Just a definition by a mathematician? But there should be some good reasons for adopting the definition.
Topics
About: junior high school mathematics
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Why Are Negative Numbers Required?
- 2: Let Us Construct the Integral Numbers Set
- 3: Some Notes
- 4: In the Axiomatic Theory, the Arithmetic Is Just Defined So, but
- 5: Addition of Integral Numbers
- 6: Subtraction of Integral Numbers
- 7: Multiplication of Integral Numbers
- 8: Division of Integral Numbers
- 9: Order of Integral Numbers
Starting Context
- The reader knows the background of this site.
Target Context
- The reader will know the reason why negative numbers are required, what integral numbers really are, the reasons for the arithmetic of integral numbers, and the reason for the order of integral numbers.
Orientation
There is an article on becoming a benefactor of humanity by being a conduit of truths
Main Body
1: Why Are Negative Numbers Required?
Special-Student-7-Hypothesizer
Japanese students seem to learn negative numbers first at junior high school.
Special-Student-7-Rebutter
Well, do Japanese elementary school students not know minus degrees Celsius?
Special-Student-7-Hypothesizer
In Japan, temperatures are usually called in Celsius, and elementary school students in northern areas daily experience minus degrees Celsius in winters.
Special-Student-7-Rebutter
But they do not know what minus degrees Celsius is ...
Special-Student-7-Hypothesizer
At least officially.
In the ZFC set theory, the natural numbers set is first constructed, the integral numbers set is constructed from the natural numbers set, the rational numbers set is constructed from the integral numbers set, and the real numbers set is constructed from the rational numbers set.
In Japanese schools, positive rational numbers seem to be learned before integral numbers.
Special-Student-7-Rebutter
I wonder why.
Special-Student-7-Hypothesizer
That is an interesting point. Probably, the concept of negative number is supposed to include an abstraction that is rather harder to be grasped.
In fact, a 1 / 3 apple can be shown to students as a physical object, but how about a -1 apple?
Special-Student-7-Rebutter
Students have to escape the notion that any number should correspond to some physical objects.
Special-Student-7-Hypothesizer
We begin to need negative numbers when we begin to consider differences.
For example, an apple tree had 17 apples the last year and had 10 apples this year, then, what is the increase of the crops of the tree from the last year to this year?
Special-Student-7-Rebutter
It is 7 decrease.
Special-Student-7-Hypothesizer
I asked the "increase".
Special-Student-7-Rebutter
I seem to be expected to say -7.
Special-Student-7-Hypothesizer
Note that getting a difference is different from getting a remain.
For example, the apple tree had 10 apples, and 3 of them have been stolen by some crows, then, '10 - 3 = 7' apples remain; more than 10 apples cannot be stolen, and the remain cannot be negative.
The elementary school mathematics does not need to cope with negative numbers because subtraction in it is to get remains.
Special-Student-7-Rebutter
The number 7 as a remain corresponds to the physical 7 remained apples, but the number -7 as a difference does not correspond to any specific physical objects but to the situation that the crops of the tree decreased in the year.
Special-Student-7-Hypothesizer
Someone may say that that -7 corresponds to some 7 apples of the last year, but that is not true: in fact, which 7 apples does he or she mean?
Special-Student-7-Rebutter
Probably, he or she picked up some arbitrary 7 apples, but -7 does not correspond to that specific 7 apples.
Special-Student-7-Hypothesizer
We need negative numbers as decreases when increases are asked.
In the same vein, we need negative numbers as increases when decreases are asked.
In fact, the decrease of the crops of the tree from this year to the last year is -7: we can talk also about the difference from this year to the last year instead of from the last year to this year.
Special-Student-7-Rebutter
I have seen an explanation that negative numbers are required in order to graduate a line.
Special-Student-7-Hypothesizer
In fact, any line can be graduated without negative numbers: for example, mark an origin, mark 1, 3, 5, ... in one direction from the origin, and mark 2, 4, 6, ... in the other direction from the origin.
Special-Student-7-Rebutter
In that way, certainly, the line can be graduated satisfying the requirement that each point is uniquely identified, but the expectation that the numbers should be monotonously increasing in one direction is not fulfilled.
Special-Student-7-Hypothesizer
The line can in fact be graduated by only the open real numbers interval, \((0, \pi)\), by using \(tan^{-1} (x) + \pi / 2\), and also by any positive bounded open real numbers interval, (a, b), by scaling and translating \((0, \pi)\), then, it is monotonously increasing.
Special-Student-7-Rebutter
A prevalent expectation is that the origin 0 should be at a finite location.
Special-Student-7-Hypothesizer
Then, of course, negative numbers are required, because obviously, all the numbers in a direction from the origin have to be less than 0.
But anyway, the primal purpose of graduating the line should be to uniquely identify each point with a number, and for the purpose, negative numbers are not required.
2: Let Us Construct the Integral Numbers Set
Special-Student-7-Hypothesizer
Let us think of the construction of the integral numbers set from the natural numbers set in this article, according to the arguments of the ZFC set theory, which means that we do not think of rational numbers.
An idea is to regard -7 to be the pair (10, 17), which means that -7 is the increase from 17 to 10.
Special-Student-7-Rebutter
Is (17, 10) not more natural?
Special-Student-7-Hypothesizer
Well, it is just the prevalent convention, which has been adopted probably because '10 - 17' is imagined.
Special-Student-7-Rebutter
Anyway, also ones like (9, 16) should be valid.
Special-Student-7-Hypothesizer
Yes, so, in fact, -7 = {(0, 7), (1, 8), ...}, which is an equivalent class in \(\mathbb{N} \times \mathbb{N}\) where \(\mathbb{N}\) is the natural numbers set.
The concept of 'equivalence class' may not be familiar to most junior high school students, but it is just a subset of a set when the set is divided into such subsets, where "divided into such subsets" means that any 2 different such subsets do not intersect and any element of the set is included in a subset.
Special-Student-7-Rebutter
'equivalence class' is a concept you cannot escape in higher mathematics, because it appears often there.
In fact, we see many equivalence classes even in lower mathematics, like the even numbers subset and the odd numbers subset in the natural numbers set, although they are not called equivalence classes there.
Special-Student-7-Hypothesizer
We need to define also positive integral numbers: 7 = {(7, 0), (8, 1), ...}.
Special-Student-7-Rebutter
That may be a point that some junior high school students do not easily understand: "We already have the natural number 7; why do we need such another 7?".
Special-Student-7-Hypothesizer
It is in fact an important point that integral 7 is really different from natural 7: natural 7 corresponds to 7 apples while integral 7 corresponds to the increase from 10 to 17, not to 7 apples.
Special-Student-7-Rebutter
So, we need integral 7 because integral 7 and natural 7 are different in meanings.
Special-Student-7-Hypothesizer
And we need it also in quest of symmetry: allowing (10, 17) but not (17, 10) is not symmetric. In other words, we want to treat negative integral numbers and positive integral numbers in the same way, so, we want negative integral numbers and positive integral numbers in the same shape.
Special-Student-7-Rebutter
We will understand the convenience of it when we think of the arithmetic of integral numbers.
Special-Student-7-Hypothesizer
{(0, 7), (1, 8), ...} is really denoted as [(0, 7)], because citing 1 element (0, 7) unambiguous determines the equivalence class. Instead, it can be [(1, 8)] or [(10, 17)] or something, because the purpose of identifying the equivalence class is fulfilled by any element of the equivalence class.
Special-Student-7-Rebutter
So, we have some leeway of expressing -7 as [(0, 7)] or [(1, 8)] or something, but that is not any problem, because the equivalence class is unambiguously specified all right.
3: Some Notes
Special-Student-7-Hypothesizer
In the following, just 7 is a natural number while [+7] or [-7] is an integral number.
The arithmetic of integral numbers is always done with only integral numbers.
So, '[+10] + [-7]' or '[+10] - [+7]' is allowed, but '[+10] - 7' is not allowed. You know, '[+10] - 7' is usually implicitly changed to '[+10] - [+7]', but this article does not allow such implicit changes, because such implicit changes hide exact mechanisms.
'[+10] + [-7]' and '[+10] - [+7]' are different things, although the results will turn out to be the same.
Although '[+10] - [+17]' has the answer, '10 - 17' has no answer, because the arithmetic of natural numbers has to have natural numbers as results. [-7] is [(10, 17)], not '10 - 17', which has no result.
Special-Student-7-Rebutter
Of course, we are not saying that we need to always make such distinctions in daily lives: it is just that the purpose of this article requires such distinctions.
4: In the Axiomatic Theory, the Arithmetic Is Just Defined So, but
Special-Student-7-Hypothesizer
In the following, we will examine why the arithmetic of integral numbers is as it is. For example, why [-1] + [-1] = [-2]?
Special-Student-7-Rebutter
Someone may say that the arithmetic is just defines so by a mathematician.
Special-Student-7-Hypothesizer
That stance is not logically wrong, but becomes a hindrance to deepen understandings, because any definition should have been made with some good reasons, and understanding the reasons clarifies the significance of the concept.
Any definition can be made up as far as it is well-defined, logically speaking, but we can and should ask why such a definition is wise.
Special-Student-7-Rebutter
But how can a definition of the arithmetic of integral numbers be judged to be wise?
Special-Student-7-Hypothesizer
A point is whether it accords with the interpretation of integral numbers as differences.
For example, what is the addition of 2 increases?
Special-Student-7-Rebutter
When the 2 increases are the increase of the crops of the apple tree from 2 years ago to the last year and the increase from the last year to this year, the addition of the 2 will be the increase from 2 years ago to this year.
Special-Student-7-Hypothesizer
Then, the addition of integral numbers should be wise to accord with that.
Special-Student-7-Rebutter
That requirement seems to already uniquely determine the definition of the addition of integral numbers.
Special-Student-7-Hypothesizer
Another point is whether it corresponds to the arithmetic of natural numbers when the concerned integral numbers are positive.
For example, we do not want '[+1] * [+2]' to be other than [+2], do we?
Special-Student-7-Rebutter
Probably not: we want the arithmetic of integral numbers to be an extension of the arithmetic of natural numbers in a meaning.
Special-Student-7-Hypothesizer
Another point is whether it satisfies the rules commonly expected to hold.
Special-Student-7-Rebutter
What rules exactly?
Special-Student-7-Hypothesizer
The commutativity, like '[-1] + [-2] = [-2] + [-1]', the associativity, like '[-1] + [-2] + [-3] = [-1] + ([-2] + [-3])', the distributability, like '[-1] * ([-2] + [-3]) = [-1] * [-2] + [-1] * [-3]', the reverse-ness of addition and subtraction, like '[0] - [-7] + [-7] = [0]', and the reverse-ness of multiplication and division, like '[+1] / [-7] * [-7] = [+1]', for example.
5: Addition of Integral Numbers
Special-Student-7-Rebutter
What should '[-1] + [-1]' be?
Special-Student-7-Hypothesizer
It is the accumulation of 1 decrease and 1 decrease, which should be 2 decrease, which means that '[-1] + [-1] = [-2]'.
Special-Student-7-Rebutter
That seems the only reasonable option as far as we honor the interpretation of negative numbers as decreases and the interpretation of additions as accumulations.
Special-Student-7-Hypothesizer
Also '[+1] + [+1] = [+2]' seems the only reasonable option.
Also '[-2] + [+1] = [-1]', '[-1] + [+2] = [+1]', '[+1] + [-2] = [-1]', '[+2] + [-1] = [+1]' come as only reasonable options.
As the exact definition, \([(n_{1, 1}, n_{1, 2})] + [(n_{2, 1}, n_{2, 2})] = [(n_{1, 1} + n_{2, 1}, n_{1, 2} + n_{2, 2})]\) will realize the above examples, where \(n_{i, j}\) is any natural number.
We can use that uniform definition for the entire integral numbers set because also positive integral numbers are formulated in the same shape with negative integral numbers.
Special-Student-7-Rebutter
We need to check that that definition is a valid definition.
Special-Student-7-Hypothesizer
That means that the result does not depend on the representatives of the equivalence classes. In fact, \([(n_{1, 1} + i, n_{1, 2} + i)] + [(n_{2, 1} + j, n_{2, 2} + j)] = [(n_{1, 1} + i + n_{2, 1} + j, n_{1, 2} + i + n_{2, 2} + j)] = [(n_{1, 1} + n_{2, 1}, n_{1, 2} + n_{2, 2})]\).
Special-Student-7-Rebutter
Let us check whether the commutativity and the associativity hold.
Special-Student-7-Hypothesizer
For the commutativity, \([(n_{1, 1}, n_{1, 2})] + [(n_{2, 1}, n_{2, 2})] = [(n_{1, 1} + n_{2, 1}, n_{1, 2} + n_{2, 2})] = [(n_{2, 1} + n_{1, 1}, n_{2, 2} + n_{1, 2})] = [(n_{2, 1}, n_{2, 2})] + [(n_{1, 1}, n_{1, 2})]\), because of the commutativity of natural numbers.
For the associativity, \([(n_{1, 1}, n_{1, 2})] + [(n_{2, 1}, n_{2, 2})] + [(n_{3, 1}, n_{3, 2})] = [(n_{1, 1} + n_{2, 1}, n_{1, 2} + n_{2, 2})] + [(n_{3, 1}, n_{3, 2})] = [(n_{1, 1} + n_{2, 1} + n_{3, 1}, n_{1, 2} + n_{2, 2} + n_{3, 2})] = [(n_{1, 1} + (n_{2, 1} + n_{3, 1}), n_{1, 2} + (n_{2, 2} + n_{3, 2}))] = [(n_{1, 1}, n_{1, 2})] + [(n_{2, 1} + n_{3, 1}, n_{2, 2} + n_{3, 2})] = [(n_{1, 1}, n_{1, 2})] + ([(n_{2, 1}, n_{2, 2})] + [(n_{3, 1}, n_{3, 2})])\), because of the associativity of natural numbers.
6: Subtraction of Integral Numbers
Special-Student-7-Rebutter
What should '[-3] - [-7]' be?
Special-Student-7-Hypothesizer
Let us think of '[-3] - [-7] + [-7]'.
Special-Student-7-Rebutter
That is about the reverse-ness of addition and subtraction: subtracting any number and adding back the same number should mean no change.
Special-Student-7-Hypothesizer
That means, '[-3] - [-7] + [-7] = [-3]'.
Then, the option for '[-3] - [-7]' is unique from the definition of addition of integral numbers. In fact, '\([(0, 3)] = [(n_1, n_2)] + [(0, 7)] = [(n_1, n_2 + 7)]\)' where \(n_i\) is a natural number, so, \((n_1, n_2) = (4, 0)\), for example; I mean, it can be also (5, 1) or something, but '[(4, 0)] = [(5, 1)]'.
As the exact definition, \([(n_{1, 1}, n_{1, 2})] - [(n_{2, 1}, n_{2, 2})] = [(n_{1, 1} + n_{2, 2}, n_{1, 2} + n_{2, 1})]\).
The definition is valid, because \([(n_{1, 1} + i, n_{1, 2} + i)] - [(n_{2, 1} + j, n_{2, 2} + j)] = [(n_{1, 1} + i + n_{2, 2} + j, n_{1, 2} + i + n_{2, 1} + j)] = [(n_{1, 1} + n_{2, 2}, n_{1, 2} + n_{2, 1})]\).
Special-Student-7-Rebutter
The reason of that definition may not be immediately clear for someone.
Special-Student-7-Hypothesizer
If \(0 \le n_{1, 1} - n_{2, 1}\) and \(0 \le n_{1, 2} - n_{2, 2}\), \([(n_{1, 1} - n_{2, 1}, n_{1, 2} - n_{2, 2})]\) will do, but that is not guaranteed, so, we have used \([(n_{1, 1}, n_{1, 2})] = [(n_{1, 1} + n_{2, 1} + n_{2, 2}, n_{1, 2} + n_{2, 1} + n_{2, 2})]\) in order to ensure that \(0 \le n_{1, 1} + n_{2, 1} + n_{2, 2} - n_{2, 1} = n_{1, 1} + n_{2, 2}\) and \(0 \le n_{1, 2} + n_{2, 1} + n_{2, 2} - n_{2, 2} = n_{1, 2} + n_{2, 1}\).
Special-Student-7-Rebutter
We should check that that definition is consistent with the natural numbers arithmetic.
Special-Student-7-Hypothesizer
When \(n_2 \le n_1\), \([+n_1] - [+n_2] = [(n_1, 0)] - [(n_2, 0)] = [(n_1, n_2)] = [(n_1 - n_2, 0)] = [+(n_1 - n_2)]\), which corresponds to \(n_1 - n_2 = n_1 - n_2\).
Special-Student-7-Rebutter
Some important properties to be checked are \(i_1 - [+n_2] = i_1 + [-n_2]\) and \(i_1 - [-n_2] = i_1 + [+n_2]\) where \(i_1\) is any integral number and \(n_2\) is any natural number.
Special-Student-7-Hypothesizer
\(i_1 - [+n_2] = [(n_{1, 1}, n_{1, 2})] - [(n_2, 0)] = [(n_{1, 1}, n_{1, 2} + n_2)] = [(n_{1, 1}, n_{1, 2})] + [(0, n_2)] = i_1 + [-n_2]\).
\(i_1 - [-n_2] = [(n_{1, 1}, n_{1, 2})] - [(0, n_2)] = [(n_{1, 1} + n_2, n_{1, 2})] = [(n_{1, 1}, n_{1, 2})] + [(n_2, 0)] = i_1 + [+n_2]\).
They are useful because now we can reduce any subtraction to an addition and the rules of additions can be employed.
For example, \(i_1 - [+n_2] - [+n_3] = i_1 + [-n_2] + [-n_3] = i_1 + [-n_3] + [-n_2] = i_1 - [+n_3] - [+n_2]\).
7: Multiplication of Integral Numbers
Special-Student-7-Rebutter
What should '[-3] * [-7]' be?
Special-Student-7-Hypothesizer
Let us think of '[-3] * [+7]' first.
It will be reasonable to assume that '[-3] * [+7] = [-3] + [-3] + [-3] + [-3] + [-3] + [-3] + [-3]'.
Special-Student-7-Rebutter
That is the original concept of multiplication.
Special-Student-7-Hypothesizer
So, '[-3] * [+7] = [-21]'.
Then, let us think of '[-3] * ([+7] + [-7])'.
What is it?
Special-Student-7-Rebutter
'[+7] + [-7] = [0]' is already known, so, '[-3] * ([+7] + [-7]) = [-3] * [0]'.
Special-Student-7-Hypothesizer
It is reasonably [0].
Now, let us suppose that the distribution law holds, which means that '[-3] * ([+7] + [-7]) = [-3] * [+7] + [-3] * [-7] = [0]'.
Then, as '[-3] * [+7] = [-21]', '[-3] * [-7]' has the unique option to be [+21].
Special-Student-7-Rebutter
So, any negative number multiplied by any negative number is positive, which has come from making some natural requirements, among which is the distributability.
Special-Student-7-Hypothesizer
The interpretation of '[-3] * [-7]' as [-7] times [-3] increases is probably enigmatic, but '[-3] * [-7]' has the good reason to be [+21].
As the exact definition, \([(n_{1, 1}, n_{1, 2})] * [(n_{2, 1}, n_{2, 2})] = [(n_{1, 1} * n_{2, 1} + n_{1, 2} * n_{2, 2}, n_{1, 1} * n_{2, 2} + n_{1, 2} * n_{2, 1})]\).
The motivation for that is '\((n_{1, 1} - n_{1, 2}) * (n_{2, 1} - n_{2, 2}) = n_{1, 1} * n_{2, 1} + n_{1, 2} * n_{2, 2} - (n_{1, 1} * n_{2, 2} + n_{1, 2} * n_{2, 1})\)', informally speaking.
Special-Student-7-Rebutter
"informally speaking" means that that is not really allowed because '\(n_{1, 1} - n_{1, 2}\)' is not valid when '\(n_{1, 1} \lt n_{1, 2}\)': note that it is in the natural numbers arithmetic, not in the integral numbers arithmetic.
Special-Student-7-Hypothesizer
It is just a motivation, not a valid calculation.
Anyway, the definition is valid, because \([(n_{1, 1} + i, n_{1, 2} + i)] * [(n_{2, 1} + j, n_{2, 2} + j)] = [((n_{1, 1} + i) * (n_{2, 1} + j) + (n_{1, 2} + i) * (n_{2, 2} + j), (n_{1, 1} + i) * (n_{2, 2} + j) + (n_{1, 2} + i) * (n_{2, 1} + j))] = [(n_{1, 1} * n_{2, 1} + i * n_{2, 1} + j * n_{1, 1} + i * j + n_{1, 2} * n_{2, 2} + i * n_{2, 2} + j * n_{1, 2} + i * j, n_{1, 1} * n_{2, 2} + i * n_{2, 2} + j * n_{1, 1} + i * j + n_{1, 2} * n_{2, 1} + i * n_{2, 1} + j * n_{1, 2} + i + j)] = [(n_{1, 1} * n_{2, 1} + n_{1, 2} * n_{2, 2}, n_{1, 1} * n_{2, 2} + n_{1, 2} * n_{2, 1})]\).
Special-Student-7-Rebutter
Let us check the commutativity.
Special-Student-7-Hypothesizer
'\([(n_{2, 1}, n_{2, 2})] * [(n_{1, 1}, n_{1, 2})] = [(n_{2, 1} * n_{1, 1} + n_{2, 2} * n_{1, 2}, n_{2, 1} * n_{1, 2} + n_{2, 2} * n_{1, 1})] = [(n_{1, 1} * n_{2, 1} + n_{1, 2} * n_{2, 2}, n_{1, 1} * n_{2, 2} + n_{1, 2} * n_{2, 1})] = [(n_{1, 1}, n_{1, 2})] * [(n_{2, 1}, n_{2, 2})]\)'.
Special-Student-7-Rebutter
Let us check the associativity.
Special-Student-7-Hypothesizer
'\([(n_{1, 1}, n_{1, 2})] * [(n_{2, 1}, n_{2, 2})] * [(n_{3, 1}, n_{3, 2})] = [(n_{1, 1} * n_{2, 1} + n_{1, 2} * n_{2, 2}, n_{1, 1} * n_{2, 2} + n_{1, 2} * n_{2, 1})] * [(n_{3, 1}, n_{3, 2})] = [((n_{1, 1} * n_{2, 1} + n_{1, 2} * n_{2, 2}) * n_{3, 1} + (n_{1, 1} * n_{2, 2} + n_{1, 2} * n_{2, 1}) * n_{3, 2}, (n_{1, 1} * n_{2, 1} + n_{1, 2} * n_{2, 2}) * n_{3, 2} + (n_{1, 1} * n_{2, 2} + n_{1, 2} * n_{2, 1}) * n_{3, 1})] = [(n_{1, 1} * (n_{2, 1} * n_{3, 1} + n_{2, 2} * n_{3, 2}) + n_{1, 2} * (n_{2, 1} * n_{3, 2} + n_{2, 2} * n_{3, 1}), n_{1, 1} * (n_{2, 1} * n_{3, 2} + n_{2, 2} * n_{3, 1}) + n_{1, 2} * (n_{2, 1} * n_{3, 1} + n_{2, 2} * n_{3, 2}))] = [(n_{1, 1}, n_{1, 2})] * [(n_{2, 1} * n_{3, 1} + n_{2, 2} * n_{3, 2}, n_{2, 1} * n_{3, 2} + n_{2, 2} * n_{3, 1})] = [(n_{1, 1}, n_{1, 2})] * ([(n_{2, 1}, n_{2, 2})] * [(n_{3, 1}, n_{3, 2})])\)'.
Special-Student-7-Rebutter
Let us check the distributability.
Special-Student-7-Hypothesizer
'\([(n_{1, 1}, n_{1, 2})] * ([(n_{2, 1}, n_{2, 2})] + [(n_{3, 1}, n_{3, 2})]) = [(n_{1, 1}, n_{1, 2})] * [(n_{2, 1} + n_{3, 1}, n_{2, 2} + n_{3, 2})] = [(n_{1, 1} * (n_{2, 1} + n_{3, 1}) + n_{1, 2} * (n_{2, 2} + n_{3, 2}), n_{1, 1} * (n_{2, 2} + n_{3, 2}) + n_{1, 2} * (n_{2, 1} + n_{3, 1}))] = [(n_{1, 1} * n_{2, 1} + n_{1, 2} * n_{2, 2} + n_{1, 1} * n_{3, 1} + n_{1, 2} * n_{3, 2}, n_{1, 1} * n_{2, 2} + n_{1, 2} * n_{2, 1} + n_{1, 1} * n_{3, 2} + n_{1, 2} * n_{3, 1})] = [(n_{1, 1} * n_{2, 1} + n_{1, 2} * n_{2, 2}, n_{1, 1} * n_{2, 2} + n_{1, 2} * n_{2, 1})] + [(n_{1, 1} * n_{3, 1} + n_{1, 2} * n_{3, 2}, n_{1, 1} * n_{3, 2} + n_{1, 2} * n_{3, 1})] = [(n_{1, 1}, n_{1, 2})] * [(n_{2, 1}, n_{2, 2})] + [(n_{1, 1}, n_{1, 2})] * [(n_{3, 1}, n_{3, 2})]\)'.
'\((i_2 + i_3) * i_1 = i_2 * i_1 + i_3 * i_1\)' where \(i_j\) is any integral number can be easily proved by using the commutativity, as '\((i_2 + i_3) * i_1 = i_1 * (i_2 + i_3) = i_1 * i_2 + i_1 * i_3 = i_2 * i_1 + i_3 * i_1\)'.
Special-Student-7-Rebutter
An important property to be checked is \([-n_1] = [-1] * [+n_1]\) where \(n_1\) is any natural number.
Special-Student-7-Hypothesizer
\([-n_1] = [(0, n_1)] = [(0, 1)] * [(n_1, 0)]\).
8: Division of Integral Numbers
Special-Student-7-Hypothesizer
Note that as we are thinking of only integral numbers, we do not think about '[-7] / [-3]' or something.
Special-Student-7-Rebutter
What should '[-21] / [+7]' be?
Special-Student-7-Hypothesizer
We already have '[-3] * [+7] = [-21]'.
We suppose that '[-3] * [+7] / [+7] = [-3]', which is the reverse-ness of multiplication and division.
Then, '[-3] * [+7] / [+7] = [-21] / [+7] = [-3]'.
Special-Student-7-Rebutter
What should '[-21] / [-7]' be?
Special-Student-7-Hypothesizer
'[+7] * [-3] / [-3] = [+7]', by the reverse-ness of multiplication and division, while we already have '[+7] * [-3] = [-21]', so, '[+7] * [-3] / [-3] = [-21] / [-3] = [+7]'.
Special-Student-7-Rebutter
What should '[+21] / [-3]' be?
Special-Student-7-Hypothesizer
We already have '[-7] * [-3] = [+21]'. '[-7] * [-3] / [-3] = [+21] / [-3] = [-7]', by the reverse-ness of multiplication and division.
As the exact definition, '\([(n_{1, 1}, n_{1, 2})] / [(n_{2, 1}, n_{2, 2})]\)' is '\([((n_{1, 1} - n_{1, 2}) / (n_{2, 1} - n_{2, 2}), 0)]\)' when '\(n_{1, 2} \le n_{1, 1}\)' and '\(n_{2, 2} \lt n_{2, 1}\)'; '\([(0, (n_{1, 1} - n_{1, 2}) / (n_{2, 2} - n_{2, 1}))]\)' when '\(n_{1, 2} \le n_{1, 1}\)' and '\(n_{2, 1} \lt n_{2, 2}\)'; '\([(0, (n_{1, 2} - n_{1, 1}) / (n_{2, 1} - n_{2, 2}))]\)' when '\(n_{1, 1} \le n_{1, 2}\)' and '\(n_{2, 2} \lt n_{2, 1}\)'; '\([((n_{1, 2} - n_{1, 1}) / (n_{2, 2} - n_{2, 1}), 0)]\)' when '\(n_{1, 1} \le n_{1, 2}\)' and '\(n_{2, 1} \lt n_{2, 2}\)'.
Special-Student-7-Rebutter
Does only the definition of division not look messy?
Special-Student-7-Hypothesizer
If you are talking about having to make specifications in cases, it seems just a natter of appearance.
Special-Student-7-Rebutter
Are appearances irrelevant?
Special-Student-7-Hypothesizer
At least, appearances are deceptive.
Special-Student-7-Rebutter
Anyway, an important property to be checked is \(i_1 / [-n_2] = [-1] * i_1 / [+n_2] = [-1] * (i_1 / [+n_2])\) where \(i_1\) is any integral number and \(n_2\) is any natural number.
Special-Student-7-Hypothesizer
When \(i_1 = [+n_1]\), \(i_1 / [-n_2] = [(n_1, 0)] / [(0, n_2)] = [(0, n_1 / n_2)] = [-n_1] / [+n_2] = [-1] * [+n_1] / [+n_2] = [-1] * i_1 / [+n_2] = [-1] * [(n_1 / n_2, 0)] = [-1] * ([(n_1, 0)] / [(n_2, 0)]) = [-1] * (i_1 / [+n_2])\); when \(i_1 = [-n_1]\), \(i_1 / [-n_2] = [(0, n_1)] / [(0, n_2)] = [(n_1 / n_2, 0)] = [+n_1] / [+n_2] = [-1] * [-n_1] / [+n_2] = [-1] * i_1 / [+n_2] = [-1] * [(0, n_1 / n_2)] = [-1] * ([(0, n_1)] / [(n_2, 0)]) = [-1] * (i_1 / [+n_2])\).
Special-Student-7-Rebutter
Let us check the commutativity.
Special-Student-7-Hypothesizer
Before that, let us check that \(i_1 / i_2 / i_3 = i_1 / (i_2 * i_3)\), which makes the check of the commutativity easy.
Special-Student-7-Rebutter
OK.
Special-Student-7-Hypothesizer
When \(i_j = [+n_j]\), \(i_1 / i_2 / i_3 = [+n_1] / [+n_2] / [+n_3] = [(n_1, 0)] / [(n_2, 0)] / [(n_3, 0)] = [(n_1 / n_2, 0)] / [(n_3, 0)] = [(n_1 / n_2 / n_3, 0)] = [(n_1 / (n_2 * n_3), 0)] = [(n_1, 0)] / [(n_2 * n_3, 0)] = [(n_1, 0)] / ([(n_2, 0)] * [(n_3, 0)]) = [+n_1] / ([+n_2] * [+n_3]) = i_1 / (i_2 * i_3)\); when \(i_1 = [-n_1], i_2 = [+n_2], i_3 = [+i_3]\), \(i_1 / i_2 / i_3 = [-n_1] / [+n_2] / [+n_3] = [-1] * [+n_1] / [+n_2] / [+n_3] = [-1] * ([+n_1] / [+n_2]) / [+n_3]\), which is by the above property where \(i_1\) is taken to be \([+n_1]\), \(= [-1] * ([+n_1] / [+n_2] / [+n_3])\), which is by the above property where \(i_1\) is taken to be \([+n_1] / [+n_2]\), \(= [-1] * ([+n_1] / ([+n_2] * [+n_3])) = [-1] * [+n_1] / ([n_2] * [n_3])\), which is by the above property where \(i_1\) is taken to be \([+n_1]\), \(= [-n_1] / ([n_2] * [n_3]) = i_1 / (i_2 * i_3)\); when \(i_2 = [-n_2], i_3 = [+n_3]\), \(i_1 / i_2 / i_3 = i_1 / [-n_2] / [+n_3] = [-1] * (i_1 / [+n_2]) / [+n_3]\), which is by the above property where \(i_1\) is taken to be \(i_1\), \(= [-1] * (i_1 / [+n_2] / [+n_3])\), which is by the above property where \(i_1\) is taken to be \(i_1 / [+n_2]\), \(= [-1] * (i_1 / ([+n_2] * [+n_3])) = [-1] * (i_1 / [+(n_2*n_3)]) = i_1 / [-(n_2 * n_3)]\), which is by the above property where \(i_1\) is taken to be \(i_1\), \(= i_1 / ([-n_2] * [+n_3]) = i_1 / (i_2 * i_3)\); when \(i_2 = [+n_2], i_3 = [-n_3]]\), \(i_1 / i_2 / i_3 = i_1 / [+n_2] / [-n_3] = [-1] * (i_1 / [+n_2] / [+n_3])\), which is by the above property where \(i_1\) is taken to be \(i_1 / [+n_2]\), \(= [-1] * (i_1 / ([+n_2] * [+n_3])) = [-1] * (i_1 / [+(n_2*n_3)]) = i_1 / [-(n_2*n_3)]\), which is by the above property where \(i_1\) is taken to be \(i_1\), \(= i_1 / ([+n_2] * [-n_3]) = i_1 / (i_2 * i_3)\); when \(i_2 = [-n_2], i_3 = [-n_3]\), \(i_1 / i_2 / i_3 = i_1 / [-n_2] / [-n_3] = [-1] * (i_1 / [+n_2]) / [-n_3]\), which is by the above property where \(i_1\) is taken to be \(i_1\), \(= [-1] * [-1] * (i_1 / [+n_2]) / [+n_3]\), which is by the above property where \(i_1\) is taken to be \([-1] * (i_1 / [+n_2])\), \(= i_1 / [+n_2] / [+n_3] = i_1 / ([+n_2] * [+n_3]) = i_1 / ([-n_2] * [-n_3]) = i_1 / (i_2 * i_3)\).
Now, the proof of the commutativity is: \(i_1 / i_2 / i_3 = i_1 / (i_2 * i_3) = i_1 / (i_3 * i_2) = i_1 / i_3 / i_2\), because of the commutativity of multiplication.
9: Order of Integral Numbers
Special-Student-7-Rebutter
What should the order of integral numbers be?
Special-Student-7-Hypothesizer
That naturally comes from the interpretation of integral numbers as differences.
Which is the bigger increase between the [+7] increase and the [+3] increase?
Special-Student-7-Rebutter
That should be about which case increased more.
Special-Student-7-Hypothesizer
Then, '\([+3] \lt [+7]\)'.
Which is the bigger increase between the [+7] increase and the [-3] increase?
Special-Student-7-Rebutter
The [-3] case did not increased at all.
Special-Student-7-Hypothesizer
I think that '\([-3] \lt [+7]\)' is more reasonable.
Which is the bigger increase between the [-3] increase and the [-7] increase?
Special-Student-7-Rebutter
[-7] is more far from increase.
Special-Student-7-Hypothesizer
I think that '\([-7] \lt [-3]\)' is more reasonable.
