The Unlikeliness of Numbers Sharing Factors

[@WrathofMath]

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[@tomkerruish2982]

Something you're glossing over is that you can't simply choose two positive integers "at random"; there is no uniform distribution over them. What can be done, however, is to find the probability that two integers between 1 and some large N, uniformly distributed, are relatively prime, and then consider the limit as N goes to infinity.The drawback to doing this is that it's a boring slog. I'll add that this result proves that there are infinitely many primes, for otherwise, we'd have a finite product of rational numbers equaling an irrational number. Edit: this depends on independently knowing that pi² is irrational.

[@schwingedeshaehers]

you can calculate the probility, / a series that cenverges to it

[@methatis3013]

​@@schwingedeshaehers thing is, the limit you get doesn't satisfy some important properties of probability. If you are chosing random integers, you NEED some distribution over them

[@schwingedeshaehers]

@@methatis3013 uniform from 1 to infinity

[@Mike10w848]

You can think of it as going over all the possible m and n that are in N(or Z+, whatever you prefer to call the positive integer set as) and count how many of them didn't have a common factor. It's kinda how probability works, "How much can this event occur out of all events?"

[@methatis3013]

@@Mike10w848 this doesn't work. A partitive set of natural numbers is not a σ-algebra, and thus you can't define probability this way. Furthermore, probability of choosing a number k (using this definition, assuming "uniform" distribution) would be 0, yet the sum over all integers should be 1, which is simply not true In other words, you need to be careful when defining a distribution over natural numbers. There isn't such a thing as uniform distribution on natural numbers

[@wolffang21burgers]

As a few people highlighted, this depends how we are choosing our numbers "at random": Let S_n be the set of all numbers that can be written in the form p_1^a_1 + p_2^a_2 + ... + p_n^a_n with p_i being the ith prime, and 0 inf, S_n -> Naturals, and the probability 2 numbers (chosen using the method of chosing detailed above [asuming the Axiom of Choice]) share a common factor is 1.

[@AJMansfield1]

You can also prove P=1 independent of AoC, using the "Fermi-Dirac factorization" argument I gave in my own earlier comment.

[@AJMansfield1]

Still trying to figure out if there's some 'natural feeling' way of limiting to the set of all integers that gets P=0 though. Since, sure, you _could_ just use subsequences from an artificial permutation of the integers that you've set up to produce prime numbers much faster than composites -- say, the sequence of s_n = {p_(n-pi(n)) if n is composite, otherwise c_(pi(n))} for primes p_n, composites c_n, and prime-counting function pi(n)). But, that's hardly a 'natural' way of picking an integer, the way picking it by its factorization is.

[@pxl_21]

0:36 you forgot 2 There I edited it

[@beaclaster]

15/2 is definitely an integer

[@ABlueGoomba2623]

Wow, thank you amazing person for teaching me that 15/2 equals a positive whole integer, I highly appreciate it.

[@DeJay7]

these replies... OP meant that he forgot to say 2 is a factor of 28, which is absolutely correct.

[@James2210]

0:36 better timestamp

[@dischqrge4846]

@@DeJay7 Ikr, this is peak of insufferable behavior.

[@ryewaldman2214]

a good follow up to this video is to calculate the probability that two randomly selected integers less than some value N are coprime.

[@nickronca1562]

I thought about this exact question before, and knew everything up to 8:57 but didn't know where to go from there.

[@DeJay7]

Didn't have a rabbit in your hat, did you? But seriously, yeah that's one aspect of mathematics that just sucks.

[@N____er]

​@DeJay7 maybe if you think hard enough or solve enough math you'll know the tools to coax that rabbit out of the hat... or you'll just learn a lot and not be able to. Either way, it's a win for me.

[@nickronca1562]

@@N____er Nah, I was never going to solve the product of all those terms. I would have never guessed that the product is the reciprocal of adding up the reciprocals of the squares of all numbers.

[@robharwood3538]

There's a hidden assumption in your statement of the problem, which is that the probability of picking a 'random' positive integer is *_equal_* to the probability of picking any other positive integer. In other words, you are assuming that the probability *_distribution_* over *_all_* positive integers is the Uniform Distribution. However, the Uniform Distribution is not a valid probability distribution for an infinite discrete set (like the positive integers), since you can't sum up an infinite number of things that all have the same probability (proportional to 1/N, as N -> inf.), and end up with a finite sum, unless that sum is 0. And in either case (sum is infinite, or sum is 0), this does not qualify as a valid probability distribution, since all valid probability distributions *must* sum to exactly 1. In other words, you can't 'normalize' the sum into a valid probability distribution by dividing everything by the total of the sum. Suppose the sum above was some value S, then with a valid probability distribution, you could divide each term of the sum by S to that the new normalized sum will sum up to S/S = 1. But in each case, if S = 0 or S is inf., you either get division by 0 (obviously invalid) or division by inf., making each number's probability = 0, and the sum of infinite 0s is 0, not 1, so not a valid prob. dist.! So, your solution only works for a kind of 'pseudo-probability' (i.e. using a non-normalized 'pseudo-probability' distribution). If you were to use a valid probability distribution over the positive integers, then this would change the answer to the question significantly. For example, you could have a Geometric Distribution, where the probability of each number is proportional to r^n, for some constant r. As long as r < 1, then the sum of probabilities converges, and can be normalized by dividing it by 1/(1-r). For an r value *_very_* close to 1, but still strictly less than 1, then this distribution will *approximate* a uniform distribution for 'small enough' positive integers. But it for 'large enough' integers, it will still be driven towards 0, as r^n goes to 0. You *might* be able to recover this result by considering the *limit* of *_finite_* uniform distributions on the numbers from 1 to N, as N -> inf. I'm not 100% sure on that, as it depends on some convergences which I haven't looked into. But that would still only be a *limit.* It's still not a valid probability distribution to have Uniform probability on all *infinite* positive integers.

[@AJMansfield1]

@@robharwood3538 indeed, also the result you get for that "limit" depends on _how_ you approach infinity here. If instead of picking from the integers up to N, you pick from the first N terms of OEIS A052330, the probability of two numbers sharing a factor converges to 1.

[@minerscale]

Doesn't measure theory find a way to save this idea of a non-normalised distribution?

[@wolffang21burgers]

Let S_n be the set of all numbers that can be written in the form p_1^a_1 + p_2^a_2 + ... + p_n^a_n with p_i being the ith prime, and 0 inf, S_n -> Naturals, and the probability 2 numbers (chosen using the method of chosing detailed above [asuming the Axiom of Choice]) share a common factor is 1.

[@AJMansfield1]

@@wolffang21burgers You can do that independent of AoC: Choose each random integer as a product of a subset of the Fermi-Dirac Primes (i.e. decompose each p_i^a_i into a product of p_i^2^k equivalent to representing a_i in binary). Take S_n as the first 2^n integers ordered by these subsets (i.e. the first 2^n elements of OEIS A052330). Two values in S_n share a Fermi-Dirac factor with P = 1-(3/4)^n. Two numbers share a prime factor at least as often as they share a Fermi-Dirac factor. Limiting n->inf, S_n -> naturals, P -> 1.

[@AJMansfield1]

There is NO well-defined way to uniformly pick a random integer -- and you can't even approach it either, the way you can other kinds of infinity. Otherwise, here's a proof of P=1, i.e. _almost all_ pairs of randomly selected integers share a factor: The Fermi-Dirac factorization of a number represents a number as the product of exactly 0 or 1 of every possible number of the form p^2^k for prime p and whole number k. If two numbers share a Fermi-Dirac factor, they share a factor; therefore, the probability of sharing a factor must be at least as large as the probability of sharing a Fermi-Dirac factor. By the fundamental theorem of arithmetic, every finite positive integer can be uniquely represented as a finite bitstring that encodes the presence or absence of the corresponding term of OEIS A050376 (the sequence of Fermi-Dirac factors). Two numbers share a Fermi-Dirac factor if both numbers share a 1 in some position of their Fermi-Dirac factorization bitstrings. Select our two integers a,b uniformly from the first 2^N terms of OEIS A052330 (the integers ordered by these bitstrings). The probability of that they share a factor is at least as large as the probability that they share a '1' bit in some position of these bitstrings, so P(N) >= 1 - (3/4)^N. The limit as N increases without bound is therefore P = 1 -- two randomly-selected integers share a factor with probability 1.

[@ntuneric]

it would be interesting to see how this answer changes for random numbers in a chosen interval [N1, N2]. for example, if the two numbers are between 1 and 97, chances of a number being even are 48/97, of being a multiple of 3 are 32/97 and so on, so the answer would look like (1-(48/97)^2)(1-(32/97)^2)...(1-(1/97)^2)

[@JohnDlugosz]

So ask Wolfram Alpha or ChatGPT to graph it.

[@GrandRezero]

Just start the video and my initial gut reaction says at least 25% likely that two random numbers share a common factor as every other number is even so there's a one in four chance of both numbers being even and thus sharing a common factor

[@momom6197]

You got the right idea! The starting step of the proof is to do that, but for every prime number rather than just 2.

[@craig4320]

If this problem were well specified, you could check your result by running a computer program that randomly selects pairs of positive integers and checks whether the GCD is 1. If you are right, the more pairs the program checks, the closer its result should get to your result.

[@Ki0212]

Another way to simplify the product is to remember the sum of an infinite GP: 1+x+x^2… = 1/(1-x) So (1-x) = 1/(1+x+x^2…) Thus the product becomes 1/((1+1/2^2+(1/2^2)^2+…)(1+1/3^2+(1/3^2)^2…)…) Since the expression in the denominator contains every power of every prime, upon expanding, we would retrieve every number. This it becomes 1/(1+1/2^2+1/3^2+…) Which is the reciprocal of the Basel Sum, which is 6/pi^2

[@irrelevant_noob]

How does that help tho? o.O

[@Waffle_6]

such a simple and beginner friendly video, but wow, so pretty

[@AllanChandler-n9g]

Not that it affects the outcome but you missed two as a factor of 28 when working out gcd(15,28).

[@Ki0212]

We did this in class a few days ago! (Some of us managed to get it in the product form, however we couldn’t simplify it)

[@WrathofMath]

It's a very fun exercise!

[@tangsolaris9533]

Cool use of infinite series!

[@chloe-un9cn]

the product in the pulling a rabbit section reminds me of the sieve of eratasthones algorithm for finding primes! im wondering if theres any kind of practical link there, but since you already need your list of primes there probably isnt. still cool how a product can do essentially the same computational steps as iteration through a list

[@battle00333]

Its interesting that if we have a * b = 1 where a > 1, that we end up with b = 1/a

[@jbrecken]

Why is the product of probabilities infinite? You're checking if all the prime numbers are factors of m and n, but you should stop checking when you get to the last prime less than m and n.

[@MrDannyDetail]

The numbers are randomly chosen from any of the infinity of possible integers, which means we effectively don't yet know what they will be and therefore don't know where to stop, so we calculate the infinite product. For any practical case where were are told one or both numbers we can choose to stop at the largesst prime that is still less than the smaller one of m and n, but if we do know the exact numbers that were chosen we could just factor each of then (provided they aren't too large) and get a definite answer, rather than even trying to calculate a probability that wouldn't really makes sense for two specific numbers anyway.

[@lincolnrimmer8615]

At 1:23 he says the numbers are between 1 and infinity so 'we don't know how big they are.' Therefore, the prime factors are also between 1 and infinity.

[@empathogen75]

You’re right to call this out, it’s not really rigorous. It’s more correct to say it’s limit of the odds of choosing two relatively prime numbers out of a range from 1 to N as N gets very large (goes to infinity.)

[@empathogen75]

Btw it converges relatively quickly - I did a quick python script choosing numbers from one to 10000 and checking if they were co prime 1000x and got a little over 60%

[@Censeo]

If we know m and n then it is either 0 chance or 1 chance that they are co primes

[@AjCohn]

the zelda soundtracks in these videos make my heart happy

[@janTasita]

I'm actually surprised it's as close as it is to 1/2. I expected one case to be much more likely than the other as the numbers got bigger, though it would have taken me a while to guess which one.

[@minerscale]

I guessed admittedly based on the title that the probability that two numbers are coprime was 1. I was very surprised to see the basel problem come up, but perhaps not too surprised because of the persistent link between the zeta function and the primes.

[@andrewmirror4611]

Btw there is a Matt Parker п by hand video using this theorem.

[@deltamico]

Another problem is you're not comparing the speed of conversions of (1+2^(-2)+3^(-2)+4^(-2)...)(1-2^(-2))(1-3^(-2))(1-5^(-2))... and (1+2^(-2)+3^(-2)+4^(-2)...)

[@AK-vx4dy]

Coprimes are cool, if for example money weights were choosen as coprimes you can count them by weight even if they mixed

[@gdclemo]

Up to the product of the two weights. If you want truly independent weights you need irrational ratios, assuming an infinitely accurate scale.

[@DeJay7]

As someone somewhat experienced in mathematics, at least watching a lot of problems here on KZbin, the result being the reciprocal of that famous sum (which contains π) was not necessarily surprising. Now we just need a 3Blue1Brown video on why that's the case. And, because of the discussion about random large integers having a factor of some prime p (occurring every p terms), I'm not at all surprised that it is around 0.61, I didn't expect it to be either 0 or 1. But, just intuitively, the probability being 0.61 is just so off. It's so close to 1/2 that it makes me uncomfortable.

[@htpc002Weirdhouse]

@DeJay7 : It already exists, KZbin.com/watch?v=d-o3eB9sfls .

[@odomobo]

How did you know to multiply by the infinite sum (1 + 1/2^2 + ...) ? That step seemed like magic. Even though it's easy to prove once you have it, I would have never found that particular infinite sum in a million years.

[@r-prime]

You should instead try to expand out the product, and see what you get!

[@minerscale]

I think it's just one of those things you have to be *really really* clever to find out. Euler found this trick I think. Once you've seen problems similar to it you might think to find a series you can multiply by to sieve out each element in your target series exactly once. But you're absolutely right, it's astonishing anyone came up with this at all! I wouldn't worry so much about not seeing it yourself.

[@g10royalle]

I feel like the analysis is wrong. How are you choosing your m and n? Is this from a uniform distribution?

[@chopseh]

Sorry what were you saying? I was too distracted from the observatory song from Majora’s Mask

[@McMxxCiV]

I don't understand why the total probability is (1-1/2^2)(1-1/3^2) etc and not 1 - 1/2^2 - 1/3^2 - etc. Surely you just add the probabilities for every factor and then subtract the sum from 1?

[@ntuneric]

source: "surely"

[@ntuneric]

bcs if you just add the probabilities for every prime, then you count two times the cases when the numbers are divisible by both primes. for example, what's the probability of m and n being divisble both by 2 and 3? 1/4+1/9-1/36 , hence the probability of m and n neither being divisible by 2 or by 3 is 1 - 1/4 - 1/9 + 1/36 = 1 - 1/4 - (1-1/4)/9 = (1-1/4)(1-1/9)

[@fullfungo]

Because “divisible by 2” and “divisible by 3” are not mutually exclusive. If I told you that a bag has some balls and 1/2 is red and 1/3 is green, then indeed 1-1/2-1/3 is not red and not green. Because they are mutually exclusive. But if I told you a bag has 1/2 red balls and 1/3 (independently) are small, then there are (1-1/2)•(1-1/3) non-red non-small balls. Maybe try drawing a diagram with all possible categories; this might help.

[@McMxxCiV]

@@fullfungo Thanks, got it now.

[@mier2673]

But n and m are finite and series is infinite, some p are never divide m or n, because they are less that p, why we can say that if we use any n or m that's mean we can use infinite series, it's just bottom approximation of probability, you didn't show that it's equal to real probability

[@alihaydar728]

If your random two numbers are 13 and 20 , why are you cheking if they both divide 37 and bigger primes. You should ne checking from 2 up to half of the smaller number

[@Tumbolisu]

we aren't checking whether two specific numbers are coprime; we are calculating the likelyhood of two random numbers being coprime. if we know that the numbers are 13 and 20, then literally none of this matters, as the result is simply "yes, these are coprime". try doing this calculation with random numbers that range between 1 and 10. now again with numbers between 1 and 100. and now one last time with numbers between 1 and 1000. you will need to do the same calculation as in the video, except you can limit how many prime numbers to check. but every time you increase the upper limit, you need to include more and more prime numbers. so if the upper limit were to be infinite, you would need to include every single prime number.

[@LemoUtan]

But P(p|m) = 0, not 1/p, if p > m.

[@islandfireballkill]

You have infinite m that are bigger than p and finite m less then it. So the probability for a random integer m and fixed prime p that p > m = 0.

[@oida10000]

Yes but given that there infinitely many positiv integers larger or equal to p that could be multiplies of it and only finitely many positiv integers less then it, so it doesn't really matter.

[@DeJay7]

It's also the case that P(p|m) = 0, if p does not divide m. So I don't know where you're getting at with that.

[@Danicker]

In my opinion, the presenter phrased this poorly. He said P(p|m) is the probability, given m, that an arbitrary prime p divides m. This should be flipped around. Given a prime p, P(p|m) represents the probability that an arbitrary number m is a multiple of p. (There is still a lack of rigour in the phrase "arbitrary integer m", but this can be solve by looking at a uniform distribution from 1 to N, and then considering the limit as N goes to infinity)

[@Christopher-e7o]

X,2x+5=8

[@cooltaylor1015]

Where the hell is pi coming from? Nothing approaching circles was involved...

[@WrathofMath]

That's the magic!

[@minerscale]

3Blue1Brown has an excellent video giving some intuition about where you might imagine a circle coming from when solving the basel problem (the sum of the recipricol of the squares of the integers). As for multiplying in that series in the first place, well, you'll just have to thank Euler for spotting such a neat trick.

[@S0larisPrime]

Gosh dang it, that pesky circle constant rears its ugly head again!

[@petrospaulos7736]

why do we test all primes and not only the primes up to min(m,n) ?

[@irrelevant_noob]

Because min(m,n) is not a constant.

[@SixtyStone]

7:05 isn't it just (p² - 1)/p²?

[@r75shell]

You haven't explained why sharing factor of one prime is independent of sharing of other prime

[@ABlueGoomba2623]

I am the 134th liker of this video and the 1708ish viewer :)

[@antonyisbwos]

Why do u post at midnight 😭 😭

[@lincolnrimmer8615]

Firstly, timezones exist. There are other countries outside of north America. Secondly, you're clearly awake at 'midnight' to be replying so quickly, so what's the issue here?

[@antonyisbwos]

@lincolnrimmer8615 :0

[@costa_marco]

I don't think this argument is sound due to the following reasoning: Any positive integer number can only be even or odd, so P(n|2) = ½ and P(m|2) = ½, so P(n|2 and m|2) = ¼. But if P(n coprime m) = 1, that implies that P(n∤2 or m∤2) = 1, therefore, P + ¬P > 1, which is a contradiction of one of the videos assumptions for this proof. Either my assumption that numbers can only be odd or even is false, or the conclusion of the video is false or the postulate assumed is false. I really want a mathematician input on this.

[@vincehomoki1612]

Where was it stated that P(n coprime m) = 1?

[@johnosullivan675]

🤯

[@writerightmathnation9481]

You’ve a basic error from the beginning but it’s a common error. I’ll describe the error. I have in mind, but there are probably others I didn’t catch. My concern is that you didn’t specify the notion of probability that you will use. What do you think I mean by that? Well I’m not done watching a video but I’ll hazard. I guess that somehow you made what we often refer to as “the equally likely assumption“. One problem with this and dealing with integers is that there are countably infinitely many positive integers, and the usual properties of probability that involve measures don’t apply, such as countable additivity of disjoint events, etc. No countably infinite set can support such measures. The usual adjustment is to do probability calculations on proper(hence finite) initial segments of the set of positive integers, obtain a formula dependent on a parameter n, and then compute the limit as n grows without bound. Ok, so this in some way makes sense, but you swept it under the rug. There remains another problem of practicality that I’ll cover below, but let’s analyze what this means with regard to the wording you use in the beginning of this video. You ostensibly will compute “the” probability of the occurrence of some arithmetic event among all countably infinitely many positive integers, but there is no such probability, and to justify claiming there IS such a notion, you must modify (“extend”) the definition of probability to a situation in which measures theory applied to probability would fail. Ok, so let’s say this is ok because it’s a kind of extension of the notion of probability. We now have existence, in a sense, so that if you’d said you were computing “a” probability, we can concede, but pedagogically, you left a gaping hole in the explanation of that idea. Now, there’s a problem with uniqueness related to practicality. Certain numbers are so large that a human being never will “randomly choose” two numbers of such sizes or larger. This means that for each person p, there exists a natural number N(p), such that the probability that person p chooses two numbers greater than N(p) is zero. Perhaps this is because numbers larger than N(p) are not describable in any language, formal or otherwise, by person p. Perhaps this is because numbers larger than N(p) are not describable in a finite amount of time in any language, formal or informal, by person p, or it may be due to various other reasons. If we take N to be the supremum of all N(p) over all persons p living on earth over the lifetime of the earth, then no one can “choose randomly” two numbers larger than N. That is, we should limit our sample space to positive integers less than or equal to N. In this case, we’re essentially considering a weighted measure, for each M>N, on the sample space consisting of all positive integers less than m and then taking a limit of those probabilities, in which all events (sets) that include numbers greater than N are assigned zero probability. These probabilities are all equal to the result one gets with the sample space consisting only of numbers less than or equal to N, and this will not be equal to the probability you computed in this video presentation by assuming there’s no upper bound on the numbers people can choose. …

[@psychoDon525]

Before watching any of the video, I'm going to go with my instincts. Just thinking "well what's the chance that two numbers are both even." Which would be 1 in 4. Then I figure the chance of two numbers being divisible by 3 is 1 in 9, and it looks like it's 1/2^2+1/3^2+1/4^2+1/5^2[...] Which definitely has an easier way to write out. The infinite sum of all inverse squares is something I don't know enough calculus to remember how to derive, but google tells me it's 0.6449[...] (if you subtract 1 because we're not counting the likelihood of them sharing 1 as a factor). So the likelihood that any two randomly-generated numbers share a factor is just under two thirds?

[@Bangilnel]

Nah, you'd have to subtract every case which you counted multiple times. You added 1/2^2 to 1/4^2 for example which inflates the number in your calculation. In the same way if you only added the primes p^(-2) you'd still have to subtract cases in which the numbers share more than one factor like 6 for example when both numbers share 2 and 3

[@JohnDlugosz]

I think there's a problem with your explanation: when doing the 3's for example you find the 1/6^2 has already been cancelled. But you can't just ignore it! You _still_ have to subtract that term, so it remains in the answer. After multiplying, you'll find all the composite numbers have multiple copies remaining to be subtracted.

[@ricardoortega1139]

that number won't appear because 2 is already cancelled, it's only multiples of 3 and other numbers that are not multiples of 2

[@irrelevant_noob]

@JohnDlugosz maybe you were tricked by the phrasing at 12:07... he should've said "all the *_remaining_* multiples of 3 in this sum". Because how exactly would 1/6^2 come up to get subtracted?

[@markwrede8878]

70.6% of integers are composites, while the remaining 29.4% are prime. Primes with small sequential differences continue to populate about the terminals of larger prime sequential differences.

[@DavidvanDeijk]

If you go high enough 99% of integers is composite, primes are not as common as they are amongst the first 100 integers

[@markwrede8878]

@@DavidvanDeijk No, sir. Although their abundance declines to a very low local abundance, their sequential differences are finite, so at values beyond that prime of largest difference, the incidence climbs to reach 29.4% at infinity.

[@zanti4132]

I'm not sure where your misunderstanding is, but you're wrong. Of the first 1000 integers, 168 are prime, or 16.8%. Of the first 1,000,000 integers, 78,498 are prime, so now it's down to under 8%. The ratio will continue to drop as you increase the number of integers. In fact, speaking in probabilistic terms, the probability a randomly chosen positive integer is prime is 0%. Yes, I know prime numbers exist, but there's another infinity paradox for you.

[@markwrede8878]

@@zanti4132 Do the math. If the sequential differences were endless, the abundance of primes at infinity would approach zero. I show my work in this video: kzbin.info/www/bejne/mpzWm2efoNuagK8 Math teachers are wrong.

[@junkgum]

2.5% chance fyp space takeruppers.. >.001% We each still live on earth after 100 yrs passes.

[@lordmaster6667]

but half of all numbers are even. if you chose a random number, there's a 50% its even, and the 2nd random number would have a 50% chance of being even. shouldn't that automatically mean them not sharing a factor is less than half?

[@momom6197]

That's a 25% chance they're both even, which is compatible with the end result of 61% < 75%

[@cotlim]

Well that is 25% for them both to be even (0.5*0.5)

[@ntuneric]

it depends also on the domain of random numbers. for numbers between 1 and 99, there's a 49.(49)% chance it's even and 50.(50)% chance it's odd, because there are 49 even numbers and 50 odd numbers.

[@tristanridley1601]

This is the first term in the infinite product he used. (1-1/2^2) It is kinda interesting how the majority of all coprime numbers are just both even.

[@lordmaster6667]

@@cotlim right, coin flips. Forgot that step, thanks

[@peteneville698]

Why do so many content creators insist on having barely audible yet still annoying-as-hell music in the background?

[@minerscale]

I like it!