In Part 1, we see that the building blocks of numbers start with the Natural Numbers defined through the five Peano Axioms. In this post, we ponder the invention of the Integers. Welcome to a 4 part series (this is part 2) of ‘What Are Numbers?’.

As we reflect on our own journey in learning about numbers from a young age, or if we teach young children about negative numbers and witness their initial confusion, or know that historically it wasn’t until the 17th century that negative numbers were accepted in Europe, it almost seems absurd that such a concept as a negative number exists. After all, how do you concretely show the number -45 on your fingers and toes? What does it mean that something is less than nothing?

Practical models such as associating different direction with negatives help us get over this hurdle. One can convince a class of year 7 students that going backwards means a negative number; or when you reverse a reversal (negative of a negative), something ends up facing the original way again (positive); or when you describe the change of the temperature an hour ago versus the time now as -4 degrees Celcius. However, these are merely *models* or practical use cases of the negatives – how are they actually *defined*? How do you actually *prove* that two negative integers multiply to a positive number?

So what is the algebraic view on this?

## Solving Simple Equations

With the natural numbers, it is able to solve equations such as:

\[ x + 2 = 3\]

However, it cannot solve other equations such as:

\[x + 3 = 2\]

This is because there is no concept of subtraction past 0 in the Natural Numbers: from the Peano Axioms discussed in the previous post, there are no numbers that have 0 as their successors.

However, it would be aesthetically pleasing to be able to solve all equations of the form \(x + b = a\). It also turns out that being able to solve these equations is incredibly useful as highlighted by some of the practical models mentioned previously. However, as we only know about Addition and Multiplication in the Natural Numbers, we cannot (yet) just say the answer is \( x= a-b\) and be done with it.

To do that requires a new system of arithmetic where subtraction into numbers less than 0 is defined. We could just postulate more Axioms to make it work, but it turns out that we already have enough (from the five Peano Axioms) to construct a set of objects (that we call Integers) that solve these particular equations. We talk about sufficient conditions in Mathematics, and the Peano Axioms are sufficient to construct the Integers!

## Some Definitions First

Before we can clearly define the Integers from the Natural Numbers, we need to define some key concepts first that are mostly related to Set Theory.

### Definition (Cartesian Product)

The Cartesian product of two sets \(A\) and \(B\) is

\[ A\times B = \{(a, b) | a \in A, b \in B\}\]

For example: \[\{1,2\}\times\{a, b, c\} = \{(1,a), (1,b), (1, c), (2,a), (2,b), (2,c)\}\]

### Definition (Relation)

A Relation \(R\) over sets \(A\) and \(B\) is a subset of the Cartesian product \(A \times B\). In HSC Stage 6, the concept of a Function is developed further on top of this definition.

We write \(a R b\) if \((a,b) \in R\).

For example, the relation \(<\) over the sets \(\{1,2,3,4,5\}\) and \(\{2,4\}\) is the set \(\{(1,2), (1,4), (2,4), (3,4)\}\).

We are allowed to write \(1 < 2\) since \((1,2)\) is in the Relation \(<\).

### Definition (Equivalence Relation)

An Equivalence Relation \(\sim\) on a set \(X\) is a Relation over \(X\) and \(X\), i.e. is a subset of \(X \times X\) that also obeys the following for every \(a, b, c \in X\):

- (Reflexivity) \(a \sim a\).
- (Symmetry) if and only if \(a \sim b\) then \(b \sim a\).
- (Transitivity) if \(a \sim b\) and \(b \sim c\), then \(a \sim c\).

For example:

- Equality (\(=\)) e.g. \( 6 = 2\times 3 \)
- Congruent triangles e.g. \(\triangle ABC \equiv \triangle XYZ\) means they are the same shape and size up.
- Modular arithmetic e.g. \(1 \equiv 13 \mod{12}\)
- As a non-example, \(<\) as defined previously is not an equivalence relation, as it is not a relation over the same set but other two different sets \(\{1,2,3,4,5\}\) and \(\{2,4\}\). We don’t even need to check the three conditions.
- As another non-example, \(\geq\) defined over the Natural Numbers in the normal way is not an equivalence relation, as \(3 \geq 2\) does not imply \(2 \geq 3\) and thus fails Symmetry.

### Definition (Equivalence Class)

The Equivalence Class of an element \(x \in X\) with an equivalence relation \(\sim\) defined over it is the set of all elements that \(x\) is equivalent to, and we write:

\[ [x] = \{ s \in X | s \sim x \} \]

For example, if \( X\) is the set of all triangles, and \(\equiv\) is the equivalence relation for congruency, the equivalence class of a triangle with side lengths 3cm, 4cm and 5cm would be the set containing all the triangles with side lengths 3cm, 4cm and 5cm – but could possibly be rotated, translated or reflected into different orientations. What is important is that they all preserve the properties we are interested in, namely side lengths and angles of the vertices.

Note that the set of *all* equivalence classes are:

- Disjoint: for if \([a] \cap [b] \not = \{\}\), then we have some \(x \sim a\) and \(x \sim b\). By Symmetry, \(a \sim x\), and hence by Transitivity we have \(a \sim b\). This implies \([a] = [b]\) whenever the intersection of two equivalence classes is not empty. If \([a] \not = [b]\) then their intersection is empty, i.e. disjoint.
- Every element has an equivalence class.

Hence, the set of equivalence classes make up a disjoint union of the set that the equivalence relation is defined over. We write the quotient \(\frac{X}{\sim}\) to be the set of all equivalence classes. This notation is similar to how we can interpret \(\frac{20}{5}\) as factorising the number 20 into 5 equivalent parts of the whole. The only difference here is that each equivalence class in \(\frac{X}{\sim}\) would contain more data or structure than just the division of two numbers.

## Forming The Integers

We have now got enough tools that allow us to define the Integers.

The motivating factor in this section is that we need solutions to the equations in the from \(x + b = a\). Armed with our definitions and only the Natural Numbers (and addition), let’s have a look at how we can rigorously define the Integers.

We observe that the solution to equations such as \(x + 2 = 1\), \(x + 3 = 2\), \(x + 4 = 3\), etc should have an *equivalent* solution – yet we currently do not have an object to describe this solution as \(x \not \in \mathbb{N}\). We also note that these equations depend only on two natural numbers, one to the left of the \(=\) sign, and one to the right.

We can therefore encode these equations as elements in the Cartesian product \(\mathbb{N} \times \mathbb{N}\):

The highlighted coordinates (1,2), (2,3) and (3,4) correspond to the equations \(x + 2 = 1\), \(x + 3 = 2\) and \(x + 4 = 3\) respectively. This suggests that they would belong to the same* equivalence class*! We can also see visually that the equivalence classes form diagonals in the diagram above. Cool maths makes cool patterns.

To do so, we need to construct the equivalence relation. We can do that on this set \(\mathbb{N} \times \mathbb{N}\) with the following rule:

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

This is motivated by the need that in order for \(x + b = a\) and \(x + d = c\) to have the same solution, whatever \(a – b\) means must equal \(c-d\). As we currently have not defined the concept of subtraction yet, we can only define the equivalence of solutions here with addition, so we write this as \(a + d = b + c\) instead.

We can verify that this is an equivalence relation:

- Reflexivity: \((a, b) \sim (a, b)\) if and only if \(a + b = b+a\) which is true. Hence \(\sim\) is Reflexive.
- Symmetry: \((a, b) \sim (c, d)\) if and only if \(a+d = b + c\) if and only if \(b+c = a+d \) if and only if \((c, d) \sim (a, b)\). Hence, \(\sim\) is Symmetric.
- Transitivity: \((a, b) \sim (c, d)\) and \((c, d) \sim (e, f)\) if and only if \( a + d = b + c\) and \(c + f = d + e\). Adding them together, we get \(a+d + c +f = b + c + d +e\). We can cancel \(c\) and \(d\), leaving us with \(a+f = b + e\), hence \((a, b) \sim (e, f)\).

Hence, we can now factor out the equivalence classes, and voila! We have our set of Integers:

\[ \mathbb{Z} = \frac{\mathbb{N} \times \mathbb{N}}{\sim}\]

A visual representation of this set of equivalence classes is achieved by grouping all the same coloured diagonals of coordinates in the diagram above. We can ascribe to each of those equivalence classes a *representative*. The most natural choice would be to pick the coordinates along the axes of the diagram, such as (0,2), which represents that diagonal’s equivalence class, and this equivalence class acts as the *solution* (note the singular) to all equations \(x+2 =0\), \(x+3=1\), etc. Instead of writing \([(0,2)]\) each time we want to use this equivalence class/solution, we now use new notation, and write \(-2\) instead.

This means each diagonal in \(\mathbb{N} \times \mathbb{N}\) *is* an integer!

To sum it up:

\[ 0 = \{(0,0), (1,1), (2,2), \ldots\}\]

\[ -1 = \{(0,1), (1,2), (2,3), \ldots\}\]

\[ 1 = \{(1,0), (2,1), (3,2), \ldots\}\]

etc…

The above construction almost seems circular as we have defined the Integers by basically identifying equivalent equations that may or may not have had solutions in \(\mathbb{N}\) as *the* solutions to themselves. A helpful way to see that this is what we do all the time anyway, is to consider this example:

John is 12.5 years old and his father is twice his age, how old is his father? By letting \(x\) equal the father’s age and saying that \(x\) is the solution, we’ve done something similar to the construction above. Only this time, we can only pick one representative of the equivalence class that represents the solution as it only contains one element: \(\{2(12.5) = 25\}\).

## Algebraic Properties

### Addition

We’ve now constructed the Integers, but is the definition of addition and multiplication still possible? It had better be, as we are used to already working with them.

We define addition in \(\frac{\mathbb{N} \times \mathbb{N}}{\sim}\) with the following:

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

Note that the left \(+\) is the addition between two Integers that we are defining, and the right \(+\) is the old addition between two Natural Numbers.

To check that this is well defined, we need to check that \((a+c, b+d)\) goes to the correct, expected equivalence class:

Let’s pick \((a_1,b_1)\) and \((a_2, b_2)\) from the same equivalence class and pick \((x_1, y_1)\) and \((x_2, y_2)\) from the same equivalence class. We have:

\[(a_1, b_1) \sim (a_2, b_2)\]\[(x_1, y_1) \sim (x_2,y_2)\]

Hence:

\[ a_1 + b_2 = b_1 + a_2\]\[x_1 + y_2 = y_1 + x_2\]

Adding together and regrouping:

\[ (a_1 + x_1) + (b_2 + y_2) = (a_2 + x_2) + (b_1 + y_1) \]

From this we can see that:

\[ (a_1 + x_1, b_1 + y_1) \sim (a_2 + x_2, b_2 + y_2)\]

As these Integers were chosen arbitrarily, we have shown that the addition of any two representatives of an equivalence class stay in the same equivalence class.

We can now also define subtraction among integers \(a – b\) as \(a + (-b)\). In the equivalence class form this looks like adding \([(a,0)] + [(0,b)]\) to get \([(a, b)]\).

### Multiplication

We define Integer multiplication depending only on the Natural Number multiplication as the following:

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

The motivation behind this definition is that the equivalence classes for \([a,b]\) solve equations with solution “\(a-b\)” and \([c,d]\) solve equations with solution “\(c-d\)”. Multiplying by what we want to get once we’ve defined the system properly, we expect: \((a-b)(c-d) = ac – ad – bc + bd = (ac + bd) – (bc + ad)\). This informs our definition of multiplication above.

We can similarly prove that the choice of representative does not affect the equivalence class of the result so this operation is well defined. (Try it yourself!)

### Multiplying Two Negatives

To see why two negatives multiply to a positive:

\(-a\) is the symbol representing the equivalence class \([(0,a)]\) and \(-b\) is the symbol representing the equivalence class \([(0, b)]\). Using the definition of multiplication, we see that

\[\begin{align*}

(-a)\times(-b) &= [(0,a)] \times [(0,b)]\\

&= [(0 \times 0 + a\times b, 0\times b + a \times 0)]\\

&= [(ab, 0)] \\

&= ab

\end{align*}

\]