Another Proof of Euclid's Theorem (infinite number of primes)?












-1














Here $mathbb N = {2,3,4,dots}$ with the binary operation of addition.



If $m in mathbb N$ we denote by $G_{mathbb N} (m)$ the semigroup generated by $m$.



Definition: A number $p$ is said to be prime if for all $m lt p$, $;p notin G_{mathbb N} (m) $.



We denote the set of non-empty finite subsets of $mathbb N$ by $mathcal F (mathbb N)$.



Let $mathtt E$ be a function



$quad mathtt E: mathbb N to mathcal F (mathbb N)$



satisfying the following:



$quad quadquadforall n in mathbb N$



$tag 0 mathtt E (2) = {2}$



$tag 1 text{ If } (forall text{ prime } p lt n) ; n notin G_{mathbb N} (p) text{ then } mathtt E (n) = {n}$



$tag 2 text{ If } , (exists text{ prime } p lt n) ; n in G_{mathbb N} (p) text{ then } mathtt E (n) text{ is the union of all such primes}$



$tag 3 mathtt E (n+1) cap mathtt E (n) = emptyset$



We have the following result:



Theorem 1: There exist one and only one function $mathtt E$ satisfying $text{(0)}$ thru $text{(2)}$; it will also satisfy $text{(3)}$. Moreover, for every $n$, all the numbers in the set $mathtt E (n)$ are prime (the prime 'factors').




Question: Can the theorem be proved in this $(mathbb N,+)$ setting?




If yes, we can continue.



Theorem 2: The set of all prime numbers is an infinite set.

Proof

If $a_1$ is any number, consider the 'next further out' number



$tag 4 a_2 = sum_{i=1}^{a_1+1}, a_1 = sum_{i=1}^{a_1},( a_1 + 1)$.



A simple argument using $text{(3)}$ shows that $mathtt E (a_1) subsetneq mathtt E (a_2);$ (c.f. Bill Dubuque's remark).



Employing recursion we get a sequence $a_1, a_2, a_3,dots$ with a corresponding chain of strictly increasing sets



$quad mathtt E (a_1) subsetneq mathtt E (a_2) subsetneq E (a_3) dots$



So there are sets of prime numbers with more elements than any finite set. $blacksquare$



My Work



Please see



Using the recursion theorem to implement the Sieve of Eratosthenes.



The proof of theorem 2 is along the lines found in the proof given by Filip Saidak. Also, if we set $a_1$ to $1$ in theorem 2 we get the researched OEIS sequence A007018.



Note that the proof supplied by Filip Saidak has most likely been known for many years; see Bill Dubuque's answer to the math.stackexchange.com question



Is there an intuitionist (i.e., constructive) proof of the infinitude of primes?






abstract-algebra elementary-number-theory prime-numbers alternative-proof semigroups







share|cite|improve this question
























  • Note, multiplication (of positive integers) is recursive addition. That is, we define $acdot1$ to be $a$ and $acdot(n+1)$ to be $acdot n+a$.
    – Barry Cipra
    Nov 22 '18 at 17:39












  • @BarryCipra Yes I know. But I get excited when something that appears 'bound to multiplication' - the prime numbers - can get released into a more elementary framework.
    – CopyPasteIt
    Nov 22 '18 at 17:44






  • 1




    The proof via $,n(n!+!1),$ has more prime factors than $n$ for $,nge 1$ is surely much older than Saidak's 2005 paper. The generated sequence is OEIS A007018. Note: adding $1$ yields Sylvestoer sequence $a_{n+1} = a_n^2 - a_n + 1 = $ OEIS A000058. See the OEIS notes for other connections.
    – Bill Dubuque
    Nov 22 '18 at 18:05








  • 1




    @CopyPasteIt The least factor $> 1$ of $n$ is prime (i.e. irreducible) is already quite minimal.
    – Bill Dubuque
    Nov 22 '18 at 19:08








  • 1




    @CopyPasteIt Well one can unwind everything down to Peano arithmetic, but that won't be very arithmetically enlightening. Why prefer assembly language over the beautiful higher-level language carefully crafted by number theorists over many centuries?
    – Bill Dubuque
    Nov 22 '18 at 19:37


















-1














Here $mathbb N = {2,3,4,dots}$ with the binary operation of addition.



If $m in mathbb N$ we denote by $G_{mathbb N} (m)$ the semigroup generated by $m$.



Definition: A number $p$ is said to be prime if for all $m lt p$, $;p notin G_{mathbb N} (m) $.



We denote the set of non-empty finite subsets of $mathbb N$ by $mathcal F (mathbb N)$.



Let $mathtt E$ be a function



$quad mathtt E: mathbb N to mathcal F (mathbb N)$



satisfying the following:



$quad quadquadforall n in mathbb N$



$tag 0 mathtt E (2) = {2}$



$tag 1 text{ If } (forall text{ prime } p lt n) ; n notin G_{mathbb N} (p) text{ then } mathtt E (n) = {n}$



$tag 2 text{ If } , (exists text{ prime } p lt n) ; n in G_{mathbb N} (p) text{ then } mathtt E (n) text{ is the union of all such primes}$



$tag 3 mathtt E (n+1) cap mathtt E (n) = emptyset$



We have the following result:



Theorem 1: There exist one and only one function $mathtt E$ satisfying $text{(0)}$ thru $text{(2)}$; it will also satisfy $text{(3)}$. Moreover, for every $n$, all the numbers in the set $mathtt E (n)$ are prime (the prime 'factors').




Question: Can the theorem be proved in this $(mathbb N,+)$ setting?




If yes, we can continue.



Theorem 2: The set of all prime numbers is an infinite set.

Proof

If $a_1$ is any number, consider the 'next further out' number



$tag 4 a_2 = sum_{i=1}^{a_1+1}, a_1 = sum_{i=1}^{a_1},( a_1 + 1)$.



A simple argument using $text{(3)}$ shows that $mathtt E (a_1) subsetneq mathtt E (a_2);$ (c.f. Bill Dubuque's remark).



Employing recursion we get a sequence $a_1, a_2, a_3,dots$ with a corresponding chain of strictly increasing sets



$quad mathtt E (a_1) subsetneq mathtt E (a_2) subsetneq E (a_3) dots$



So there are sets of prime numbers with more elements than any finite set. $blacksquare$



My Work



Please see



Using the recursion theorem to implement the Sieve of Eratosthenes.



The proof of theorem 2 is along the lines found in the proof given by Filip Saidak. Also, if we set $a_1$ to $1$ in theorem 2 we get the researched OEIS sequence A007018.



Note that the proof supplied by Filip Saidak has most likely been known for many years; see Bill Dubuque's answer to the math.stackexchange.com question



Is there an intuitionist (i.e., constructive) proof of the infinitude of primes?






abstract-algebra elementary-number-theory prime-numbers alternative-proof semigroups







share|cite|improve this question
























  • Note, multiplication (of positive integers) is recursive addition. That is, we define $acdot1$ to be $a$ and $acdot(n+1)$ to be $acdot n+a$.
    – Barry Cipra
    Nov 22 '18 at 17:39












  • @BarryCipra Yes I know. But I get excited when something that appears 'bound to multiplication' - the prime numbers - can get released into a more elementary framework.
    – CopyPasteIt
    Nov 22 '18 at 17:44






  • 1




    The proof via $,n(n!+!1),$ has more prime factors than $n$ for $,nge 1$ is surely much older than Saidak's 2005 paper. The generated sequence is OEIS A007018. Note: adding $1$ yields Sylvestoer sequence $a_{n+1} = a_n^2 - a_n + 1 = $ OEIS A000058. See the OEIS notes for other connections.
    – Bill Dubuque
    Nov 22 '18 at 18:05








  • 1




    @CopyPasteIt The least factor $> 1$ of $n$ is prime (i.e. irreducible) is already quite minimal.
    – Bill Dubuque
    Nov 22 '18 at 19:08








  • 1




    @CopyPasteIt Well one can unwind everything down to Peano arithmetic, but that won't be very arithmetically enlightening. Why prefer assembly language over the beautiful higher-level language carefully crafted by number theorists over many centuries?
    – Bill Dubuque
    Nov 22 '18 at 19:37
















-1












-1








-1







Here $mathbb N = {2,3,4,dots}$ with the binary operation of addition.



If $m in mathbb N$ we denote by $G_{mathbb N} (m)$ the semigroup generated by $m$.



Definition: A number $p$ is said to be prime if for all $m lt p$, $;p notin G_{mathbb N} (m) $.



We denote the set of non-empty finite subsets of $mathbb N$ by $mathcal F (mathbb N)$.



Let $mathtt E$ be a function



$quad mathtt E: mathbb N to mathcal F (mathbb N)$



satisfying the following:



$quad quadquadforall n in mathbb N$



$tag 0 mathtt E (2) = {2}$



$tag 1 text{ If } (forall text{ prime } p lt n) ; n notin G_{mathbb N} (p) text{ then } mathtt E (n) = {n}$



$tag 2 text{ If } , (exists text{ prime } p lt n) ; n in G_{mathbb N} (p) text{ then } mathtt E (n) text{ is the union of all such primes}$



$tag 3 mathtt E (n+1) cap mathtt E (n) = emptyset$



We have the following result:



Theorem 1: There exist one and only one function $mathtt E$ satisfying $text{(0)}$ thru $text{(2)}$; it will also satisfy $text{(3)}$. Moreover, for every $n$, all the numbers in the set $mathtt E (n)$ are prime (the prime 'factors').




Question: Can the theorem be proved in this $(mathbb N,+)$ setting?




If yes, we can continue.



Theorem 2: The set of all prime numbers is an infinite set.

Proof

If $a_1$ is any number, consider the 'next further out' number



$tag 4 a_2 = sum_{i=1}^{a_1+1}, a_1 = sum_{i=1}^{a_1},( a_1 + 1)$.



A simple argument using $text{(3)}$ shows that $mathtt E (a_1) subsetneq mathtt E (a_2);$ (c.f. Bill Dubuque's remark).



Employing recursion we get a sequence $a_1, a_2, a_3,dots$ with a corresponding chain of strictly increasing sets



$quad mathtt E (a_1) subsetneq mathtt E (a_2) subsetneq E (a_3) dots$



So there are sets of prime numbers with more elements than any finite set. $blacksquare$



My Work



Please see



Using the recursion theorem to implement the Sieve of Eratosthenes.



The proof of theorem 2 is along the lines found in the proof given by Filip Saidak. Also, if we set $a_1$ to $1$ in theorem 2 we get the researched OEIS sequence A007018.



Note that the proof supplied by Filip Saidak has most likely been known for many years; see Bill Dubuque's answer to the math.stackexchange.com question



Is there an intuitionist (i.e., constructive) proof of the infinitude of primes?






abstract-algebra elementary-number-theory prime-numbers alternative-proof semigroups







share|cite|improve this question















Here $mathbb N = {2,3,4,dots}$ with the binary operation of addition.



If $m in mathbb N$ we denote by $G_{mathbb N} (m)$ the semigroup generated by $m$.



Definition: A number $p$ is said to be prime if for all $m lt p$, $;p notin G_{mathbb N} (m) $.



We denote the set of non-empty finite subsets of $mathbb N$ by $mathcal F (mathbb N)$.



Let $mathtt E$ be a function



$quad mathtt E: mathbb N to mathcal F (mathbb N)$



satisfying the following:



$quad quadquadforall n in mathbb N$



$tag 0 mathtt E (2) = {2}$



$tag 1 text{ If } (forall text{ prime } p lt n) ; n notin G_{mathbb N} (p) text{ then } mathtt E (n) = {n}$



$tag 2 text{ If } , (exists text{ prime } p lt n) ; n in G_{mathbb N} (p) text{ then } mathtt E (n) text{ is the union of all such primes}$



$tag 3 mathtt E (n+1) cap mathtt E (n) = emptyset$



We have the following result:



Theorem 1: There exist one and only one function $mathtt E$ satisfying $text{(0)}$ thru $text{(2)}$; it will also satisfy $text{(3)}$. Moreover, for every $n$, all the numbers in the set $mathtt E (n)$ are prime (the prime 'factors').




Question: Can the theorem be proved in this $(mathbb N,+)$ setting?




If yes, we can continue.



Theorem 2: The set of all prime numbers is an infinite set.

Proof

If $a_1$ is any number, consider the 'next further out' number



$tag 4 a_2 = sum_{i=1}^{a_1+1}, a_1 = sum_{i=1}^{a_1},( a_1 + 1)$.



A simple argument using $text{(3)}$ shows that $mathtt E (a_1) subsetneq mathtt E (a_2);$ (c.f. Bill Dubuque's remark).



Employing recursion we get a sequence $a_1, a_2, a_3,dots$ with a corresponding chain of strictly increasing sets



$quad mathtt E (a_1) subsetneq mathtt E (a_2) subsetneq E (a_3) dots$



So there are sets of prime numbers with more elements than any finite set. $blacksquare$



My Work



Please see



Using the recursion theorem to implement the Sieve of Eratosthenes.



The proof of theorem 2 is along the lines found in the proof given by Filip Saidak. Also, if we set $a_1$ to $1$ in theorem 2 we get the researched OEIS sequence A007018.



Note that the proof supplied by Filip Saidak has most likely been known for many years; see Bill Dubuque's answer to the math.stackexchange.com question



Is there an intuitionist (i.e., constructive) proof of the infinitude of primes?






abstract-algebra elementary-number-theory prime-numbers alternative-proof semigroups




abstract-algebra elementary-number-theory prime-numbers alternative-proof semigroups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 '18 at 12:11







CopyPasteIt

















asked Nov 22 '18 at 17:10









CopyPasteItCopyPasteIt

4,0651627




4,0651627












  • Note, multiplication (of positive integers) is recursive addition. That is, we define $acdot1$ to be $a$ and $acdot(n+1)$ to be $acdot n+a$.
    – Barry Cipra
    Nov 22 '18 at 17:39












  • @BarryCipra Yes I know. But I get excited when something that appears 'bound to multiplication' - the prime numbers - can get released into a more elementary framework.
    – CopyPasteIt
    Nov 22 '18 at 17:44






  • 1




    The proof via $,n(n!+!1),$ has more prime factors than $n$ for $,nge 1$ is surely much older than Saidak's 2005 paper. The generated sequence is OEIS A007018. Note: adding $1$ yields Sylvestoer sequence $a_{n+1} = a_n^2 - a_n + 1 = $ OEIS A000058. See the OEIS notes for other connections.
    – Bill Dubuque
    Nov 22 '18 at 18:05








  • 1




    @CopyPasteIt The least factor $> 1$ of $n$ is prime (i.e. irreducible) is already quite minimal.
    – Bill Dubuque
    Nov 22 '18 at 19:08








  • 1




    @CopyPasteIt Well one can unwind everything down to Peano arithmetic, but that won't be very arithmetically enlightening. Why prefer assembly language over the beautiful higher-level language carefully crafted by number theorists over many centuries?
    – Bill Dubuque
    Nov 22 '18 at 19:37




















  • Note, multiplication (of positive integers) is recursive addition. That is, we define $acdot1$ to be $a$ and $acdot(n+1)$ to be $acdot n+a$.
    – Barry Cipra
    Nov 22 '18 at 17:39












  • @BarryCipra Yes I know. But I get excited when something that appears 'bound to multiplication' - the prime numbers - can get released into a more elementary framework.
    – CopyPasteIt
    Nov 22 '18 at 17:44






  • 1




    The proof via $,n(n!+!1),$ has more prime factors than $n$ for $,nge 1$ is surely much older than Saidak's 2005 paper. The generated sequence is OEIS A007018. Note: adding $1$ yields Sylvestoer sequence $a_{n+1} = a_n^2 - a_n + 1 = $ OEIS A000058. See the OEIS notes for other connections.
    – Bill Dubuque
    Nov 22 '18 at 18:05








  • 1




    @CopyPasteIt The least factor $> 1$ of $n$ is prime (i.e. irreducible) is already quite minimal.
    – Bill Dubuque
    Nov 22 '18 at 19:08








  • 1




    @CopyPasteIt Well one can unwind everything down to Peano arithmetic, but that won't be very arithmetically enlightening. Why prefer assembly language over the beautiful higher-level language carefully crafted by number theorists over many centuries?
    – Bill Dubuque
    Nov 22 '18 at 19:37


















Note, multiplication (of positive integers) is recursive addition. That is, we define $acdot1$ to be $a$ and $acdot(n+1)$ to be $acdot n+a$.
– Barry Cipra
Nov 22 '18 at 17:39






Note, multiplication (of positive integers) is recursive addition. That is, we define $acdot1$ to be $a$ and $acdot(n+1)$ to be $acdot n+a$.
– Barry Cipra
Nov 22 '18 at 17:39














@BarryCipra Yes I know. But I get excited when something that appears 'bound to multiplication' - the prime numbers - can get released into a more elementary framework.
– CopyPasteIt
Nov 22 '18 at 17:44




@BarryCipra Yes I know. But I get excited when something that appears 'bound to multiplication' - the prime numbers - can get released into a more elementary framework.
– CopyPasteIt
Nov 22 '18 at 17:44




1




1




The proof via $,n(n!+!1),$ has more prime factors than $n$ for $,nge 1$ is surely much older than Saidak's 2005 paper. The generated sequence is OEIS A007018. Note: adding $1$ yields Sylvestoer sequence $a_{n+1} = a_n^2 - a_n + 1 = $ OEIS A000058. See the OEIS notes for other connections.
– Bill Dubuque
Nov 22 '18 at 18:05






The proof via $,n(n!+!1),$ has more prime factors than $n$ for $,nge 1$ is surely much older than Saidak's 2005 paper. The generated sequence is OEIS A007018. Note: adding $1$ yields Sylvestoer sequence $a_{n+1} = a_n^2 - a_n + 1 = $ OEIS A000058. See the OEIS notes for other connections.
– Bill Dubuque
Nov 22 '18 at 18:05






1




1




@CopyPasteIt The least factor $> 1$ of $n$ is prime (i.e. irreducible) is already quite minimal.
– Bill Dubuque
Nov 22 '18 at 19:08






@CopyPasteIt The least factor $> 1$ of $n$ is prime (i.e. irreducible) is already quite minimal.
– Bill Dubuque
Nov 22 '18 at 19:08






1




1




@CopyPasteIt Well one can unwind everything down to Peano arithmetic, but that won't be very arithmetically enlightening. Why prefer assembly language over the beautiful higher-level language carefully crafted by number theorists over many centuries?
– Bill Dubuque
Nov 22 '18 at 19:37






@CopyPasteIt Well one can unwind everything down to Peano arithmetic, but that won't be very arithmetically enlightening. Why prefer assembly language over the beautiful higher-level language carefully crafted by number theorists over many centuries?
– Bill Dubuque
Nov 22 '18 at 19:37












1 Answer
1






active

oldest

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It is valid, but seems to be fundamentally the same as the classical proof; both hinge upon the fact that $p_1times p_2times cdotstimes p_n+1$ is divisible by some prime not in ${p_1,p_2,dots,p_n}$.






share|cite|improve this answer

















  • 1




    @Copy It is poor form to change the question in a way that invalidates existing answers.
    – Tobias Kildetoft
    Nov 23 '18 at 17:08











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It is valid, but seems to be fundamentally the same as the classical proof; both hinge upon the fact that $p_1times p_2times cdotstimes p_n+1$ is divisible by some prime not in ${p_1,p_2,dots,p_n}$.






share|cite|improve this answer

















  • 1




    @Copy It is poor form to change the question in a way that invalidates existing answers.
    – Tobias Kildetoft
    Nov 23 '18 at 17:08
















1














It is valid, but seems to be fundamentally the same as the classical proof; both hinge upon the fact that $p_1times p_2times cdotstimes p_n+1$ is divisible by some prime not in ${p_1,p_2,dots,p_n}$.






share|cite|improve this answer

















  • 1




    @Copy It is poor form to change the question in a way that invalidates existing answers.
    – Tobias Kildetoft
    Nov 23 '18 at 17:08














1












1








1






It is valid, but seems to be fundamentally the same as the classical proof; both hinge upon the fact that $p_1times p_2times cdotstimes p_n+1$ is divisible by some prime not in ${p_1,p_2,dots,p_n}$.






share|cite|improve this answer












It is valid, but seems to be fundamentally the same as the classical proof; both hinge upon the fact that $p_1times p_2times cdotstimes p_n+1$ is divisible by some prime not in ${p_1,p_2,dots,p_n}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 22 '18 at 17:22









Rafay AsharyRafay Ashary

83618




83618








  • 1




    @Copy It is poor form to change the question in a way that invalidates existing answers.
    – Tobias Kildetoft
    Nov 23 '18 at 17:08














  • 1




    @Copy It is poor form to change the question in a way that invalidates existing answers.
    – Tobias Kildetoft
    Nov 23 '18 at 17:08








1




1




@Copy It is poor form to change the question in a way that invalidates existing answers.
– Tobias Kildetoft
Nov 23 '18 at 17:08




@Copy It is poor form to change the question in a way that invalidates existing answers.
– Tobias Kildetoft
Nov 23 '18 at 17:08


















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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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