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Posted (edited)

We wonder the primality of prime numbers and why they are located in specific locations.

I discovered a way of why this is the way it is. Primes and their locations are cause by this multiplicative-odd-number series process:

First the odd numer 3 is multiplied to itself and above 3 odds:

3*3=9

3*5=15

3*7=21

3*9=27

and so on...

Second the next odd of 3 which is 5:(same process like above)

5*5=25

5*7=35

5*9=45

5*11=55

and so on

Third, fouth and so on.This process will continue unto infinity.

let's find out using the above process why 2,3,5,7 are primes below 10.

3*3= 9

5*5=25

here, the odd numbers except 1 are automatically primes below 9 because there is no other way the odd 3 can be multiplied except to itself and above it. Since 3*3= 9 therefore all the odd numbers except 1 are prime numbers. 5*5 also can't accomodate that because it equals to 25.

 

Note: Below 10 is only an example. Actually it can explain all prime numbers. Sorry for my bad English if you don't understand what I'm saying.

Edited by Randolpin
Posted

So what you are saying is that prime numbers smaller than any number are not the products of any primes below that number. That is not surprising. It is what "prime" means. And the basis of Eratosthenes Sieve. (But I suppose a small amount of credit is due for working out what "prime number" means.)

Posted
2 hours ago, Randolpin said:

let's find out using the above process why 2,3,5,7 are primes below 10.

I couldn't see where that shows 2 is prime?

  • 1 month later...
Posted

In a small way, this dialogue highlights a prevalent issue in mathematics.  Many mathematicians are very talented and fluent in the received wisdom of their discipline;  but they lack a proper understanding of its philosophical issues.  In particular, they often do not fully understand the concept of 'proof'.  For example, certain middle-level theorems in mathematics require that the number 0 be treated as an even number.  Some mathematicians consider this to be a proof that 0 is in fact an even number, without any further reference to number theory or the axioms of arithmetic.  No competent second-year philosophy student would fall into such an error;  why then is it so prevalent in mathematics?

Posted
17 minutes ago, amplitude said:

In a small way, this dialogue highlights a prevalent issue in mathematics.  Many mathematicians are very talented and fluent in the received wisdom of their discipline;  but they lack a proper understanding of its philosophical issues.  In particular, they often do not fully understand the concept of 'proof'.  For example, certain middle-level theorems in mathematics require that the number 0 be treated as an even number.  Some mathematicians consider this to be a proof that 0 is in fact an even number, without any further reference to number theory or the axioms of arithmetic.  No competent second-year philosophy student would fall into such an error;  why then is it so prevalent in mathematics?

The only thing this thread highlights is a misunderstanding of the word cause.

Posted
21 hours ago, amplitude said:

In a small way, this dialogue highlights a prevalent issue in mathematics.  Many mathematicians are very talented and fluent in the received wisdom of their discipline;  but they lack a proper understanding of its philosophical issues.  In particular, they often do not fully understand the concept of 'proof'.  For example, certain middle-level theorems in mathematics require that the number 0 be treated as an even number.  Some mathematicians consider this to be a proof that 0 is in fact an even number, without any further reference to number theory or the axioms of arithmetic.  No competent second-year philosophy student would fall into such an error;  why then is it so prevalent in mathematics?

    You say "some mathematicians consider this to be a proof that 0 is in fact an even number" but your "this" appears to refer to "certain middle-level theorems in mathematics require that the number 0 be treated as an even number".  I don't know of any mathematician who would think that.  I am not sure what you mean by "mathematician" but I would expect any undergraduate math major to know that "0 is even" because 0= 2*0.

Posted
22 minutes ago, HallsofIvy said:

I would expect any undergraduate math major to know that "0 is even" because 0= 2*0.

Or that adding up even numbers gives an even number.

2+0 =2

 

It's laughable that you need anything complicated to prove it.

Posted
On 10/14/2017 at 9:53 AM, amplitude said:

In a small way, this dialogue highlights a prevalent issue in mathematics.  Many mathematicians are very talented and fluent in the received wisdom of their discipline;  but they lack a proper understanding of its philosophical issues.  In particular, they often do not fully understand the concept of 'proof'.  For example, certain middle-level theorems in mathematics require that the number 0 be treated as an even number.  Some mathematicians consider this to be a proof that 0 is in fact an even number, without any further reference to number theory or the axioms of arithmetic.  No competent second-year philosophy student would fall into such an error;  why then is it so prevalent in mathematics?

I am sure, given your response, that you do not know many mathematicians.

 

The closest your response comes to reality is if mathematicians were answering a different question - of why our definition of even numbers should include 0. We could have decided to care only about positive even integers. There are even some situations where that would be useful. But in nearly any useful situation where positive even integers satisfy a condition, so do all even integers. In some sense, evenness "carves nature at the joints"; "positive evenness" does not. So why should we restrict ourselves to looking at positive even integers, rather than all even numbers?

Posted (edited)

I figured out when I was much younger that the sum of any of the digits of a prime number never come to factors of 9. Some random examples

 

953 = 9 + 5 + 3 = 17 = 1 + 7 = 8

1087 = 1 + 8 + 7 = 16 = 1 + 6 = 5

3187 = 3 + 1 + 8 + 7 = 19 = 1 + 9 = 10

3691 = 3 + 6 + 9 + 1 = 19 = 1 + 9 = 10

7793 = 7 + 7 + 9 + 3 = 26 = 2 + 6 = 8

 

And I am quite sure the rule continues. I thought in my younger years it was a discovery, but it turned out there was statement of it somewhere and no doubt has a name which I certainly don't remember now. I think the jury is out, but most mathematicians tend to think it is likely there is some rule that describes them because they do show patterns from time to time, such as this one above - I think there is a rule out there since it is obeying certain principles within its structure. Who knows?

Why it avoids factors of nine is interesting, for me at least, because the sum of its components in the multiplicative table show interesting anomalies, such as a palindrome made of prime numbers

[18][27][36][45][54][72][81]

The difference of 1 with 8 is 7. The difference of 2 and 7 is 5... and so on, produces:

75311357

Which is a palindrome constructed of prime numbers (if you take 1+1) = 2. For some reason, this always interested me, probably for naive reasons. 

What can be said of it, is that this is the first four prime numbers without running into double digits - again, if and only if you consider the factors of 1 portraying the prime number 2.

Edited by Dubbelosix
  • 1 month later...
Posted

  "Casting out nines' says that if you add the digits of any  number, and, if the result has more than one digit, add digits again, repeating until you have a single digit, the result will be the remainder when the original number is divided by 9.  For example, the sum of digits of 3242747 is 3+ 2+ 4+ 2+ 7+ 4+ 7= 29, 2+ 9= 11, 1+ 1= 2.  9 divides into 3242747  360305 times with remainder 2.  As you can see from the Wikipedia article studio links to, this method is over a thousand years old.

Posted
On 11/19/2017 at 8:29 AM, DrKrettin said:

Does this all work for other bases? Would it be casting out of 7s in octal, or casting out of Fs in hexadecimal?

0, 0; 1, 1;

10, 1; 11, 10, 1;

100, 1; 101, 10, 1; 110, 10, 1; 111, 11, 10, 1;

The binary version definitely prefers 1 more than 0, surprise.

0, 0; 1, 1; 2, 2;

10, 1; 11, 2; 12, 10, 1;

100, 1; 101, 2; 102, 10, 1; 110, 2; 111, 10, 1; 112, 11, 2; and we appear to be alternating between 1 and 2, but all even numbers above 2 are non-primes.

0, 0; 1, 1; 2, 2; 3, 3;

10, 1; 11, 2; 12, 3; 13, 10, 1;

20, 2; 21, 3; 22, 10, 1; 23, 11, 2;

30, 3; 31, 10, 1; 32, 11, 2; 33, 12, 3; and our only 3, which is the analogue of 9, Did not appear again after three itself.

Grunt work complete.

Posted (edited)

In response to the OP before reading through the entire thread, your premise has already been covered in the thread that discussed my discovery of a prime factor harmonic matrix, though your misunderstanding concept of what causes the locations of primes numbers to be where they are.

The locations of primes are not directly distributed by the periodicity of their prime factors.

They are indirectly determined by the positions of the composite numbers that the periodicity of prime factors create.

Another words, Primality or the positions of prime numbers are only detemined by what’s left over after all composite number positions within a given range have already been determined. Which is exactly how number siev’s work. Note, that since a number sieve was the first image I used in that post, you can see it in the link I provided which is making me rethink the first image that I put in posts from now on, since it gives the impression that I’m describing a number seive and not a PFHM

That’s one of the reasons, though not the main reason, why it takes so long to calculate their positions when they get large and also why they seem to have a random like property in their distribution.

Edited by TakenItSeriously
Posted
On ‎11‎/‎19‎/‎2017 at 6:29 AM, DrKrettin said:

Does this all work for other bases? Would it be casting out of 7s in octal, or casting out of Fs in hexadecimal?

Presuming you mean if casting out works in other bases without any regard to the OP and Primes, the answer is yes. Note that in base ten the digital root is congruent mod 9. (Note that numbers divisible by 9 return 9 for the digital root and 0 for mod 9, but the results are congruent) Similarly, casting out 7s in octal is congruent mod 7, e.g. 97 in octal is 141 and the digital root in octal is 1+4+1=6 and 97 mod 7 = 6. In octal, a number with a digital root of 7 divides evenly by 7. This holds for all bases. I have no proof but recall seeing one some years ago.

As to binary, all digital roots = 1.

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