# Is it an abuse of language to say “*the* integers,” “*the* rational numbers,” or “*the* real numbers,” etc.?

I'm finding that the more math I learn, the more concepts I thought were well-defined seem to be intuitive and naive. Here I'm asking about whether it's an abuse of language to refer to "the integers," "the rational numbers," or "the real numbers". This is what I mean:

Suppose, for the sake of starting somewhere, that we have a set which we call "the integers" and denote by $\mathbb Z$.

1. I can define "the rational numbers," which I denote by $\mathbb Q$, as the set of equivalence classes of ordered pairs of integers, where $(a,b)$ is equivalent to $(c,d)$ if $ad = bc$. But now when I say "the integers," do I mean the original set $\mathbb Z$, or the image of this set in $\mathbb Q$, i.e., the collection of equivalence classes of ordered pairs of the form $(a,1)$ where $a\in \mathbb Z$?

2. I can define "the real numbers," which I denote by $\mathbb R$, as the set of equivalence classes of Cauchy sequences of rational numbers, where $(a_n)$ is equivalent to $(b_n)$ if $a_n - b_n \to 0$. But now when I say "the rational numbers," do I mean the original set $\mathbb Q$, or the image of this set in $\mathbb R$, i.e., the collection of equivalence classes of sequences of the form $(q,q,q,\dotsc)$ where $q\in \mathbb Q$?

3. I can define "the complex numbers," which I denote by $\mathbb C$, as the set of equivalence classes of the polynomial ring $\mathbb R[X]$, where $f(x)$ is equivalent to $g(x)$ if $(x^2 + 1) \mid (f(x) - g(x))$. But now when I say "the real numbers," do I mean the original set $\mathbb R$, or the image of this set in $\mathbb C$, i.e., the collection of equivalence classes of constant polynomials in $\mathbb R[X]$?

And then of course I can take sets further up on the list and try to ask where they fit in lower down, e.g., I can ask whether or not by $\mathbb Z$, I mean a specific subset of $\mathbb R$, or a specific subset of $\mathbb C$, etc. Going further, I can also ask, given an arbitrary integral domain $D$, for example, and given its field of fractions $F$, whether or not when I say "$D$" I am referring to the original integral domain or its image in $F$, etc.

What is going on? Which set I should think of when I hear someone say "the integers" or "the rational numbers"? Is this a matter of learning to no longer think of these sets or structures as unique, but rather unique up to isomorphism? Or unique up to unique isomorphism? So when someone says "the rational numbers," should I be thinking of any field isomorphic to the rational numbers? Feel free to modify tags or provide comments.

• It is a "philosophical issue" still open... See Paul Benacerraf, What Numbers Could Not Be (1965). – Mauro ALLEGRANZA Jun 9 '16 at 14:35
• I'm not sure that this is an exact duplicate of Is "$a + 0i$" in every way equal to just "$a$"?, but I'm also not sure it isn't a duplicate of that. Barry Mazur's paper “When is one thing equal to another thing” discusses this in great detail, and argues strongly in favor of the "unique up to unique isomorphism" view. – MJD Jun 9 '16 at 14:54
• I don't understand why you think this is an interesting question... – reuns Jun 9 '16 at 16:32
• your question, as stated, is as inane as me saying, "when I'm talking about the Chicago Bears, am I allowed to call them the Bears? Or, like, since there are other types of bears, can I not do that?" – dbliss Jun 9 '16 at 19:04
• @user1952009 It's not so much that I think this is an interesting question per se; rather, I'm asking it in order to resolve my confusion arising from the fact that it's possible to model in different ways the same seemingly primitive mathematical objects. – justin Jun 9 '16 at 21:15

How you probably should think about this, most of the time -- and certainly the approach that seems to be most productive for most ordinary mathematics -- is that the integers, real numbers and even complex numbers exist in and of themselves in some "Platonic" sense, independently of what we think of them.

The set-theoretic constructions you find in textbooks construct models of the Platonically existing numbers within a pure set theory. Knowing how to do this -- and in particular knowing that it can be done -- is important and brings with it many technical advantages, but you should not let it trick you into thinking that these set-theoretic models are "what numbers really are". Doing so would make a mockery of the centuries and millennia where mathematicians reasoned about numbers without set theory even having been invented. It would be absurd that Dedekind, Cantor, Zermelo, and other mathematicians working in the 19th and 20th centuries were for the first time discovering what Euclid, Euler, Gauss and so forth had really been talking about.

Thus for ordinary mathematics the most useful approach is to continue thinking that all of the real numbers (and the complex ones too, if you can bring yourself to it) just exist, that you can put them all into a set, and that the rationals and integers are particular subsets of them.

Of course this "naive" view is not sufficient in order to work in axiomatic set theory, logic, or similar foundational areas of mathematics. There you're interested in how well the set-theoretic formalism can capture our naive ideas about integers and real numbers, and perhaps even in to which extent the usual Platonic belief in the existence of numbers can be upheld when one really thinks about it.

It it good to be able to ask such philosophical questions, and to know enough to begin answering them.

All I'm saying here is that it is not productive to let such philosophical uncertainty paralyze you when doing mathematics outside the foundational domain. One of the benefits about knowing about foundational theories is that it allows you to relax and know that whichever philosophical objections one might have to such-and-such argument in everyday mathematics can be handled another day, because we have a good model of the fundamental assumptions that everyday mathematics depends on.

• I appreciate the point about not letting philosophical uncertainty get in the way of productivity. This applies outside of math as well. For instance, it is interesting to consider the philosophical argument that sensory experiences may be illusions, but it's not productive to become paralyzed by this uncertainty and neglect to eat, or look both ways before crossing the road, etc. – justin Jun 9 '16 at 16:43
• @JohnHughes: Eudoxus and Dedekind did share a fundamental insight (all we need to know about a real number is which rationals are smaller than it, and which are greater), but as relevant for this discussion there is the important difference that Eudoxus did not (at least not as his ideas are known today through Euclid) think that the continuum was something that needed constructing; he was figuring how to reason in a uniform way about proportions between quantities (lengths, areas, angles, etc.) that were supposed already to exist. (...contd) – Henning Makholm Jun 9 '16 at 19:59
• @CharlesStaats: Yes that would be a typing error, strictly speaking. However, one can get away with considering integers as both real numbers and p-adic numbers in two ways. The first way is to have automatic type-conversion, which arguably is what people actually have in mind. The second way is unnatural but is standard in set theory, namely to first construct a structure that has the properties desired of reals, and then construct an isomorphic copy by replacing the embedding of the integers with the integers themselves, suitably modifying all functions. – user21820 Jun 10 '16 at 0:59
• Funny. I see numbers and sets both as models/constructs that do not inherently exist on their own. Just because humans discussed and used numbers before the idea of sets was developed, does not, in my mind, make numbers any more fundamental or "real". But I didn't write this comment to argue, merely to throw a different philosophy into the mix. I also don't see the question of whether numbers are wholly artifacts or exist on their own outside of human thought as having an objective "right" answer. Not a fan of Plato, either. – Todd Wilcox Jun 10 '16 at 2:04
• Hmm... Do they "exist in a Platonic realm" in such a fashion that there is a set of intermediate cardinality between $\mathbb{N}$ and $\mathbb{N}^{\mathbb{N}}$, or in such a fashion that there does not exist a set of intermediate cardinality between $\mathbb{N}$ and $\mathbb{N}^{\mathbb{N}}$? – Mikhail Katz Jun 10 '16 at 8:41

Is this a matter of learning to no longer think of these sets or structures as unique, but rather unique up to isomorphism? Or unique up to unique isomorphism?

Yes. And depending on what you mean by "the", it's not even an abuse of language!

(this post assumes some familiarity with the ideas of category theory)

The "uniqueness" of the real numbers is a theorem that says if $A$ and $B$ are any two complete, ordered fields, then there is a unique isomorphism $A \to B$.

Now, consider the following two full subcategories of the category of rings:

• the category $\mathcal{A}$ of all complete ordered fields and isomorphisms between them
• the category $\mathcal{B}$ consisting just of the complete ordered field $\mathbb{R}$ and its identity map

The uniqueness theorem above can be rephrased as saying that the categories $\mathcal{A}$ and $\mathcal{B}$ are equivalent.

There is a general philosophy in category theory that equivalence is the "right" notion, rather than equality or isomorphism. If we had a suitable language for doing mathematics, there wouldn't be any distinguishing between $\mathcal{A}$ and $\mathcal{B}$: it would be perfectly correct to speak of the complete ordered field.

From this perspective, that that is not the current state of affairs and instead we have to juggle transporting the mathematics we do along unique isomorphisms is simply an unfortunate accident of history that elevated the notion of identity over that of equivalence and that had an impact on how mathematics was developed.

• In other words: "the category of all complete ordered fields is equivalent to a point." If every two objects were isomorphic, but not necessarily uniquely so, we'd only get to say: "the category of all complete ordered is equivalent to a group." – goblin Jun 10 '16 at 5:27
• @William: There's something called an "anafunctor" which is a functor whose values on objects is only defined up to unique isomorphism. If you take the notion that an object of a category is a functor from the terminal category and replace functor with anafunctor, you would get my example that $\mathcal{A}$ is an "object" of the category of rings. – Hurkyl Jun 10 '16 at 6:44
• I think the idea of weakening equality to equivalence should appear in anything on higher category theory, but definitely at the nlab and the nforum. The best candidate we have for the "suitable language" is homotopy type theory. – Hurkyl Jun 10 '16 at 6:45
• @goblin: Yeah, which is actually kind of interesting now that I think about it. It seems to suggest things like the complexes, viewed as a -ring, really should not be thought of as a *set of elements, but instead as a $(\mathbb{Z}/2)$-set of objects. – Hurkyl Jun 10 '16 at 6:55

It all hangs on the isomorphism theorems at each level. They show that any two sets with the properties of N, Z, Q, R, C are isomorphic. Of course, those are isomorphisms that preserve sums, products, order , completeness, whenever that structures are defined.

Answering your question more directly: Yes, it is an abuse of language, but it is justified because of the isomorphism theorems. Proving those will be an excellent exercise in your understanding of the subject.

• I don't know I would attribute the justification specifically to the isomorphism theorems. The isomorphism theorems (assuming we are talking about the same thing) are theorems that can be used to construct specific isomorphisms. – Thomas Jun 9 '16 at 14:38
• Maybe I have seen them with another name, since all my classes are not in English. I refer to theorems like " any two linearly ordered conmutative rings, without extremes, complete, with a dense subset isomorphic to Q, are isomorphic (to R)". What are you refering to? – user346735 Jun 9 '16 at 14:48
• See en.wikipedia.org/wiki/Isomorphism_theorem for what was meant, I think. – quid Jun 9 '16 at 15:33
• oh! you meant those. Thank you for the terminology acclaration. – user346735 Jun 9 '16 at 15:45
• Personally I find it an abuse of language to speak of isomorphisms without specifying isomorphisms of which category. In the above cases it is not always clear which categories are involved, although one can probably invent proper categories for the occasion. – Marc van Leeuwen Jun 10 '16 at 14:58

The central issue here is not that these structures are merely "unique up to isomorphism," but that they are unique up to unique isomorphism: any two models for each of the structures you mention are isomorphic to each other in a unique way. Therefore (using common sense) there is really nothing to freak about over calling $\mathbf R$ "the" real numbers even though it has multiple models. It'd be like getting worried that those who write $\mathbf R$ and those who write $\mathbb R$ might in some way be describing different things.

None of the other answers are giving a compelling example where use of a definite article would not be a good idea, so let me do that: if you ever construct an algebraic closure $\overline{k}$ of a field $k$, you should only speak of $\overline{k}$ as "an algebraic closure" of $k$ rather than "the algebraic closure" of $k$ since two different algebraic closures of $k$ are isomorphic over $k$ but not uniquely so (if $k$ is not itself already algebraically closed). This is a real difference between constructing algebraic closures and the examples you give.

Another example: all cyclic groups of the same size are isomorphic, but not uniquely so (if the size is greater than $2$). Therefore you should speak of "a cyclic group of order $n$" rather than "the cyclic group of order $n$" (if $n > 2$). The computational significance of this issue comes up in the way some cyclic groups are used in cryptography: for prime numbers $p$, the groups $\mathbf Z/(p-1)\mathbf Z$ and $(\mathbf Z/p\mathbf Z)^\times$ are both cyclic of order $p-1$ and thus are isomorphic as abstract groups. However, computationally the groups are quite different: the first one has a "canonical" generator $1$ and given two generators $a$ and $b$ it's very easy to solve $ax \equiv b \bmod p-1$ for an $x$, while there is no simple formula for a generator of $(\mathbf Z/p\mathbf Z)^\times$ and if you know two generators $g$ and $h$ it appears to be a hard problem to solve $g^x \equiv h \bmod p$ for an $x$ (this is basically the discrete log problem).

Or you could do math in Russian and then your whole question disappears since Russian doesn't have articles and yet Russians seem to be able to conduct math at a high level quite successfully.

• To emphasize a point related to your first example, "the complex numbers" is quite different from the integers, rationals, or reals, because there are now two isomorphisms instead of a unique one. – Noah Snyder Jun 12 '16 at 9:42
• @NoahSnyder, and similarly for the quaternions. – KCd Jun 12 '16 at 11:58
• @NoahSnyder: And yet, nobody bats an eye at saying "the complex numbers", either. – Nate Eldredge Jun 12 '16 at 16:40
• @NateEldredge: Two is very close to one. – Noah Snyder Jun 12 '16 at 20:36

As explained already, this is, strictly speaking, an abuse of notation, one justified implicitly (or EXplicitly, if you ask) by reference to those structure-preserving maps. Computer programming has provided an excellent metaphor for what is going on here: implicit type coercion. This is what lets you write, e.g, 'X + 1', where 'X' is of type 'double', and have the machine understand that you really mean 'X + 1.0'.

So, for example, mathematicians have noted that there is a natural map from Z to Q - i.e 'n --> (n,1)' - such that the objects on the right side of '-->' behave the same arithmetically as those on the left side (I'm glossing over a bunch of stuff here). Based on this, they have learned - and expect you in turn to learn - that when they take an 'n' from Z and try to reason about it within Q, that 'n' should be translated to '(n,1)' in your head automatically, by habit. Of course no one ever explains this explicitly during your education, they expect you to just 'pick it up'.

• FWIW, I program in OCaml, and it entirely refuses to let you write x + 1 if x is a float, forcing you to write x +. 1.0 instead :) I sometimes wish that standard mathematical practice (at least at the higher levels) was a little more pedantic about inserting explicit coercions where necessary. – Ben Millwood Jun 14 '16 at 15:11

Good questions all. When I say "the integers," I may mean one of several things. For instance, I might say that $f(x) = \sin(\pi x)$ is zero on the integers, in which case I'm probably saying that $x$ is a real number, and that for certain real numbers --- the integers --- $f$ happens to take the value $0$. When I say "We do induction over the positive integers," I mean the integers themselves, not embedded in the real line.

Mathematicians are often sloppy about this, and it almost never causes confusion. Context usually makes things clear.

There are cases when careful distinctions matter. For instance, the integers can be embedded in the integers as a proper subset (via $n \mapsto 2n$, for instance), although this embedding is not a ring homomorphism. We could then refer to the codomain as the integers, and the image of the map as "a copy of the integers", even though we'd typically call it "the even integers". So we have, in this situation, two different things that we could call the integers. When this happens, you have to be more careful and give things individual names. I'd write something like this:

"Let $D$ and $C$ both be the integers, and consider the function $f : D \to C: n \mapsto 2n$. Then the image of $f$, $f(D)$, is set-isomorphic to the integers, but $f$ is not a homomorphism of rings. In particular, the identity element for multiplication in $C$ is not in the image of $f$."

The categoricity claim for each of the number systems $\mathbb{Z,Q,R}$ depends on fixing a background set theory. In this sense the answer to your question is affirmative: the definite article used in this way is a bit of an abuse of language, but a pardonable one in many contexts.

This is a point that has been widely discussed in the literature, for instance by Joel David Hamkins; see e.g., here and here. To see that this could not be otherwise, perform the following thought experiment.

Naively assumed "uniqueness" of $\mathbb{N}$ naively speaking would imply the uniqueness of the collection of sequences of natural numbers $\mathbb{N}^{\mathbb{N}}$. Now consider the following question.

Question CH. Does there exist a subset of $\mathbb{N}^{\mathbb{N}}$ of strictly intermediate cardinality between $\mathbb{N}$ and $\mathbb{N}^{\mathbb{N}}$?

Thus the naive uniqueness of the natural numbers would seem to imply that there is a determinate answer to Question CH. However, the latter is thought by many to be a question that does not have a determinate answer, and moreover given a set-theoretic universe where the answer to CH is affirmative, there are well-understood techniques for constructing a set-theoretic universe where the answer to CH is negative, and vice versa.

Those who advocate for a realist/platonist position as the surest best way to do well in mathematics overlook the fact that such a position is a fairly recent one, dating from the end of the 19th century.

The great masters of the past: Leibniz, Euler, Cauchy, and others all did just fine without believing in triangles fluttering in a rarefied realm of the abstracta. Telling the OP that he should believe this whether or not there is a basis for such a belief is apparently not the practice of a scientist.

• Could you point me in the direction of some of Joel David Hamkins' discussions of this? I'm quite new to the set theory literature. – justin Jun 9 '16 at 14:54
• worldscientific.com/worldscibooks/10.1142/8997 You might want to start by looking at his discussion of the question whether we have an absolute notion of finiteness. – Mikhail Katz Jun 9 '16 at 15:10
• jdh.hamkins.org/question-for-the-math-oracle Actually I meant this post of his, but the one above is also interesting. – Mikhail Katz Jun 9 '16 at 15:12
• I don't find it super convincing that being sure of what $\mathbb N$ is must lead you to being sure of what $\mathbb N^\mathbb N$ is, and I'm definitely unconvinced that you can't know what both of those are and still be unable to resolve CH. CH feels like it's about the existence of sets that aren't necessarily directly related to either $\mathbb N$ or $\mathbb N^\mathbb N$. On a more superficial level, CH is about sets, whereas I feel like I can talk about natural numbers and sequences of them without necessarily pinning myself to a particular set theory. – Ben Millwood Jun 10 '16 at 17:24
• @BenMillwood, this is an interesting viewpoint but I am not entirely sure what you mean. Of course I also like to talk about natural numbers and sequences of them without pinning myself to a particular set theory. However, the continuum hypothesis in this case is equivalent to the question whether there is a subset of $\mathbb{N}^{\mathbb{N}}$ of strictly intermediate cardinality, as I mentioned in my answer. So if natural numbers and sequences thereof have a definite existence for you, then so should be an answer to CH. Do we agree so far? – Mikhail Katz Jun 13 '16 at 6:52

Yes it is technically abuse of language/notation. There are several ways to construct/define the integers/rational numbers/ ... But all these constructions will lead to things that are "isomorphic". If we think about the integers as an Abelian group, then the different groups will be isomorphic.

This type of "abuse" isn't something to worry about. Just imagine the chaos we would have if every time we talked about the integers we had to make clear exactly how it was constructed. We use the same "abuse" when we in algebra will say things like: There is only one group of order $3$. We should, of course, say that up to isomorphism, there is only one group of order $3$, but it is not uncommon to exclude the "up to isomorphism" part. Likewise we can talk about the dihedral group of a certain order even though there are several says of constructing these.

• The assumption of uniqueness up to isomorphism is problematic as I explained in my answer. – Mikhail Katz Jun 9 '16 at 15:42

There are many ways to define a thing. As long as we have a consistent way to declare the definitions as equivalent, we can choose our preferred definition and use that as the "operating definition". As long as we have machinery for declaring equivalency of definitions, we just say, if only as a matter of practicality, that the things being defined are indeed the same thing.

This doesn't mean that in some deep existential/ontological/philosophical sense they are actually the same thing. It just means they can be identified with each other in a logically consistent and helpful way.

As was noted in the original post, a real number can be thought of as an equivalence class of Cauchy sequences of rationals (once we accept the existence of the rationals). Call these equivalence classes $\mathbb R_{Q.seq}.$ But this is only one way to define the real numbers.

We may posit the existence of a continuum and define each real number as a specific point on that continuum. Call these $\mathbb R_\frak c.$ Or we might say that a real number is identical to an equivalence class of sequences natural numbers upon which the concatenation of all but the first with placement of the decimal according the the first (as long as we accept the existence of the naturals, a decimal point, and our ability to concatenate and place such a decimal, etc.). Call these $\mathbb R_{N.cat}.$

Do we have $\mathbb R_{Q.seq}=\mathbb R_{\frak c}=\mathbb R_{N.cat}?$

Consider $x,y,z$ in $\mathbb R_{Q.seq}$, $\mathbb R_{\frak c}$, $\mathbb R_{N.cat}$ respectively. $x$ is an equivalence class of sequences of rationals. $y$ is a point on a continuum. $z$ is an equivalence class of sequences of natural numbers. (Note that although we might like to think of $z\in\mathbb R_{N.cat}$ as being a collection of sequences of rationals as well, those sequences are not necessarily Cauchy.)

How can we even check if $x=y=z$ when, structurally, they are three distinct types of objects? Certainly they are all pair-wise non-identical (again since they are from three distinct classes of objects).

However, we are able to define one-to-one and onto mappings $f:\mathbb R_{\frak c}\rightarrow\mathbb R_{Q.seq}$ and $g:\mathbb R_{N.cat}\rightarrow\mathbb R_{Q.seq}$ so that when $f(y)=x$ and $g(z)=x$, we say that $x\overset{f}{=}y$ and $x\overset{g}{=}z$ hence $y\overset{g^{-1}\circ f}{=}z.$ Of course, we have to have some "belief" the existence of such mappings!

In practice, we drop the mappings indicated in the equivalences and just say $x=y=z=r$ where $r$ is our chosen object for defining a specific "real number" (potentially from none of the given definitions of $\mathbb R$).

A simpler example:

To distill the argument into much simpler terms, consider: $\frac{1+2}{3}=1$ and $10-9=1.$ I'll call attention to two ideas:

Quantitative equivalence/identity.

$$\frac{1+2}{3}=10-9$$

Specific 'structural' identity.

$$\text{Equation}\left[\frac{1+2}{3}\right]\neq\text{Equation}\left[10-9\right]$$

Where the latter is calling attention to the fact that both sides are distinct collections of mathematical operations. (Note: I'm just making up the terminology.)

We have two classes:
$[1]_3=$ a specific equivalence class of arithmetic operations on $3$ integers. Specifically, $(n,m,k,o_1,o_2)\in[1]_3$ where $n,m$ are finite sequences of $1$'s, $k$ is a natural number, $o_1$ is removal or concatenation, and $o_2$ is dividing the result into $k$ equivalent finite sequences of ones.
$[1]_2=$ an equivalence class of operations on $2$ integers defined similarly with elements $(n,m,o_1)$ where $n,m$ are finite sequences of ones and $o_1$ is either concatenation or removal.

We say $[1]_3=[1]_2$ because we have a mapping which allows us to calculate/evaluate them and get the number $1$ as the result. Of course, I'm likely brushing aside many foundational and technical issues.

Very often notation is abused in order of brevity. This only causes a problem if one does not know the original meaning anymore. Hence abusing the notation is fine as long as it is clear to everyone what is meant.

I do not agree with your examples: 1. and 2. rather describe that $\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}$. The usual way is to construct $\mathbb{Q}$ as an extension to $\mathbb{Z}$ using exactly the equivalence classes that you described. Similarly $\mathbb{R}$ is usually constructed as an extension to $\mathbb{Q}$ to make it complete. In this sense, it does not make much sense to define for instance $\mathbb{Z}$ as a subset of something that has been originally built on $\mathbb{Z}$. Similarly, 3. is kind of a circular reasoning, describing $\mathbb{C}$ as some equivalence class over a polynomial ring over $\mathbb{R}$ and then defining $\mathbb{R}$ as a subset of this set of equivalence classes.

So I would say that your examples rather describe connections between sets, or in the third case an isomorphism. The set $\mathbb{R}$ for example is constructed by postulating certain axioms and then showing that there exists a unique set up to isomorphism that satisfies these axioms, and just because one finds such an isomorphism it does not mean that the notation is not clear.

But there are many other examples, where one omits things that are clear. For example, for the dimension, a span, or similar constructions over a field $K$ one usually omits $K$ in the notation when it is clear which field is meant. One defines a function directly by $f(x)=x$, when it is clear what is its domain and range, and that this describes the assignment for all elements of the domain. Also, the integers are actually an example in the way that there is several kinds of integers. I have seen for instance p-adic integers, Gaussian integers and Eisenstein integers.

So when someone talks to you about the integers or the rational numbers, then the context should make it clear how this is meant, and usually it will not cause any problem that there are actually several isomorphic such spaces.

• Well, the "uniqueness up to isomorphism" that you mentioned is problematic, as I explained in my answer. – Mikhail Katz Jun 9 '16 at 15:13
• After construction the rationals as equivalence classes, I often see $\mathbb{Z} \subset \mathbb{Q}$ forced, either by modifying $\mathbb{Q}$ or changing what we mean by $\mathbb{Z}$. – Hurkyl Jun 9 '16 at 15:31