Euler's Proof of the Infinitude of Primes

[@WrathofMath]

CORRECTION: 12:46 is a logical error. The convergence of the sum on the left does not imply finitely many primes, because it is certainly possible the product on the right consists of infinitely many factors yet still converges. Only that remark was incorrect, the logic of the actual argument made in the video is correct. It is the case that finitely many primes would imply convergence of the product, which thus implies convergence of the sum. Hence if the sum diverges it must be that the product diverges, which is only possible if there are infinitely many primes.

[@eefaaf]

Ah, I was exactly looking for this correction.

[@hugh081]

This is such a chaotic channel. Infinite products over the primes one week, short division the next...

[@WrathofMath]

Its a day by day thing 🤣

[@RobertShadowLee]

8:50 The proof strongly relies on the Fundamental Theorem of Arithmetic, and one might ask whether the infinitude of primes is required to prove the FTA, as this could make the proof circular. However, it turns out that you can prove the FTA without assuming the infinitude of primes, so it all works out. Still, it would be helpful to clarify this, as brief details like this can nullify the validity of a proof.

[@feliksporeba5851]

Yes. Was hoping for the inclusion of that detail

[@PotentialDevGcimOgism.]

laughs in analytic continuation

[@ronhoffman1008]

Really liked the method for deriving the equality. Very interesting.

[@markwrede8878]

Multiplication delivers a vast cycle of integers, eventually failing to distinguish values. These are distinguished not by primes, but by the sequential differences between primes. Phi describes this slope value for all twin primes: 5, 7, 13, 19, 31, 43, .... Other such values exist as well, with the same properties for the square primes (square root of 11 divided by 4), sexy primes (square root of 29 divided by 6), and so on.

[@crazyape968]

Strangely, "sieve" is pronounced "siv", not "seev". 8:25 Great proof, BTW. I can't believe how simple it was.

[@MH-sf6jz]

The way of reasoning does not work by arguing for s>1, because it is never established the product is continuous in s.

[@liledw13]

The mario 64 music is a nice touch. 🤌

[@AsiccAP]

This feels non-rigorous to me.The proof the equality for s>1 seems sound to me, but why can we assert that it also works with s=1? The entire principal of the proof for s>1 is multiplying infinite series and subtracting them, which doesnt make sense if they diverge in the first place. Even the "equality" of s=1 case doesn't make sense, since the Left hand side diverges to infinity, you basically are saying something "equals infinity", which doesn't make sense in the real number system, as infinity is not a real number.

[@THEDeathWizard87]

It can be shown that the infinite sum converges if s>1. Once you prove that, you can add and subtract and reorder the terms without changing the answer. After showing that the sum and product are equal, you can let s approach 1 from above, which will cause the sum, and therefore the product also, to approach positive infinity. Yes this video isn’t rigorous, but it’s only meant to be an entry-level explanation that shows the outline of a proof, not the full thing

[@AsiccAP]

@THEDeathWizard87 Yes, this is exactly the rigorous proof I was looking for. I understand that this video isn't meant to be rigorous, but I guess a video can't entertain all the audience. I just wish there were addendums at the end of the video for mathematical rigour.

[@Iponamann]

@@THEDeathWizard87 Though it seems obviously true, I do not know how to prove that the limit as s approaches 1 from above diverges. We can’t simply reorder the summation and limit because the series of functions Sum(1 to N) 1/n^s does not converge uniformly to the function Sum(1 to infinity) 1/n^s. It should be enough to show that Sum(1 to infinity) 1/n^s is unbounded but I haven’t managed to do that either.

[@atrus3823]

This would have been a lot clearer if you wrote the product as 1/(1 - 1/p^s). You could then right away see the relationship to the seemingly arbitrary multiplying you were doing. In fact, why not start with the equality and multiply on both sides.

[@12_5tech]

If s = 2 then it we have primpes products on right

[@Iponamann]

I found this proof unsatisfactory because of how you let s=1. The equality of the infinite sum and product was proved assuming that s is strictly greater than 1. It would be enough to show that the function Sum(1 to infinity) 1/n^s is unbounded for s in (1,infinity), but this was not done. If someone can provide me a proof of this please let me know.

[@rockapedra1130]

11:03 wait ... At this point we still don't know the multiplication is infinite, right?

[@OfficialStickPM]

What purpose would anything under 3 and over 5 would ever be needed to understand? Can you explain that to me in another video or a reply so I can grasp why you want all these ?

[@hdthor]

Seems much easier to just do a proof by contradiction. Suppose there were a finite set of primes P, then the number that is (the product of the elements of P) + 1 is therefore is congruent to 1 modulo p for each p in P. Because it’s never congruent to 0 modulo any prime, it is not a composite, and so it is a prime, and therefore is an element of P. But it is bigger than any element in P. Contradiction. So P must be infinite.

[@DerekRoss1958]

Simpler but not so much fun! 😉

[@talastra]

But that wasn't how Euler did it, which is the point of the video. Yeah?

[@suatozcan1813]

Wonderful.

[@atrus3823]

I don’t find this proof convincing. Don’t you need to assume there are infinite primes to be able to wipe out the infinite terms of the sum? But you need the fact that they are equal to show there are infinitely many primes, so it’s circular.

[@zinc_magnesium]

This proof is not presented very rigorously but it is correct. You don’t need to assume infinite primes, as each prime has infinite multiples. All you need to assume is that each number >1 can be represented as a product of prime numbers. In this way, each prime number accounts for an infinite number of terms, and after you have removed all multiples of all prime numbers, whether there are a finite or infinite number of primes, you will have removed all integers greater than 1.

[@theupson]

no. your argument is materially defective, coming from you playing extremely fast and loose with the ellipsis and what it represents. ANY INDIVIDUAL TERM will eventually be canceled, but after any number of iterations, there always remain an infinite number of terms on the right. i havent read euler's version of this, but the natural logic is a proof by contradiction: IF the primes be finite, THEN the harmonic sum (over N) is equal to the given product (over a finite support). the seive ("siv", not "seeve") of eratosthenes argument you are trying to employ here works in this structure- 1/n will have been canceled by the time its largest prime factor has been included on the right- so in this hypothetical you really would be left with only "1" after a finite number of steps. the right product is trivially convergent, the harmonic sum is divergent, contradiction. did you notice how bringing the siv into it gets you within a gnat's whisker of just using euclid's proof? i love me some euler, but this result is just a toy.

[@WrathofMath]

I don’t see any disagreement between us aside from you wishing I had been more explicit about the limits.

[@CosmicHase]

Hallo.

[@CosmicHase]

Im not german, it's a typo.

[@inutamer3658]

Guten tag. Wie gehts