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











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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 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 at 17:44










  • Agreed. It's like discovering that gold isn't just valuable stuff that giltters, but can be dealt with profitably as heavy stuff that settles to the bottom.
    – Barry Cipra
    Nov 22 at 17:48












  • @BarryCipra Yes, sieving for gold seems like fun!
    – CopyPasteIt
    Nov 22 at 17:50






  • 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 at 18:05

















up vote
0
down vote

favorite












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 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 at 17:44










  • Agreed. It's like discovering that gold isn't just valuable stuff that giltters, but can be dealt with profitably as heavy stuff that settles to the bottom.
    – Barry Cipra
    Nov 22 at 17:48












  • @BarryCipra Yes, sieving for gold seems like fun!
    – CopyPasteIt
    Nov 22 at 17:50






  • 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 at 18:05















up vote
0
down vote

favorite









up vote
0
down vote

favorite











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 at 12:11

























asked Nov 22 at 17:10









CopyPasteIt

3,9541627




3,9541627












  • 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 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 at 17:44










  • Agreed. It's like discovering that gold isn't just valuable stuff that giltters, but can be dealt with profitably as heavy stuff that settles to the bottom.
    – Barry Cipra
    Nov 22 at 17:48












  • @BarryCipra Yes, sieving for gold seems like fun!
    – CopyPasteIt
    Nov 22 at 17:50






  • 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 at 18:05




















  • 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 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 at 17:44










  • Agreed. It's like discovering that gold isn't just valuable stuff that giltters, but can be dealt with profitably as heavy stuff that settles to the bottom.
    – Barry Cipra
    Nov 22 at 17:48












  • @BarryCipra Yes, sieving for gold seems like fun!
    – CopyPasteIt
    Nov 22 at 17:50






  • 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 at 18:05


















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 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 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 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 at 17:44












Agreed. It's like discovering that gold isn't just valuable stuff that giltters, but can be dealt with profitably as heavy stuff that settles to the bottom.
– Barry Cipra
Nov 22 at 17:48






Agreed. It's like discovering that gold isn't just valuable stuff that giltters, but can be dealt with profitably as heavy stuff that settles to the bottom.
– Barry Cipra
Nov 22 at 17:48














@BarryCipra Yes, sieving for gold seems like fun!
– CopyPasteIt
Nov 22 at 17:50




@BarryCipra Yes, sieving for gold seems like fun!
– CopyPasteIt
Nov 22 at 17:50




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 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 at 18:05












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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





















  • @Copy It is poor form to change the question in a way that invalidates existing answers.
    – Tobias Kildetoft
    Nov 23 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





















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















up vote
1
down vote













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





















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













up vote
1
down vote










up vote
1
down vote









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 at 17:22









Rafay Ashary

80118




80118












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


















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
















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




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


















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