Number theory (for example) defines a prime number excluding 1, which is allowed by pure algebra. The number 1 used to be a prime number until recently, about 1950, for algebra reasons.

Dear Juan Weisz ,

I try to understand what do you want to say but I can't.

" repel"? How?

There is a strong relationship between algebra and number theory, they cooperate together in a unified important math subject:

Algebraic Number Theory. Can you explain more?

Best regards

Issam

Rebel, not repel. To seem to be at odds with. Im sure this is just an apparent feature, but it is hard to see a reason.

For some reason number theory is very hard, people make very little progress there compared to algebra. However

the Z should also be subject to algebraic laws.

Certainly any positive real number is a product of another two, different from zero or 1.(although this is not an axiom of R that I know about), so you can understand factoring. It is just that the factoring is much more complex when it comes to Z.

The illustration I gave , I think indicates that sometimes algebra indicates one thing, number theory another.

Ed. The unique factoring theorem for Z is a result of number theory, not of algebra.

The fundamental theorem of algebra is algebra, different. But also addresses factors.

Methinks that number theory repels algebra especially when the question owner defines willy-silly what is number theory and what is algebra. The former, however, includes collective effects that the latter ignores on purpose. Also, "rebels" is the incorrect word in English, they complement each other and overlap.

Here is another example of algebra not quite making it in number theory territory.

Take the identity

xxx+yyy = (x+y)(xx+yy-xy)

and think about the Fermat problem with cubes.

Obviously we could replace the left hand side by n for x and m for y, for all n,m of interest. On the right hand side we could never get a cube, based on this identity.

If we square (x+y) we get an expression differing by the other factor by a 3xy term. And since (x+y) could be anything

this seem OK.

However this is not considered a valid proof. Why? Because each factor on the right hand side could be a non prime

integer, and we could borrow some factor from one and pass it to the other, giving yet some hope of getting a cube.

Incredibly people have never really understood this sort of barrier well enough to make progress.

There are two distinct, and for the most part disjoint, areas of number theory: algebraic number theory and analytic number theory. The first of these certainly doesn't rebel algebra; it's based heavily on field theory. However, it hasn't been able to produce much in the way of improving our understanding of prime numbers. This is where analytic number theory has played a significant role, introducing such powerful tools as Dirichlet series and Riemann Zeta function.

I wouldn't say it rebels algebra as much as it identifies its limitations. Every field has limitations, and the strength of mathematics is most felt when one is able to apply tools from other fields to circumvent these limitations. The use of analytic number theory to produce results about the integers is only one example of many.

In fact one could argue that the theory of prime numbers shouldn't be regarded as a subfield of algebra in the first place. Algebra is based on abstraction, which means it strips an object, such as the integers, of all of its properties except those which are most basic to it, namely its operations. The integers only serve as one basic model for such abstraction. Apart from that, they quickly lose their relevance.

A7. Uniqueness of prime decomposition of Gaussian integers

Answer 7, posted on April 16, 2019

========================

Thanks to the uniqueness of the prime decomposition

of Gaussian integers, we may observe that

the two decompositions of 10 in the original question

are the same:

10 = [3 + i][3 – i] = [(1 + i)(2 – i)][(1 – i)(2 + i)]

and

10 = [2][5] = [(1 + i)(1 – i)][(2 + i)( 2 – i)].

================

With best regards, Jean-Claude

Number theory is the systematic theoretical study of numerical structures which by definition is a part in the larger portion of algebraic structures. Prime numbers for instance, numbers with specific divisibility behaviors that are located in particular locations, are part and parcel of algebraic structures. In fact, parts of mathematics, although growing large within their territory, but creating more overlaps with other parts of mathematics. Rebellion in this context does not have meaning and purpose.

Jean,

that seems to do it, thanks, Juan

A10. Polynomials whose graphs pass through a given point

Answer 10, posted on April 18, 2019

========================

The polynomials

p(z) = (z + i)(z – i) and q(z) = (z - 1)(z + 2)

given in the original question are monic quadratic

polynomials, whose roots are Gaussian integers,

and they take the value 10 at z = 3, that is to say,

their graphs pass through the point (3, 10).

=====================

This suggests the problem of finding all such polynomials,

that is find all polynomials f(z) = (z – r)(z – s)

such that r and s are Gaussian integers and f(3) = 10.

=====================

We have seen in answer 7 that, up to multiplication

by any of the four units 1, -1, i, -i

of the ring Z[i] of Gaussian integers,

the prime decomposition of 10 in Z[i] is

10 = (1 + i)(1 – i)(2 + i)(2 – i).

We can make the prime decomposition in Z[i] unique

by choosing associates of primes whose arguments lie

in the interval [0, pi/2). In particular

1 – i = -i(i + 1) = -i(1 + i) is an associate of (1 + i)

and the argument of (1 + i) is pi/4,

which is in the interval [0, pi/2),

so that the standard prime power decomposition of 10 is

10 = -i(1 + i)^2(2 + i)(2 – i).

=====================

We want to find polynomials f(z) = (z – r)(z – s)

such that f(3) = 10 that is, such that

f(3) = (3 – r)(3 – s) = 10.

Let A = 3 – r and B = 3 – s.

Then finding Gaussian integers r and s

such that (3 – r)(3 – s) = 10

is equivalent to finding Gaussian integers A and B

such that AB = (1 + i)(1 – i)(2 + i)(2 – i).

Thanks to the uniqueness of the prime decomposition

of Gaussian integers up to multiplication by units,

this is equivalent to choosing some of the prime factors

(1 + i), (1 – i), (2 + i), (2 – i) for A,

all the others for B,

and possibly multiplying the product AB

by (-1)(-1) = 1 or by i(-i) = 1,

for example,

AB = (-A)(-B) = -A(-B) = -(-A)B

or AB = i(-iA)B = (iA)(-iB) = … .

=====================

If we choose A = (1 + i)(2 – i) = 3 + i

and B = (1 – i)(2 + i) = 3 – i, we obtain

3 – r = A = 3 + i, which implies that r = -i

and 3 – s = B = 3 – i implies that s = i.

With this choice of A and B, we obtain

f(z) = (z – r)(z – s) = (z + i)(z – i) = p(z),

which is the first polynomial given in the original question.

=====================

If we choose A = (1 + i)(1 – i) = 2

and B = (2 + i)( 2 – i) = 5, we obtain

3 – r = A = 2, which implies that r = 1,

and 3 – s = B = 5 implies that s = -2.

With this choice of A and B, we obtain

f(z) = (z – r)(z – s) = (z - 1)(z + 2) = q(z),

which is the second polynomial given in the original question.

=====================

If we choose A = -(1 + i)(1 – i) = -2

and B = -(2 + i)( 2 – i) = -5, we obtain

3 – r = A = -2, which implies that r = 5,

and 3 – s = B = -5 implies that s = 8.

With this choice, we obtain

f(z) = (z – r)(z – s) = (z – 5)(z – 8),

which is different from the two polynomials

given in the original question.

=====================

With best regards, Jean-Claude

Jean,

Your four gaussian number factors of a non prime integer, here 10

also gives for x=3, written as polynomial with roots, gives

[xx-2x +2] [xx-4x+5] giving 5 times 2

For x=0 you also get, the other way around, did not expect this. You understand?

Here algebra is a sort of many for one affair. I wonder how far up in degree of polynomial

you can get. Your examples are for N=2

One can also try x=3,5,7,...all nonprimes with a factor of 2 from xx+1

Regards, Juan

ALGEBRA vs. NUMBER THEORY

Allow a late comer to answer to several persons at the same time. The subject of the debate, about « the apparent misfit of number theory with algebra », seems void to me. There is no misfit whatsoever. At the surface, the arguments given by the OP make no sense (excuse me for my frankness), see a detailed refutation in my first appended file. At the core, the two sides of the alleged « rebellion » must be defined before any further discussion. What are numbers, what is algebra ?

It’s Kronecker, I think, who said that "God made the integers, all else is the work of man." Fractions were known to the Babylonians, the Egyptians, etc. but let us jump directly to the ancient Greeks, because of their geometric vision of the universe. They viewed fractions as « proportions » (this word is still used in common language) via Thales’ theorem which people nowadays learn in elementary school. But for want of adequate notions/notations, they could hardly manipulate their « proportions » (at least before Diophantus, and non-Greek mathematicians as Al-Khawarizmi or Brahmagupta) in what we now usually call algebraic calculations, i.e. calculations taking place in rings and fields. Yet they knew about prime factorization and the existence of an infinite number of primes (see Euclid’s « Elements »). The discovery of the irrationality of sqrt 2 by the school of Pythagoras came as an earthquake. Even nowadays, there are pseudo-mathematicians who continue to divagate about irrationality (see a thread neighbouring this one). The next step, the field of real numbers, was already hidden behind the famous paradoxes of Zeno (around -450 B.C.) : the arrow never hitting its target, Achilles never catching up with the tortoise… Mathematically speaking, these paradoxes are lifted by the convergence of the series 1/2+1/4+...+1/2n+ … in R. Metaphysically speaking, we must accept that Time can be assimilated with R. Anyway the notion of convergence had to wait for Cauchy to be firmly established, and one way to constuct R nowadays is to topologically complete Q by way of equivalence classes of Cauchy sequences. In my opinion it’s the best way, because when replacing the archimedean (usual) metric by the p-adic metric (p being a prime number), one gets the p-adic completion of Q, which is the field Qp of p-adic numbers. These completions are compatible with the field structure, and the astonishing phenomenon is that they are the only existing ones (Ostrowski’s theorem). This is how analysis (classical or p-adic) makes its entry in the world of number theory.

I conclusion, I roughly agree with the opinion of @Andrew J. Woldar : the goal of algebra is to recognize structures behind operations (such as rings and fields), to organize them and build machineries to deal with them. These machineries, if ubiquitous enough, can be applied to many different domains (number theory, algebraic geometry, differential geometry, algebraic topology, etc.). Conversely, analytical tools are needed in case the underlying structure is not existing, or not clearly distinguished (just think of the set of prime numbers). But I disagree with too neat a distinction between the algebraic and analytic approaches. I’ll take just two examples. The first is the explosion, starting from the middle of the last century, of the theory of modular forms which combines complex analysis with symmetry groups. Applied to number theory, it’s now nicknamed « the 5-th operation of arithmetic », the power and versatility of which can be witnessed in Wiles’ proof of Fermat’s Last Theorem, and its subsequent developments. The second example is about the mystery of the « special values » of the Zeta function, see my second appended file. Thank you for your attention.

@Juan Weiz, Issam Kaddoura, Jean-Claude Evrard, Andrew J. Woldar Sorry, there was a confusion between two files.

Dear Thong Nguyen Quang Do ,

Regardless of the files, your answer is excellent and inspiring.

Best regards

A15. A standard prime decomposition in Gaussian integers

Answer 15, posted on April 19, 2019

========================

In answer 10, I gave a wrong “standard” prime decomposition

of the integer 10 in Gaussian integers.

================

My personal definition of “standard” is an attempt

to make the decomposition unique, that is,

without “up to multiplication by units”

and

without “up to the order of the factors”.

================

For this, I do the following:

1. For every Gaussian prime p in the decomposition,

I choose the associate of p whose argument

(in the sense of argument of a complex number)

is in the interval [0, pi/2).

2. I order the primes in the decomposition

in increasing order of their modules

(in the sense of modules of complex numbers).

3. I order the primes that have the same module

in increasing order of their arguments.

4. I collect the identical primes into a power of a prime,

so that the prime decomposition

becomes a prime power decomposition.

================

Let us find the standard prime decomposition

of the integer 10 in Gaussian integers. We have

10 = [2][5]

= [(1 + i)(1 – i)][(2 + i)( 2 – i)]

= (1 + i)[(-i)(1 + i)](2 + i)[(-i)(1 + 2i)]

= (-i)(-i)(1 + i)(1 + i)(2 + i)(1 + 2i)

= (-1)(1 + i)^2(2 + i)(1 + 2i),

so that the standard prime decomposition of 10 is

10 = (-1)(1 + i)^2(2 + i)(1 + 2i).

================

I am very interested in the observations and questions

posted by Juan in answers 5 and 11.

I will try to find some answers very slowly.

I will be very happy if anyone else finds some answers before me.

At the present time, I am in a terrible battle

to try to simplify the proof that every solution

of the cubic Fermat equation in quadratic integers

can be reduced to conjugate form.

I would be very happy if anyone finds a simpler proof.

================

The current simplest proof is the following:

================

Why the solutions of cubic Fermat equation

can be written in conjugate form –

A detailed explanation of Alexander Aigner's finding

https://www.researchgate.net/publication/331111922_Why_the_solutions_of_cubic_Fermat_equation_can_be_written_in_conjugate_form_-_A_detailed_explanation_of_Alexander_Aigner's_finding

================

I have no problem with the title of this thread.

It is frequent in sciences

that we are surprised by an observation

and we have a first feeling in very imprecise shaky terms.

================

With best regards, Jean-Claude

A16. The integers 3 and 10 are fundamentally related

Answer 16, posted on April 19, 2019

========================

I am taking a short break from my nightmare

about the cubic Fermat equation

that I described in answer 15,

and I have found the following answer

to one of the questions asked by Juan in answer 11.

========================

One of the prime decompositions of 10 into Gaussian primes,

up to multiplication by units, and up to the order of the factors, is

10 = [2][5]

= [(1 + i)(1 – i)][(2 + i)( 2 – i)]

= (1 + i)(1 – i)(2 + i)(2 – i).

========================

Let a = 1 + i, b = 1 – i, c = 2 + i, and d = 2 – i,

so that 10 = abcd.

========================

The two polynomials considered by Juan in the answer 11 are:

p(z) = z^2 – 2z + 2 = (z – a)(z – b)

and

q(z) = z^2 – 4z + 5 = (z – c)(z – d).

Not surprisingly, we have

p(0)q(0) = (0 – a)(0 – b)(0 – c)(0 – d)

= (-a)(-b)(-c)(-d) = abcd = 10,

so that p(0)q(0) = 10.

========================

Following the original question of this thread,

Juan made the following stupefying observation:

p(3)q(3) = 10. Why?

========================

We can look for a stupefying fundamental relation

between 3 and 10 for billions of years and never find any,

and we risk a heart attack if we find any.

========================

So, if you have heart problems,

do not read what follows.

========================

The stupefying fundamental relation between 3 and 10

is a + d = 3 and b + c = 3.

========================

It follows that

p(3)q(3) = (3 – a)(3 – b)(3 – c)(3 – d)

= dcba = abcd = 10,

so that p(3)q(3) = 10.

========================

With best regards, Jean-Claude

Juan Weisz

J. W. : xx+1 = xx-ii = (x+i)(x-i)

This is unique, fundamental theorem of algebra.

No! This gets disproved.

Do not listen to the advocates of the wrong theory.

See: www.mathe-neu.de and have a view at the fundamental new math.

Peter

It is good for a given and fixed polynomial, but not the set of polynomials that can pass through a given fixed point.

regards, JW

Thong, Issam

Maybe what I mean is that perhaps you can ultimately succesfully join the algebra and number theory, but that

this would require a lot of work, depending on the specific problem, and is a fertile ground to explore new results.

I dont think you can yet argue that all these are already solved.

best regards, Juan

Dear Juan Weisz ,

The subjects evoked in your new post and your original post are totally different in content. In your new post, no genuine number theorist can agree with your first sentence, because algebra and number theory have already been "successefully joined" since the times of Kummer and Dedekind, the founders of algebraic number theory. Regardless of this initial disagreement, no genuine scientist can dispute the pertinence of your general considerations on the development of knowledge, because they lie at the core of scientific research.

What is disputable is your view of the "apparent misfit between algebra and number theory" exposed in your initial post. I took great care (but with great bluntness, I recognize) to refute point by point your erroneous conceptions on the existence and uniqueness of prime factorization in number rings. Again, this goes back to Kummer and Dedekind, and everyone entering a discussion on this subject should know it. If you disagree somewhere with my refutation, please let we know and we'll discuss. Otherwise, I don't know what I can do.

Best regards, Thong

Cannot immediately answer, because I was not familiar with some of the results the specialists in number theory

already incorporate, I am learning something. A priori your answer sounds a bit too grandious, but cannot say for sure now.

You know what you studied, but perhaps have not asked sufficient questions yet. I admit that my questions were posed on shaky grounds. Am not a mathematician, handle much of algebra rather well, but am out of depth in number theory.

Algebraic number is simply a root of a polynomial with rational coeficients.? Then what is different from these numbers

with simply real or complex numbers?(without inventing new abstract numbers such as i ?)

An algebraic number is a root of a polynomial with coefficients in Q. But saying this implies already that this "root" lives somewhere. For simplicity, suppose we work inside C (although this is not very economical), where C has been constructed from R by adding i, with i2 = -1. An algebraic integer is a root of a polynomial with coefficients in Z s.t. the coefficient of the term of highest degree is 1. A set of algebraic numbers which constitute a field is called a number field , say K. The subset of algebraic integers in a number field K constitutes a ring AK, called its ring of integers. Elementary examples are (Q, Z) and, in the case you consider, (Q(i), Z[i]) where Q(i) (resp. Z[i]) is constructed as C, just replacing R by Q (resp. Z). But note that general (K , AK ) are not so simple.

To do interesting arithmetic, working in a ring of integers AK , you must define prime elements. The usual definition in Z cannot be just translated to any AK because, as I said, primes in Z can decompose in AK , and even worse, uniqueness of decomposition is not automatic. The most famous counter-example is the erroneous proof of Fermat's Last Theorem given by Lamé around 1850 . Kummer re-established uniqueness by replacing prime elements with prime ideals, but I'll not go further into this because, in the case of the gaussian integers Z[i] , prime ideals are just formed of multiples of prime elements, up to units. This short introduction is enough, I think, to understand what I was talking about.

PS: What do you mean by "grandious" ?

Dear Juan Weisz , your question is not clear to me. Can you please explain it in more simpler form?

Tapas, from general algebra I have it that roots of polynomials are either real or complex in pairs.

So when I hear that someone is seeking "algebraic numbers", I ask myself what is all this new gun

powder about. It is not about old fashioned real or complex numbers?

Thong,

Will take me a while , so from xx-2=0, you conclude that sqrt(2) is an algebraic integer?

I know integers are a ring and not a field.

Yes, I quite appreciate that certain decompositions are not unique ie. 10 =2*5=(3+i)(3-i) if Gausian integers are allowed.

Grandious would be too general a statement.

Aparently you see the importance of working with the irreducible or prime elements, can be prime in one system and not in another.

I think certain numbers can only really be understood from matrices,

as in

0 -1

1 0

for i. Then you appreciate the difference between i and -i.

The prime number 5 decomposes in gaussian as (2+i)(2-i).

Regards, Juan

Juan Weisz

What is good for a given and fixed polynomial? For what is it good?

You hadn’t a look at my homepage, had you? So you don’t understand anything about `i´ .

The imaginary unit is the unsolved problem to take the square-root (of minus one).

The problem stays unsolved because of wrong definitions. I gave the solution!

There are many other problems in classical math. All about `i´ makes no sense, except its counterprove. Are you interested in any explanation / discussion?

Regards, Peter

Peter

Well, the general form of a power expansion has polynomial form, and is central to the discussion of analytic functions.

Ideal tools to teach derivatives and such. Then you can also expand in Complex variables, or in ring domains.

You can form special functions in mathematical physics using polynomials, orthonormal systems used in QM and solutions to problems in electromagnetic theory.

Why wouldent I understand anything about i? Square my matrix and see what you get.

Work out the algebra of

x -y

y x

and see if you do not get the same rules as (x1+iy1)(x2+iy2) by multiplying such indexed matrices together.

Best regards, Juan

Juan Weisz

It´s not you alone. No-one knows the truth about i until it was revealed (uncovered) by me in 2008.

You only repeat what is taught wrong. So my assertion was right.

1. If you square i it is defined to get -1. But the same by -i . So ambiguous.

2. If you take the square-root of anything you should have four results! By i you have none. [√4 = +2 and +2 or -2 and -2; think about; `or´: no product comes by only one factor]

3. Definition to i goes: i2 = -1 (i2 + 1 = 0) => y = f(x) = f(i)

i is variable: x = +i or -i => y = 0; x = 0i => y = +1; x = +2i or -2i => y = +3 …

Plot it and you get a well known, classical graph: y = -x2 + 1 ! [nothing with imaginary!]

4. J. W.: The prime number 5 decomposes in gaussian as (2+i)(2-i)

Don’t do that foolish thing again. Both the 2´s are real, both the i´s are imaginary. So of different unit by definition! You also won’t add apples to pears, length to width, pressure to area … . First you have to define a set in which both units are subsets (fruit for apples and pears, for example). Then you may add.

Above the contradictions inside the theory. More sophisticated to think about the right formalism.

What is measure, what is unit? Mathematicians as well as physicists went the wrong way. Have a look at my homepage. Step by step whole the new formalism gets presented. Mathematics get paramounted to a nature-science. Most is already done in German. Translation will come soon as well as the missing parts.

I hope you are strong enough to discuss this topic. There are more faults than you point on.

Regards, Peter

Dear Peter Kepp ,

To show new accepted results one should announce them and submit the theory to some indexed journal to share his findings with a serious scientific community. Otherwise, you will stay alone with your results that seem to be weird to everyone. Frankly, you are talking about something vague and not clear. Best regards

Dear Issam Kaddoura

thanks for `weird´.

In German that means `to be dignified for remembering´.

What is vague at `an imaginary value is not balanceable to a real value´?

I represented it at the deepest level.

I am not `alone´. Have a look at the counter at www.mathe-neu.de .

The inner core of my assertions to the imaginary unit was published in a double peer reviewed journal [European Scientific Journal, Vol. 9, Nr. 33, Nov. 2013, http://eujournal.org/index.php/esj/issue/archive   or direct:

http://eujournal.org/index.php/esj/issue/view/103   (16-th articel)].

This is also on my homepage (so far in German only). See you.

Regards, Peter

Issam Kaddoura, Peter Kepp,

What is surely clear and not vague at all, is that all of Peter's abrupt statements are non sense.

1. If you square i it is defined to get -1. But the same by -i . So ambiguous.

Wy "ambiguous" ? If you square +2 and -2, you get 4. More generally, sqrt x is only defined of to a sign, wherever x may live. In more scholarly word, the function sqrt x cannot be defined without adding conditions.

2. If you take the square-root of anything you should have four results! By i you have none.

Completly wrong. Given x, you have at most two values for sqrt x . For example, +sqrt 2 and -sqrt 2 are the two real square roots of 2, and also the two complex ones. But there is no sqrt of -1 in R, and two in C, which are i and -i . And there are two sqrt of i, wich are cos(pi/4)+i sin (pi/4) and cos(pi/4) - i sin (pi/4), which make exactly four 4-th roots of 1. More generally, any complex number z admits exactly two sqrt., any complex polynomial of degree n admits n roots (counted with multiplicities).

3. Definition to i goes: i2 = -1 (i2 + 1 = 0) => y = f(x) = f(i)

i is variable: x = +i or -i => y = 0; x = 0i => y = +1; x = +2i or -2i => y = +3 …

Plot it and you get a well known, classical graph: y = -x2 + 1 ! [nothing with imaginary!]

This is hardly understandable. What is your function ? If it is f(x)= - x2 +1, then f(i)=2 , f(2i)=5, ... not what you say. And even if the values values were correct , what do you mean by nothing with imaginary ? The value f(x) at an imaginary x has no need to be imaginary, since by definition i2 =-1. Besides you cannot plot a graph with an imaginary x axis and a real y axis. This does not make any sense.

4. J. W.: The prime number 5 decomposes in gaussian as (2+i)(2-i)

Don’t do that foolish thing again. Both the 2´s are real, both the i´s are imaginary. So of different unit by definition! You also won’t add apples to pears, length to width, pressure to area … . First you have to define a set in which both units are subsets (fruit for apples and pears, for example). Then you may add.

But Juan is doing exactly what you wish. It's you who don't understand what you're doing. Here you have not 2 distinct types of fruits (apples and pears), but only one type, because a real is just a complex with null imaginary part (and an imaginary is a complex with null real part). So you may say for example that there are green and red apples, which you can eat without changing your menu.

You suggested to Juan to take a look at your home page. After reading all this non sense, I know what I would answer if I were at his place.

On Google you can find at least 12 lists of "Predatory (fake) Scientific Journals" warning against the Eastern so called European Sci. J. See e.g. https://www.researchgate.net/...predatory.../Bealls-list-of-potential-...

Dear Thong Nguyen Quang Do

Thanks for your explanations. But you are definitely wrong! What I hardly learned is: the higher the graduation the lower the understanding of fundamental facts.

Bad thinkers always shelter bad thinkers; because of saving the wrong theory.

If you are able to argue, then argue. If you are only able to slander ... poor Thong!

1. ambiguous

The imaginary unit gets defined by: i2 = -1

This leads to the wrong understanding that this is a bijective (one to one) definition. But also -i2 = -1 is valid. Both give the full version of the expert opinion of the definition of imaginary, which got ruled out by my papers.

2. Square-root

The radicand must be the result of the multiplication of two (same in (absolute) value) factors. No quantity could be squared! If math does, it goes wrong.

Math, as the formalism of calculation for real things (nature) has to follow an isometry to the calculated reality — otherwise it is useless.

So in classical math √4 looks for +2 and +2 as well as for -2 and -2. End of explanation to √ (please think about before you answer)!

3. Definition is function

You mixed the end with the middle! In f(x) = -x2 + 1 (= 0 if x = 1) you should not take any measure `i´ as variable x. This is exclusively real. It is equal to the plot of the function I declared. I named the graph, not its units. Please read precise.

Imaginary function (again):

Definition (complete) goes: +i2 + 1 = 0 as well as -i2 + 1 = 0

(y = ax2 + bx + c; a = 1, b = 0, c = 1 0 = (+1i)2 + 1 as well as 0 = (-1i)2 + 1)

[already two points at the real / imaginary area].

Here you must take a logical step. y = f(x) => you take any real value for x.

If i is the measure you have to take the product of any value as a factor to i .

Third point of the graph comes by 0i : y = (0i)2 + 1 = 1

The other points by various value for the factor x to i .

Than you may look at both the graphs, the exclusively real (y = -x2 + 1) and the imaginary one (y = (xi)2 + 1). They are identical!

4. Addition of a real number and an imaginary number

What did you read? The gaussian was named as: (2 + i)(2 – i). Inside the brackets there are reals (2) and imaginaries (i). The rules of combination are addition and subtraction. Please do that with any two quantities which are different in unit.

There is no relation to red or green apples which may be my menu.

Regards, Peter

Dear Thong Nguyen Quang Do,

I think you are wasting your time. There is no way to convince Peter Kepp with your words, although you have made a sincere and admirable attempt. I salute you for this. - Andrew

I know, there have been many craps of this type on RG, especially about "elementary" proofs of FLT or the (non)trancendence of Pi. But this case is at such a low level that I had to react. With no effect, visibly. Thank you for your concern, Thong NQD

@Peter Kepp I've just read your "mathematical" reply , which is on the same non sensical level as before, so I'll quit here. But I'd really (in the ordinary, non mathematical sense) like to have an answer concerning the "predatory" European Scientific Journal.

Dear Peter Kepp ,

Such replies are not good for your scientific reputation.

RG is a handy platform to share knowledge with full respect to the experts in their fields. You should learn and benefit from their experience.

Do you think the reviewers of your articles are more expert than Professor Thong!!

You should know with whom you are talking. And then reply.

Issam Kaddoura

You are treated as you treat others. First reply came from Thong.

My reputation comes by the truth.

For knowledge no title doesn´t make a difference!

Since 2008 I heard such ... from established `professors´. No matter. All of them don´t (or won´t) understand. Students are interested. A new theory needs time.

Do you have to say anything on the facts?

Juan Weisz

Do you know the other `rebels´ on classical math?

2500 years old problem:

Binoms of the Babyloniens.

Do you know theseone which produces:

-a2 – 2ab – b2 as well as -a2 + 2ab – b2 ?

250 years old problem:

`i´ – we just talk about; but following the Euler-equivalence, new definition of √ ?

150 years old problem:

Continuum hypothesis of Cantor?

Also the disproving of the mathematical field?

Regards, Peter

A39. Language: Two very different cases

Answer 39, posted on April 22, 2019

========================

In the art of written communication,

there are two very different cases:

========================

Case 1: The writer is not attacking anyone.

In this case, there is no problem to write a first draft

in an imprecise shaky language.

A splendid example is the original question posted by Juan.

He wrote a first imprecise and shaky draft

with a wonderful observation of two monic quadratic polynomials

whose roots are Gaussian integers

and whose graphs pass through a given integer point.

This leads to the wonderful problem of finding all such polynomials.

In this case, it does not matter that the language used by Juan

was imprecise and shaky. Scientists are overflowed

with too much work, and try to save time with drafts

written at very high speed. When they are not attacking anyone,

it does not matter if they write a first draft above speed limit

with high risks of mistakes. Nobody will be hurt by such mistakes.

========================

Case 2: The writer is attacking some people.

In such a case, it is essential to use a very correct

and precise language.

In answer 27, Peter attacked the complex number i,

implying that all users of complex numbers are primitive idiots,

including all mathematicians, physicists, and engineers.

We all know for sure that this is false:

Every day, there are hundreds of jumbo jets that fly safely,

huge buildings do not collapse,

huge bridges do not collapse,

Internet, phones, and television are working perfectly.

This shows that all those scientists are not primitive idiots,

and they do use complex numbers every day.

To attack complex numbers, it is essential to start

from the precise definition of complex numbers:

https://en.wikipedia.org/wiki/Complex_number#Formal_construction

========================

With best regards, Jean-Claude

Thong Nguyen Quang Do:

So you may say for example that there are green and red apples, which you can eat without changing your menu.

The reds as well as the greens are subsets (minor-sets) in all the apples (major-set). If you distinguish between the colour (red or green) you sort. Then the major-set `apples´ gets resolved (disbanded) into its constituents (subsets / minor-sets).

[the higher the gaduation ...]

Why excluding two parts of whole the area when looking for the square-root?

The graphic definition of a product comes by the (flat) area.

In analytical geometry the origin is the position (see Constantin Frantzen, Philosophie der Mathematik) of both the vectors which represent the units for the factors (their position in relation to the origin and their orientation). The four possible variations of prefixes of the factors stay fix in each quarter of this geometrical construct.

By searching (by square-root) for the two factors (equal in absolute value), which produce the radicand, classical definition excludes the areas `plus by minus´ as well as `minus by plus´.

That´s to complete. I did it! Follow the reform.

Having more enemies — getting more popular.

Having more followers — getting more popular.

Both are no indicator for being right or being wrong.

A42. The Cayley–Dickson construction

Answer 42, posted on April 23, 2019

========================

Concerning the well-known equivalent definition of complex numbers

by means of 2 by 2 matrices with real entries,

that was mentioned by Juan in the answer 26,

here is the definition on the encyclopedia Wikipedia,

followed by some generalizations:

========================

Matrix representation of complex numbers

https://en.wikipedia.org/wiki/Complex_number#Matrix_representation_of_complex_numbers

========================

Complex structure of C^n:

https://en.wikipedia.org/wiki/Linear_complex_structure#Cn

========================

Generalizations and related notions

https://en.wikipedia.org/wiki/Complex_number#Generalizations_and_related_notions

========================

Cayley–Dickson construction

https://en.wikipedia.org/wiki/Cayley–Dickson_construction

========================

With best regards, Jean-Claude

36 could be the product of 36 and 1, 18 and 2, 12 and 3, 9 and 4 or 6 and 6.

The square-root looks for the last pair. The sums of the pairs are: 36, 20, 15, 13 or 12. The minimum of the sums also is the last. By that the square-root comes by a construct `minimum´.

In words to take the square root may be defined as:

Half the value of the minimum of the absolute of the sum of the factors which could produce the radicand.

I√cI = 1/2 min I(a + b)I; a * b = c

But this fails for negative radicand. So reform to:

Half the value of the minimum of the sum of the absolutes of the factors which could produce the radicand.

I√cI = 1/2 min (IaI + IbI); a * b = c

This is for classical, single prefix. So ambiguous (the absolute as double-result). The complete reform allows double prefixes (the sources complete documented).

[very short representation, but well understood — also by some higher graduated persons, even professors]

Peter,

That sort of stuff is well known, for example when they prove the inequality between the arithmetic and geometrical mean

for positive elements.

Other people

The polynomial in u

(u-x)(u-x)=0 may be interpreted as having a solution u=x+ey, where e is another abstract number where ee=0, similar to i.

The system x+ey is known as the ring of dual numbers, whoes algebric properties are easily deduced.

It can be given as the matrix

x 0

y x

The transpose of e , e(T) is equally valid for this purpose, as -i is to i, in matrix form -i = i(T).

In terms of algebra i=e -e(T), and e e(T)+e(T)e =I, the unit matrix.

In still another system, the Hyperbolic, is x+ty, where tt=1. That is all that is necesary, without assuming t=1 or -1.

So this t is also abstract.

This can be given in terms of the matrix

x y

y x

Solution to (u-x)(u-x)-yy=0

These two are rings with closure , commutative. Similar to the complex, except the complex is also a field.

They are invertible over almost all the parameter space of the reals.

I suppose over Z they are also number rings, to play around with.

I do not think that all these possibilities have properly been explored, especially with more variables

ie. x+ty+ttz, where ttt=1 or -1? Or have they?

But no problem extending the Complex like over rings (not fields)

regards, Juan

Juan,

you are free to name the source where my reform

I√cI = 1/2 min (IaI + IbI); valid for all a, b: a * b = c

already was represented. Where? When?

`i´ represents the question (exercise, task) to take the square-root. But it wasn´t done! Repeating old definitions doesn´t disproof my theory (knowledge).

For addition you may take ` ç´ (or every strange type) as a not solved problem:

4 + 3 = ç => (4 + 3) + 2 = ç + 2 (no solution in real numbers base 10)

=> ç – 3 + 2 = ((4 + 3) – 3) + 2 = 4 + 2 = 6

[the same in addition like i in multiplication]

It´s very much deepest level clarification. I am prepared.

Peter,

Frankly, I didn't understand what are you talking about.

If one has any specific proposition, he states and proves it.

Where is your question? What is your claim?

There is nothing clear in between your lines!

Show me any clear mathematical statement, then one can judge

if it is true or not.

I√cI = 1/2 min (IaI + IbI); valid for all a, b: a * b = c

What is this? It has no meaning.

(*) as not solved problem!!

In fact, there is no problem.

Are you trying to define an operation over some set?

(*) `i´ represents the question (exercise, task) to take the square-root.

No, the complex i, is not to take the square root. Who said that?

Issam,

so I can´t help you.

Peter,

I'm trying to help you.

Dear Issam,

thanks for your offer to help. Please visit my homepage first. Then we may talk about to make you ready to see the propositions.

Dear Peter,

Yes, I visited the site www.mathe-neu.de . Your message: The current mathematics is full of contradictions. Especially, with the complex i. But this is not the case. Can you show one contradiction?

Dear Issam,

yes, at my homepage most about imaginary — as far as being worked out yet — only is represented in German. German papers are ready for last preparing and will come soon.

Now a little bit in English:

One could solve the exercise `square-root´. Even if the radicand is negative. I did. This is part of Part 13. (in German and link to ESJ) of my homepage.

If the square-root looks for the — anyway equal in value — two factors which could produce the radicand, then it is not necessary to exclude these with different prefixes. Take them also and whole the area gets source.

The so called `unit´ i never could lead to -1 because there couldn’t exist a square of any unit. Contradiction

If i should define a special unit, the radicand has to depend on two units. Because the radicand is taken from the reals (as best one unit), we have the second contradiction.

To square a real number leads to positives exclusively by definition of algebra. The radicand, which has to be a product (by inverse operation to square-root), is negative. It couldn’t exist by definition of algebra (rules of multiplication on binoms of the Babyloniens). Contradiction

So the definition — if one takes it as a proof — comes as a circular argument.

Regards, Peter

Dear Peter,

What is the complex number i ?

I think you have a misunderstanding with the quantity i. What is i?

Recall that:

The story started with the solution of quadratic equations that have negative discriminant. For example:

Is there any real number such that x^2 + x + 1= 0,

or x^2 + 1= 0?. The answer is NO.

One can prove that x^2 + x + 1 > 0 and x^2 + 1 > 0 for all real numbers x.

In general, If b^2 -4ac < 0 , then ax^2 +bx + c ≠ 0 for any real x.

In fact, If a>0 and b^2 -4ac < 0 , then ax^2 +bx + c > 0 for any real x

and If a < 0 and b^2 -4ac < 0 , then ax^2 +bx + c < 0 for any real x.

Now, the mathematicians think to change (extend) the field of real numbers R so that they can find solutions for the equation

ax^2 +bx + c =0, where b^2 -4ac < 0 in the new field.

For this purpose, they defined a new set C = { (a,b), a,b belongs R } with new addition and multiplication operations (+), (.) on the set C such that

(a, b) + (c, d) = ( a+c, b+d)

(a, b).(c, d) = ( ac-bd, ad+bc)

C equipped with (+), (.) (C , (+), (.) ) is a field.

(The proof is direct by showing all axioms of the field) This new field is called:

the field of complex numbers. You can observe that R is a subset of C

where R={(a,0), a belongs R}.

We set a for (a,0) to simplify notations.

Now, every complex number z, is an ordered pair, i.e z = (x,y)

where x is called the real part and y is called the imaginary part

( names by convention).

Go back to the equation x^2 + x + 1= 0, it has negative discriminant so, It has no solutions in R. Can we solve this equation in C?

We want to find (a,b) in C such that

(a,b)^2 + (a,b) + (1,0) = (0,0)...*

Apply formula the multiplication operation: (a,b)^2=(a,b).(a,b)= (a^2 - b^2 , 2ab)

Substitute in the equation *

(a^2 - b^2 , 2ab) + (a,b) + (1,0)= (0,0).

It splits into two equations in R

a^2 - b^2 +a + 1 = 0.

2ab +b = b(2a+1) = 0.

from the second equation, we have b = 0 or a = -1/2.

For b=0, we get a^2 +a + 1 = 0, the original equation in R

which is not solvable.

But if a = -1/2, we obtain b^2 = 1/4 -1/2 +1 = 3/4, with real solutions

are b = sqrt(3)/2 or b = - sqrt(3)/2

hence, (-1/2, sqrt(3)/2) and ( -1/2, - sqrt(3)/2 )

are two solutions for x^2 + x + 1 = 0 , in the field C.

Similarly, x^2 + 1= 0 has the solutions

(0, 1) and (0, -1) .

You can observe (0,1).(0,1) = (-1,0) .

To simplify the notations, we named (0,1) = i.

i is an ordered pair (0,1).

The statement sqrt(-1) = i is a popular notation, used probably by engineers to carry calculations. The exact mathematical meaning is

(0,1).(0,1) =( -1,0)

So, the imaginary unit i = (0,1), is logical and live in the mentioned field C,

called the field of complex number.

Best wishes

Dear Issam,

that´s recalling old stuff. I know all of that (by studying). But you didn´t give any commant on the contradictions. Which way you represent the unsolved problem (√-1) doesn´t matter. Look at the definition of `circular argument´. x^2 + 1 = 0 only has solutions by that! You can´t stop the reform. Truth ever will succeed.

Regards, Peter

What I think about i, is that ii=-1, is a bit too sketchy on its own., although enough to get the addition and multiplication rules.

Numbers also operational properties, as well as intrinsic number properties.

To ilustrate consider

i(x+iy) = -y +ix

So i transforms a vector (x,y) into (-y,x); you eventually find i and -i rotating (x,y) in opposite directions, when you draw the points on a graph.

The number 2 operationally transforms a nonzero number, to one double its magnitude, as well as a counting meaning, or just a real number.

Dear Juan,

I agree that using, ii = -1, one can get the addition and multiplication rules. This is what the students do in their calculations of complex numbers, but we should keep in mind what is the exact mathematical meaning of complex numbers, to show and prove new results and to make advances in the topic. Also, to avoid confusion with other different algebraic structures over the plane.

This usually happened with many students and instructors who are not mathematicians. They make fatal errors all the time.

Concerning your example:

Transformations are one of the important applications on the field

of complex numbers. As an example f(z): C → C

defined by f(z) = k.z, z ≠ 0 , k is complex constant,

is a complex mapping called transformation. In your example k = i = (0,1)

applying the multiplication rule over complex numbers,we have

f(z)= k.z = iz = (0,1).(x, y)= (0×x-1×y, 0×y+1×x) = (-y, x),

exactly the same as you did (equivalent statements)

iz = i(x+iy)= ix + ii(y) = -y + ix .

f(z) = iz acts as transformation, it rotates (x,y)

( which can be thought as a vector) by the angle (π/2) anticlockwise,

we obtain a new complex number (or vector) (-y, x).

Details are available in any elementary complex textbook.

Such applications are geometric consequences of the

algebra of the complex field.

Best regards

Dear Peter,

√(z₀)=(z₀)1/2 is equivalent to say z² = z₀,

and using the same approach as in my previous answer, we

solve z² = z₀, to determine the roots for

z² = -1 ⇒ x² - y² + 2ixy = -1,

then x² - y² = -1 and 2xy = 0,

solving the two equations we obtain

x=0 or y = ±1

therefore, √(-1) =(-1)1/2 has two values

√(-1) = (0,1) =i or √(-1) = (0,-1) = - i.

If we interested in the principal value, we select

one value √(-1) = i.

You may have another interpretation of the complex i.

Why not, you can build your model that is equivalent to the field

of the complex numbers with different interpretations!

Your new algebraic postulates should be consistent if so,

this new proposed model should be equivalent to the algebraic model of the complex field. Otherwise, it is a wrong one simply because the complex field is consistent.

Best regards

A57. First part of a new document

Answer 57, posted on April 26, 2019

========================

I have started to rewrite and improve some of my answers

to some of the questions of this thread

in the following document:

========================

Polynomials whose graphs intersect at integer points

https://www.researchgate.net/publication/332686211_Polynomials_whose_graphs_intersect_at_integer_points

========================

With best regards, Jean-Claude

Dear Jean-Claude Evard ,

I noticed that you used the statement (the graph of f(z) ) many times in your article: Polynomials whose graphs intersect at integer points ). I prefer to change the title. For the following reason:

There is is no graphs for w=f(z) over the complex field.

f(z) the complex map is considered as a transformation from

(x,y) plane into (u,v) plane.

You said:

P(z) = (z-i)(z-i) and q(z)=(z-1)(z+2) have the same value 10 at at z = 3,

and you deduced that the graph of p(z) and q(z) intersects at ( 3,10).

Mathematically, this is not accepted.

You can paraphrase as the following:

the mapping P (z) transforms the point (3,0) into the point (10, 0)

and the mapping q(z) transforms the point (3,0) into the point (10, 0).

That is both p(z) and q(z) transform the point (3,0) in the xy-plane into

the point (10,0) in the uv - plane.

In fact, there is an infinite number of transformations that have the same property.

For example the family of complex functions g(z) = (z- 3)h(z) +10, where h(z) is any arbitrary complex function. g(z) has the same property, it transforms (3,0) into (10,0).

Best regards

Dear Issam Kaddoura ,

I will not say some thinks to everything. Field is special topic. No command on the contradictions?

But you did (disproved) it by yourself.

I. K.: √(-1) = (0,1) = i or √(-1) = (0,-1) = - i.

√(-1) = i => i = √(-1) [commutativity by `=´]

i = √(-1) = √(-1) = -i [your equations in line]

i = √(-1) = -i [reflexivity]

i = -i [transitivity]

2i = 0 [addition of i]

No command.

Regards, Peter

Dear Peter Kepp ,

There are no contradictions at all.

You have confusion with fundamental concepts.

NO one says (-1)^1/2 = i = -i !!!! this is a wrong claim.

One can say (-1)^1/2 equals i or - i.

In the complex field, (-1)^1/2 is a multiple - valued quantity.

I said in the previous answer:

the expression (-1)^1/2 is equivalent to solve z^2 = -1,

(The notations are done by convention), and then we

have two distinct roots i and - i.The principal value of (-1)^1/2 = i .

I think you need to distinguish between the different concepts,

the exact definition, the setup of notations of the complex variables and the corresponding relations.

Best regards

Jean-Claude Evard and Issam Kadoura,

By my knowledge, the graph is the mathematical notion which is defined for any function, and generally, for any relation!!!

However, in the answer (a day ago), Issam Kadoura wrote:

“There is is no graphs for w=f(z) over the complex field.”

Best regards

Romeo Meštrović ,

This is your wrong knowledge.

It is better to sketch the graph of p(z)=( z-i)(z-i) if you can,

rather than showing unprofessional statements.

As usual, aggressive respond by you against my answers!!

P(z) is a transformation which sends regions(domain) from the xy -plane into regions in the uv - plane.

Probably you have a misunderstanding of some graphs to show domain coloring by using the tools of complex functions !!!.

Peter, why stop there?

Why not 1 = √(1) = -1 ?

Then all numbers are zero, and everyone can go home.

This is an example of bastardizing notation.

By definition, i is the principle root of x^2 + 1.

You cannot use "magic" to assert that the two roots of this quadratic are equal.

See list (Part A) available at RG page https://www.researchgate.net/publication/332353557_List_of_publications_by_topics

of my 24 papers in Complex and Functional Analysis! Moreover, my PhD Dissertation concerns the topics of Complex, Functional and Real Analysis.

Dear Issam,

I will not waste my time in arguing against the statements of poeple who are not well-informed about the logic of proofs. You ignored `transitivity´ (don´t you?), so you are out.

Last help: By √(-1) = i as well as by √(-1) = -i no operation is represented; both are declarations, which should be written by the use of a triple line instead of the equal-sign (not your fault; special type is missing here).

If the roots of the equations were to handle as operations, I wouldn´t see a problem by two different results. But that insight comes by logic!

Extension, some hours later:

The full result of an ambiguous radicand (only one prefix) must be written as a logically complete (single) result by `or´:

√(+1) = +1 and +1 or -1 and -1

√(-1) = +1 and -1 or -1 and +1

Because both (by and) are the factors the root is looking for.

The multi-prefix-rule (by my reform) brings products like:

++1, +–1, –+1, or ––1 (the complete prefixes of both the factors)

Their roots are well defined. No problem by `ë.

@ all

Because there are no well founded arguments against the disproved theory on `i´. We may turn the pages over and focus on the first part of another problem in classical math; the binoms.

The Babylonians used two. We were teached by three. By them there are two different values (a and b) and two different prefixes (+ and –). Full the variation leads to 16 different equations. Why are only three of them teached?

If they should represent the rules of multiplication of factors including prefixes, the binom

(+a – b )(+a – b) = ((+a)(+a)) + ((+a)(-b)) + ((-b)(+a)) + ((-b)(-b))

fulfills all different combinations.

The left side needs three operations for a result. The right side seven. What is the advantage of the right side in operating? Shouldn´t there be any advantage?

Who finds (ad hoc) the missing ones (excluding logical repeating)?

Dear Peter,

According to your web site. It may be you have to build a model for the complex field via a special Functor F: (usual complex field C) →(peter model of C)

F should preserve all properties of the first field, the elements, and operations, etc. otherwise, the model is not accepted. Riemann, the great mathematician, redefined the domain of some mappings over the complex plane and constructed the famous Riemann surfaces, interchanged the complex plane by the new surfaces to get rid of the multiple values of some mappings.

Hope you will correct the conflicts and contradictions of your model to stay consistent with the current complex space.

Best regards

Romeo Meštrović

No doubt, you are a good mathematician with good experience.

Our conflict is with the style of how to show our answers.

In your previous answer, you declared that my answer is wrong.

But you didn't show any evidence. You posted a quick and biased answer without careful reading to the contents. I asked you to sketch the graph, this will provide a better, useful and helpful reply for all followers. Guessing is not the right way in a serious scientific discussion.

Really, I wish the best for you and all researchers who seek knowledge.

wish you more academic achievements.

Best regards

There is an interesting connection with calculus from the ring of dual nos.

in that

f(x+ey) = f(x) +ey (df/dx)

where ee=0

It is easily shown from the cut off of the Taylor expansion after second term.

The log is alternative from Complex variables.

ln(x+ey)= ln [(x)(1+ey/x)] = ln(x) + ln(1+ey/x) =ln(x) +ey/x

which shows that 1/x is the derivative of ln(x) with rspect to x, algeraically.

Differently from ln(z), we are constrained to x>0 and do not have the complicationn of Riemann sheets.

I will bet on the circumstance that nobody who is not able to see the theory about `i´ as ruled out would be able to extend the isometry of:

1) A refers to B as (a + b)(a + b) refers to a^2 + 2ab + b^2

C refers to D as ————— refers to -a^2 – 2ab – b^2

2) A refers to B as (a – b)(a – b) refers to a^2 – 2ab + b^2

C refers to D as ————— refers to -a^2 + 2ab – b^2

How to represent C by 1) and 2); wich stay for the 4th and 5th binom?

If you are able, why not teaching?

Dear Juan Weisz ,

Dual numbers is another example of an embedding of the real number in a commutative ring. They constitute an algebraic ring structure which is useful in many applications, in particular, the quantum mechanics.

On the other hand, complex numbers provide powerful tools and theorems that help to study and solve almost all problems.

Dual numbers have excellent simple behavior but are limited to particular cases.

In general, the analysis of dual numbers (RING) is more much complicated than that of the complex numbers that have the structure of a complete field.

I think that any embedding(algebraic structure) of the real numbers by a complete field should be isomorphic to the complex field.

Best regards

Dear Peter Kepp ,

Your last question in the thread is not clear!

I won the bet! Even the question isn´t understood.

A75. Definition of the graph of a function

Answer 75, posted on April 28, 2019

========================

To answer the comments of Issam Kaddoura in answers 58, 62, and 68,

and the comment of Romeo Meštrović in answer 61,

according to the definition of the graph of function:

https://en.wikipedia.org/wiki/Graph_of_a_function#Definition

if f is a function from Z[i] into Z[i],

and if f(3) = 10, then the ordered pair (3, 10)

of Gaussian integers 3 and 10 belongs to the graph of f.

========================

So, there is no mistake about this

in my document:

Polynomials whose graphs intersect at integer points

https://www.researchgate.net/publication/332686211_Polynomials_whose_graphs_intersect_at_integer_points

========================

It is well-known that we cannot “draw the graph”

of a function from C into C. For this, we would need

to draw it in a four-dimensional space: See:

Tetraview:

https://en.wikipedia.org/wiki/Tetraview

========================

In spite of the fact that we live in a universe

containing particles living in higher dimensions,

we are locked into a 3-dimensional space,

and therefore, we cannot “draw the graph” of a function from C to C.

========================

Concerning the problem of “dimensions” of our universe, see:

Extra dimensions:

https://en.wikipedia.org/wiki/String_theory#Extra_dimensions

“In bosonic string theory, spacetime is 26-dimensional”,

“in superstring theory, it is 10-dimensional”,

“in M-theory it is 11-dimensional”.

========================

With best regards, Jean-Claude

Dear Jean,

You have arranged my answer in a new form:

you said that p,q: Z[i] → Z[i]

exactly what I have said p(3,0) =(10,0)

q(3,0)= (10,0) both transforms (3,0) into (10,0).

it is a transformation which transforms numbers into numbers

that is F( region or set) = region or set.

Your answer was:

(https://www.researchgate.net/publication/332686211_Polynomials_whose_graphs_intersect_at_integer_points)

You said that (3,10) belongs to the graphs of both polynomials p and q,

p(3), q(3), (3,10), notice that all are not well defined!

Even you want to adjust and change your document,

you need to use 4- dimensional space and set ( (3,0),(10,0) ), to stay consistent with the accepted definition.

In any elementary textbook of complex variables, one can read and study

the complex functions as transformations and following some absurd document out of billion facts doesn't give one the right to defend wrong arguments.

Your new answer:

(It is well-known that we cannot “draw the graph”

of a function from C into C. For this, we would need

to draw it in a four-dimensional space)

agree with my previous answer, in the presence of a domain(Region)

that show ( Region, f(Region) ), which is another topic (manifolds) different than what we are talking about!! it is visualized as a real manifold subset of R4, where

all ( x,y, u(x,y) ,v(x,y)) all are real components.

To read the following textbook is useful:

(1) https://math.unice.fr/.../Brown-Churchill-complex%20Variables%20and%20Application..

The chapters 2 (page 35, 8 (page 311), 9( page(355),

this will help to understand the complex mappings.

PS.

The web site: https://en.wikipedia.org/wiki/Graph_of_a_function#Definition

doesn't support your answer or the used notations.

Best regards

Dear Peter Kepp ,

Your answer:

( I won the bet! Even the question isn´t understood. )

Do you think that was a good reply about our serious discussion concerning your ideas about the complex number i.

It seems we were wasting time. And all your ideas were jokings!!

Issam

My understanding is that x+ey is a commutative ring but not a field, in that not all nonzero elements are invertible.

For example here the inverse of ey alone does not exist, and that ey is an ideal of the ring.

The inverse can be taken if x is not zero.

I do not know if there is any intermediate clasification for this kind of ring and the ring of integers, where the inverse cannot be taken for any element.

The only metric I can see here is x itself.

I tend to agree that the only two varible commutative field of continuous variables is the Complex numbers, where all nonzero elements are invertible.

Probably Im a minority, but I see more usefull here the more general concept of ring, than field, in grouping together

the complex with a host of other cases (rings). ie, dont see that much difference.

My last questions were meant rhetorical — no exercises. The answers could be done by copy and paste because all the 16 combinations of types (a, b) and prefixes (+, –) are represented at my homepage.

Now the second part of this problem (binoms of the Babyloniens).

Did anyone, who reads here, had read Martinez (negative math)?

The binoms are not an unambiguous proof for the multiplication and division of prefixed values (+a, +b, -a, -b).

To take the inverse rule also is plausible (circular argument). Reform is needed.

I am looking for the lions of oz. Who wants to be courageous?

"Pay no attention to that man behind the curtain. The Great Oz has spoken."

I believe tha discussions related to the proposed question would be extended to the set of quaternions, i.e., the expressions of the form:

a + bi + cj + dk

where a ,b,c and d are real numbers and i, j and k are symbols that can be interpreted as unit vectors pointing along the three spatial axes. Under usual operations addition and scalar multiplication, the set of quaternions is made a 4 dimensional vector space over the field of real numbers, with {1, i, j, k} as a basis. Moreover, under the so-called Hamilton product which follows the distributive law (in which 1 is the multiplicative unit, i2= j2 = k2 = -1, ij = -ji = k, jk = - kj = i, ki = -ik = j), the set of quaternions becomes a division algebra.

It is interesting to notice that the equation

z2 + 1 = 0,

has infinitely many quaternion solutions, which are the quaternions bi + cj + dk such that b2 + c2 +d2 = 1. Hence, these "roots of –1" form a unit sphere in the three-dimensional space of vector quaternions.

Romeo Meštrović

The disproved field as well as the right use of any unit in math should be called later by me at this topic. If you want to be informed know, you may have a look at my homepage (www.mathe-neu.de).

A 83 On the ambiguity of the result of prefixed factors

Answer 83, posted on April 29, 2019

=============================

Continnuing the 2nd part to binoms:

classic: (a – b)(a – b) = a^2 – 2ab + b^2

`Martinez´: (a – b)(a – b) = -a^2 + 2ab – b^2

classic: (2 – 3)(2 – 3) = (-1)(-1) = +1

`Martinez´: (2 – 3)(2 – 3) = (-1)(-1) = -1

Why is Martinez correct? You only will ask if you transport the solution of `classic´ to `Martinez´. Do it complete and it seems well.

`Martinez´:

plus by plus as well as minus by minus => minus

plus by minus as well as minus by plus => plus

`Martinez´:

(+2)(+2) = -4

(+2)(-3) = +6

(-3)(+2) = +6

(-3)(-3) = -9

Complete: (2 – 3)(2 – 3) = (-1)(-1) = -1

By `binom-way´: (2 – 3)(2 – 3) = -4 + 6 + 6 – 9 = -1

Why isn’t there any differentiaton between the prefix (as the position) and the symbol (as the order to combine; addition, subtraction) by the binoms?

What about the first material contact to visiting friendly aliens who use the way to calculate like Martinez does? Their manoeuvre may base on our informations about coordinates. Their calculations could bring the crash.

Reform shows unambiguous solution to that problem (be friendly to every alien).

A84. From the definition (a, b) to the usual notation a + bi

Answer 84, posted on April 29, 2019

========================

To answer the comments of Issam Kaddoura in answer 76,

about the first version of my document

https://www.researchgate.net/publication/332686211_Polynomials_whose_graphs_intersect_at_integer_points

posted on April 26, 2019,

the notation z = (a, b) is used only at the beginning

of most books of complex analysis,

in the definition of complex numbers,

and just after this definition.

Immediately after this, these books say that

the tradition is to use the notation

z = a + bi = a(1, 0) + b(0, 1) = (a, b).

Immediately, after this, these books say that

when b = 0, the tradition is to write z = a

instead of z = a + 0i or z = (a, 0).

Consequently,

the sentence “z = 3” is correct,

the sentence “z = 3 + 0i” is correct but unusual,

the sentence “z = (3, 0)” is correct but unusual.

Furthermore, if we have a function from C into C

such that f(3) = 10,

then

the sentence “f(3) = 10” is correct and usual,

the sentence “f(3 + 0i) = 10 + 0i” is correct but unusual,

the sentence “f((3, 0)) = (10, 0)” is correct, but unusual.

If G denotes the graph of f, then

the sentence “(3, 10) is in G” is correct and usual,

the sentence “(3 + 0i, 10 + 0i) is in G" is correct, but unusual,

the sentence “((3, 0), (10, 0)) is in G” is correct, but unusual.

========================

With best regards, Jean-Claude

Dear Romeo Meštrović ,

Nice observation in your post about the quaternions q = a +ib +jc + kd with the notations mentioned

in your answer ( i2= j2 = k2 = -1, ij = -ji = k, jk = - kj = i, ki = -ik = j ) that extends the complex numbers.

I have a little bit to add:

The real numbers have an extension into the complex numbers ( a +ib) and the complex numbers

have an extension into the quaternion numbers (q = a +ib +jc + kd ) and the quaternions has an

extension into the octonions ( noncommutative and nonassociative ) normed division algebra

and one can determine many solutions for z2 = -1.

Octonions are helpful to study the black holes entropy and quantum information science.

A duplication of octonions generate the sedenions.

It is the magic of Abstract Algebra.

Best regards

Dear Jean-Claude Evard ,

I am not a referee for your article. It was a bit of advice that helps to put things in better form.

You can reject or accept advice.

The usual concept is that no graphs for complex functions, they are transformations.

If one for some reason needs to duplicate complex values into R4 , then the usual notation

is to set ( x,y,u,v) belongs to dimension four to show the case.

If p(z) = (z- i)(z + i): Z[i] into Z[i]. It transforms (3,0) into (10,0).

You insist to use (3,10) which is mathematically wrong!!!

The right notation is ( (3,0), (10,0)) belongs to Z[i]xZ[i] in dimension 4.

Now, if we transform (3,1): p(3,1) = (9,6)

It is a must to use ( (3,1), (9,6)) belongs to Z[i]xZ[i]

We can't reduce the number of tuples from (x,y,u,v) into (a,b),

and how you will fix this case in your article?

Wish you to produce nice articles free of any wrong notations.

Best regards

Dear Andrew J. Woldar ,

Thank you for the advice. Sometimes one needs to have a (big) heart and listen to all.

Best regards

Personally I do not see quaternions as the best way to expand the Complex (NON COMMUTATIVE) You want comutative

so you go via rings.

JW

Yes, with the quaternions you lose commutativity, and with the octonions you further lose associativity. But why is it important to preserve these properties? Is it for Nature's sake? I fear that Nature would vehemently disagree with you.

For example, Lie algebras are anti-commutative and non-associative, yet they play a vastly important and extensive role in mathematics and quantum mechanics. Octonions are crucial to string theory and quantum computing.

Besides, when it comes to extensions over the reals, you don't have many choices. I believe that was the original point that Professor @Issam Kaddoura was making.

Dear Juan,

You are right; we prefer the commutative and associative operations for the algebraic structure.

But this was far-fetched in the algebraic structures with higher dimensions.

Notice that we extend the real numbers into complex numbers to gain a new dimension, the new structure preserves the Field axioms, but we lost the ( order property).

So to gain more degree of freedom, we should sacrifice some (friendly) properties to stay with an accepted well defined algebraic structure.

The great importance of the new structures was mentioned by professor Andrew J. Woldar where higher dimensions are essential in the applied physics.

See for example the text

www.theworld.com/~sweetser/quaternions/ps/book.pdf

Best regards

Dear Professor Issam Kaddoura,

Thank you very much for the link you provided to Professor Sweetser's book. It looks extremely interesting. I hope to eventually get around to reading it!

Best regards,

Andrew

"The great importance of the new structures was mentioned by professor Andrew J. Woldar where higher dimensions are essential in the applied physics". Not only in physics, but in geometry and number theory too ! After the discovery of the quaternions by W.R. Hamilton on 16 October 1843 during a walk with his wife ("presumably doomed to silence"), a vast theory of non commutative K-algebras has been developped and has been flourishing until now. By a K-algebra we mean a finite dimensional vector space over a field K, equipped with a not necessarily commutative but associative K-linear multiplication. For example, a K-algebra of matrices with fixed dimension.

The theme of central simple K-algebras A (the center, i.e. the set of elements of A which commute with all A, coincides with K, and A admits no sub-algebra other than A and K) is particularly prominent. In geometry, quaternion algebras (of dimension 4, generalizing Hamilton's) are associated with the study of conics, and central simple algebras (of higher dimensions) with that of the so-called Severi-Brauer varieties. In number theory, the classification of central simple algebras has given birth to the so called Brauer group, the study of which culminated in the monument of the 20-th century known as "class-field theory". As stressed by @Andrew J. Woldar, "Nature vehemently disagrees with commutativity".

@Juan Weisz : A field is just a ring in which every non null element is invertible. So commutativity is a specificity of neither rings nor fields. You encounter non commutative rings every day, for instance when multiplying matrices. Not to say anything of non commutative groups !

A94. Traditions can cause heart attacks

Answer 94, posted on April 30, 2019,

on the comments of Issam Kaddoura in answer 86,

about my document:

https://www.researchgate.net/publication/332686211_Polynomials_whose_graphs_intersect_at_integer_points

========================

If f is a function from C into C,

then by definition of the graph of a function

https://en.wikipedia.org/wiki/Graph_of_a_function#Definition

the graph of f is the set of all ordered pairs (z, f(z))

such that z belongs to C.

========================

Therefore, if f(3) = 10, then

(3, f(3)) = (3, 10) belongs to the graph of f.

========================

In most books of complex analysis,

complex numbers are defined as ordered pairs (a, b)

of real numbers a and b:

https://en.wikipedia.org/wiki/Complex_number#Construction_as_ordered_pairs

========================

Immediately after the definition,

the following usual notation is introduced:

(a, b) = a(1, 0) + b(0, 1) = a + bi,

and the following simplification is introduced:

(a, 0) = a.

========================

Consequently, in the above example where f(3) = 10,

we have

3 = 3 + 0i = (3, 0)

and

10 = 10 + 0i = (10, 0).

========================

At this point, mathematicians risk a heart attack,

because the equalities 3 = (3, 0) and 10 = (10, 0)

look horribly wrong, and they are obtained

by applying the definitions 100% correctly.

========================

Furthermore, the point (3, 10) of the graph of f

is (3, 10) = ((3, 0), (10, 0)),

which causes a second heart attack to mathematicians,

because it seems that R^2 = R^4,

and all scientists know that this is wrong.

========================

If we want a third heart attack, we may observe

that when we construct the set of rational numbers,

the pair (2, 3) represents the rational number

2/3 = 4/6 = 6/9 = 8/ 12 = … = (-2)/(-3) = … ,

while when we work with complex numbers,

we have (2, 3) = 2 + 3i.

From this, we are horrified to obtain

2/3 = (2, 3) = 2 + 3i,

which gives us a third heart attack.

========================

Therefore, to be rigorous,

we should use a subscript Q when we work in Q

and a subscript C when we work in C:

(2, 3)_Q = 2/3 and (2, 3)_C = 2 + 3i,

but the tradition is to omit these subscripts.

========================

We may also observe that

3 is “essentially” the same as (3, 0)_C,

but it is not “absolutely” the same.

The set R is embedded into the set C,

but it is not contained in C.

We have a one-to-one correspondence

between 3 and (3, 0).

To be rigorous, we should add the subscript C:

3_C = (3, 0)_C

========================

Thus, in my above example, I should write

(3_C, 10_C) is in the graph of f,

but the tradition is to omit the subscript C.

========================

With best regards, Jean-Claude

of course, being mostly in Physics, do not deny the noncommutative properties coming through operators in QM,

angular momentum and rotations.

However looking into the case of numbers , there is some kind of slogan coming via math, to the effect that since only fields are good, one could not go beyond the Complex. In the case of numbers, noncommutative properties seem to give these a property

of unreal, because you are never on solid constructive ground.

So when I found out that with commutative rings, invertible over most of the parameter space, you could go to any number of independent

variable, including 3, even giving generalized Cauchy Riemann, I decided that this line of propaganda that physics seemed to be recieving from math was , intentionally or not, wrong, I have decided to combat it a bit. Do not impovrish math just because phyicists are using something else.

Thong, your remark

That is what I also said.

There are categories of matrices, that if you stay within the special kind, the elements of whih are commutative with each other.

The simplest examples are

x -y

y x

and

x y

y x

and circulant in general.

Dear Jean,

A heart attack is created with defending wrong claims! According to me, your notations are completely wrong; the whole article is full of unaccepted mathematical facts and notations. This is my right to evaluate what I have read. It seems you have a problem with the higher dimensions!! To talk about graphs for (z,f(z)), you should use four real components. It is not a tradition!! It is a must.

I agree that in the dimension two one can consider the real x as (x,0),

and the imaginary y as (0,y), because there are no other choices.

In dimension 4. mathematicians will never use (a,b) to represent numbers in Z[i]xZ[i] because it is wrong notation and one can think of several possibilities:

(a,b,0,0), (0,a,0,b), ((0,0,a,b), ..etc.

Your statement:

>

My answer is: your suggested notations are not well defined.

(3,10) has the following possible representations:

(3,10,0,0), (0,0,3,10), (3,0,0,10),(3,0,10,0), (0,3,0,10),(0,3,10,0)

So, the notation (3,10) has no meaning in Z[i]xZ[i].

You think that one should use (3,0,10,0) out of all, why?

By the way, you answered yourself, when you wrote:

>

Best regards

@Juan Weisz

Your latest reply is full of confusion. You write:

1) However looking into the case of numbers , there is some kind of slogan coming via math, to the effect that since only fields are good, one could not go beyond the Complex.

* Who says that ? Certainly not number theorists ! Hamilton invented the quaternions in 1843, and the descendants of these (called "skew fields") could go beyond the complex numbers precisely because they abandoned commutativity. In number theory, as I stressed many times, arithmetic only exists in rings of integers. The recourse to fields, or K-algebras, or analytic tools, or whatever, is one fascinating characteristic of number theory. Not only for technical reasons, but for conceptual reasons which I've tried to evoke in a previous post recalling the history of the notion of "number".

2) In the case of numbers, noncommutative properties seem to give these a property of unreal, because you are never on solid constructive ground.

* Totally wrong. This is a complete misunderstanding of the nature of the development of math. You seem to believe that it's based on random introduction of new exotic axioms. These must of course be logically coherent (whatyou call "solid constructive ground"), but this is not sufficient. They deserve consideration only when they produce interesting developments (tools or ideas), either to help to solve anciently unassailable problems, or to open new unsuspectable domains. The history of the final demonstration of Fermat's LT is a perfect illustration. To give an example outside the strict domain of math., recall the 5-th axiom of euclidian geometry: "Through a point outside a line, you can draw one and only one parallel to that line". Euclid himself was not convinced of its necessity. Only after it was abandoned could the so called non-euclidian geometries flourish... and give birth to the theories of Relativity (Special and General) in physics.

3) Thong, your remark. That is what I also said. There are categories of matrices, that if you stay within the special kind, the elements of which are commutative with each other.

* I fear you misunderstand the notion of categories. Loosely speaking, a category consists of objects endowed with a structure (say, groups) and of applications between these which respect this structure (say, group homomorphisms). If you consider e.g. matrices of given size (nxn), with coefficients in a fixed field, they form a non commutative ring w.r.t. addition and multiplication, but nothing prevents this ring from containing a subring which is commutative (your example). The notion of commutativity does not belong to the axioms of rings. The notion of fields gives a similar example : a field is a commutative field; a non commutative field is called a skew field. This somewhat confusing terminology is of course due to historical reasons (see above).

A98. Definition of the graph of a function

Answer 98, posted on May 1, 2019,

on the comments of Issam Kaddoura in answer 96,

about my document:

https://www.researchgate.net/publication/332686211_Polynomials_whose_graphs_intersect_at_integer_points

========================

In my above document, I am using the usual definition

of the graph of a function:

https://en.wikipedia.org/wiki/Graph_of_a_function#Definition

========================

Definition:

Let E and F be non-empty sets.

Let f be a function from E into F.

Then the graph of f is the set of all ordered pairs (x, f(x))

such that x belongs to E.

========================

Therefore, if f is a function from C into C

such that f(3) = 10,

then (3, f(3)) = (3, 10) belongs to the graph of f.

========================

With best regards, Jean-Claude

Thong

Im glad number theorist dont think that.

However the weight of number theory in math applied to science , mostly math. analysis and such is very relative.

Typically we know very little about specific topics in number theory, so therefore you and I dont appreciate

how others, in very different areas may think.

Im almost sure that somewhere in math there is , or was some theorem saying you cannot go beyond Complex numbers.

(I would guess the condition is that you cannot continue with commutative and field at the same time)

Perhaps our confusion is that the term number is not a technical one in math.

If I say that I cannot work with noncommutative numbers as a tool of measurement in science, there is a reason, which

I tried to explain. The concept of area would go up in smoke. Is area xy or yx?

It is quite possible that I have misused the term category as you might use it, but I think you understoood me.

Again, some matrices within a subset of given form commute among themselvs; their eigenvalues can be treated as numbers.

Think about this, then answer.

Best regards, Juan

A 100 posted on May 2, 2019

Jean-Claude Evard

Thanks for giving numbers to your answers. By including the observation of all other answers you help to have a structure for reverse looking. Best!

I asked RG for that, but self was too lazy to do it at the topics I am involved.

If everyone would do as you there may come a confusion because some more answers could reach before posting self.

What Issam and Thong wouldn’t accept is, that there has to be a reform.

I consequently (as often as mighty) write `number (of amount)´ if meaning how many of what is meant. By the reform every mathematical quantity (which could be combined with another quantity) has to be expressed by three expressions:

1) number (of amount) [element of the naturals excluding zero or rationals]

2) measure [in square-brackets, to display incommensurability; 1, π, e, √2, …]

3) unit [pure number, length, width, height, area, volume, arc, …, distance, time, speed, accelaration, ..., pressure, …]

Regards and thank you very well, Peter