## Building the $~\mathbb{Z}~$ from $~\mathbb{N}$

Intuitively, to go from $\mathbb{N}$, we just add in $-n$ for each $n \in \mathbb{N}$ with $n > 0$. We will see two different ways of handling this.

**Grothendieck construction**. We start with $\mathbb{N}$ and consider the set $\mathbb{N} \times \mathbb{N}$ of ordered pairs of natural numbers. On this, we impose a relation. We write $(n_1,m_1) \sim (n_2,m_2)$ if \(n_1 + m_2 = n_2 + m_1\)

**Theorem**. This is an equivalence relation on $\mathbb{N} \times \mathbb{N}$.

## **Proof**. (Expand to view)

We check that $\sim$ is reflexive. For any $(n,m)$, we have $(n,m) \sim (n,m)$ since $n + m = n + m$. Next we check symmetry. Assume that $(n_1,m_1) \sim (n_2,m_2)$. Then, by definition, we have $$ n_1 + m_2 = n_2 + m_1 $$ Flipping the equality we get $$ n_2 + m_1 = n_1 + m_2 $$ so $(n_2,m_2) \sim (n_1,m_1)$. Now we check transitivity. Assume that $(n_1,m_1) \sim (n_2,m_2)$ and $(n_2,m_2) \sim (n_3,m_3)$. Then, we have $$ n_1 + m_2 = n_2 + m_1 \\ n_2 + m_3 = n_3 + m_2 $$ So $$ n_1 + m_3 + n_2 = n_1 + n_3 + m_2 = n_1 + m_2 + n_3 = n_2 + m_1 + n_3 = n_3 + m_1 + n_2 $$ We know that we cancel addition in $\mathbb{N}$ so we can conclude that $$ n_1 + m_3 = n_3 + m_1 $$ ■

Since we have an equivalence relation, we can form the quotient by that equivalence relation.

**Definition**. The *integers*, denoted $\mathbb{Z}$, are the set of equivalence classes of $\mathbb{Z} := \mathbb{N} \times \mathbb{N}/\sim$.

Ok, maybe this does not look like our familiar integers. Where is $-1$? Where are the $\mathbb{N}$’s even?

**Definition**. We will let \(\begin{aligned} i : \mathbb{N} & \to \mathbb{Z} \\ n & \mapsto [(n,0)] \end{aligned}\) and let \(\begin{aligned} j : \mathbb{N} & \to \mathbb{Z} \\ n & \mapsto [(0,n)] \end{aligned}\)

**Lemma**.

- Both $i$ and $j$ are injective.
- Every element of $\mathbb{Z}$ is in the image of $i$ or of $j$.
- If $i(n) = j(m)$, then $n=m=0$.

## **Proof**. (Expand to view)

Assume that $[(n,0)] = [(m,0)]$. Then we have seen that this is equivalent to $(n,0) \sim (m,0)$. So $n + 0 = m + 0$ or $n = m$. A similar argument shows that $j$ is injective.

Let $[(n,m)]$ be an equivalence class. Assume that $n \geq m$. Then $$ n + 0 = (n-m) + m $$ So $(n,m) \sim (n-m,0)$ and $[(n,m)] = [(n-m,0)] = i(n-m)$. Now assume that $n \leq m$. Then $$ n + m-n = 0 + m $$ so $(n,m) \sim (0,m-n)$ and $[(n,m)] = [(0,m-n)] = j(m-n)$.

Assume that $i(n) = j(m)$. Then $[(n,0)] = [(0,m)]$. So $(n,0) \sim (0,m)$. Thus, $$ n + m = 0 + 0 = 0 $$ Since $n,m \in \mathbb{N}$, we have $n = m = 0$. ■

Despite the unfamiliar definition, we see that $\mathbb{Z}$ is exactly two copies of $\mathbb{N}$ with their two $0$’s identified.

Furthermore, we can add elements of $\mathbb{Z}$! First, we start with componentwise addition on $\mathbb{N} \times \mathbb{N}$. \((n_1,m_1) + (n_2,m_2) := (n_1+n_2,m_1+m_2)\)

**Theorem**. The following is a well-defined function. \(\begin{aligned} \mathbb{Z} \times \mathbb{Z} & \to \mathbb{Z} \\ ([(n_1,m_1)],[(n_2,m_2)]) & \mapsto [(n_1+n_2,m_1+m_2)] \end{aligned}\)

## **Proof**. (Expand to view)

The possible problem with this as a function is that the output might depend on the representatives of the equivalence. Certainly the function $$ ([(n_1,m_1)],[(n_2,m_2)]) \mapsto (n_1+n_2,m_1+m_2) $$ is not well-defined. For example, $[(0,0)] = [(1,1)]$ but we get different outputs for $([(0,0)],[(0,0)])$ and $([(1,1)],[(1,1)])$.

Assume that $[(n_1,m_1)] = [(n_1^\prime,m_1^\prime)]$ and that $[(n_2,m_2)] = [(n_2^\prime,m_2^\prime)]$. Then $$ n_1 + m_1^\prime = n_1^\prime + m_1 \\ n_2 + m_2^\prime = n_2^\prime + m_2 $$ The two outputs we get are $(n_1+n_2,m_1+m_2)$ and $(n_1^\prime + n_2^\prime, m_1^\prime + m_2^\prime)$. We need to check these give the same equivalence class. It is equivalent to checking that $(n_1+n_2,m_1+m_2) \sim (n_1^\prime+n_2^\prime,m_1^\prime+m_2^\prime)$. We have $$ n_1+n_2+m_1^\prime + m_2^\prime = n_1^\prime+n_2^\prime +m_1+m_2 $$ using the equalities above. ■

The function $i$ now satisfies \(i(n+m) = [(n+m,0)] = [(n,0)] + [(m,0)] = i(n) + i(m)\) so it intertwines addition in $\mathbb{N}$ and addition in $\mathbb{Z}$. Similarly $j(n+m) = j(n) + j(m)$.

**Example**.

- We have \([(n,m)] + [(0,0)] = [(0,0)] + [(n,m)] = [(n,m)]\) for any $(n,m)$.
- We have \([(n,0)] + [(0,m)] = \begin{cases} [(n-m,0)] & \text{ if }n \geq m \\ [(0,m-n)] & \text{ if } n \leq m \end{cases}\) In particular, $[(n,0)] + [(0,n)] = [(0,0)]$.

So $[(n,0)]$ is really $n$ while $[(0,n)]$ is $-n$ (though we could reverse the roles here).

**Theorem**. The function \(\begin{aligned} \mathbb{Z} \times \mathbb{Z} & \to \mathbb{Z} \\ ([(n_1,m_1)],[(n_2,m_2)]) & \mapsto [(n_1n_2 + m_1m_2,n_1m_2+n_2m_1)] \end{aligned}\) is well-defined.

We omit the proof of this theorem. Let’s look at the outputs of this function on simple entries.

- What if we input $i(n)$ and $i(m)$? Then, $i(nm) = [(nm,0)]$ comes out.
- What if we inpute $j(n)$ and $j(m)$? Then we get $[(nm,0)] = i(nm)$.
- How about $i(n)$ and $j(m)$? Then we get $[(0,nm)] = j(nm)$.

If you think about this for a little bit, you will recognize it as *multiplication* on $\mathbb{Z}$.

Next we will compare this to Lean’s `Int`

.