Rational Numbers

# The Set of Rational Numbers $$\mathbb{Q}$$

## Construction of $$\mathbb{Q}$$

Assume $$\mathbb{Z}$$, the integers, their arithmetic and order.

An equivalent class is a set whose members satisfy an equivalent relation. For example, $$(1, 3) \sim (2, 6)$$ are equivalent ordered pairs from the class “$$\frac{1}{3}$$”.

Informally, $$\mathbb{Q}$$, the rational numbers, is the set of all such equivalent classes. Formally, $$\mathbb{Q} = \left \{\frac{m}{n} : m, n \in \mathbb{Z}, n \neq 0 \right \}$$, where $$\frac{m}{n}$$ is an equivalent class of ordered pairs $$(m, n)$$ with relation $$(p, q) \sim (m, n)$$ if $$pn = qm$$ and $$q, n \neq 0$$.

Check that $$\sim$$ is an equivalent relation:

1. Reflexivity: Since $$mn = nm$$ for all $$m, n \in \mathbb{Z}$$ and $$n \neq 0$$, $$(m, n) \sim (m, n)$$ is true.
2. Symmetry: Assume $$(p, q) \sim (m, n)$$ where $$p, q, m, n \in \mathbb{Z}$$ and $$q, n \neq 0$$. We have $$pn = qm$$, which is the same as $$mq = np$$, so $$(m, n) \sim (p, q)$$. Thus, $$(p, q) \sim (m, n) \implies (m, n) \sim (p, q)$$.
3. Transitivity: Assume $$(a, b) \sim (c, d)$$ and $$(c, d) \sim (e, f)$$ where $$a, b, c, d, e, f \in \mathbb{Z}$$ and $$b, d, f \neq 0$$. We have $$ad = bc$$ since $$(a, b) \sim (c, d)$$ and $$cf = de$$ since $$(c, d) \sim (e, f)$$. We also have \begin{align} & ad = bc \\ \implies & adf = bcf & \text{(since } f \neq 0 \text{)} \\ \implies & adf = bde & \text{(since } cf = de \text{)} \\ \implies & af = be & \text{(by cancellation law on } \mathbb{Z} \text{)} \\ \implies & (a, b) \sim (e, f) \end{align} Thus, $$(a, b) \sim (c, d)$$ and $$(c, d) \sim (e, f) \implies (a, b) \sim (e, f)$$.

## Properties of $$\mathbb{Q}$$

The chosen definition of addition should not depend on the representative chosen, i.e. it should be well-defined. The following definition is not well-defined. $\frac{a}{b} + \frac{c}{d} = \frac{a + c}{b + d}$ Check: $$\frac{1}{2} + \frac{2}{3} = \frac{3}{5}$$ but $$\frac{2}{4} + \frac{2}{3} = \frac{4}{7}$$.

A well-defined but useless definition is as follows. $\frac{a}{b} + \frac{c}{d} = \frac{0}{1}$

The correct definition is as follows. $\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$ To show that the definition is well-defined, we must show that if $$(a, b) \sim (a', b')$$ and $$(c, d) \sim (c', d')$$ then $$(ad + bc, bd) \sim (a'd' + b'c', b'd')$$.

Proof. Assume $$(a, b) \sim (a', b')$$ and $$(c, d) \sim (c', d')$$. So, $$ab' = ba'$$ and $$cd' = dc'$$. Since $$ab' = ba'$$, $$ab'dd' = ba'dd'$$, which is the same as $$adb'd' = bda'd'$$. Since $$cd' = dc'$$, $$cd'bb' = dc'bb'$$, which is the same as $$bcb'd' = bdb'c'$$. Hence, $$adb'd' + bcb'd' = bda'd' + bdb'c'$$, which means $$(ad + bc)b'd' = bd(a'd' + b'c')$$. So, $$(ad + bc, bd) \sim (a'd' + b'c', b'd')$$. Thus, if $$(a, b) \sim (a', b')$$ and $$(c, d) \sim (c', d')$$ then $$(ad + bc, bd) \sim (a'd' + b'c', b'd')$$. QED.

### Multiplication

Definition of multiplication on $$\mathbb{Q}$$: $\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$

Proof of well-definedness. Assume $$(a, b) \sim (a', b')$$ and $$(c, d) \sim (c', d')$$. So, $$ab' = ba'$$ and $$cd' = dc'$$. Hence, $$ab'cd' = ba'dc'$$, which is the same as $$acb'd' = bda'c'$$. So, $$(ac, bd) \sim (a'c', b'd')$$. Thus, if $$(a, b) \sim (a', b')$$ and $$(c, d) \sim (c', d')$$ then $$(ac, bd) \sim (a'c', b'd')$$. QED.

### $$\mathbb{Q}$$ is an Extension of $$\mathbb{Z}$$

$$\left\{\frac{n}{1} : n \in \mathbb{Z} \right\}$$ behaves the same as $$\mathbb{Z}$$. The correspondence is $$\frac{n}{1} \longleftrightarrow n$$.

Proof. Assume $$m, n \in \mathbb{Z}$$. Note that $$\frac{n}{1} + \frac{m}{1} = \frac{n \cdot 1 + m \cdot 1}{1 \cdot 1} = \frac{n + m}{1}$$ and $$\frac{n}{1} \cdot \frac{m}{1} = \frac{n \cdot m}{1 \cdot 1} = \frac{nm}{1}$$. QED.

### Order on $$\mathbb{Q}$$

An order on a set $$S$$ is a relation, denoted $$\lt$$, satisfying

1. (law of trichotomy) if $$x, y \in S$$, exactly one of these is true: $$x < y$$, $$x = y$$, $$y < x$$, and
2. (transitivity) if $$x, y, z \in S$$, $$x < y$$, and $$y < z$$, then $$x < z$$.

A set $$S$$ is called an ordered set if there is an order on it.

For example, in $$\mathbb{Z}$$, $$m < n$$ if $$n - m$$ is positive. Another example is, in $$\mathbb{Z} \times \mathbb{Z}$$, $$(a, b) < (c, d)$$ if $$a < c$$ or $$a = c$$ and $$b < d$$ (dictionary order).

$$\frac{m}{n}$$ is positive if $$mn > 0$$.

Proof of well-definedness. Assume $$(a, b) \sim (a', b')$$ and $$\frac{a}{b}$$ is positive. Since $$\frac{a}{b}$$ is positive, $$ab > 0$$. Since $$(a, b) \sim (a', b')$$, $$ab' = ba'$$, so $$a'b'b = ab'b'$$, which implies $$a'b'b > a$$ since $$b'b' > 0$$ for all $$b' \neq 0$$. Because $$a'b'b > a$$, $$a'b'bb > ab$$ since $$b \neq 0$$. This implies $$a'b'bb > 0$$ and thus $$a'b' > 0$$ because $$bb > 0$$ for all $$b \neq 0$$. Hence, $$\frac{a'}{b'}$$ is positive. Therefore, if $$(a, b) \sim (a', b')$$ and $$\frac{a}{b}$$ is positive then $$\frac{a'}{b'}$$ is positive. QED.

Order on $$\mathbb{Q}$$: $$\frac{m}{n} < \frac{m'}{n'}$$ if $$\frac{m'}{n'} - \frac{m}{n}$$ is positive (subtraction is defined by addition on the negation of the second operand).

Proof of well-definedness. Assume $$\frac{a}{b} < \frac{c}{d}$$, $$(a, b) \sim (a', b')$$, and $$(c, d) \sim (c', d')$$. Since $$\frac{a}{b} < \frac{c}{d}$$, $$\frac{c}{d} - \frac{a}{b}$$ is positive, which means $$(cb - ad)bd > 0$$, so either $$cb - ad > 0$$ and $$bd > 0$$ or $$cb - ad < 0$$ and $$bd < 0$$. Since $$(a, b) \sim (a', b')$$, $$ab' = ba'$$. Since $$(c, d) \sim (c', d')$$, $$cd' = dc'$$. There are four cases to consider.

Case 1: $$cb - ad > 0$$, $$bd > 0$$, and $$b'd' > 0$$. We have \begin{align} & cb - ad > 0 \\ \implies & cb > ad \\ \implies & cbb'd' > adb'd' & \text{(since } b'd' > 0 \text{)} \\ \implies & c'bb'd > a'dbd' & \text{(since } cd' = dc' \text{ and } ab' = ba' \text{)} \\ \implies & c'b' > a'd' & \text{(since } bd > 0 \text{)} \\ \implies & c'b' - a'd' > 0 \\ \implies & (c'b' - a'd')b'd' > 0 & \text{(since } b'd' > 0 \text{)} \\ \implies & \frac{c'b' - a'd'}{b'd'} \text{ is positive} \\ \implies & \frac{c'}{d'} - \frac{a'}{b'} \text{ is positive} \\ \implies & \frac{a'}{b'} < \frac{c'}{d'} \end{align}

Case 2: $$cb - ad > 0$$, $$bd > 0$$, and $$b'd' < 0$$. We have \begin{align} & cb - ad > 0 \\ \implies & cb > ad \\ \implies & cbb'd' < adb'd' & \text{(since } b'd' < 0 \text{)} \\ \implies & c'bb'd < a'dbd' & \text{(since } cd' = dc' \text{ and } ab' = ba' \text{)} \\ \implies & c'b' < a'd' & \text{(since } bd > 0 \text{)} \\ \implies & c'b' - a'd' < 0 \\ \implies & (c'b' - a'd')b'd' > 0 & \text{(since } b'd' < 0 \text{)} \\ \implies & \frac{c'b' - a'd'}{b'd'} \text{ is positive} \\ \implies & \frac{c'}{d'} - \frac{a'}{b'} \text{ is positive} \\ \implies & \frac{a'}{b'} < \frac{c'}{d'} \end{align}

Case 3: $$cb - ad < 0$$, $$bd < 0$$, and $$b'd' > 0$$. We have \begin{align} & cb - ad < 0 \\ \implies & cb < ad \\ \implies & cbb'd' < adb'd' & \text{(since } b'd' > 0 \text{)} \\ \implies & c'bb'd < a'dbd' & \text{(since } cd' = dc' \text{ and } ab' = ba' \text{)} \\ \implies & c'b' > a'd' & \text{(since } bd < 0 \text{)} \\ \implies & c'b' - a'd' > 0 \\ \implies & (c'b' - a'd')b'd' > 0 & \text{(since } b'd' > 0 \text{)} \\ \implies & \frac{c'b' - a'd'}{b'd'} \text{ is positive} \\ \implies & \frac{c'}{d'} - \frac{a'}{b'} \text{ is positive} \\ \implies & \frac{a'}{b'} < \frac{c'}{d'} \end{align}

Case 4: $$cb - ad < 0$$, $$bd < 0$$, and $$b'd' < 0$$. We have \begin{align} & cb - ad < 0 \\ \implies & cb < ad \\ \implies & cbb'd' > adb'd' & \text{(since } b'd' < 0 \text{)} \\ \implies & c'bb'd > a'dbd' & \text{(since } cd' = dc' \text{ and } ab' = ba' \text{)} \\ \implies & c'b' < a'd' & \text{(since } bd < 0 \text{)} \\ \implies & c'b' - a'd' < 0 \\ \implies & (c'b' - a'd')b'd' > 0 & \text{(since } b'd' < 0 \text{)} \\ \implies & \frac{c'b' - a'd'}{b'd'} \text{ is positive} \\ \implies & \frac{c'}{d'} - \frac{a'}{b'} \text{ is positive} \\ \implies & \frac{a'}{b'} < \frac{c'}{d'} \end{align}

Therefore, if $$\frac{a}{b} < \frac{c}{d}$$, $$(a, b) \sim (a', b')$$, and $$(c, d) \sim (c', d')$$ then $$\frac{a'}{b'} < \frac{c'}{d'}$$. QED.

Write $$x > y$$ for $$x < y$$. Write $$x \leq y$$ for $$x < y$$ or $$x = y$$.

### $$x^2 = 2$$ has No Solution in $$\mathbb{Q}$$

Proof (by constradiction). Assume that $$x^2 = 2$$ has a solution in $$\mathbb{Q}$$, i.e. let $$x = \frac{p}{q}$$, where $$p, q \in \mathbb{Z}$$, $$q \neq 0$$, and $$p$$ and $$q$$ are in lowest terms, i.e. they have no common factors.

So, $$\left(\frac{p}{q}\right)^2 = 2$$, hence $$p^2 = 2q^2$$. Then $$p^2$$ is even (divisible by 2). Then $$p$$ is even (because if $$p$$ were odd, $$p^2$$ would be odd).

So, $$p = 2m$$ for some $$m \in \mathbb{Z}$$, hence $$p^2 = 4m^2$$ and $$4m^2 = 2q^2$$. Then $$q^2 = 2m^2$$.

Then $$q^2$$ is even, hence $$q$$ is even. This contradicts the fact that $$p$$ and $$q$$ are in lowest terms.

Therefore, $$x^2 = 2$$ must have no solution in $$\mathbb{Q}$$. QED.

### $$\mathbb{Q}$$ is a Field

A field is a set $$F$$ with two operations $$+$$, and $$\times$$, satisfying the following axioms:

• (A1). $$F$$ is closed under $$+$$.
• (A2). $$+$$ is commutative.
• (A3). $$+$$ is associative.
• (A4). $$F$$ has an is additive identity. Call it $$0$$.
• (A5). Every element has an additive inverse.
• (M1). $$F$$ is closed under $$\times$$.
• (M2). $$\times$$ is commutative.
• (M3). $$\times$$ is associative.
• (M4). $$F$$ has an is multiplicative identity. Call it $$1$$.
• (M5). Every element except $$0$$ has a multiplicative inverse.
• (D). $$\times$$ distributes over $$+$$.

In $$\mathbb{Q}$$, $$0$$ element is $$\frac{0}{1}$$. $$1$$ element is $$\frac{1}{1}$$.

$$\mathbb{Z}$$ is not a field because it does not satisfy (M5).

$$\mathbb{Q}$$ is an ordered field, which is a field with an order such that the order is preserved by the field operations, i.e.

1. $$y < z \implies x + y < x + z$$, and
2. $$y < z, x > 0 \implies xy < xz$$.