1

I'm studying Analysis from the book of Terence Tao and I have a problem with a paragraph where Tao establishes that we can set the natural number $n$ equal to integer $n—0$. He starts checking that $(n—0) + (m—0) = (n + m)—0$, $(n—0)×(m—0) = nm—0$ and $(n—0) = (m—0) \iff n = m$. From this argument he says that we can identify the natural numbers with integers by setting $n \equiv n—0$ and I don't know why that argument justify that we can give that "identification". What means exactly "$\equiv$" in this context? Finally, how we can compare two different objects when he sets that $n = n—0$?

I would appreciate some bibliographical references that reinforce why, in effect, we can set that $n = n—0$.

  • Are the very long dashes meant to be something other than subtraction? If not, I would suggest that you use a normal subtraction symbol, which is typeset with a hyphen (-). If so, can you please explain what these long dashes are? Otherwise, it is quite confusing to me. – Xander Henderson Jul 13 '23 at 01:29
  • 1
    @XanderHenderson Tao says "we will temporarily write integers not as a difference $a − b$, but instead use a new notation $a−!!!−,b$ to define integers, where the $−!!!−$ is a meaningless place-holder, similar to the comma in the Cartesian co-ordinate notation $(x, y)$ for points in the plane." – Henry Jul 13 '23 at 01:47
  • Yikes. Okay. I see what he is doing. – Xander Henderson Jul 13 '23 at 01:48
  • Tao does not say "we can set the natural number $n$ equal to integer $n−!!!−,0$" since he starts with the natural numbers and then defines the integers using an equivalence relation on an ordered pair of natural numbers. What he does say is "The integers $n−!!!−,0$ behave in the same way as the natural numbers $n$; indeed one can check" that addition and multiplication work as expected as you quote. So in that sense the integer $n−!!!−,0$ is equivalent to (but not identical to) the natural number $n$ and he uses $\equiv$ as a symbol to show this. – Henry Jul 13 '23 at 01:54
  • 1
    This is a special case of a general construction that completes monoids (or semigroups) into groups, e.g. see the links here about groups of differences. – Bill Dubuque Jul 13 '23 at 03:18

1 Answers1

1

Preliminaries

I find the notation $m \mathbin{—} n$ uncomfortable—it looks a little too much like a subtraction symbol (but isn't), and is difficult to typeset (I've copy-pasted it from the question, where it is some kind of unicode entity; I find this unwieldy). Because of this, I am going to use a slightly more standard notation: an integer is an equivalence class of ordered pairs of natural numbers, where two pairs $(a,b)$ and $(c,d)$ are equivalent if $a+d = b+c$.

In order to avoid writing "the equivalence class of $(m,n)$" over and over again, adopt the notation $[(m,n)]$ for that equivalence class. In more annoying set notation, $$ [(m,n)] = \{ (a,b) : m+b = n+a \}. $$ It appears that this is what Tao means by $m\mathbin{—}n$.

There is some work to do in order to ensure that this relation really is an equivalence relation, and there is a little more work to do in order to define addition and multiplication, but the short version is that it is possible to add and multiply: suppose that $x$ and $y$ are two integers. Each integer is an equivalence class of pairs, so choose representatives from each class, say $(x_1, x_2) \in x$ and $(y_1, y_2) \in y$. Then

  1. $x + y$ is defined to be $[(x_1 + y_1, x_2 + y_2)]$, and
  2. $x \cdot y$ is defined to be $[(x_1y_1 + x_2y_2, x_1y_2 + x_2y_1)]$.

Again, there is quite a lot that is being swept under the rug here. For example, it is not obvious that these operations are well defined (e.g. if we choose different representatives from each class, does this addition give the same result?).

Embedding the Naturals into the Integers

The next argument is that the naturals can be "canonically" embedded into the integers. That is, there is a function from the naturals to the integers which "preserves" everything good about the naturals (i.e. the additive and multiplicative structures). The embedding is as follows: if $n$ is a natural number, then map $n$ to the equivalence class represented by $(n,0)$.

Note that the statement is not (or should not be) that $n$ is equal to $(n,0)$. Rather, the statement says that there is a function which identifies $n$ with $(n,0)$. In more abstract notation, there is function $$ \iota : \mathbb{N} \to \mathbb{Z} : n \mapsto [(n,0)]. $$

The claim, then, is that $\iota$ preserves the additive and multiplicative structures of $\mathbb{N}$. That is, $\iota(m+n) = \iota(m)+\iota(n)$, and $\iota(mn) = \iota(m)\iota(n)$. This is what Tao is checking in the quoted section:

  1. Addition is Preserved: Let $m,n\in \mathbb{N}$. Then $$\iota(m+n) = (m,0) + (n,0) = (m+n, 0+0) = (m+n, 0) = \iota(m+n). $$

  2. Multiplication is Preserved: Let $m,n\in \mathbb{N}$. Then $$ \iota(m\cdot n) = (m,0)\cdot (n,0) = (m\cdot n + 0\cdot 0, m\cdot 0 + 0\cdot n) = (m\cdot n, 0) = \iota(mn). $$

To summarize, if $n$ is a natural number, then there is a natural or canonical way of equating $n$ to an integer—specifically, identify $n$ with the equivalence class of $(n,0)$. This identification preserves addition and multiplication, which means that, for all intents and purposes, we can treat the equivalence class of $(n,0)$ as though it were the same thing as $n$. These two things are not equal, but they behave the same way, so we can kind of ignore the fact that they aren't quite equal.

The notation for this is $n \equiv [(n,0)]$ (or, in Tao's notation, $n \equiv n \mathbin{—} 0$).

References?

This is not a topic which I learned out of a book—I originally learned it out of Donald Pfaff's lecture notes when I took a class on set theory from Dr Pfaff when I was an undergraduate. My impression is that this is, essentially, how most people learn the topic—everyone who teaches a class in which this material comes up has their own set of lecture notes, which they pass on to their students. At least, that is what a quick Google search seems to indicate—there are lots and lots and lots of lecture notes.

Looking through the books on my shelf, the only one I can find which seems to cover the topic at an elementary level is

  • Hamilton, Norman; Landin, Joseph, Set theory and the structure of arithmetic, 3rd printing. Boston: Allyn and Bacon, Inc. XI, 264 p. (1963). ZBL0127.27901.

I have neither studied, nor taught, from this text, so I cannot really vouch for it (it is one of the books I picked up when some professor or another retired), but a quick skim of the relevant chapter convinces me that it is probably a totally reasonable text.

  • Thanks!... I'm studying Analysis by my own and I have few preliminaries of equivalence relations, and by the way of tacle my question, I see that is necessary to have a some strong background of equivalence relations and concepts related. – José Guzmán Jul 13 '23 at 14:05
  • Furthermore, I don't understand why Tao establishes the equality. Here is the part of the paragraph of this fact: "For instance the natural number $3$ is now considered to be the same as the integer $3—0$, thus $3 = 3—0$. Of course, if we set $n$ equal to $n—0$, then it will also be equal to any other integer which is equal to $n—0$." – José Guzmán Jul 13 '23 at 15:34
  • The point is that $3$ (the natural number) and $[(3,0)]$ (the integer represented by $(3,0)$) behave identically. You can add it to other numbers, and multiply it by other numbers. The two objects are equivalent (i.e. $3 \equiv [(3,0)]$), so Tao takes the extra step and says "Hey, since they are basically the same object, let's ignore the fact that they are different, and say that they are not just equivalent, but equal". – Xander Henderson Jul 13 '23 at 15:37