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My source BBFSK

I need to add that natural numbers in this context are defined as starting with 1. I didn't think that would impact the answer, but apparently it does. $n-0$ provides a "bridge" between the integers and the natural numbers.

This question pertains to the extension of the natural numbers to the integers. The expression $a-\bar{a}$ is not yet called subtraction.

Given the arbitrary natural numbers $a$ and $\bar{a},$ the expression $a-\bar{a}$ is defined to be the solution of the equation

$$ a=x+\bar{a}, \tag{1} $$

and is called an integer. In the case where $a>\bar{a},$ we could just as well say that $x$ is a natural number. We were able to do that before contemplating integers. It was the case where $a<\bar{a}$ that induced us to extend the domain of numbers.

Our goal is to show that certain integers are equal to natural numbers. To this end we define equality between integers and natural numbers by setting \begin{equation} \left(c+\bar{a}\right)-\bar{a}=c\text{, and }c=\left(c+\bar{a}\right)-\bar{a}, \tag{2} \end{equation}

with the stipulation that these and only these equations are to hold between natural numbers and integers. Reflexivity remains unaffected by this definition, but to show comparability, that is

$$ \alpha=\gamma\land\beta=\gamma\implies\alpha=\beta, $$

we must show it holds for all cases of $\alpha,\beta,\gamma$ as integers or natural numbers. In total there are eight such cases. The two cases in which all three variables are of the same kind have already been dispensed with. Due to the defined symmetry of (2) we need not treat the two cases which arise from another by substituting $\alpha$ with $\beta$.

The first of the remaining four cases is $$ \alpha=a\in\mathbb{N},\beta=b-\bar{b},\gamma=c-\bar{c}\in\mathbb{Z}. $$

This is where things get a bit murky for me.

First we treat $\alpha=\gamma,$ which written explicitly is $a=c-\bar{c}.$ But we are required to somehow relate this to (2). So, perhaps we should write $a=\left(a+\bar{c}\right)-\bar{c}.$ Here I'm not sure what is legal. It seems that since we have asserted $a$ to be a natural number, we should be free to set $$ \left(a+\bar{c}\right)-\bar{c}=c-\bar{c}. $$

Since both sides are known to be natural numbers dressed up like integers. Is that permissible?

Also, since $a=c-\bar{c}$ is defined to be the solution of the equation $c=a+\bar{c},$ and $a$ is a natural number, it follows from the definition of $>$ for natural numbers that $c>\bar{c}.$ It therefore seems reasonable that we could immediately conclude that $a=c-\bar{c}\in\mathbb{N}\implies c=a+\bar{c}$ without explicitly appealing to (2). But that feels like cheating. Is it cheating?

The gist of my question seems to boil down to: given the definition (from the book) of equivalence between an integer and a natural number:

$$\left(c+\bar{a}\right)-\bar{a}=c\text{, and }c=\left(c+\bar{a}\right)-\bar{a},$$

how does the result (from the book) $a=c+\bar{a}$ follow?

Notice that $a$ does not appear in the above definition. The only clue I have as to where $a$ comes from is that every prior appearance of $\bar{a}$ is in the pair $\left(a,\bar{a}\right)$ used in the mapping $\left(a,\bar{a}\right)\to{a-\bar{a}}.$ Apparently I am expected to look at $\left(c+\bar{a}\right)$ and see $a$.

The nature of BBFSK is that such conclusions are often left to the reader.

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  • $\begingroup$ Re: $0$, see my edit. $\endgroup$ – Noah Schweber Jan 21 at 18:23
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EDIT: The issue with $0$ is trivial. If we don't want to consider $0$ as a natural number, just replace $(n,0)$ with $(n+1, 1)$ everywhere below. $0$ provides a convenient, but inessential, bridge from $\mathbb{N}$ to $\mathbb{Z}$, and we can easily do without it.

(In my opinion we shouldn't omit $0$ - in particular I don't think history always matches up with the ideal, and this is a case in point: although $0$ was a relatively late introduction, I think it definitely should be included in the naturals - but this doesn't matter mathematically.)


Aside: I'm going to phrase things in terms of reflexivity-symmetry-transitivity rather than reflexivity-comparativity, since that's the more common language in my experience.

I think this is made more mysterious than it needs to be due to the phrasing of the definitions. Ultimately you're overloading the symbol "$=$," and this leads to a sense of slipperiness.

Instead, the right approach in my opinion is (something like) the following:

  • First, we define a pre-integer to be a pair $(a,b)\in\mathbb{N}$. Intuitively my $(a,b)$ is your $a-b$, and my pre-integers are your integers, but $(i)$ using ordered pairs avoids suggestive (and worrying) notation and $(ii)$ I'm going to reserve "integer" for the resulting equivalence classes (see next point).

  • Second, we define an equivalence relation between pre-integers as follows: $$(a,b)\sim (a',b')\iff a+b'=b+a'.$$ This avoids overloading "$=$," and puts everything in terms of already-understood operations on naturals (so no mixing naturals and integers as such). We now say that integers are $\sim$-classes of preintegers.

    • ... Of course, that presupposes that $\sim$ is in fact an equivalence class, but that's easy to prove:

      • Reflexivity: If $a=a'$ and $b=b'$ then $a+b'=a+b=a'+b$.

      • Symmetry: If $a+b'=a'+b$ then $a'+b=a+b'$.

      • Transitivity (the only nontrivial one, and a large part of what you're asking): Suppose $(a,b)\sim(a',b')$ and $(a',b')\sim (a'',b'')$; we want to show $(a,b)\sim (a'', b'')$, which amounts to showing that $a''+b=a+b''$. To do this, consider the (trivially true) equation $$a+a'+a''+b+b'+b''=a+a'+a''+b+b'+b''.$$ This converts to $$(a+b')+(a'+b'')+(a''+b)=a+a'+a''+b+b'+b'',$$ and we can simplify the left hand side using our assumptions to get $$(a'+b)+(a''+b')+(a''+b)=a+a'+a''+b+b'+b''$$ or better yet $$[a'+a''+b+b']+(a''+b)=[a'+a''+b+b']+(a+b'').$$ Now cancelling appropriately - remembering that cancellativity is a property of natural numbers which we already know, despite its "subtraction-y" flavor - we get $$a''+b=a+b'',$$ which is what we wanted. (Also note that I've used associativity of addition of naturals a couple times above; again, this is fine since associativity of addition of naturals is something we've presumably already established.)

  • Finally, again avoiding overloading "$=$" we say that a natural $n$ is equivalent to the integer $[(a,b)_\sim]$ if and only if $(a,b)\sim(n,0)$ (technically I need to show that this definition is well-defined, but that's trivial). The point is that your eight cases don't emerge since we've defined the embedding of naturals into integers after establishing the integers "on their own."

  • And the approach above also makes it completely straightforward to compare operations on integers with the corresponding operations on naturals. For example, we define addition of pre-integers as $(a,b)\underline{\oplus} (c,d)=(a+c,b+d)$. We can show quite easily that if $(a,b)\sim (a',b')$ and $(c,d)\sim(c',d')$ then $(a,b)\underline{\oplus}(c,d)\sim(a',b')\underline{\oplus}(c',d')$, and so $\underline{\oplus}$ extends to a well-defined binary operation $\oplus$ on integers. Basic properties of $\oplus$ are now easy to establish, e.g. commutativity comes from the "mod $\sim$" version of the trivial pre-integer identity $$(a,b)\underline{\oplus}(c,d)=(a+c,b+d)=(c+a,d+b)=(c,d)\underline{\oplus}(a,b).$$

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  • $\begingroup$ Actually, my source claims we only need reflexivity and comparativity, to establish the traditional three properties of an equivalence relation. I will need to sleep on the rest of it. My overall approach is similar to yours. I used $\mu\left[a,\bar{a}\right]$ for $a-\bar{a}$. $\endgroup$ – Steven Hatton Jan 20 at 17:57
  • $\begingroup$ @StevenHatton That's true, a relation satisfying comparativity (I've never heard it called that, but you're right it is not the same as transitivity) and reflexivity is an equivalence relation: symmetry holds because, if $aEb$, then by reflexivity we have $bEb$ and $aEb$, so by comparativity we get $bEa$, and symmetry + comparativity clearly gives you transitivity. $\endgroup$ – Noah Schweber Jan 20 at 18:02
  • $\begingroup$ @StevenHatton Yes, the approach I've outlined above is essentially the same as yours, but as I said I think the notational shift - in particular, avoiding overloading "$=$" from the start - makes things much clearer. $\endgroup$ – Noah Schweber Jan 20 at 18:03
  • $\begingroup$ (Incidentally, the proof that comparativity + reflexivity implies transitivity does in fact require reflexivity. Consider the relation on $\{1,2, 3\}$ consisting of: $1E1, 1E2, 2E1, 2E2, 2E3$. Then this is clearly nontransitive since $1E2, 2E3$, but $1\not E3$, yet it is comparative: the only way $\alpha,\beta,\gamma$ could be a failure of comparativity would be if at least one of $\alpha,\beta$ were $3$, but then no $\gamma$ would have $\alpha E\gamma$ and $\beta E\gamma$.) $\endgroup$ – Noah Schweber Jan 20 at 18:12
  • $\begingroup$ My approach which I was comparing to yours is the one in my notes. I use $\mu\left[a,\bar{a}\right]$ in place of $a-\bar{a}$ to keep my mind form "pre-processing" the notation. The book, and I also use ordered pairs with your equivalence relation. I use over-bars to avoid confusion with successor notation from the previous section. The book uses primes, as do you. I failed to specify $\mathbb{N}=\left\{1,2,\dots\right\},$ which precludes $\left(n,0\right)$. The book calls the integers "residue classes" of $\sim$. I like the organization of your presentation. $\endgroup$ – Steven Hatton Jan 21 at 13:48
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I think this is the correct answer.

In order to keep track of what is certainly a natural number and what may or may not be a natural number I use a Roman font for the unknown in $a=\mathrm{c}+\bar{a},$ where the relative order of $a$ and $\bar{a}$ is unspecified. We have defined the solution to be $\mathrm{c}\equiv a-\bar{a},$ which is an element of our extended domain.

We assert that our commutative, associative and cancellation laws of addition developed in $\mathbb{N}$ are valid in the extended domain. Applying these to a bit of minor algebraic acrobatics give the following

$$a=\mathrm{c}+\bar{a}=\left(a-\bar{a}\right)+\bar{a}\text{, and}$$

$$\mathrm{c}=\left(\mathrm{c}+\bar{a}\right)-\bar{a}=a-\bar{a}.$$

Which includes the form used in the definition of equality between integers and natural numbers. But in the case where $c$ is a natural number, we don't use the Roman font. In this case all of the variables are natural numbers. Beginning with the definition and working backwards

$$c=\left(c+\bar{a}\right)-\bar{a}=a-\bar{a}\text{, and}$$

$$a=c+\bar{a}\text{. So}$$

$$c=\left(c+\bar{a}\right)-\bar{a}\iff c=a-\bar{a}\iff a=c+\bar{a}.$$

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