Proof of infinitely many prime numbers

Question

Here's the proof from the book I'm reading that proves there are infinitely many primes:

We want to show that it is not the case that there only finitely many primes. Suppose there are finitely many primes. We shall show that this assumption leads to a contradiction. Let $p_1, p_2,...,p_n$ be all the primes there are. Let $x = p_1...p_n$ be their product and let $y=x+1$. Then $y \in \mathbb{N}$ and $y \neq 1$, so there is a prime $q$ such that $q \mid y$. Now $q$ must be one of $p_1,...,p_n$ since these are all primes that there are. Hence $q \mid x$. Since $q \mid y$ and $q \mid x$, $q \mid (y-x)$. But $y-x=1$. Thus $q \mid 1$. But since $q$ is prime, $q \geq 2$. Hence $q$ does not divide 1. Thus we have reached a contradiction. Hence our assumption that there are only finitely many primes must be wrong. Therefore there must be infinitely many primes.

I have a couple of questions/comments regarding this proof. I will use a simple example to help illustrate my questions:

Suppose only 6 primes exist: $2, 3, 5, 7, 11, 13$

Let $x = p_1p_2p_3p_4p_5p_6=30,030$

Let $y = x + 1 = 30,030+1 = 30,031$

Questions/Comments:

The proof states there is a prime $q$ such that $q \mid y$ and that $q$ must be either $p_1, p_2, p_3, p_4, p_5,$ or $p_6$. However, none of the 6 primes listed, $(2,3,5,7,11,13)$, divides $30,031$. In fact, the only divisors for $30,031$ are $1, 59, 509$ and $30,031$. Doesn't the proof then break down here since there is no prime $q$ that divides $y$?
The prime factorization of $30,031$ is $59 \times 509$. These two numbers are factors of $30,031$ and are in fact primes themselves since they are only divisible by themselves or by 1. Have I shown that there exists $\gt 6$ primes? If so, what can I conclude now that I have shown this?
I don't understand why the contradiction $q$ divides $1$ and $q$ does not divide $1$ lead us to the assumption that the finitely many primes must be wrong. I understand how we reached the contradiction. I don't understand why contradiction leads us to the conclusion shows that are assumption that there is only finitely many primes is wrong.

My apologies for the long post. Thanks for any and all help.

unique factorization requires a unique prime decomposition. If no $p$ exists, there must be a new prime... — Eleven-Eleven, Oct 08 '16 at 23:50
The point of the argument is that there must be some prime not in the list that divides the number you produce. Your example confirms this! You found two such primes $59,509$ neither of which is on the list. Thus, your list can't have been complete. — lulu, Oct 08 '16 at 23:51
The proof makes an assumption that there are finitely many primes, But it then goes on to show, given the conditions, this actually can't be the case. Therefore, the flaw is in your assumption, since all arguments in the proof are mathematically valid. — Eleven-Eleven, Oct 08 '16 at 23:53
Thanks for the quick replies everybody. Could you expound a bit more about #3?. Why does this contradiction lead us to the fact that the assumption that there are finitely many primes must be wrong? — Kurt Mueller, Oct 08 '16 at 23:53
As stated above, all of your mathematical arguments are correct when you started from your original assumption. Yet you are led to a point in the proof that is impossible. If the mathematical arguments are sound and correct, what is left for there to be wrong...? The original assumption must therefore be incorrect. — Eleven-Eleven, Oct 08 '16 at 23:55

score 46 · Accepted Answer · answered Oct 08 '16 at 23:56

46

This is an excellent example of a proof that is traditionally phrased as a proof by contradiction but is much better understood constructively. From a constructive viewpoint, the proof shows that given any list of primes $p_1, \ldots, p_n$ there is a prime $q$ (any prime divisor of $p_1p_2\ldots p_n + 1$) that is distinct from each $p_i$. So given any finite set of primes we can find a prime that is not in that set.

answered Oct 08 '16 at 23:56

Rob Arthan

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+1 for emphasizing the constructive aspect of this proof. – P Vanchinathan Oct 09 '16 at 00:44
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Except that this process doesn't actually construct a new prime (the example $y = 30031$ being composite). – Daniel R. Collins Oct 09 '16 at 02:20
@DanielR.Collins So how does the process that gives a prime divisor of $p_1p_2\ldots p_n+1$ fail to construct a new prime? – Rob Arthan Oct 09 '16 at 02:28
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@Rob: I think Daniel’s point is that the proof constructs an integer $y>1$ that is not divisible by $p_1$, $p_2$, …, $p_n$, and then it *asserts* that there must be a prime $q$ such that $q \mid y$, but it *doesn’t construct* $q$. – Scott - Слава Україні Oct 09 '16 at 04:10
^ Yes, that's it. – Daniel R. Collins Oct 09 '16 at 04:14
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This is not a constructive proof. It is an existence proof that is best understood by contradiction. Downvote downvote downvote downvote downvote – djechlin Oct 09 '16 at 07:13
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@Scott But there's a simple algorithm to generate a prime divisor of any given integer. – Jack M Oct 09 '16 at 08:05
@JackM yep, but it's horrendously slow. $O(10^l)$ specifically. – John Dvorak Oct 09 '16 at 09:19
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@JanDvorak: yes, the proof also leads to a very poor estimate for the growth rate of the primes (I seem to recall this is discussed in Hardy and Wright's book on number theory). Nonetheless, it is still constructive. – Rob Arthan Oct 09 '16 at 10:51
@JackM: quite so! Thank you. – Rob Arthan Oct 09 '16 at 10:52
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@djechlin This is a constructive proof and you're utterly wrong. The set of divisors of a positive integer is finite. Now just take the minimal element greater than $1$ in the set of divisors; this is obviously prime. Would you say that finding the minimal element in a finite set of integers is not constructive? By the way, the proof by contradiction seems to imply that $p_1p_2\dots p_n+1$ is prime, which is the main source for confusion in newbies. – egreg Oct 09 '16 at 11:20
@egreg I don't see an intuitive reason to call that constructive. Suppose I wrote a proof that there must be a counterexample of the Frankl's conjecture that has at most $1000$ elements, without any hint on what that counterexample might be. This would mean enumerating all elements in $\mathcal{P}(\mathcal{P}([1000]))$, a finite set, and testing them would give the counterexample. But I wouldn't call this proof constructive. However I do understand the notion that it is more constructive than some other proofs, so I ask: Is there a widely-accepted meaning for the word constructive? – JiK Oct 10 '16 at 10:02
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@JiK Given any (finite) family of prime numbers, the procedure finds a prime number not belonging to the list. What more do you need to call this constructive? – egreg Oct 10 '16 at 10:05
@egreg A polynomial-time algorithm for finding that prime given the input might be nice! I don't know, that's why I asked what the word means. – JiK Oct 10 '16 at 10:12
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@JiK Why should it be polynomial time? That's completely irrelevant. – egreg Oct 10 '16 at 10:13
@egreg Is there a widely-accepted meaning for the word constructive? – JiK Oct 10 '16 at 10:14
@JiK: Would "each existence statement and case distinction can be encoded as a recursive function" be enough for you? – Johannes Kloos Oct 10 '16 at 11:34
@JohannesKloos I don't know. Luckily my opinion doesn't matter because I don't decide what's widely accepted. – JiK Oct 10 '16 at 11:37
It might help to think about the proof in purely set-theoretic terms. Let $\textbf P$ denote the set of prime numbers (this is a set thanks to the axiom schema of specification). Euclid's theorem states that $\textbf P$ is a countably infinite set. Since $\textbf P$ is contained in the natural numbers $\textbf N$, it suffices to show that $\textbf P$ is simply infinite. To show this, it suffices to show that for every finite subset $F\subseteq\textbf P$ of the prime numbers $\textbf P$, the complement $\textbf P\setminus F$ is a non-empty set. – SurfaceIntegral Sep 25 '21 at 20:11
I would say that overall, the proof is direct/constructive, but there are many smaller instances within the proof that utilize a quick proof by contradiction. – SurfaceIntegral Sep 25 '21 at 20:16
You may use little lemmas that involve proof by contradiction in your proof, of the infinitude of primes, but they aren't necessary. – Rob Arthan Sep 25 '21 at 20:32

score 14 · Answer 2 · answered Oct 08 '16 at 23:55

The proof does break down in a sense. You have reached a contradiction, which means the hypothesis that there are finitely many primes can not possibly be true.
So you thought you had 6 primes, you found 2 extra. Can you just add these two to your list and get all of the primes? If you repeat the process on these 8 primes, you will find that you will have even MORE primes by considering the product + 1. You can keep going and going and you will never run out of primes. This is the idea of the proof. It uses contradiction because you can do it all in one step and avoid the potential issue of doing the process infinitely many times.

score 9 · Answer 3 · edited Jun 12 '20 at 10:38

Suppose only 6 primes exist: 2,3,5,7,11,13

However, none of the 6 primes listed, (2,3,5,7,11,13), divides 30,031.

Then we already have a contradiction. Since there are only 6 primes (we supposed that at the beginning) and none of them divide 30,031, then 30,031 must be prime. However, 30,031 is not one of the only 6 primes that exist. So 30,031 cannot be prime, yet it must be prime.

So the proof works. In fact, it works precisely the same regardless of what set of numbers we suppose are the only primes that exist. Thus no finite set of numbers can include all the primes that exist. Thus there are an infinite number of primes.

I think you mean "... no finite set of numbers ..." – origimbo Oct 10 '16 at 10:48 — origimbo, Oct 10 '16 at 10:48

score 5 · Answer 4 · answered Oct 09 '16 at 00:02

Point 1: It's a theorem that any natural number $n>1$ has a prime factor. The proof is easy: for any number $n>1$, the smallest natural number $a>1$ which divides $n$ is prime (if it were not prime, it would not be the smallest).

Point 2: Yes, you have proved there are more than six primes. So what? The proof by contradiction doesn't suppose there are only six, but that there are a finite number of them.

Point 3: Actually, it's not really a proof by contradiction stricto sensu. It is proved that any finite list of primes in incomplete.

score 5 · Answer 5 · answered Oct 09 '16 at 01:51

You ask in #1 if the proof breaks down. No. What breaks down is the assumption that there are no more primes. You assumed you had a complete list of primes. Then you constructed a number that is not divisible by any of the primes in your list. Only two possibilities remain: Either the number you constructed is prime, or it is divisible by a number that is not in your list. Either way your list is not complete, which contradicts your initial assumption.

Simpy put, you assumed you had a complete list and by using that assumption you proved that the list was not complete. The assumption must therefore be wrong.

The answer to #3 is basically the same. By finding a contradiction when you made an assumption, you proved the assumption to be incorrect.

The answer to #2 is no, you did not prove you anything by noting that 59 and 509 are both prime. You are trying to prove that there is a finite list of primes. If you choose a particular set of primes as you did {2, 3, 5, 7, 11, 13} and show that that particular set doesn't hold all the primes, a skeptic would just say that you need to add more primes (like 59 and 509) because you didn't make your set big enough. The proof has to be more general which is why it doesn't say how many primes are in the finite set. The proof is written so that it works no matter how big the finite set is.

Eleven-Eleven · Answer 6 · 2016-10-09T00:22:41.837

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Here is a non-mathematical approach to the logic behind the proof by contradiction...

Suppose your assumption is that a suspect in a brutal murder is innocent because he said he couldn't have been at the location the person was murdered. You feel as though he is possibly telling the truth, but what you do know is that there are 5 other suspects. You know with 100% certainty due to the detail of the case there can't be anymore than 5 other suspects via proven information. But, the police can verify using obvious physical evidence such as video cameras that the 5 suspects could not possibly be guilty. Then there is either another suspect or the one assumed innocent is guilty. But we said that it was impossible for any other suspects to be considered. Therefore our original suspect is guilty.

This is exactly what is happening in a proof by contradiction.

edited Oct 09 '16 at 00:22

answered Oct 09 '16 at 00:03

Eleven-Eleven

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This is good. But I think a better analogy is that you have 6 suspects and you assume that these are the only six possible suspects in the world. Then cameras show each and every one is innocent. Thus our assumption that we have all the suspects is wrong, and therefore there is a seventh possible person who could have (and did) do it. – fleablood Oct 09 '16 at 00:58
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Damn.... you are right. It made logical sense when i wrote it... but yours is better. – Eleven-Eleven Oct 09 '16 at 01:02
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Yours makes sense. And is a proof by contradiction. I just thought for this particular proof this is a closer analogy. I wouldn't say yours was wrong or anything. – fleablood Oct 09 '16 at 01:05
That was my original goal; understand why contradition works. But yours does that AND is analogous to the problem at hand. Thanks – Eleven-Eleven Oct 09 '16 at 01:07
If you remove the extraneous details, your example does not involve proof by contradiction. What you say about "your assumption" and what "you feel" is irrelevant to the outcome: if one out of six people is known to be guilty and five of those people are known to be innocent then there is only one possibility for the guilty party. The same objection applies to fleablood's variant. In a proof by contradiction, you make use of the false hypothesis. – Rob Arthan Oct 09 '16 at 01:49
If you assume that your suspect is innocent, and you know that there are a finite number of suspects, then you ARE using the fact that the assumption is innocent because then you are trying to show that the remaining suspects are guilty on the assumption that yours is innocent. I was trying to "humanize" and "legitimize" the example in a way that the OP would understand on a non-mathematical level which was why I chose the word feel. When proving infinitude of primes, you may "feel" or intuit the path to help logically show the way. – Eleven-Eleven Oct 09 '16 at 01:54
@Eleven-Eleven: let me say it again: if it is known that one out of a given set of six people is guilty and if five of those people are known to be innocent, then there is only one possibility for the guilty party. This argument does not involve proof by contradiction. – Rob Arthan Oct 09 '16 at 02:57
Lets assume that one specific person of our six suspects is innocent and that it is impossible for any other suspect to be involved. This implies the person is not guilty. This implies that one of the remaining suspects must be guilty. But it's not suspect 1 because of evidence..and it's not suspect 2 because of evidence... and... and it's not suspect 5 because of evidence. Therefore none of the 5 remaining suspects is guilty which means that the original suspect must be guilty. I just used implication of the assumption so show that "P" and "not P" are valid under the assumption... – Eleven-Eleven Oct 09 '16 at 03:37
The statement $(A \lor B \lor C \lor D \lor E \lor F) \land (\lnot A \land \lnot B \land \lnot C \land \lnot D \land \lnot E) \Rightarrow F$ is valid in intuitionistic propositional logic. Hence it can be proved without the principle of proof by contradiction. You are getting ex falso quodlibet mixed up with the law of the excluded middle. – Rob Arthan Oct 09 '16 at 20:56
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I will concede. I get it. – Eleven-Eleven Oct 09 '16 at 23:01

String · Answer 7 · 2017-03-12T08:06:20.933

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The proof is essentially overexplaining. The point of contradiction would have been much clearer if the author had:

Reached it earlier.
Introduced less notation.

Since $p_1\cdots p_n+1$ cannot have any of $p_1,...,p_n$ as a prime factor we already have a contradiction.

The rest is just explaining to death why $p_1,...,p_n$ cannot be prime factors of $y=p_1\cdots p_n+1$ which should be clear since $y$ is $1$ off from being a multiple of either of those primes.

edited Mar 12 '17 at 08:06

answered Oct 10 '16 at 09:12

String

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I think the gif dosn't really fit this site, though I can't say if there is a specific rule against such things – Yuriy S Oct 10 '16 at 11:27
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@YuriyS: Hmm ... I am sorry to learn that people may find it inappropriate! That was by no means my intention! Do you think it would work better by removing the gif an leave the link and the figure of speach, or are all of those opfuscating the points I tried to make? – String Oct 10 '16 at 12:21
@String: your link to a cartoon movie of a dead animal being bludgeoned is disproportionate and offensive. – Rob Arthan Mar 12 '17 at 01:27
@RobArthan: Sorry, in some parts of the world such a cartoon would be considered merely a funny way to illustrate the saying about beating a dead horse. No offense intended, only a light tone. I cannot help that people do take offense, so I have removed it. Still it puzzles me how a cartoon matching the content of a saying would offend. I am from Denmark, after all. – String Mar 12 '17 at 08:11
@String: on MSE it's easier just to use neutral language. As I am English (after all), may I point out that the cliched phrase is actually "flogging a dead horse" and it doesn't have the connotations you think it does (it's not "explaining to death", it's using a tired old argument that has lost all interest or relevance). Your cartoon doesn't help with that. – Rob Arthan Mar 13 '17 at 01:55

score 1 · Answer 8 · edited Jun 12 '20 at 10:38

Well... yes... it does break down. You assumed that there are only 6 primes and reached a contradiction. You've successfully proved that there aren't only 6 primes.
Under the assumption that there are only 6 primes 30,031 isn't factorizable.
In proof by contradiction you prove proposition A by assuming A is not true, and through a series of logical steps reach an impossibility, thus proving that A must be true. Concurrently, in this proof, we assumed there is a finite number of primes. After a series of logical inferences we reached a contradiction. Since the only assumption we made in the process is that there are a finite number of primes, that assumption must be wrong.

score 1 · Answer 9 · answered Oct 10 '16 at 12:39

Either $ 6 $ primes you considered are not all the primes, or $ 30031 $ is a new prime. Either way assumption is false. Proof by contradiction.
No, you haven't shown that two more primes exist. In this example yes, but not in general
Again, based on your assumption you reached something impossible that $q $ divides $ 1$. There is no such $q $ except $1$.

Start with $ 2,3,5 $ being the only prime numbers. Then you would have $ 31 $ as the new number. $ 2,3,5 $ do not divide $ 31$. But unlike your case, $31$ is a prime number. So the assumption is wrong.

score 0 · Answer 10 · answered Mar 19 '24 at 16:08

Another way to arrive at a contradiction from your initial set-up is related to Rob Arthan's answer.

Given some arbitrary finite set of primes, say $\{p_1, p_2, ..., p_k\}$, define $N$ to be 1 more than their product, namely $$N = p_1 p_2 ... p_k\ + 1.$$

Now every positive integer is uniquely determined (up to ordering) by its prime divisors, and so $N$ must have at least one prime divisor, say $q$.

But $q$ cannot be one of the $p_i$s: if it was, then it would divide $N$ (which we just deduced) and also $p_1p_2...p_k$ (because $q$ is one of the $p_i$s). But then if $q$ divides $N$ and $N+1$, it must also divide their difference, $N - p_ 1p_2...p_k = 1$. This is a contradiction (since 1 is the empty product and does not have any prime divisors) and so $q$ must be a prime not in the finite set $\{p_1, p_2, ..., p_k\}$.

But the finite set we started with was arbitrary! This means that in any finite set of primes there exists an integer (one of the same form as $N$), at least one of whose prime divisors is not in that set. We conclude that there must be infinitely many primes.

Proof of infinitely many prime numbers

10 Answers10