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My interpretation of the sub-paragraph in the book which I am asking about is this:


BEGIN SUMMARY

Note that in this context 0 is not considered to be a natural number.

The pairs of natural numbers $\left\langle a,a^{\prime}\right\rangle \in\mathbb{N}\times\mathbb{N}$ representing integers are transformed by the mapping $\left\langle a,a^{\prime}\right\rangle \mapsto a-a^{\prime}\in\mathbb{Z}.$

The pairs of integers $\left\langle \alpha,\alpha^{\prime}\right\rangle \in\mathbb{Z}\times\left(\mathbb{Z-}\left\{ 0\right\} \right)$ representing rational numbers are transformed by the mapping $\left\langle \alpha,\alpha^{\prime}\right\rangle \mapsto\alpha/\alpha^{\prime}\in\mathbb{Q}.$

The elements $c\in C\subset\mathbb{N}$ consist of all “stand-alone” natural numbers appearing in equations such as $c=\left(c+a^{\prime}\right)-a^{\prime}=a-a^{\prime}$ necessary for the proofs. In particular, all of those needed for the application of the cancellation law $x+c=y+c\iff x=y$; provided that the subtrahends $a^{\prime}$ in the corresponding $\left\langle a,a^{\prime}\right\rangle$ are also $a^{\prime}\in C$. We must also require $a^{\prime},b^{\prime}\in C\implies a^{\prime}+b^{\prime}\in C.$

When we introduce the neutral element $a-a^{\prime}\equiv0\iff a=a^{\prime},$ this requirement $a\in C$ must be extended to include the minuends $a$ of $\left\langle a,a^{\prime}\right\rangle$.

These restrictions must not impact the validity of the proofs regarding the arithmetic properties of all integers, and will exactly correlate to a subset $\mathcal{Z}\subset\mathbb{Z}$ such that $\alpha^{\prime}=0\notin\mathcal{Z}$ for all denominators in $\alpha/\alpha^{\prime}$ where $\left\langle \alpha,\alpha^{\prime}\right\rangle \in\mathbb{Z}\times\left(\mathbb{Z-}\left\{ 0\right\} \right);$ and the cancellation law is $\left(\xi\ne0\land\xi\left(\alpha-\beta\right)=0\right)\iff\alpha=\beta.$

This will enable us to immediately apply all of the results established in deriving the laws of integer arithmetic to rational number multiplication.

Apparently, this means there will be some element or equivalence class in $\overline{C}\equiv\mathbb{N}-C$ corresponding to $0\in\mathbb{Z}$ under the respective mappings $\left\langle a,a^{\prime}\right\rangle \mapsto a-a^{\prime}\in\mathbb{Z}$ and $\left\langle \alpha,\alpha^{\prime}\right\rangle \mapsto\alpha/\alpha^{\prime}\in\mathbb{Q}.$

END SUMMARY


The book is now available on archive.org so if you are interested in looking at the original discussion, here it is:

Behnke, Bachmann, Fladt & Suss [Eds.] - Fundamentals of Mathematics Vol.1, The Real Number System and Algebra[MIT Press, 555p]

Please be aware that I have read through the relevant material multiple times. I am not asking the easy questions, such as why is $0$ excluded as a second element in the pairs representing rational numbers? The restriction discussed below is not logically necessary for the development of integer arithmetic. It is placed on that development so that the results are applicable to the development of rational number arithmetic without exception.

My question is: what does the quoted text mean? It is evident that in the introduction of integers the authors want to avoid using combinations of natural numbers and ordered pairs of natural numbers which would be analogous to dividing by zero in the introduction of the rational numbers. But it is not clear what, exactly the set $C$ is to be.

The following is from BBFSK V-I, page 108, etc.:

Properties (6),(7),(11) are respectively the associative, commutative, and cancellation laws of natural number addition. Natural numbers are defined with $1$ as the non-successor.

It is to be noted that for the following developments we require only the properties (6),(7),(11) of the natural numbers and their addition. This fact will become important in section 3.1, where the procedure described here is applied to multiplication instead of addition (with $a/a^{\prime}$ instead of $a-a^{\prime}$). Since in that section we shall be introducing only pairs $\left(a,a^{\prime}\right)$ with $a^{\prime}\ne0,$ we make the further remark that in what follows (as can easily be seen from the proofs) the rule (11) is needed only for natural numbers $c$ restricted to a proper subset $C,$ provided that the pairs $\left(a,a^{\prime}\right)$ are restricted to $a^{\prime}\in C$ and for $a^{\prime},b^{\prime}\in C$ we have $a^{\prime}+b^{\prime}\in C;$ the only exception is the proof of the existence of the inverse element at the beginning of section 2.3, where we must also require $a\in C.$

Since this paragraph refers to several pages of text, it is difficult to provide as much context as I would like. The section where the paragraph appears presents the extension of the domain of natural numbers to the domain of integers. In this section the individual elements of the pairs $\left(a,a^{\prime}\right)$ are natural numbers. I am guessing the reference to natural numbers $c\in C$ means natural numbers appearing as terms in expressions involving both integers $\left(a,a^{\prime}\right)\mapsto\left(a-a^{\prime}\right)$ and "stand-alone" natural numbers.

Integers are introduced to provide solutions to equations of the form

$$ a=c+a^{\prime}\text{ where }a\le a^{\prime}. $$

A relation between pairs is defined as

$$ \left(a,a^{\prime}\right)\bumpeq\left(b,b^{\prime}\right)\iff a+b^{\prime}=a^{\prime}+b. $$

Reflexivity and comparativity

$$ \left(a,a^{\prime}\right)\bumpeq\left(a,a^{\prime}\right), $$

$$ \left(a,a^{\prime}\right)\bumpeq\left(c,c^{\prime}\right)\land\left(b,b^{\prime}\right)\bumpeq\left(c,c^{\prime}\right)\implies\left(a,a^{\prime}\right)\bumpeq\left(b,b^{\prime}\right), $$

are then shown. Reflexivity is obvious. and comparativity is show as follows:

$$ \left(a,a^{\prime}\right)\bumpeq\left(c,c^{\prime}\right)\iff a+c^{\prime}=a^{\prime}+c\text{ and,} $$

$$ \left(b,b^{\prime}\right)\bumpeq\left(c,c^{\prime}\right)\iff b+c^{\prime}=b^{\prime}+c $$

$$ \implies a+c^{\prime}+b^{\prime}=a^{\prime}+c+b^{\prime} $$

$$ =a^{\prime}+b+c^{\prime}=a^{\prime}+b^{\prime}+c $$

$$ \implies a+b^{\prime}=a^{\prime}+b\text{ [by (11)]}. $$

Equality between integers is defined by

$$ a-a^{\prime}=b-b^{\prime}\iff a+b^{\prime}=a^{\prime}+b. $$

Integer addition is defined by

$$ \left(a-a^{\prime}\right)+\left(b-b^{\prime}\right)=\left(a+b\right)-\left(a^{\prime}+b^{\prime}\right). $$

Equality between natural numbers and (positive) integers is defined by

$$ \left(c+a^{\prime}\right)-a^{\prime}=c\text{ and }c=\left(c+a^{\prime}\right)-a^{\prime}. $$

The authors prove the consistency of this definition of equality by explicitly showing that comparativity

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

holds for all possible combinations of integers and natural numbers.

Case: $\alpha=a,\beta=b-b^{\prime},\gamma=c-c^{\prime};$

$$ \alpha=\gamma\implies c=a+c^{\prime}\text{ and }\beta=\gamma\implies b+c^{\prime}=b^{\prime}+c $$

$$ \implies b+c^{\prime}=b^{\prime}+a+c^{\prime} $$

$$ \implies b=b^{\prime}+a\text{ [by (11)]} $$

$$ \implies a=b-b^{\prime}. $$

Case: $\alpha=a-a^{\prime},\beta=b-b^{\prime},\gamma=c;$

$$ \alpha=\gamma\implies a=c+a^{\prime}\text{ and }\beta=\gamma\implies b=c+b^{\prime} $$

$$ \implies a+c+b^{\prime}=b+c+a^{\prime} $$

$$ \implies a+b^{\prime}=a^{\prime}+b\text{ [by (11)]} $$

$$ \implies a-a^{\prime}=b-b^{\prime}. $$

Case: $\alpha=a,\beta=b,\gamma=c-c^{\prime};$

$$ \alpha=\gamma\implies c=a+c^{\prime}\text{ and }\beta=\gamma\implies c=b+c^{\prime} $$

$$ \implies a+c^{\prime}=b+c^{\prime} $$

$$ \implies a=b\text{ [by (11)]}. $$

Case: $\alpha=a,\beta=b-b^{\prime},\gamma=c;$

$$ \alpha=\gamma\implies a=c\text{ and }\beta=\gamma\implies b=c+b^{\prime} $$

$$ \implies b=a+b^{\prime}\implies a=b-b^{\prime}. $$

And finally, the neutral element is defined by $0\equiv a-a,$ where $a\in\mathbb{N}.$ It is from this that they arrive at

$$\left(a-a^\prime\right) + \left(a^\prime-a\right)=0,$$

which is the closest thing they have to a definition of the inverse of an integer. All other discussion of the inverse deals with Abelian groups in general.

In the section introducing rational numbers, the elements of $\left(a,a^{\prime}\right)$ may be any integer with the restriction that $a^{\prime}\ne0$.

Perhaps there is a hint in the observation that addition in $\mathbb{N}$ is not a surjective self-mapping. That is $\forall_{a,b}\left[a+b\ne 1\right],$ for example.

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  • $\begingroup$ The proper subset $C$ where $c∈C$ refers to a proper subset of natural numbers which include only "stand-alone" natural numbers which does not include the "extended" symbols of the form $a - a', b - b'$ that were introduced to solve $x + a' = a$ and such. With reference to (11) what we have is,$a = b$, if $a + c = b + c$, where $a,b,c \in N$. What the author claims I think is that this is required only for the "stand-alone" natural numbers and not for $a-a'$ kind. But, for inverse you need to include the whole set and atleast one extra symbol '$0$' or like $a-a$. since $a-a=b-b$ from sec.2.3 $\endgroup$ – Gopal Anantharaman May 31 at 22:45
  • $\begingroup$ Lord Shark the Unknown cannot answer this question. $\endgroup$ – Steven Thomas Hatton Jun 1 at 17:34
  • $\begingroup$ I'm pretty sure that $C$ is never to be extended to include $0$. There will, however, be some $z\in\mathbb{N}-C$ corresponding to $0\in Q$. I added a summary to the original question. $\endgroup$ – Steven Thomas Hatton Jun 1 at 19:25
  • $\begingroup$ I suspect you have an in-joke with Shark? Also +1 purely for the sheer effort involved in assembling this question. $\endgroup$ – The Count Jun 2 at 16:04
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It looks like a considerable investment of time and eyeballs would be required to get through the archive.org book/photo images and give the exact/detailed answer the OP is looking for. But we'll skim over it and make some statements/comments that might be helpful.

Idea 1: When you apply the construction in the book,

$$(\Bbb N^{ \gt 0},+) \mapsto (\Bbb Z,+)$$

you prove that $(\Bbb Z,+)$ is a group with a natural injective morphism

$$(\Bbb N^{ \gt 0},+) \hookrightarrow (\Bbb Z,+)$$

So you don't have to prove that (11) (the cancellation law) is true, since it is true for any group.

I'm fairly confident that the construction transforms any commutative semigroup $C$ into a group $G$. But in general, $C$ might 'collapse' so there is no injection from $C$ into $G$.

Another example of a commutative semigroup are the integers with $0$ removed, $(\Bbb Z^*,\times)$;
it satisfies (11).

Idea 2: One thing that I found interesting/puzzling in the book was the introduction of

$\text{(32)} \quad (c + a') - a' = c\text{,} \quad c = (c + a') - a'$

It looks like the authors are using combinatorial/symbolic arguments to cut down on the checking, and if you understand $\text{(32)}$ you are following the argument better than I am. I would guess that with $\text{(32)}$ they are selecting the forms to represent a natural number $c$ from the equivalence relation block.


When you are examining only positive numbers $P$ you have fewer axiomatic properties to check if you want to extend them. One powerful axiom is

$$\tag 1 \text{For every } x,y,z \in P \text{, if } z = x + y \, \text{ then } \, z \ne x $$

So if you want to 'go from' $\Bbb N^{ \gt 0}$ to the positive rational numbers there is less number property checking to be concerned with if you don't have an additive identity. And then you can get all of $\Bbb Q$ with your 'number property' checklist.

The standard approach to constructing the integers $\Bbb Z$ is found here, where you start with the natural numbers $\Bbb N^{ \ge 0}$ with both addition and multiplication.

But the same technique can be applied to the counting numbers $\Bbb N^{ \gt 0}$ with only addition, extending it to the integers with addition defined. You can easily check the technique does not alter the validity of the properties (6),(7) and (11). When you do it this way you get, free-of-charge, the additive identity to go along with the fact that you can now solve equations of the form (22),

$$ x + a' = a$$

When you solve the equation $x + a = a$ you will get the same answer every time, and you call it $0$.

In section 2.4 of the book they take the next step, defining multiplication on $\Bbb Z$. Later they construct the rational numbers using the same technique.

Since the authors know that properties (6),(7) and (11) get 'passed through' the construction, they won't recheck them again. But they will have to check that multiplication still distributes over addition in $\Bbb Q$.

The OP mentioned that over $\Bbb N$, $\forall_{a,b}\left[a+b\ne 1\right]$:

Proof: By $\text{(1)}$, $\;1 \ne 1 + b$ and by definitions of the inequality $1 \lt 1 + b$. Since $1$ is the minimum element, $1 \le a$. So

$$ 1 \lt 1 + b \le a + b$$

and $a + b \ne 1$.

In general, before extending your number system, you list out all the core/independent axioms that describe the system up to isomorphism. Then, when you construct as extension, using say an equivalence relation on a subset of a Cartesian product, you drop some of the axioms or 'revamp them' and use your logic flowchart to see what properties are satisfied in the extension.

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  • $\begingroup$ I've already gone through the development of the real numbers. This is the only part of the chapter that I am still unsure of. These are my (work in progress) notes drive.google.com/open?id=1HRn2OJJV8OiJNBvDBTpsSjDL0VHdLvbE $\endgroup$ – Steven Thomas Hatton Jun 2 at 15:21
  • $\begingroup$ If my answer doesn't add anything I'll be happy to delete it... $\endgroup$ – CopyPasteIt Jun 2 at 15:26
  • $\begingroup$ You can also work out your own theory that you have total confidence in. You might have to redundantly prove some number properties - take it as your medicine. Then as time permits go back to the book material and notes. It may make perfect sense! $\endgroup$ – CopyPasteIt Jun 2 at 15:32
  • $\begingroup$ In BBFSK the authors outline multiple approaches to different stages in constructing the real number system, and compare their merits and consequences. Near the end of the chapter they discus the approach you suggested as a way of simplifying the remaining proofs. Regarding my question, I would never suspect a need for this exception had they not mentioned it. The chapter has no exercises, per se. The exercises are implied by such comments as the sub-paragraph in question. Mine is not a question of having a development I fully understand. I want to fully understand this development. $\endgroup$ – Steven Thomas Hatton Jun 3 at 16:05
  • $\begingroup$ It may interest you to know that the chapter under discussion takes a very similar overall philosophy to the approach you are advocating when extending from one domain to the next. I would be interested in any suggestions you have for books or other discussions of number theory you may have to offer. You seem to know what you are doing. $\endgroup$ – Steven Thomas Hatton Jun 3 at 18:13

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