Welcome to a four part series on ‘What Are Numbers?’. In the previous part, we constructed the Integers by using the equivalence classes of Natural Number ordered pairs that represent equations in the form \(x + b = a\). For example, the Integer we write down in the usual way as \(-2\) describes the set \(\{(0,2), (1,3), (2,4),\ldots\}\), which correspond to the equations \(x+2=0, x+3=1, x+4=2,\ldots\) respectively.

In this part, we aim to extend the Integers even further to the set of Rational Numbers.

## Solving More Equations

We learn about Rational Numbers in high school as numbers we can write as fractions, i.e. in the from \(\frac{a}{b}\) where the numerator \(a\) and denominator \(b\) are integers, separated by a bar (called a vinculum). Of course, we can’t divide by zero so the denominator cannot be \(0\).

We are familiar with this concept by the time we hit high school because we learn about fractions in primary school, and we’ve always been taught to *notate* it as \(\frac{a}{b}\). We are taught from a young age that this means \(a\) lots of \(b\) parts of the whole. However, why does it have to be like that? Why can’t we just write it as an ordered pair \((a,b)\)?

In fact, that is what we will do in this part as we examine solutions to the equations in the form:

\[ bx = a\]

The Integers can solve equations such as \(2x = 6\), but there is no \(x \in \mathbb{Z}\) that solves \(2x = 5\).

We can encode all equations of the form \(bx = a\) as ordered pairs \((a,b)\) where \(a\) and \(b\) are Integers. However, the equation \(0x = a\) where \(a \not = 0\) never yields any solutions as \(0\) multiplied onto another integer is always \(0\) (sets with this property are called Integral Domains). Moreover, the equation \(0x = 0\) will always be true for any value of \(x\in\mathbb{Z}\) and is therefore indeterminate. We thus avoid these equations with \(b = 0\).

## Forming The Rational Numbers

We now wish to form equivalence classes on the set of ordered pairs of Integers \((a,b)\) where \(b \not = 0\) (i.e. the set \(\mathbb{Z} \times \mathbb{Z}\backslash \{0\}\)) that each represent an equation in the form \(bx = a\).

To do this, we are really asking what rule needs to be applied to two equations \(bx = a\) and \(dx = c\) so they are described to have the same solution.

We identify two equations having the same solution with the equivalence relation:

\[(a,b) \sim (c,d) \mbox{ if and only if } ad = bc\]

To show that this is an equivalence relation:

- Reflexivity: \((a,b) \sim (a,b)\) if and only if \(ab = ba\) which is true. Hence it is reflexive.
- Symmetry: \((a,b) \sim (c,d)\) if and only if \(ad = bc\) if and only if \(bc = ad\) if and only if \((c,d) \sim (a,b)\). Hence it is symmetric.
- Transitivity: \((a,b) \sim (c,d)\) and \((c,d) \sim (e,f)\) if and only if \(ad = bc\) and \(cf = de\). Multiplying these equations together, we get \(adcf = bcde\). We can cancel \(d\) and \(c\) from both sides to get \(af = be\). Hence \((a,b) \sim (e,f)\).

Hence, we have successfully formed the set of rational numbers when we factor out the equivalence classes!

\[ \mathbb{Q} = \frac{\mathbb{Z} \times \mathbb{Z}\backslash\{0\}}{\sim}\]

Let’s now take a look at what members of \(\mathbb{Q}\) look like, say for the equation \(2x = 3\). This equation is represented by the ordered pair \((3,2)\). Equivalent equations represented by \((c,d)\) as per the equivalence relation are such that \(3d = 2c\). Just listing a few out that satisfy this, we see that \((3,2), (-3,-2), (6,4), (-6,-4), (9,6), (-9,6), (12,8), \ldots\) are equivalent. Hence:

\[ [(3,2)] = \{(3,2), (-3,-2), (6,4), (-6,-4), (9,6), (-9,6), (12,8), \ldots\} \]

If we want a visual representation of what this looks like, we can look at the lines that pass through the origin that also pass through integer coordinates. Each line represents an equivalence class, i.e. one rational number.

In the diagram above, we can see that between any two lines (rational numbers), the gap between them increases as they splay out more towards infinity. These gaps eventually widen enough so that an integer coordinate appears. Hence, between any two rational numbers, is another rational number!

There’s a kind of indescribable beauty in the symmetry of all this!

## Simplest Fractions

Notice that each equivalence class has a member \((a, b)\) such that the greatest common factor (also known as the highest common factor and notated \(\gcd\)) of \(a\) and \(b\) is \(1\). This is what we know in our elementary/secondary study of fractions to be the ‘simplest fraction’. We now choose this ordered pair \((a, b)\) and notate it as \(\frac{a}{b}\) to represent its equivalence class.

In the example above, the equivalence class of \([(3,2)]\) would therefore be represented by \(\frac{3}{2}\). The equivalence class of \([(4,8)]\) would be represented by \(\frac{1}{2}\).

This then leads to the usual way Rational Numbers are presented:

\[ \mathbb{Q} = \left\{ \left. \frac{a}{b} \right| a, b \in \mathbb{Z}\mbox{ and } b \not = 0\mbox{ and } \gcd (a, b) = 1 \right\} \]

If we left out the \(\gcd (a, b) = 1\) condition out, it would imply the following:

- Equivalent fractions are not equivalent. In listing the set of Rational Numbers, we would eventually write down: \(\{\ldots, \frac{1}{2}, \frac{2}{4}, \ldots\}\). These two fractions will therefore represent different equivalence classes, i.e. it will mean that the solution to \(2x = 1\) and \(4x=2\) are not the same. By our construction, this is a contradiction.

Of course in our use of Algebra in a standard context, we always use unsimplified fractions. For example, we can most certainly write \(\frac{4}{8}\) and use it, especially when adding two fractions we’ve been taught to change them to the same common denominator.

What we really mean when we write down an unsimplified fraction such as \(\frac{4}{8}\) is its *equivalence class* \([(4,8)]\) which is represented by the simplest fraction \(\frac{1}{2}\), but no one does this as there isn’t any real benefit using the notation \([(4,8)]\) over \(\frac{4}{8}\) outside the context of understanding the algebraic science behind the Rational Numbers – in fact, we use notation such as the fraction to make working with them easier than if we were to write them as ordered pairs.

## Algebraic Properties

For the following, we should prove the operations are well-defined, i.e. the result of addition and multiplication always ends up in the same equivalence class regardless of which ordered pair of the equivalence class we choose. I’ll leave that as an exercise for the reader.

### Addition

Addition is defined in \(\mathbb{Q}\) by the following:

\[ [(a, b)] + [(c, d)] = [(ad + bc,bd)] \]

Here, the \(+\) to the left of the equality sign is the addition we are defining between two rational numbers. The use of addition and multiplication on the right are the operations defined in the previous part for Integers, which were themselves defined depending on the addition and multiplication of Natural Numbers.

### Multiplication

Multiplication is defined in \(\mathbb{Q}\) by the following:

\[ [(a, b)] \times [(c, d)] = [(ac, bd)] \]

Here, the \(\times\) symbol on the left of the equality sign is the multiplication we are defining between two rational numbers and the multiplications on the right are between two integers.

We see that the structure of rational numbers depend on the structure of the integers, and the structure of integers depend on the structure of the natural numbers.

## Monoids, Groups, Rings and Fields

So far, I haven’t really discussed the operations of addition and multiplication in the contexts of the sets they’re defined in. Here is a good time to do so before moving onto the next part about Real Numbers.

### Monoids

A Monoid is a set such that:

- it has a binary operation that is associative, i.e. if \(M\) was a set of elements and an operation \(*\) was defined on it, then \(m_1 * (m_2 * m_3) = (m_1 * m_2) * m_3\) for every \(m_1, m_2, m_3 \in M\).
- there exists an identity element, i.e. there exists \(e \in M\) such that \(e * m = m * e = m\) for every \(m \in M\).

A commutative operation \(*\) is such that \(a*b = b*a\).

We can see that the set of Natural Numbers \(\mathbb{N}\) is a monoid under addition with \(0\) as the identity, and a monoid under multiplication with \(1\) as the identity. We also note that addition and multiplication are commutative.

### Groups

A Group is a Monoid such that every element has an inverse, i.e. there exists \(m^{-1}\) such that \(m * m^{-1} = e\).

If the operation \(*\) is commutative, we call the group an abelian group.

Essentially, an inverse of an element brings back that element to the identity. For example, the additive inverse of \(5\) is \(-5\). We can see that the set of Natural Numbers is not a Group, but the set of Integers are a Group under addition.

The study of Group Theory runs deep and is often labelled the study of symmetry. Applications of Groups permeate deep into many parts of the sciences: Physics such as Quantum Mechanics (see work done by Hermann Weyl), Chemistry such as counting number of isomers of a molecular formula, just to name a few. We will not go too deep into their theory in this post.

### Rings

A Ring is a set equipped with two operations, namely Addition (\(+\)) and Multiplication (\(\times\)) such that it is an abelian group under addition, a monoid under multiplication, and multiplication distributes over addition, i.e. \(a \times (b+c) = a\times b + a\times c\).

A commutative ring is one where multiplication is also commutative.

We see that the set of Integers is a commutative ring as it is an abelian group under addition, a monoid under multiplication which is commutative and multiplication distributes over addition. Hence, we often call it the *Ring* of Integers.

### Fields

A Field is a commutative ring with \(1 \not = 0\) (this is saying the additive and multiplicative identities are not the same element) such that all non-zero elements have a multiplicative inverse.

We see that the set of Integers are not a Field as take for example the integer \(3\). There is no other integer that multiplies \(3\) to get a result of \(1\). However, the set of Rational numbers are a field:

\[\frac{a}{b} \times \frac{b}{a} = \frac{ab}{ab} = \frac{1}{1}\]

This brings us to a close of this post. In the next part, we focus on constructing the real numbers.