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paulunderwood 2020-10-10 13:40

Prime Numbers Book
 
13 Attachment(s)
Enjoy and criticize. It ain't no C&P but might help some. :book:

After 20 downloads, I have made the changes to typos and Mersenne data. A new copy is now uploaded.

2 More downloads and I have uploaded a new version without the Euler Phi Function nonsense -- Thanks Robert.

24 more downloads and a new version is up with accumulated corrections and suggested text.

13 more downloads. The replacement (12th Oct) version corrects some grammatical errors, includes suggestions by Nick and a Lucas Sequence algorithm.

12 more downloads. The replacement (Oct 15th) makes the suggestions by Nick and corrects some typos.

4 more downloads. The replacement builds on suggestions by Nick. There is now a section on Gaussian primes. Replacement (18 Oct,2020) uploaded.

Dr Sardonicus 2020-10-10 14:10

There is a caveat in using the term "number" to mean "natural number" or "positive integer."

Euclid distinguished between 1 ("the unit") and integers greater than 1, which he called "numbers." Thus, a "prime number" was, [i]by definition[/i] an integer greater than 1.

The unit, 1, is neither prime nor composite.

kriesel 2020-10-10 14:28

spell check; grammar
 
Legendre, J[B]o[/B]cobi and Kronecker (should be Jacobi) (page 3)

[SIZE=2]"The determinant of a A is" (bottom of page 9)[/SIZE]
The determinant of a [B]matrix[/B] A is

nothing else conspicuous through page 10 on a first pass.

Pretty clean.

Viliam Furik 2020-10-10 14:52

Page 3 - Typo in "Jocobi"
Page 32 - Fig. 7.1 contains n = 2 twice, lots of unknown where it shouldn't be (n= 61,89,107)

R. Gerbicz 2020-10-10 15:59

On page 39:
"The Euler Phi Function is useful since for all a
a^(1+φ(n)) ≡ a (mod n)."

This is very false, a small counter-example is a=2, n=4.

bhelmes 2020-10-10 20:46

"It is important to analize the the exponentiating" page 49.


Greetings
Bernhard

Knut 2020-10-10 20:53

You write i were the square root of -1.


As far as I know, the square root r of a number x is the non-negative solution of the equation r^2 = x as far as real numbers are concerned.



As i is neither positive nor negative and i^2 = -1 and (-i)^2 = -1 in my opinion it is not correct to say i would be the square root of -1.


Best regards


Knut

paulunderwood 2020-10-10 22:40

[QUOTE=Knut;559490]You write i were the square root of -1.


As far as I know, the square root r of a number x is the non-negative solution of the equation r^2 = x as far as real numbers are concerned.



As i is neither positive nor negative and i^2 = -1 and (-i)^2 = -1 in my opinion it is not correct to say i would be the square root of -1.


Best regards


Knut[/QUOTE]

I have read [url]https://en.wikipedia.org/wiki/Imaginary_unit[/url] What would you say I should write? [I]i[/I] is fixed to be one of the two square roots of [I]-1[/I]?

paulunderwood 2020-10-10 22:43

I have tidied up what I want write about "The Fermatian Child".

I will refrain from reposting the paper until more input from folks.

Knut 2020-10-11 05:28

I think there is no square root of -1.


Instead, define the "imaginary unit" i with i^2 = -1. And then define complex numbers.


This, in my opinion, could help to avoid disturbances.



E.g.


x^2 = 3*i
x^2 = - 3*i
x^2 = 2 + 3*i
x^2 = 2 - 3*i
x^2 = - 2 + 3*i
x^2 = - 2 - 3*i



There are two (complex) solutions of each equation, of course. Which of the two solutions is to be named the "square root"? I am not aware of a definition.


I apologize fpr my poor command of the English language and hope you can see my point.


Best regards


Knut

Nick 2020-10-11 07:44

There are a number of errors at the start, some just details of definitions but others hiding more important points.
How it would be best to resolve these depends strongly on who you are writing this for.
What is the intended audience?

paulunderwood 2020-10-11 08:40

[QUOTE=Nick;559539]There are a number of errors at the start, some just details of definitions but others hiding more important points.
How it would be best to resolve these depends strongly on who you are writing this for.
What is the intended audience?[/QUOTE]

The man in the street with a little interest in primes and testing :grin:

Please detail what the errors are so that I can fix them.

Nick 2020-10-11 22:26

[QUOTE=paulunderwood;559547]The man in the street with a little interest in primes and testing :grin:

Please detail what the errors are so that I can fix them.[/QUOTE]
Feedback on version dated 11 October 2020 (wow you've written a lot!)
p5 The proof that root 2 is irrational relies on the uniqueness of prime factorization.
As this is a special property of the integers (not true in all number systems), I would at least mention it.
p6 Ordinary sets are unordered and may not have repeating elements (this is so that they correspond with properties - if you select all objects satisfying a certain condition, you want what you get to be a set). Orderings and multisets can be constructed from ordinary sets if needed (in fact, so can everything in mathematics).
p6 "subtracting all elements of one set from another" sounds confusing to me, as if you are calculating x-y for each x in the first set and y in the second. I would consider "removing"
p7 The group axioms \(e\circ g=g\) needs to be \(g\circ e=g\), or state it and the next one both ways around.
p7 under multiplication the units of a commutative ring with 1 form an abelian group, but these are not all the elements of the ring (except in the trivial case where the ring has just 1 element).
p8 If you want to be able to relax condition 11 then you need to write condition 12 both ways around!
p8 A field is also required to have 1 not equal to 0.
I'm out of time now - I'll look again in the week.

paulunderwood 2020-10-12 05:33

[QUOTE=Nick;559610]Feedback on version dated 11 October 2020 (wow you've written a lot!)
[/quote]
Thank you very much for taking tome to read the text. I have made some changes to my local copy but will upload after more changes.
[quote]
p5 The proof that root 2 is irrational relies on the uniqueness of prime factorization.
As this is a special property of the integers (not true in all number systems), I would at least mention it.
[/quote].
I mention it now in the both the paragraph on integers and when arguing about the irrationality of root 2.
[quote]
p6 Ordinary sets are unordered and may not have repeating elements (this is so that they correspond with properties - if you select all objects satisfying a certain condition, you want what you get to be a set). Orderings and multisets can be constructed from ordinary sets if needed (in fact, so can everything in mathematics).
[/quote]
I have added the word "not".
[quote]
p6 "subtracting all elements of one set from another" sounds confusing to me, as if you are calculating x-y for each x in the first set and y in the second. I would consider "removing"
[/quote]
I replaced the word "subtract" with "remove",
[quote]
p7 The group axioms \(e\circ g=g\) needs to be \(g\circ e=g\), or state it and the next one both ways around.
[/quote]
I have subtended e rather than prepending it and in all subsequent instances where identity existence in mentioned including the section on rings,
[quote]
p7 under multiplication the units of a commutative ring with 1 form an abelian group, but these are not all the elements of the ring (except in the trivial case where the ring has just 1 element).
[/quote]
I found this difficult. I have made mention of the center, although it might be out of context:
[quote]
\item multiplicative identity : $(\exists 1\in {\cal S})(\forall a\in{\cal S})\hspace{.1in}a\times 1=a$.
The units form an Abelian group apart from a ring with one element, called the {\em center}.
[/quote]
[quote]
p8 If you want to be able to relax condition 11 then you need to write condition 12 both ways around!
[/quote]
Done!
[quote]
p8 A field is also required to have 1 not equal to 0.
[/quote]
I have added "where $1\ne 0$".
[quote]
I'm out of time now - I'll look again in the week.[/QUOTE]
Again, many thanks,

Edit I have uploaded t[URL="https://mersenneforum.org/showthread.php?t=26067"]he latest copy -- 12 Oct[/URL].

Nick 2020-10-15 09:09

Some feedback on the version dated 12 October 2020:

2.1 Primes.
I find the first sentence confusing (though that may just be me).
Your explanation of what prime numbers are at the start of chapter 1 was clearer.

2.2 Euclid's algorithm
If you are going to use both "greatest common divisor" and "highest common factor",
perhaps saying that they are 2 names for the same thing would make it clearer.
You say the gcd is the last non-zero remainder but don't explain what to do if the 1st remainder was 0 already.
You explain why the last non-zero remainder divides a and b but not why it must be the highest number to do so.

3. Pythagoras
You show that a+b and a-b are both even and that no prime except 2 divides them both,
and you also have that their product is a square.
You conclude that a+b and a-b are each 2 times a square.
But the prime factorizations of a+b and a-b could each contain an even number of 2's
- you need to rule that out as well before writing a+b=2s² and a-b=2t².
(You are of course making it harder for yourself by working in the integers instead of the Gaussian integers here.
As you introduced complex numbers in chapter 1, you could consider using Gaussian integers.)

paulunderwood 2020-10-15 15:48

I have tried to correct and clarify the text and arguments as Nick pointed out in the previous post. I have uploaded the latest version -- Oct 15th, 2020 -- to [URL="https://mersenneforum.org/showthread.php?p=559426#post559426"]the original post.[/URL]

I have not written about Gaussian integers, trying to restrict the text to natural primes,

Nick 2020-10-17 10:00

Feedback on version dated 15 October 2020:

7 Mersenne Numbers
Formula for factorization of \(M_{pq}\) is almost right!

9.1 Legendre symbol
Say p is an odd prime or p not equal to 2.
Typo bottom of page 37: elements of A should go up to 10 instead of 11.

12 Frobenius
Example: taking the polynomial \(x^2+1\), you get the Gaussian integers modulo n after all!

Obviously it's up to you, but it might be worth including something on the Chinese Remainder Theorem.
It would make it easier to explain your formula for the Euler phi function.
Also, as you have a nice emphasis on the computational side of things in this, you could show
how it is used in practice to speed up RSA decryption, for example.
Just a thought, anyway.

paulunderwood 2020-10-17 23:57

[QUOTE=Nick;560123]

12 Frobenius
Example: taking the polynomial \(x^2+1\), you get the Gaussian integers modulo n after all!

Obviously it's up to you, but it might be worth including something on the Chinese Remainder Theorem.
It would make it easier to explain your formula for the Euler phi function.
Also, as you have a nice emphasis on the computational side of things in this, you could show
how it is used in practice to speed up RSA decryption, for example.
Just a thought, anyway.[/QUOTE]

There is now a section on Gaussian primes, but not mod n.

CRT: I find that quite difficult.

RSA: I have written that s or t may be chosen to be small for quick encoding or quick decoding respectively.

Thanks again for your input. [URL="https://mersenneforum.org/showthread.php?t=26067"]18th Oct, 2020 version uploaded[/URL]

a1call 2020-10-18 00:35

I love the book Paul.
Thank you very much.


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