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
asked Nov 22 at 17:10
CopyPasteIt
3,9541627
|
show 5 more comments
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
asked Nov 22 at 17:10
CopyPasteIt
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
|
show 5 more comments
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
asked Nov 22 at 17:10
CopyPasteIt
3,9541627
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
asked Nov 22 at 17:10
CopyPasteIt
3,9541627
asked Nov 22 at 17:10
CopyPasteIt
3,9541627
edited Nov 24 at 12:11
asked Nov 22 at 17:10
CopyPasteIt
3,9541627
asked Nov 22 at 17:10
CopyPasteIt
3,9541627
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
|
show 5 more comments
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
|
show 5 more comments
1 Answer
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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 ${p_1,p_2,dots,p_n}$.
answered Nov 22 at 17:22
Rafay Ashary
80118
@Copy It is poor form to change the question in a way that invalidates existing answers.
– Tobias Kildetoft
Nov 23 at 17:08
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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oldest
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up vote
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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 ${p_1,p_2,dots,p_n}$.
answered Nov 22 at 17:22
Rafay Ashary
80118
@Copy It is poor form to change the question in a way that invalidates existing answers.
– Tobias Kildetoft
Nov 23 at 17:08
add a comment |
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 ${p_1,p_2,dots,p_n}$.
answered Nov 22 at 17:22
Rafay Ashary
80118
@Copy It is poor form to change the question in a way that invalidates existing answers.
– Tobias Kildetoft
Nov 23 at 17:08
add a comment |
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 ${p_1,p_2,dots,p_n}$.
answered Nov 22 at 17:22
Rafay Ashary
80118
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 ${p_1,p_2,dots,p_n}$.
answered Nov 22 at 17:22
Rafay Ashary
80118
answered Nov 22 at 17:22
Rafay Ashary
80118
answered Nov 22 at 17:22
Rafay Ashary
80118
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
add a comment |
@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
add a comment |
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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