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i don't have much background in set theory and mathematical logic besides isomorphisms thus i can't quite understand(justify) the way of "identifying" integers with naturals in Tao's analysis.

That's how i interpret what i have read so far about integers: he constructs integers from naturals(integers are elements of the set $N×N$ so they are not the same objects as naturals he defined earlier, he uses a notation $(a,b):=a-b$ for them). He defines equality $"="$ relation between integers(based on equality between naturals), and operations of additions and multiplication for integers(again in terms of natural numbers) After that I have a problem with understanding his next paragraph:

The integers $n−0$ behave in the same way as the natural numbers n; indeed one can check that $(n−0) + (m−0) = (n + m)−0$ and $(n−0) × (m−0) = nm−0$.

I know it should mean something like this$:$ if we map $(n,0)$ with $n$, then we have a function $f:A⊆N×N↦N$, where $A$ consists of integers of the form $(n,0)$ that has properties that if $a+b=c$ ($"+"$ and $"="$ are those defined for integers), then $f(a)+f(b)=f(c)$($"+"$ and $"="$ are those defined for naturals) and if $a×b=c$ then $f(a)×f(b)=f(c)$(same thing with $"×"$ and $"="$)

Furthermore, $(n−0)$ is equal to $(m−0)$ if and only if $n = m$.

I guess that means that $f$ is injection. Though it's a surjection too.

(The mathematical term for this is that there is an isomorphism between the natural numbers $n$ and those integers of the form $n−0$). Thus we may "identify" the natural numbers with integers by setting $n ≡ n−0$;

From here it begins: What exactly does $"≡"$ sign mean? Does it stand for my function $f$? Or is it some new relation for integers like our already defined relation $"="$ but he just doesn't want to overload the sign $"="$ or something?

this does not affect our definitions of addition or multiplication or equality since they are consistent with each other. Thus for instance the natural number $3$ is now considered to be the same as the integer $3−0: 3 = 3−0$.

Hey now i wonder what $"="$ sign means, because we have $"="$ for naturals, $"="$ for integers but we don't have $"="$ for integer-naturals(did he define it implicitly?)(is it the same sign as $"≡"$?)

In particular $0$ is equal to $0−0$ and 1 is equal to $1−0$. Of course, if we set $n$ equal to $n−0$, then it will also be equal to any other integer which is equal to $n−0$, for instance $3$ is equal not only to $3−0$, but also to $4−1$, $5−2$, etc.

We can now define incrementation on the integers by defining $x++ := x + 1$ for any integer $x$; this is of course consistent with our definition of the increment operation for natural numbers.

How does this operation work? It looks like it has only one argument from $N×N$ but then computing an output it uses $1$ from $N$ so operation $"+"$ has one argument from $N×N$ and the other from N and how it supposed to react to this?!

So basically all my questions are about what we can do with isomorpisms and why we can do it.

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Your basic understanding is correct. In your paragraph under the first colored box we need $f$ to be a bijection. He claims that in the second colored box with the if and only if. The sign $\equiv$ is read "equivalent to". Once you have shown that $f$ is an isomorphism we can consider the two things related by $f$ as the same for whatever purpose we have in mind. Constructing the integers as ordered pairs of naturals is formally nice, but representing the integers as equivalence classes of naturals is very clumsy. We would like to get back to the notation we are used to with the positive integers and zero using the same symbols as naturals and the negative integers using the naturals with a minus sign prefixed. The equivalence sign shows which integer as an ordered pair corresponds to which natural. In the part you quote he never points out that the integers as ordered pairs are really equivalence classes of ordered pairs, though it is implied when he talks about $3-0$ being equal to $4-1, 5-2, $etc. I am sure that point is made in the article. I am not crazy about writing $3=3-0$ as he does because (as you say) this is asserting equality between two different sorts of objects. What he has done is give the traditional names to the integers as single numbers which may be preceded by a minus sign instead of the equivalence classes of ordered pairs. He asserts that the formal incrementation operator works on the integers just as you would expect. The basic point is that once you have an isomorphism you can think of it as having two different descriptions of the same object. You have the formally defined integers as equivalence classes and you have the informally defined integers as the naturals plus the negatives. He is showing that the informally defined ones with the rules we are used to work the same as the formally defined ones, then will say that it is much easier to write the informally defined ones so we will use that notation in the future.

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  • $\begingroup$ Yes, you are right, i have omitted that $a−b$ is the space of all pairs equivalent to $(a, b): a−b := {(c, d) ∈ N × N : (a, b) ∼ (c, d)}$. And i just now realized, that a-b is a set, not an object.(The thing is, Tao decided to distinguish some sets and objects, and functions can only map objects from some sets to objects in other sets, so how do we define addition and multiplication on sets? Because the way that he defines them needs some correction i guess >The sum of two integers, $(a−b) + (c−d)$, is defined by the formula $(a−b) + (c−d) := (a + c)−(b + d)$. $\endgroup$ – famesyasd May 6 '17 at 16:02
  • $\begingroup$ does it get to notify you if i don't mention your name in my comment? also i typed your name in comment and after that it disappeared, weird @Ross Millikan $\endgroup$ – famesyasd May 6 '17 at 16:56
  • $\begingroup$ @famesyasd: I get notified if you comment on one of my posts. You can also notify somebody with at before their user name like I did here for you. I do that to respond to someone in a comment chain who is not the poster of the answer or question. I think you would have been notified anyway because it is your question. You can only use one at per comment. $\endgroup$ – Ross Millikan May 6 '17 at 19:30
  • $\begingroup$ What makes this hard is we all know where this is going. We learn about naturals, integers, rationals and reals before we learn the word axiom. We all know the naturals are a subset of the integers, but in the formal construction they are not because the integers are equivalence classes of pairs of naturals. Then the integers are not a subset of the rationals because the rationals are equivalence classes of pairs of integers. We put a lot of work into the construction, then throw it all away and use the integers and rationals we are used to. $\endgroup$ – Ross Millikan May 6 '17 at 19:35
  • $\begingroup$ The point of the isomorphism discussion is to define a subset of the integers which we can consider the naturals. We can then use the naturals for those equivalence classes of ordered pairs of naturals. This suggests the usual notation for the negative integers and we are back to the notation we grew up with. $\endgroup$ – Ross Millikan May 6 '17 at 19:37
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Lots of questions here. I will suggest an answer to the last one

So basically all my questions are about what we can do with isomorphisms and why we can do it.

in hopes that it helps with the rest.

What Tao is doing is to show you how to construct formally what you intuitively understand as the integers $\{ \ldots, -3, -2, -1, 0, 1, 2, \ldots \}$ assuming that you know everything necessary about the natural numbers $\{0, 1, 2, \ldots \}$,

Since he wants to write with formal set theory vocabulary, unravelling the notation can be challenging.

The idea he wants to capture is that a "missing" negative number like $-5$ can be defined by the expression "$2-7$" even though that expression has no meaning in the natural numbers. But you must be careful, because "$2-7$" and "$96-99$" and "$0-5$" all capture the essence of the missing $-5$. So he tells you just when two of those expressions (each constructed from a pair of natural numbers) should count as the same integer, using the ordinary arithmetic properties of the natural numbers. The $\equiv$ sign between two such expressions says the represent that "same integer". In formal terms, $\equiv$ is an equivalence relation and Tao defines the integer represented by any of those pairs as the equivalence class - the set of all the pairs. It's how you might define the rational numbers once you know the integers as pairs of integers, where $(1,2)$, $(2,4)$ and $(75, 150)$ all represent the same rational, one half. (Tau may well do this next.)

Having done this and checked all the arithmetic facts about these new things called "integers" using only the properties of the natural numbers. he wants to back away from the formal structure. To that end he shows you that there's a faithful copy of natural numbers inside the "integers" he's constructed. That's the essence of the function $f$.

Once that's done you can forget that $5$ is (formally) the set of all the pairs that are equivalent ($\equiv$) to $(5,0)$ and that $-5$ is (formally) the set of all the pairs that are equivalent ($\equiv$) to $(0,-5)$. Then can go about business as usual with $\{ \ldots, -3, -2, -1, 0, 1, 2, \ldots \}$.

Edit:

In a comment you ask if this is "overloading" the natural numbers. That's a formal term from computer science, describing a situation where (say) the meaning of an operator symbol like "$+$" depends on the context. That operator is overloaded here, since it's used both for adding natural numbers and for adding the integers defined as sets of pairs of natural numbers. I'm not sure whether you'd say the natural numbers are themselves overloaded. It's more the opposite: two different representations of the same thing. The $5$ and the $0$ in the integer represented by $(5,0)$ are natural numbers. When you identify that pair with $5$ the symbol "5" stands for both $5$ and the equivalence class of $(5,0)$. Since the embedding $f$ is injective you won't get into trouble reusing the name. You've actually done this kind of thing before. When you work with polynomials you unthinkingly use the natural embedding of the integers into the ring of polynomials. "5" can mean the constant polynomial $5$ or the coefficient in the polynomial $5x$.

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    $\begingroup$ Shouldn't $-5$ be $2-7$? (etc) $\endgroup$ – ancientmathematician May 6 '17 at 15:37
  • $\begingroup$ @ancientmathematician yes it should, but whatever :p $\endgroup$ – famesyasd May 6 '17 at 16:08
  • $\begingroup$ @Ethan Bolker >The $≡$ sign between two such expressions says the represent that "same integer". Do you mean by $"≡"$ here actually $"=" $ defined for integers in terms of natural numbers, right? And then you refer to it in >$5$ is (formally) the set of all the pairs that are equivalent $≡$ to $(5,0)$ have you overloaded it for natural-integers? or does isomorphism allow some sort of substitution, is there some kind of axiom for that? i think i grasped the idea, but i want to understand on formal level why we can rename integers as $5$, $-5$ due to that isomorphism. $\endgroup$ – famesyasd May 6 '17 at 16:16
  • $\begingroup$ @ancientmathematician Fixed, thanks. You could have edited yourself and made the correction. $\endgroup$ – Ethan Bolker May 6 '17 at 18:09
  • $\begingroup$ @famesyasd See my edit on overloading. $\endgroup$ – Ethan Bolker May 6 '17 at 18:09

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