Error in Twin Prime ConjectureCan the twin prime conjecture be solved in this way?What is wrong with this proposed proof of the twin prime conjecture?How to pigeonhole the primes between $p_n$ and $p_n+1^2$ for twin prime conjecture?Possible method to prove infinite twin prime conjectureTwin prime conjecture proof errorWould Brun's constant being transcendental prove the twin prime conjecture?Can't understand the logical structure of Euclid's infinitely many primes proof in Rosen's book.A twin prime theorem, and a reformulation of the twin prime conjectureTwin prime conjecture and gaps between primesIterated Twin Prime conjecture

Why Choose Less Effective Armour Types?

Why do passenger jet manufacturers design their planes with stall prevention systems?

Why would a flight no longer considered airworthy be redirected like this?

Time travel from stationary position?

Sailing the cryptic seas

How Could an Airship Be Repaired Mid-Flight

Error in Twin Prime Conjecture

How do I hide Chekhov's Gun?

Look at your watch and tell me what time is it. vs Look at your watch and tell me what time it is

Did Ender ever learn that he killed Stilson and/or Bonzo?

Min function accepting varying number of arguments in C++17

How to create the Curved texte?

Employee lack of ownership

Is there a higher dimension analogue of Noether's theorem?

how to write formula in word in latex

Can I use USB data pins as power source

How could a scammer know the apps on my phone / iTunes account?

How can I track script which gives me "command not found" right after the login?

Is it possible to upcast ritual spells?

Bach's Toccata and Fugue in D minor breaks the "no parallel octaves" rule?

SOQL: Populate a Literal List in WHERE IN Clause

Do I need life insurance if I can cover my own funeral costs?

Why did it take so long to abandon sail after steamships were demonstrated?

The difference between「N分で」and「後N分で」



Error in Twin Prime Conjecture


Can the twin prime conjecture be solved in this way?What is wrong with this proposed proof of the twin prime conjecture?How to pigeonhole the primes between $p_n$ and $p_n+1^2$ for twin prime conjecture?Possible method to prove infinite twin prime conjectureTwin prime conjecture proof errorWould Brun's constant being transcendental prove the twin prime conjecture?Can't understand the logical structure of Euclid's infinitely many primes proof in Rosen's book.A twin prime theorem, and a reformulation of the twin prime conjectureTwin prime conjecture and gaps between primesIterated Twin Prime conjecture













3












$begingroup$


Euclid's theorem states:




Consider any finite list of prime numbers $p_1, p_2, ..., p_n$. It will be shown that at least one additional prime number not in this list exists. Let $P$ be the product of all the prime numbers in the list: $P = p_1p_2...p_n$. Let $q = P + 1$. Then $q$ is either prime or not.



If $q$ is prime, then there is at least one more prime that is not in the list. If $q$ is not prime, then some prime factor $p$ divides $q$. If this factor $p$ were in our list, then it would divide $P$ (since $P$ is the product of every number in the list); but $p$ divides $P + 1 = q$. If $p$ divides $P$ and $q$, then $p$ would have to divide the difference of the two numbers, which is $(P + 1) − P$ or just $1$. Since no prime number divides $1$, $p$ cannot be on the list. This means that at least one more prime number exists beyond those in the list. This proves that for every finite list of prime numbers there is a prime number not in the list, and therefore there must be infinitely many prime numbers.





My question:



Does this theorem also hold if you let $q = P - 1$?



Wouldn't $P-1$ also be necessarily a new prime number? And if so, it and $P+1$ would be a set of twin primes.




So the proof would be:



Assume there are a finite number of twin primes such that $p_n+1 - p_n = 2$.



Then, from the final set of twin primes, choose the larger of these two primes $p_n+1$. Calculate $S=p_1p_2...p_n+1$. So you now have a product of all primes up to $p_n+1$. Call this $S$. $S + 1$ is a prime number and so is $S - 1$. This is a new set of twin primes not in our original list, thus there cannot be a finite list of twin primes.



Of course, if $S - 1$ is not prime, then this falls apart.










share|cite|improve this question









New contributor




Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 6




    $begingroup$
    There’s absolutely no reason why S+1 or S-1 should be a prime number though, it merely has an unlisted prime factor.
    $endgroup$
    – Noe Blassel
    4 hours ago










  • $begingroup$
    By the way, take a look at the edits. It's a courtesy to other contributors to use MathJax to format your posts. If you're not familiar with it, it's not hard to learn -- I've been on this site for less than two months and it has become second nature.
    $endgroup$
    – Robert Shore
    4 hours ago















3












$begingroup$


Euclid's theorem states:




Consider any finite list of prime numbers $p_1, p_2, ..., p_n$. It will be shown that at least one additional prime number not in this list exists. Let $P$ be the product of all the prime numbers in the list: $P = p_1p_2...p_n$. Let $q = P + 1$. Then $q$ is either prime or not.



If $q$ is prime, then there is at least one more prime that is not in the list. If $q$ is not prime, then some prime factor $p$ divides $q$. If this factor $p$ were in our list, then it would divide $P$ (since $P$ is the product of every number in the list); but $p$ divides $P + 1 = q$. If $p$ divides $P$ and $q$, then $p$ would have to divide the difference of the two numbers, which is $(P + 1) − P$ or just $1$. Since no prime number divides $1$, $p$ cannot be on the list. This means that at least one more prime number exists beyond those in the list. This proves that for every finite list of prime numbers there is a prime number not in the list, and therefore there must be infinitely many prime numbers.





My question:



Does this theorem also hold if you let $q = P - 1$?



Wouldn't $P-1$ also be necessarily a new prime number? And if so, it and $P+1$ would be a set of twin primes.




So the proof would be:



Assume there are a finite number of twin primes such that $p_n+1 - p_n = 2$.



Then, from the final set of twin primes, choose the larger of these two primes $p_n+1$. Calculate $S=p_1p_2...p_n+1$. So you now have a product of all primes up to $p_n+1$. Call this $S$. $S + 1$ is a prime number and so is $S - 1$. This is a new set of twin primes not in our original list, thus there cannot be a finite list of twin primes.



Of course, if $S - 1$ is not prime, then this falls apart.










share|cite|improve this question









New contributor




Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 6




    $begingroup$
    There’s absolutely no reason why S+1 or S-1 should be a prime number though, it merely has an unlisted prime factor.
    $endgroup$
    – Noe Blassel
    4 hours ago










  • $begingroup$
    By the way, take a look at the edits. It's a courtesy to other contributors to use MathJax to format your posts. If you're not familiar with it, it's not hard to learn -- I've been on this site for less than two months and it has become second nature.
    $endgroup$
    – Robert Shore
    4 hours ago













3












3








3


1



$begingroup$


Euclid's theorem states:




Consider any finite list of prime numbers $p_1, p_2, ..., p_n$. It will be shown that at least one additional prime number not in this list exists. Let $P$ be the product of all the prime numbers in the list: $P = p_1p_2...p_n$. Let $q = P + 1$. Then $q$ is either prime or not.



If $q$ is prime, then there is at least one more prime that is not in the list. If $q$ is not prime, then some prime factor $p$ divides $q$. If this factor $p$ were in our list, then it would divide $P$ (since $P$ is the product of every number in the list); but $p$ divides $P + 1 = q$. If $p$ divides $P$ and $q$, then $p$ would have to divide the difference of the two numbers, which is $(P + 1) − P$ or just $1$. Since no prime number divides $1$, $p$ cannot be on the list. This means that at least one more prime number exists beyond those in the list. This proves that for every finite list of prime numbers there is a prime number not in the list, and therefore there must be infinitely many prime numbers.





My question:



Does this theorem also hold if you let $q = P - 1$?



Wouldn't $P-1$ also be necessarily a new prime number? And if so, it and $P+1$ would be a set of twin primes.




So the proof would be:



Assume there are a finite number of twin primes such that $p_n+1 - p_n = 2$.



Then, from the final set of twin primes, choose the larger of these two primes $p_n+1$. Calculate $S=p_1p_2...p_n+1$. So you now have a product of all primes up to $p_n+1$. Call this $S$. $S + 1$ is a prime number and so is $S - 1$. This is a new set of twin primes not in our original list, thus there cannot be a finite list of twin primes.



Of course, if $S - 1$ is not prime, then this falls apart.










share|cite|improve this question









New contributor




Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Euclid's theorem states:




Consider any finite list of prime numbers $p_1, p_2, ..., p_n$. It will be shown that at least one additional prime number not in this list exists. Let $P$ be the product of all the prime numbers in the list: $P = p_1p_2...p_n$. Let $q = P + 1$. Then $q$ is either prime or not.



If $q$ is prime, then there is at least one more prime that is not in the list. If $q$ is not prime, then some prime factor $p$ divides $q$. If this factor $p$ were in our list, then it would divide $P$ (since $P$ is the product of every number in the list); but $p$ divides $P + 1 = q$. If $p$ divides $P$ and $q$, then $p$ would have to divide the difference of the two numbers, which is $(P + 1) − P$ or just $1$. Since no prime number divides $1$, $p$ cannot be on the list. This means that at least one more prime number exists beyond those in the list. This proves that for every finite list of prime numbers there is a prime number not in the list, and therefore there must be infinitely many prime numbers.





My question:



Does this theorem also hold if you let $q = P - 1$?



Wouldn't $P-1$ also be necessarily a new prime number? And if so, it and $P+1$ would be a set of twin primes.




So the proof would be:



Assume there are a finite number of twin primes such that $p_n+1 - p_n = 2$.



Then, from the final set of twin primes, choose the larger of these two primes $p_n+1$. Calculate $S=p_1p_2...p_n+1$. So you now have a product of all primes up to $p_n+1$. Call this $S$. $S + 1$ is a prime number and so is $S - 1$. This is a new set of twin primes not in our original list, thus there cannot be a finite list of twin primes.



Of course, if $S - 1$ is not prime, then this falls apart.







proof-verification prime-numbers prime-twins






share|cite|improve this question









New contributor




Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









David G. Stork

11.1k41432




11.1k41432






New contributor




Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 4 hours ago









Jeffrey ScottJeffrey Scott

161




161




New contributor




Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 6




    $begingroup$
    There’s absolutely no reason why S+1 or S-1 should be a prime number though, it merely has an unlisted prime factor.
    $endgroup$
    – Noe Blassel
    4 hours ago










  • $begingroup$
    By the way, take a look at the edits. It's a courtesy to other contributors to use MathJax to format your posts. If you're not familiar with it, it's not hard to learn -- I've been on this site for less than two months and it has become second nature.
    $endgroup$
    – Robert Shore
    4 hours ago












  • 6




    $begingroup$
    There’s absolutely no reason why S+1 or S-1 should be a prime number though, it merely has an unlisted prime factor.
    $endgroup$
    – Noe Blassel
    4 hours ago










  • $begingroup$
    By the way, take a look at the edits. It's a courtesy to other contributors to use MathJax to format your posts. If you're not familiar with it, it's not hard to learn -- I've been on this site for less than two months and it has become second nature.
    $endgroup$
    – Robert Shore
    4 hours ago







6




6




$begingroup$
There’s absolutely no reason why S+1 or S-1 should be a prime number though, it merely has an unlisted prime factor.
$endgroup$
– Noe Blassel
4 hours ago




$begingroup$
There’s absolutely no reason why S+1 or S-1 should be a prime number though, it merely has an unlisted prime factor.
$endgroup$
– Noe Blassel
4 hours ago












$begingroup$
By the way, take a look at the edits. It's a courtesy to other contributors to use MathJax to format your posts. If you're not familiar with it, it's not hard to learn -- I've been on this site for less than two months and it has become second nature.
$endgroup$
– Robert Shore
4 hours ago




$begingroup$
By the way, take a look at the edits. It's a courtesy to other contributors to use MathJax to format your posts. If you're not familiar with it, it's not hard to learn -- I've been on this site for less than two months and it has become second nature.
$endgroup$
– Robert Shore
4 hours ago










1 Answer
1






active

oldest

votes


















8












$begingroup$

Remember, your original argument doesn't show that $P+1$ is itself prime. It shows that $P+1$ has a prime factor that you haven't already accounted for. So while you could make the same argument for $P-1$, you'd also reach the same conclusion, not that $P-1$ is itself necessarily prime, but only that it has some prime factor not in your original list. So that's of no help in proving the Twin Prime Conjecture.



Similarly, you don't know that $S+1$ or $S-1$ is prime. You just know that they have prime factors that aren't on your original list of twin primes, but that doesn't help you.






share|cite|improve this answer









$endgroup$








  • 3




    $begingroup$
    Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
    $endgroup$
    – Jeffrey Scott
    4 hours ago










  • $begingroup$
    Glad I could help. Acceptances of answers that you find useful are always welcome.
    $endgroup$
    – Robert Shore
    4 hours ago










Your Answer





StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);






Jeffrey Scott is a new contributor. Be nice, and check out our Code of Conduct.









draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3149966%2ferror-in-twin-prime-conjecture%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

Remember, your original argument doesn't show that $P+1$ is itself prime. It shows that $P+1$ has a prime factor that you haven't already accounted for. So while you could make the same argument for $P-1$, you'd also reach the same conclusion, not that $P-1$ is itself necessarily prime, but only that it has some prime factor not in your original list. So that's of no help in proving the Twin Prime Conjecture.



Similarly, you don't know that $S+1$ or $S-1$ is prime. You just know that they have prime factors that aren't on your original list of twin primes, but that doesn't help you.






share|cite|improve this answer









$endgroup$








  • 3




    $begingroup$
    Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
    $endgroup$
    – Jeffrey Scott
    4 hours ago










  • $begingroup$
    Glad I could help. Acceptances of answers that you find useful are always welcome.
    $endgroup$
    – Robert Shore
    4 hours ago















8












$begingroup$

Remember, your original argument doesn't show that $P+1$ is itself prime. It shows that $P+1$ has a prime factor that you haven't already accounted for. So while you could make the same argument for $P-1$, you'd also reach the same conclusion, not that $P-1$ is itself necessarily prime, but only that it has some prime factor not in your original list. So that's of no help in proving the Twin Prime Conjecture.



Similarly, you don't know that $S+1$ or $S-1$ is prime. You just know that they have prime factors that aren't on your original list of twin primes, but that doesn't help you.






share|cite|improve this answer









$endgroup$








  • 3




    $begingroup$
    Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
    $endgroup$
    – Jeffrey Scott
    4 hours ago










  • $begingroup$
    Glad I could help. Acceptances of answers that you find useful are always welcome.
    $endgroup$
    – Robert Shore
    4 hours ago













8












8








8





$begingroup$

Remember, your original argument doesn't show that $P+1$ is itself prime. It shows that $P+1$ has a prime factor that you haven't already accounted for. So while you could make the same argument for $P-1$, you'd also reach the same conclusion, not that $P-1$ is itself necessarily prime, but only that it has some prime factor not in your original list. So that's of no help in proving the Twin Prime Conjecture.



Similarly, you don't know that $S+1$ or $S-1$ is prime. You just know that they have prime factors that aren't on your original list of twin primes, but that doesn't help you.






share|cite|improve this answer









$endgroup$



Remember, your original argument doesn't show that $P+1$ is itself prime. It shows that $P+1$ has a prime factor that you haven't already accounted for. So while you could make the same argument for $P-1$, you'd also reach the same conclusion, not that $P-1$ is itself necessarily prime, but only that it has some prime factor not in your original list. So that's of no help in proving the Twin Prime Conjecture.



Similarly, you don't know that $S+1$ or $S-1$ is prime. You just know that they have prime factors that aren't on your original list of twin primes, but that doesn't help you.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









Robert ShoreRobert Shore

2,990219




2,990219







  • 3




    $begingroup$
    Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
    $endgroup$
    – Jeffrey Scott
    4 hours ago










  • $begingroup$
    Glad I could help. Acceptances of answers that you find useful are always welcome.
    $endgroup$
    – Robert Shore
    4 hours ago












  • 3




    $begingroup$
    Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
    $endgroup$
    – Jeffrey Scott
    4 hours ago










  • $begingroup$
    Glad I could help. Acceptances of answers that you find useful are always welcome.
    $endgroup$
    – Robert Shore
    4 hours ago







3




3




$begingroup$
Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
$endgroup$
– Jeffrey Scott
4 hours ago




$begingroup$
Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
$endgroup$
– Jeffrey Scott
4 hours ago












$begingroup$
Glad I could help. Acceptances of answers that you find useful are always welcome.
$endgroup$
– Robert Shore
4 hours ago




$begingroup$
Glad I could help. Acceptances of answers that you find useful are always welcome.
$endgroup$
– Robert Shore
4 hours ago










Jeffrey Scott is a new contributor. Be nice, and check out our Code of Conduct.









draft saved

draft discarded


















Jeffrey Scott is a new contributor. Be nice, and check out our Code of Conduct.












Jeffrey Scott is a new contributor. Be nice, and check out our Code of Conduct.











Jeffrey Scott is a new contributor. Be nice, and check out our Code of Conduct.














Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3149966%2ferror-in-twin-prime-conjecture%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

名間水力發電廠 目录 沿革 設施 鄰近設施 註釋 外部連結 导航菜单23°50′10″N 120°42′41″E / 23.83611°N 120.71139°E / 23.83611; 120.7113923°50′10″N 120°42′41″E / 23.83611°N 120.71139°E / 23.83611; 120.71139計畫概要原始内容臺灣第一座BOT 模式開發的水力發電廠-名間水力電廠名間水力發電廠 水利署首件BOT案原始内容《小檔案》名間電廠 首座BOT水力發電廠原始内容名間電廠BOT - 經濟部水利署中區水資源局

Prove that NP is closed under karp reduction?Space(n) not closed under Karp reductions - what about NTime(n)?Class P is closed under rotation?Prove or disprove that $NL$ is closed under polynomial many-one reductions$mathbfNC_2$ is closed under log-space reductionOn Karp reductionwhen can I know if a class (complexity) is closed under reduction (cook/karp)Check if class $PSPACE$ is closed under polyonomially space reductionIs NPSPACE also closed under polynomial-time reduction and under log-space reduction?Prove PSPACE is closed under complement?Prove PSPACE is closed under union?

Is my guitar’s action too high? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Strings too stiff on a recently purchased acoustic guitar | Cort AD880CEIs the action of my guitar really high?Μy little finger is too weak to play guitarWith guitar, how long should I give my fingers to strengthen / callous?When playing a fret the guitar sounds mutedPlaying (Barre) chords up the guitar neckI think my guitar strings are wound too tight and I can't play barre chordsF barre chord on an SG guitarHow to find to the right strings of a barre chord by feel?High action on higher fret on my steel acoustic guitar