}{y} \cdot \dfrac{ (u \cdot q) + (p \cdot v)}{q \cdot v} \\ a \cdot b + a \cdot c = \dfrac{x}{y} \cdot \left( \dfrac{u}{v} \cdot \dfrac{p}{q} \right) \\ a \cdot b + a \cdot c = a \cdot ( b \cdot c ) \\$

14. Define the ordering on $\mathbb{Q}$ as

$\dfrac{a}{b} < \dfrac{c}{d} \text{ iff } a \cdot d < c \cdot b$.

Confirm that this definition forms a total ordering on $\mathbb{Q}$.

We show that $<$ fulfills the three requirements of an order relation.

1. Comparability: Consider $\dfrac{a}{b}, \dfrac{c}{d} \in \mathbb{Q}$, where $\dfrac{a}{b} \neq \dfrac{c}{d}$. By comparability on $\mathbb{Z}$, either $a \cdot d < c \cdot b$, or $c \cdot b < a \cdot d$. Therefore either $\dfrac{a}{b} < \dfrac{c}{d}$ or $\dfrac{c}{d} < \dfrac{a}{b}$.

2. Nonreflexivity: Consider $\dfrac{a}{b} \in \mathbb{Q}$. By nonreflexivity on $\mathbb{Z}$, it is impossible that $a \cdot b < b \cdot a$. Therefore nonreflexivity holds on $\mathbb{Q}$.

3. Transitivity: Consider $\dfrac{a}{b}, \dfrac{c}{d}, \dfrac{e}{f} \in \mathbb{Q}$, where $\dfrac{a}{b} < \dfrac{c}{d}$ and $\dfrac{c}{d} < \dfrac{e}{f}$. Then $a \cdot d < c \cdot b$ and $c \cdot f < e \cdot d$. Multiplying the first equation by $f$ and the second by $b$ gives us $a \cdot d \cdot f < c \cdot b \cdot f$ and $c \cdot f \cdot b < f \cdot e \cdot b$. Then $a \cdot d \cdot f < f \cdot e \cdot b$. By the cancellation law on $\mathbb{Z}$, we can remove $d$ to see that $a \cdot f < e \cdot b$, and therefore that $\dfrac{a}{b} < \dfrac{e}{f}$.