# Building the integers from scratch (and multiplying negative numbers)

Now I understand that what I am about to ask may seem like an incredibly simple question, but I like to try and understand math (especially something as fundamental as this) at the deepest level possible. And for the life of me, I can't shake this feeling I have that something is not quite right.

Let me begin:

First of all, I was reading Terry Tao's discussion about the construction of the standard number system and was very pleased with the way in which $\mathbb{C}$ can be systematically constructed from the natural numbers through a system of homomorphisms (as a direct limit): $$\mathbb{N} \hookrightarrow \mathbb{Z} \hookrightarrow \mathbb{Q} \hookrightarrow \mathbb{R} \hookrightarrow \mathbb{C}$$

In particular, he talks about how the integers can be constructed as the space of equivalence classes of formal differences between natural numbers, where $$[a-b]\sim[c-d ] \iff a+d=b+c$$ with $a,b,c,d\in\mathbb{N}.$ He then goes on to say "with the arithmetic operations extended in a manner consistent with the laws of algebra."

Now, for $(\mathbb{N},+,\cdot)$ addition is straight forward, and $$n\cdot k:=\underbrace{k+\cdots+k}_{n\;\text{times}}$$ is well defined with $n\cdot k=k\cdot n$. Moreover, the distributive law, $n\cdot(m+k)=n\cdot m+n\cdot k$, also holds.

However, when I want to construct $(\mathbb{Z},+,\cdot)$ as above, I run into some trouble. Addition of two equivalence classes seems pretty straightforward, given by $$[a-b] + [c-d] := [ (a+c)-(b+d) ],$$ and is well defined. This makes sense as we simply "add up" the positive and negative amounts together. This definition also behaves nicely with the map $\varphi:\mathbb{N}\hookrightarrow\mathbb{Z}$, given by $n\mapsto[ n-0 ]$; with $\varphi(n+k)=\varphi(n)+\varphi(k)$, where the second addition is the addition of equivalence classes. Moreover, we have commutativity of addition; the existence of an additive identity, namely $[0-0]$; and the existence of additive inverses, $[a-b]+[b-a]\sim[0-0].$

But now, when I try to define the multiplication of classes, I am unsure how to proceed:

If we consider our usual algebraic rules, we get that $$(a-b)\cdot(c-d)=a\cdot c-a\cdot d-b\cdot c+b\cdot d,$$ which might lead us to define the multiplication of two equivalence classes as $$[ a-b ]\cdot[ c-d ]:=[ (a\cdot c+b\cdot d)-(a\cdot c+b\cdot c) ].$$ Now it's easy to check that this is indeed well defined, and obeys the distributive law with the definition of addition we've given above. However I feel that we've simply gone in a circle (logically) as we've assumed a priori that $(-b)\cdot c=-(b\cdot c)$ and $(-b)\cdot(-d)=-(b\cdot d)$.

Now, I am very familiar with the fact that if your set is a ring (with unity) then these results (particularly that negative times negative is positive) come as simply a result of playing around with additive inverses and the distributive laws (see here). However my problem is that we've only gotten this ring structure on our set of equivalence classes by appealing to the ring structure of the integers, which is exactly the thing we are trying to construct from scratch! And I do not want to simply define multiplication in this way "because it works"; it has always been my feeling that results like $(-1)\cdot(-1)=1$ should come as consequences of the structure, and not as properties we impose.

So I'm curious if there is a way around this issue. Is this particular definition of multiplication the only one that:

1. is well defined?
2. satisfies the distributive laws?
3. is associative?
4. has a multiplicative identity? $$[1-0]\cdot[a-b]=[a-b],\;\text{for all a,b\in\mathbb{N}}$$
5. gives a homomorphism which splits over multiplication? $$\varphi:(\mathbb{N},+,\cdot)\hookrightarrow(\mathbb{Z},+,\cdot),\; n\mapsto[ n-0 ]$$ $$\varphi(n\cdot k)=\varphi(n)\cdot\varphi(k)=[ n-0 ]\cdot[ k-0 ]$$

If our deffiniton of of multiplication has all these properties, then we'll have a ring structure on our set of equivalence classes, and all the familiar properties of the integers will be established. However, it is not immediately clear to me that this is our only option. Can someone shed some light on this?

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Related: math.stackexchange.com/a/62868/622 (and the rest of the thread). – Asaf Karagila Jun 9 '12 at 22:44

You've answered your own question. You say you're familiar with the fact that $(-b)c = -bc$ and $b(-d) = -bd$ in a ring. Now, if you write down a definition of multiplication that satisfies all of those properties, you get a ring structure on the set of equivalence classes, and $(-b)c = -bc$ and $b(-d) = -bd$ in a ring...!

Perhaps some extra generality will make this argument seem less circular. You have a rig $R$. This rig has an underlying additive monoid $M = (R, +)$. This abelian monoid has a Grothendieck group $G(M)$, which is the universal group into which $M$ maps. More precisely, there is a forgetful functor $F : \text{Ab} \to \text{CMon}$ from the category of abelian groups to the category of commutative monoids, and the Grothendieck group functor is its left adjoint. $G(M)$ consists of equivalence classes of formal expressions $m - n$ where $m, n \in M$ with addition defined as expected.

Now you want to put a ring structure on $G(M)$ compatible with the multiplication on $R$. It turns out that what you are actually constructing is the universal ring into which $R$ maps; more precisely, there is a forgetful functor from the category of rings to the category of rigs and you are constructing its left adjoint. Well, in any such ring we need $$(a - b)(c - d) = ac - bc - ad + bd$$

as you well know, so this is the unique possible ring structure on $G(M)$ compatible with the multiplication on $R$ (and distributivity, inverses, etc.) if it is well-defined. This is a general property of adjoint functors (they are unique up to unique isomorphism if they exist). The only thing left is to verify that it actually works.

Another way to say the above is the following. Let $R$ be a rig and let $S$ be a ring, and let $\phi : R \to S$ be any rig homomorphism whatsoever. Then $$(\phi(a) - \phi(b))(\phi(c) - \phi(d)) = \phi(ac) - \phi(bc) - \phi(ad) + \phi(bd).$$

There is nothing circular about this because $S$ is a ring and the above computation takes place in $S$.

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Sorry but my algebra knowledge is nowhere near this level. I haven't seen any category theory or semi-rings. – Patch Jun 10 '12 at 0:37
@Patch: okay, so ignore the category theory. Semirings are not so bad; just think of $\mathbb{N}$ (the only axiom that is dropped is additive inverses). The basic point here is that if you want to embed $\mathbb{N}$ into a ring, $(a - b)(c - d) = ac - bc - ad + bd$ needs to be true in that ring for all $a, b, c, d \in \mathbb{N}$. Does this make sense? – Qiaochu Yuan Jun 10 '12 at 0:50
If what you've said about that being the unique way to turn $\mathbb{N}$ into a ring, then that is exactly the answer I was looking for. I guess it's still a bit hazy to me, though, why that would be the only way to extend the definition of multiplication. But thanks for getting at the heart of my confusion. – Patch Jun 10 '12 at 1:00
@Patch: what is hazy about it? You have $\mathbb{N}$, you have some ring $R$, and you have a homomorphism $\phi : \mathbb{N} \to R$. In $R$ you have $(\phi(a) - \phi(b))(\phi(c) - \phi(d)) = \phi(ac) - \phi(bc) - \phi(ad) + \phi(bd)$. Everything is forced on you by distributivity and the uniqueness of additive inverses. – Qiaochu Yuan Jun 10 '12 at 1:02
@Patch: perhaps your confusion comes from the fact that the argument can be interpreted as working in some ring (namely the universal ring admitting a map from $\mathbb{N}$) conditional on its existence. This is actually not an unusual thing to do in mathematics. The goal of that part of the argument is to prove uniqueness, not existence. These are often done separately and there's no reason to prove existence first, especially since proving uniqueness often helps a great deal in proving existence. – Qiaochu Yuan Jun 10 '12 at 1:05

The easiest way to convince oneself that there's no circularity is to notice that the entire discussion leading up to the law $$\tag{*} [a-b][c-d]=[(ac+bd)-(ad+bc)]$$ is superfluous from a completely rigorous point of view. The discussion makes it look like $(*)$ is a deduced truth, but from a formal point of view it is just a definition. We could just have pulled it out of a hat, or found it in the lost notebooks of a crazy deceased genius. All that matters from a formal standpoint is that we can prove that this definition allows us to prove the laws we want to be true (such as the associative and distributive laws).

The "circular" discussion leading up to the definition is just there to satisfy the student's curiosity why anyone would get the idea of trying out such a complicated definition in the first place. Circularity there doesn't matter; one is allowed to use all sorts of stupid tricks, iffy intuitions, and clairvoyant cheating to decide what one wants to prove -- as long as the proof itself (which comes later) is solid.

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I realize that this is legit, but it still feels a bit like cheating to me. * I start off asking "why is $-a\cdot-b=a\cdot b?$". * To prove this I need a ring structure on my system of numbers. * I start constructing $\mathbb{Z}$ only to find that the ring structure I gave them already uses $-a\cdot -b=a\cdot b$. (It's like saying $(-1)(-1)=1$ because I said so.) – Patch Jun 10 '12 at 0:33
@Patch: you're asking the wrong question. The question is "how do I turn $\mathbb{N}$ into a ring?" and the answer is "formally adjoin additive inverses, then the distributive law forces a unique definition of multiplication on me." – Qiaochu Yuan Jun 10 '12 at 0:51
@Patch: If you ask "why is $(-a)(-b)=ab$?", then the particular construction of $\mathbb Z$ does not matter at all -- it has to be like that in any ring. First, $-a=(-1)\cdot a$ for all because $$a+(-1)\cdot a = 1\cdot a+(-1)\cdot a = (1+(-1))\cdot a = 0\cdot a = 0.$$ Then $$(-a)\cdot(-b)=(-1)\cdot a\cdot (-b) = a\cdot(-1)\cdot(-b)=a\cdot(-(-b))=a\cdot b.$$ This is independent of how we decide give meaning to the ring operations, as long as the meaning we select satisfies the ring axioms! – Henning Makholm Jun 10 '12 at 4:31

There is no circularity. You should maybe remember that $-a$ and $-b$ don't mean anything in $\mathbb{N}$, so their product inside $\mathbb{N}$ is also meaningless. The way to approach this, is to say the following

$"$Let us define a function $m:(\mathbb{N}\times \mathbb{N})\times(\mathbb{N}\times \mathbb{N})\rightarrow\mathbb{N}\times\mathbb{N}, ((a,b),(c,d))\mapsto (ac+bd,ad+bc)$. This, you will agree, is a well defined function. Next we notice, as you have, that this function is compatible with the equivalence relation $\sim$, i.e. $\forall a,a',b,b',c,c',d,d'\in\mathbb{N}$, $$(a,b)\sim (a',b') \mathrm{~and~} (c,d)\sim (c',d') \mathrm{~imply~} m((a,b),(c,d))\sim m((a',b'),(c',d,))$$ so whenever we "multiply" (read : apply $m$ to) two ordered pairs, the equivalence class of the result only depends on the equivalence classes of the pairs we "multiplied".

This being the case, we get a well defined map (which I'll still call $m$) on the set of equivalence classes $\mathbb{Z}:=(\mathbb{N}\times\mathbb{N})/\sim$ of the equivalence relation $\sim$ $$\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{Z},([a,b],[c,d])\mapsto [m((a,b),(c,d))]=[ac+bd,ad+bc]"$$

where I wrote $[a,b]$ for the equivalence class of the ordered pair $(a,b)$ (you wrote $[a-b]$, which is fine, but may lead one to believe there is some kind of negation used and to "expect" it to behave somehow like negation, for which there is a priori no reason). The fact that $(-1_{\mathbb{Z}})\times(-1_{\mathbb{Z}})=1_{\mathbb{Z}}$ and so forth are then properties you can easily verify.

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What you've said makes perfect sense, and I've already played around on paper with all of this before. I started off trying to build $\mathbb{R}$ from $\mathbb{N}$ (as the natural numbers are the only numbers we have at our disposal a priori) but I got a bit philosophical and began to run in circles. Multiplying natural numbers makes intuitive sense as repeated addition, as does multiplying a negative integer by a positive one. But this intuition fails if you try to say things like "$-3\cdot-2$ equals $-3$ added up $-2$ times." – Patch Jun 10 '12 at 0:54
@Patch: it's a silly intuition. You're much better off thinking of multiplying real numbers as composing scalings of the real line. This attitude towards multiplication, among other things, extends naturally to the complex numbers and matrices whereas "repeated addition" fails miserably. See also my comments at math.stackexchange.com/questions/56663/… . – Qiaochu Yuan Jun 10 '12 at 1:06
It might be a bit silly, or limited, but I find diving into the real numbers without some kind of deep understanding of their construction can be confusing. And I've seen time and again people unable to wrap their minds around the fact that $0.999...=1$ (which is immediate if you've formally constructed the real numbers as limits of Cauchy sequences in $\mathbb{Q}$). – Patch Jun 10 '12 at 1:33