The Prime Constant - Numberphile

[@numberphile]

Get your signed copy of Love Triangle at mathsgear.co.uk/products/love-triangle-by-matt-parker-signed

[@The.171]

I agree

[@klaxoncow]

Mind you, if someone is able to generate the Prime Constant in a different way, they've just nailed how to find primes without searching.

[@Myndale]

And a great read it is. I've read my copy, and I'm now tempted to donate it to my local library (yes, they still exist) so that other people can read it too.

[@abigailcooling6604]

I've already got mine 🙃

[@Little-pluto-behind-neptune]

Yay

[@fullfungo]

Matt, you wrote the binary representation of 0.3 instead of 1/3. I shall now call it “the Parker third”™️.

[@mekkler]

Or 'Biblical π'.

[@Ms.Pronounced_Name]

Cut the guy some slack, he runs a Minecraft channel, not a maths channel

[@deinauge7894]

yea 1/3 with a 4-digit cycle looked very suspicious. The length of the repeating cycle is always smaller than the denominator...

[@AnotherPointOfView944]

@@Ms.Pronounced_Name no slack.

[@roberttalada5196]

Don’t you round? Lol

[@pokerformuppets]

This constant is really close to sqrt(2) - 1. I suggest we just make the constant *equal* to sqrt(2) - 1 for simplicity, and then determine the primes from there.

[@Meuszik]

Remarkably close. Not absurdly, but it makes you think if there is a reason for this.

[@janaki3829]

Nreaking news! The first few new prime numbers are 2, 3, 5,7,13,16,17,18,19...

[@morismateljan6458]

@@Meuszik Oh, a semi-prime constant is even closer to the sqrt(10)

[@JuusoAlasuutari]

I suggest we also redefine π = ∛Prime[Prime[Prime[Prime[Prime[1]]]]]

[@programmingpi314]

I think this is the most popular comment that doesn't talk about the Parker Third. So I am just going to bring it up in the replies.

Small mistake, 1/3 is 0.010101 repeating in binary. The decimal aproximation after 6 binary digits is 21/64, which makes a lot more sense.

[@GreylanderTV]

this was nagging at me too

[@Blocksetter63]

Yes, the binary fraction in the video 0.010011001... , with the last 4 digits repeating, represents 0.3 in decimal not 1/3.

[@mapwiz-sf5yt]

Yes. It has the same digits as 1/11 in base 10, because 3 is one more than 2 and 11 is more than 10.

[@Criz454]

parker binary

[@TheArizus]

That's a bit more than a small mistake...

[@_toomas]

3:10 The Parker Third, also known as 3/10 :D

[@Gabbobox]

EXACTLY

[@alandouglas2789]

@@GabboboxWhat? He was correct

[@juandesalgado]

Brilliant minds allow themselves to fumble

[@Henrix1998]

At 3:10 nonetheless

[@TechnocratiK]

@@alandouglas2789 1/3 = 0.01010101... (because 1/3 = 1/4 + 1/16 + 1/64 + ...), not 0.01001100110011

[@forthrightgambitia1032]

Haha, I started calculating 1/3 in binary myself and was confused where I went wrong. But turns out Matt is wrong.

[@carloslaue1236]

That's a Parker third

[@mandolinic]

No. Matt is correct. It's the _universe_ that's wrong.

[@pi2infinity]

I love this concept of The Parker Third. In my head, my calculus was nagging me: “One-third can be represented by summing (1/4)^n, which has the really pleasant binary expansion of .0101010101…” I pay ~30% of my wages to taxes as an American schoolteacher. Yes that’s right- a full Parker Third of my teacher paycheck goes to the government!

[@mrjava66]

Federal income tax. State income tax. State sales(vat) tax. Property tax. Special extra Vat taxes(wine, gasoline, tires, some other items). Are you sure it’s just 30%

[@pi2infinity]

@@mrjava66Yes, I’m sure. All those numbers you’ve described are less than 0.3 in the manners in which they interface with me, and those numbers smaller than 0.3 do, in fact, add up to 0.3 when combined in the manners relevant to me and my unique circumstances. I assure you and everyone else reading this comment that, in general, a list of small numbers can add up to a larger number without having to add to a number larger than that larger number.

[@KuK137]

@@mrjava66 Turn brain on and fox lies off, it will help...

[@disgruntledtoons]

@@mrjava66 And don't forget the taxes that are passed along by the producers of everything you buy. Ultimately, all taxes are paid by working people.

[@johnchessant3012]

Fun fact: The "factorial constant" (the nth digit is 1 if n is some number factorial and 0 otherwise) was the first number proven to be transcendental! Roughly speaking, Liouville was able to show that rational approximations to the "factorial constant" converge faster than it's possible for rational approximations can to any irrational algebraic number.

[@jammasound]

Wow, that is a strange fact indeed.

[@stenzenneznets]

😮

[@stenzenneznets]

Thank you so much

[@stenzenneznets]

I had a proper epiphany

[@maxonmendel5757]

i vany find yhis. help

[@trummler4100]

Fun Fact: In a very recent Snapshot (24w37a), the Boat Bug (mentioned at 4:57) has been fixed!

[@charliethunkman]

Im curious how the ‘fix’ was implemented, if it was a very minute change to the gravity system, if they went case by case and canceled out the issue, or if they changed the update order inside of the entity-block collisions section.

[@YunxiaoChu]

@@charliethunkmanhmm

[@SmileyMPV]

Quite the Parker bits in that 1/3 binary expansion ngl

[@Carriersounds]

6:25 the dauge just chillin in the back

[@imveryangryitsnotbutter]

The dhowgg

[@Carriersounds]

@@imveryangryitsnotbutter butter dog, dog w the butter on em

[@AndrewConnolly-c9k]

Can I pet that dªẅg

[@James_3000]

what the dog doin

[@abigailcooling6604]

Everyone loves Skylab 🐶😊

[@xtieburn]

Speaking of numbers between 0 and 1. This reminds me of my favourite number Champernownes constant which is all positive integers. 0.12345678910111213... Its an evenly distributed, transcendental number, containing all strings, that has actually seen some use in random number generation and testing. (It can fool naive tests, despite its obvious lack of randomness.) Something tickles me about how incredibly simple it is while being so expansive and having all these interesting properties.

[@jamesknapp64]

shows that "randomness" is a very complicated thing.

[@radadadadee]

wouldn't the digits of that number be distributed according to Benford's Law? At least for the few 1000's digits, it seems 1 will be the most frequent, 2 the second most, etc.

[@landsgevaer]

​@@radadadadee Nah, the fact that zeros do occur should be a clue. All digits, in the limit, occur equally probably in the limit (including that zero even).

[@MarcusCactus]

Well, in this context, "expansive" means "expensive". Too much so!

[@MichaelRothwell1]

​@@radadadadeeWell, you'd soon get to large numbers with many digits so the particular distribution of the first digit according to Benford's law would pale into insignificance.

[@RichardHolmesSyr]

Using continued fractions, you could turn this constant back into a sequence of integers. Which isn't a monotonic sequence, but its partial sums are. So you could then turn that into a real constant, and then do its continued fractions. Hours of fun for the whole family.

[@MrSilami]

That dog sleeping in the bg cracks me up

[@lo1bo2]

What I want to know is where does the secret door lead to?

[@ChemicalVapors]

Matt forgot to check his math in 1/3. The decimal/binary expansion of a fraction 1/N cannot contain a period longer than N. (And 0011 is a period of 4, which is bigger than 3.)

[@tweer64]

Yeah, and if you calculate what it actually is, it's 0.3, not 1/3.

[@nicolasmaldonado1428]

No, it's the Parker Third

[@lachlancooke]

1:17 Matt trying to contact his home planet

[@matthewziemba7526]

I was trying so hard to figure out what that was at first! 😂

[@treepoder]

what is lachlan cooke doing over here

[@aok76_]

This comment aged like fine wine xD

[@eliasmochan]

I thought this was going to be about the continued fraction a=1/(2+1/(3+1/(5+1/(7+1/(11+... which has the property that the primes are generated by repeatedly inverting the fractional part and taking the integral part of the result. 1/a=2+a1 1/a1=3+a2 1/a2=5+a3 1/a4=7+a5 etc. (all the a's are smaller than 1)

[@dragandraganov4384]

If you think about it, this encoding can be done for an arbitrary subset of the naturals, hence we have proved that the cardinalities of the power set of the naturals and the interval (0,1) are equal.

[@TabooGroundhog]

10:12 the aliens will just think it’s the monkey typewriter planet again

[@Jiglias]

isn't it though

[@juandesalgado]

I wonder how little sense the sequence of bits will make, if they fail to catch it from the beginning... They may notice clues, like an odd number of consecutive zeroes, or (if the Twin Prime Conjecture is true) the repeated occurrence of 101

[@abigailcooling6604]

Yes, surely it will just look like random noise as the primes are a random sequence?

[@deinauge7894]

the averade distance grows logarithmically. the 1s will become more and more lonesome in the sea of 0s

[@john_hunter_]

We are kind of the monkey typewriter planet when you think about it.

[@liamroche1473]

I guessed this was going a different way, and defined a different real number containing all the primes as: 1/(2+(1/(3+1/(5+1/(7+1/(11+1/(13+1/(17+1/(19+1/(23+...)))))))))) I make this number 0.4323320871859029... Note that this construction works for a larger class of sequences of integers.

[@fullfungo]

Yeah, this is called a continued fraction

[@liamroche1473]

@@fullfungo Yes, I didn't explicitly mention the term.

[@fakenullie]

But can you recover prime numbers from your constant?

[@liamroche1473]

@@fakenullie Yes, the algorithm to turn a real number into a continued fraction is very straightforward. Of course, in the real world we can only ever do this with an approximation to the real number, giving a chosen number of the primes. You need infinite space to store arbitrary real numbers, of course.

[@wmpowell8]

I thought that too. The larger class of sequences of integers is integers that are not in increasing order. For example you could do 1/(1+1/(1+1/(1+1/(2+1/(1+1/(2+1/(1+1/(5+1/...)))))))) (which is A000001 on the OEIS btw) that is approximately 0.634049724209852

[@robko87]

funny thing is that this video can be exported and transformed to binary file and if you put "0." at the start of this file, you will again have a number between 0 and 1 :D

[@lonestarr1490]

Which means there is a monotonic sequence of natural numbers representing this video.

[@LoganKearsley]

That's the basic idea of arithmetic coding in data compression.

[@hqTheToaster]

I can't wait for you to make a Universal Scene Description that is just this video in glorious reformated 90 sound samples per second, 7p (7:5), 3 frames per second, from left to right, top to bottom, with a 3x3 pixel png file meant as a cypher for what colors and neighbors of colors to modularly find, and zip the two together in a .zip file, and then try to list the number between 0 and 1 describing that .zip file.

[@abigailcooling6604]

@@hqTheToaster With the amount of nerds who watch these videos, someone will surely try this.

[@ChrisShawUK]

A rational number as well

[@sashagornostay2188]

"If you wanna yell "we're pretty clever" - that's your number" (c) Parker

[@HangarQueen]

Ya, I loved this ending -- to an overall interesting and light-hearted episode. :-)

[@GoodBrownBear]

I don’t get this. We are just rewriting base 10 prime numbers in base 2 but why would aliens get it? They don’t use base 10. Are there really prime numbers in base 2?

[@sashagornostay2188]

@@GoodBrownBear Numbers exist outside of base. You can think of numbers as piles of little pebbles, prine numbers would be ones that you can't make rectangles of, only lines. And bases are ways of arranging these piles, properties of piles don't change if you shift them around.

[@Qbe_Root]

This is a neat way to encode _sets_ of numbers, not sequences, which is why it works out neatly with primes. In order to extend it to monotonically increasing sequences, you have to rely on the separate assumption that the bits are to be read in positional order, which is kinda weird since positions already encode the elements of the set. If you read the bits from the end instead, you'd get monotonically decreasing sequences! The only thing you can't do with this encoding and an arbitrary reading order is have the same number twice, since it makes no sense for a set to contain the same element twice, it either contains it (1) or it doesn't (0). So the "Fibonacci constant" shown in the video doesn't properly encode the Fibonacci sequence because it would need 1 twice; it encodes the set of numbers that appear in the Fibonacci sequence. (Also 0 is a Fibonacci number so the constant should go 1.11101001...) Fun fact: this idea of encoding a set of fixed elements using bits in a specific order has been used quite a bit in programming, such as with MySQL's SET type, Java's EnumSet class, or manual bitfields/flags in languages that didn't have built-in support for that.

[@RobinDSaunders]

To be pedantic, it encodes sets of numbers which satisfy excluded middle. It's sometimes useful (especially in computer science) to consider the possibility that not all sets are like this.

[@vsm1456]

regarding your fun fact, this idea was also used in the best attempt to improve Matt's code for the Wordle problem. instead of storing words as a string of letters, a, b, c, d, etc., each word is coded in bits where 1 means this letter is present in the word, 0 means this letter is absent. then, to compare if two words have the same letter, you perform bitwise-AND on them. since this operation is hardwired in x86 CPUs, it works extremely fast, so fast that full brute-force comparison of all 5-letter words takes a tiny fracfion of a second

[@alansmithee419]

"If you read the bits from the end instead..." Aren't these supposed to be infinitely long binary numbers? Are you referring to a subset of sequences that are finite in length here?

[@RobinDSaunders]

@@alansmithee419 sequences are often allowed to have terms "coming in from infinity" as well as / instead of "going off to infinity", although reading terms "from the end" might not be the clearest way to refer to this.

[@ulob]

This is how you encode all primes on a stick, using a knife. Just make a cut on the stick in the right place. In fact, you can encode all human knowledge this way (on a single stick). Good to know in case you need to prepare for a nuclear apocalypse.

[@ffggddss]

⅓ in binary is .[01]; where the bracketed part repeats forever, not .[0011], which is ⅕. Writing each as an infinite geometric series will show this. Even easier, multiply the first by 11 binary (= 3), and the second by 101 binary (= 5). Both will give .111111111... which is =1. Correction: What Matt wrote wasn't .[0011], it was .0[1001], which is .3 (decimal). Fred

[@FloydMaxwell]

You can add even more "unmistakable order" to the prime constant 'beaming' by adding a pause after each embedded prime, with the pause length equal to the number of the prime.

[@zfighter3]

19/64 is the Parker Approximation. Great video though!

[@NigelJohns]

Surprising that neither realised that it had to be 21/64. Instant red flag for me.

[@connorohiggins8000]

I got a prime number sequence accepted a few years ago (A328225) after one of these videos. This just reminded me that I never figured out why my sequence looked the way it did when it was plotted. I would love to hear some thoughts. I am not a mathematician in any form, so it could be absolutely nothing.

[@MrCheeze]

Of course, you could also do it backwards, taking a specific number and convert it into an integer sequence. For example pi would be -1, 0, 3, 6, 11, 12, 13, 14, 15, 16, 18, 19, 21... I don't know why you would, but you can. (And I just checked, it's OEIS A256108.)

[@therealax6]

I mean, when you encode a number in binary, you're essentially doing this. And encoding fractional numbers in binary is something that computers do pretty darn often...

[@claytonarg5947]

Clicking on a Numberphile and finding Matt Parker makes me so happy.

[@mitchkovacs1396]

One interesting property of this encoding scheme for increasing natural number sequences is that it lexicographically orders all of the sequences, i.e. given two such sequences A and B, we determine A '

[@IceMetalPunk]

This reminds me a bit of arithmetic coding, where given a frequency table and infinite decimals at your disposal, you can compress any data -- any file -- of any length into a single decimal number, and decompress it losslessly. That's always fascinated me, and been one of the main reasons I'm frustrated at the nonexistence of infinite precision 😂 (That and, of course, precision errors in my code...)

[@galaxygur]

It explains very well, why 0.1 + 0.3 is not equal to 0.4 in most programming languages, - because binary representation of those numbers is infinitely repeating, and therefore it must approximate it at some point.

[@toimine8930]

3:08 bruh

[@MartinPHellwig]

Only problem for the receiver is, that if they don't know they are receiving the constant of prime and start listening after our known greatest prime,, it is indistinguishable from random.

[@alexsimpson2970]

It is meaningless if there's any noise. Or if the listener hears from the middle.

[@HeroDarkStorn]

Well, you would distinguish it from random by noticing that that chance of receiving "1" lowers over time.

[@BenAlternate-zf9nr]

You could send a repeating signal of on/off pulses where the length ratio of on:off was this constant. Or transmit continuous sine waves on two different frequencies that have this ratio between them.

[@JavSusLar]

Not exactly... The distribution of beeps would become logarithmically less dense, which should awaken the suspicion of any attentive listener. However, since sending a sequence that becomes progressively more scarce can be quite impractical, it would probably be better to just send a few terms, the fewest that can give enough evidence to discard a non intelligent origin.

[@violetfactorial6806]

You would send it on repeat it with some clearly unique signal, like a pause in the transmission, to indicate the reset point.

[@oscarfriberg7661]

There’s also the Parker Prime constant, which is the binary representation of an infinitely long video where Matt Parker writes down every prime number on the brown paper.

[@diddykong3100]

Binary 1/three is not .0100110011..., it's 0.0101010101... as is easily seen by multiplying it be three = 11 to get 0.1111111111... = 1. Its successive approximations taking even numbers of digits are 1/4, 5/16, 21/64, 85/256, always of form n/(3.n +1). Multiplying 0.0100110011... by 101 = five, we get 1.0111111111... = 1.1 = three halves, so 0.0100110011... is three tenths, not a third.

[@corlinfardal]

Interestingly, with the sequence-to-real-number conversion, you can re-express a problem like the Twin Prime conjecture as whether the number corresponding to that sequence is rational or goes on forever (the primes are too spread out to allow for repeats), or the Collatz conjecture as whether the real number corresponding to a sequence of 0 if the collatz function reaches 1 and 1 otherwise equals 0.

[@leonschroder2970]

I like this new and improved Parker Third

[@markwrede8878]

Whereas primes are without end, their capacity to determine slopes is finite, and superseded by Pythagorean triplets. These are given on two parabolas: y = (x^2 + 1)/2 and y = (x^2 + 2)/4.

[@PaulBennett]

"Five is not a factor of two". That alone was worth opening KZbin for the day.

[@orena932]

I love the idea of beaming out the prime constant in binary and getting to really big numbers where you just get a crazy amount of zeroes with the occasional one sent out as well when you reach a prime

[@dielaughing73]

Perhaps aliens are huddling around their primitive radio sets somewhere waiting for that next '1' to come through

[@Topakhok]

This exact idea actually shows that there are more real numbers than integers. Because every set of integers is uniquely mapped that way to a sequence of 0 and 1, which in turn can be mapped to all real numbers in [0; 1]. And there are more sets of integers then integers themselves :)

[@iangreenhoe6611]

Look up “Gödel diagonalization”. It demonstrates not only are there more real numbers than integers, but that there are an infinite number of real numbers for each integer.

[@Topakhok]

@@iangreenhoe6611 what do you mean by “infinite number for each integer”? I just meant that |ℝ| > |ℕ|

[@iangreenhoe6611]

@@Topakhok , infinite number of real numbers for each integer.

[@Topakhok]

@@iangreenhoe6611 I mean mathematically :) What does “there is infinite number of real numbers for each integer” mean? Do you mean some exact mapping between real and integer numbers? Because it’s not true for every mapping, f(x) = 0 gives exactly zero real numbers for each non-zero integer

[@iangreenhoe6611]

@@Topakhok , no. I mean that there are different sizes of infinity. Here we are talking about aleph 0 (for N, Z, and Q) and aleph 1 (R, C, and other finite dimensional imaginary systems). Aleph 1 = (Aleph 0) ^ (Aleph 0). Gödel showed that you can map N onto Q (bijective mapping). Specifically he used N onto Q for Q between 0 and 1 inclusive, mapping to the rest of Q is easy. Doing this allowed him to construct an ordered list of rational numbers. By doing the decimal expansion of each item of this list, you can create a real number that is distinct from any item in the list of rational numbers. Further, there is literally an infinite number of ways to do so.

[@newTellurian]

In Zemeckis' Contact (1997) prime number sequence (basically a chunk of this binary prime constant) is what aliens sent us to make it clear it's an artificial signal.

[@dielaughing73]

Because of course some sci-fi writer already thought of it. There's really nothing new under the sun

[@philrichards7240]

Try reading the book "Contact" written by the late great Carl Sagan on which the film was based...

[@JamesDavy2009]

From what I researched, they actually skipped a number in the prime sequence.

[@jimmyzhao2673]

10:05 Aliens *still* using dial up. lol

[@alan2here]

Constant positive or negative curvature (pick exactly 1), then PI is a real to real function "def pi(radius_relative_to_curvature : float): -> float". Constant zero curvature, then pi is a real.

[@oneeyejack2]

I've spotted an error.. the closest number tor 1/3 over 64 is 21, not 19..so that should be 0.010101... and in fact 1/3 is 0.010101[01]...

[@lyrimetacurl0]

Yes and later it shows the odd constant 0.101010... = 0.666... So the even constant 0.010101.. must equal 0.333... :)

[@88porpoise]

Did you consider that it may be a Parker Third?

[@AnthonyCassidy50]

Its not a very efficient way of storing the prime numbers though - from a comp-sci perspective. The best way might be to take advantage of the gap between prime numbers. The largest prime gap for values under a million is 113, the largest prime gap under a billion is 287. There are 78,498 primes under a million, so even if (wastefully) storing the same amount of bits for each delta (eg for storing the largest prime under a million, only 7 bits of delta are needed) then only 7* 78,498 = 549,886 bits are needed. Much less than 1 million bits. The cost saving increases as we go higher.

[@rogercarl3969]

Don't blame Matt for miscalculating the binary approximation of 1/3. The dog ate his homework.

[@liam3284]

There are as many digits after the decimal point as there are before the decimal point. In other words, there are as many finite decimal numbers as there are integers. However there are numbers between zero and 1 that cannot be represented in finite decimal, and infinitely many of those as well. So, there are more real numbers between any two integers than there are integers.

[@voyageintostars]

THE DOG SLEEPING 😭

[@johnsteenbruggen5718]

The bit at the end about how the bitstrings correspond to monotonic increasing sequences of naturals: I find it easier to imagine each of the numbers as corresponding to a subset of naturals (so the nth bit is 1 if n is included in the subset or not). This was you can see that there is a one to one correspondence between the binary encodings of the real numbers from 0 to 1 and the set of all subsets of naturals -- i.e., the powerset. This is a nice way to see that the powerset of the countable set of naturals is uncountable, like the interval [0,1].

[@QuantumHistorian]

So there's a bijection between reals in [0, 1] and strictly monotonic positive integer sequences? Not something I would have guessed but, the way it's explained makes it seem obvious in hindsight

[@lonestarr1490]

They have to be strictly monotonic, though. So no repetitions either. But yeah, if I were confronted with that claim and asked to prove it, it would have probably stumped me quite a bit. But presented in this order it becomes completely obvious.

[@JohnnyDigital27]

It's a bijection between the reals in [0, 1] and the (binary encoding for strictly monotic sequences) represented in base 10. That detail is important, otherwise the statement doesn't make sense.

[@QuantumHistorian]

@@srenvitusthyregodlandkilde4800 But not than countably infinite large sets of countable infinites. Which is what an infinite strictly monotonic sequence is.

[@fullfungo]

@@JohnnyDigital27No it’s not base 10.

[@TheBasikShow]

While your statement is true, the map in the video is not an example of such a bijection: The set containing just 7 corresponds to the same real number as the set containing all integers bigger than 7, since 0.000000100000… = 0.000000011111111… in binary. There are, however, cleverer things you can do to get actual bijections between even more impressive sets. For example, using simple continued fractions you can biject every irrational number in [0,1] to an arbitrary infinite sequence of positive integers, whether increasing or not! In fact, by fiddling with finite sequences and rational numbers, you can biject everything in the interval [0,1) to an infinite-or-finite sequence of positive integers. And I think that’s neat!

[@josephrissler9847]

A simple transform maps positive sequences to positive monotonic sequences: Add to each term the sum of all prior terms.

[@heathrobertson2405]

I love that matt has the Parker square in a frame

[@oz_jones]

So its in a square. Would it be parker squared?

[@kephalopod3054]

0.4323320871859028689... also encodes all the primes from the continued fraction, see A084255.

[@jivejunior8753]

As has been stated by others, there is a glaring error in this video... he says pi is the circle constant, not tau :P

[@hoebare]

Pi is the Parker Tau

[@theadamabrams]

τ is not "the circle constant" either. Each of π and τ and π/2 = τ/4 could reasonably be called "*a* circle constant".

[@hoebare]

@@theadamabrams That's entirely true, but I think it's more fun to argue that τ is the best of all the circle constants.

[@rubyswolf9767]

@@theadamabrams Pi may be a circle constant but its the semicircle constant rather than a full circle

[@renerpho]

He's contractually obliged to use π. After all, it's asteroid (314159) Mattparker, not (628318)...

[@MarcusCactus]

10:00 They are too enthousiastic about sending the number to aliens. Their number having infinite composition, it is impossible to send in a finite time. And... we don't know an infinity of prime numbers. So there is a twist at the "end" of what we send. And... we don't know how many primes the aliens have managed to discover.

[@aikumaDK]

Excellent cameo work by Skylab

[@bertofnuts1132]

Not the first one to sleep during math class...

[@Snakeyes244]

This was an amazing numberphile!!! Not only a very cool constant, but I have never ever known that pi could be any other constant. It depends on the curvature of the space the circle in embedded in. That's huge! Pi isn't even constant on the surface of the Earth!

[@smylesg]

Brady: why'd I bring all this paper?

[@EricJohnson-ls1rh]

The binary representation of 1/3 is .0101... with repetition period two, not the four-digit repetition period shown in the video. The repetition period for a fraction cannot be longer than the denominator. To see this, consider the sequence of remainders as you do long division. As soon as the sequence comes back to the same remainder, the expansion must repeat. So the distinct remainders cannot be more number itself, and indeed must be strictly less, because once you get 0, the repetition terminates. As an example, the remainders in the long division of 1/7 are 3, 2, 6, 4, 5, 1, and then repeat. In binary, the inverse of 1/11 is .0101.... This is an example of the general pattern in any base notation that the inverse of 11 is .0m0m... where m is the base minus one. This is because when you multiply 11 * 0.0m0m... you get 0.mm..., which is indistinguishable from 1.

[@hammerth1421]

It took me way to long to realize that it's essentially the concatenation of the truth table of primeness.

[@danieldare2640]

Yes I think that's a good way of describing not only the concept but the video is that it is a novelty but time not wasted... it's always interesting and gets you thinking so thank you.

[@trummler4100]

10:38 The _better_ what if would be "how many Civilizations got 10 fingers?"

[@lafingman100]

"What if somewhere else in the universe the curvature of space is different and they got a different pi, whereas primes are always primes" Somehow this is incredibly profound

[@vsm1456]

prime numbers are still primes no matter the base. an example: you have a pile of rocks; if the number of rocks is compound, you can arrange this pile in a complete grid A × B size where A and B are factors. if the number is prime, you would only be able to arrange them in a single row or column. it doesn't matter how you write the number down

[@almightysapling]

A sequence does not need to be increasing in order to be encoded. In fact it doesn't even need to be a sequence. *Any* ordering on the natural numbers (or any other countable set) can be encoded by a real numbers via a similar mechanism

[@lafingman100]

10:35 "What if somewhere else in the universe the curvature of space is different and they got a different pi, whereas primes are always primes" Somehow this is incredibly profound

[@bjornmu]

Except it is wrong, pi does not depend on the curvature of space. It's a fundamental constant of pure mathematics, independent of any physical reality.

[@rmsgrey]

@@bjornmu It depends whether you're talking about pi or about pi (or about pi, or...) - there is a pi which is a fundamental constant of pure mathematics, and there is a pi which is half the ratio of a circle's circumference to its radius. In (near) Euclidean space like where we live, the two are close enough to the same that experimental determinations of pi by measuring actual drawings of circles give the same value as the purely theoretical construct to well within the margins of error, but it's not hard to come up with theoretical spaces where empirical pi is significantly different from theoretical pi.

[@theadamabrams]

@@bjornmu Indeed, highly intelligent aliens would probably stumble across the number 3.14159... (or 11.001001000... in binary) even the curvature of space caused their circumference÷diameter ratios were not always that number. π has a lot of uses beyond just circles (for example, the area under a non-normalized bell curve y = e^(-x²) is exactly √π).

[@xakaryehlynn4749]

i love that this episode was "idk, it's a cool number" and i can't find any reason this is actually *useful* (though i agree it's cool). Then it ends with "yell this out to say human civilization is smart!" and i love it

[@RoyalNexusAcademy]

Great Video! We really enjoyed your insights and creativity. KEEP UP THE AWESOME WORK!

[@hyperium007]

9:21 the voice sent me

[@Mathijs_A]

Yeah lol wth was that

[@danielrhouck]

2:59 This is incorrect; the number you are writing is 3/10 not 1/3. 1/3 in binary is 0.01010101… You can tell because repeating decimals mean you divide the repeating part by that many “nine” digits, by which I mean the base minus one. So 0.010011001… is 1001/1111 in binary or 9/15 = 3/5 in decimal, and the leading 0 after the -decimal- binary point means half, so 3/10. 01/11 is directly 1/3, so 0.0101… is 1/3.

[@Xboxiscrunchy]

I want to see that game of life simulation that generates primes. That sounds very interesting. Maybe you could do a video that explains it?

[@Tumbolisu]

the game of life is turing complete, so you can make a computer within that simply goes through every number, checks if its prime, and then display it.

[@RobinDSaunders]

@@Tumbolisu In fact you don't need to use Turing completeness here: a simple sieve works. The first published pattern that works is called "Primer" - you can find it e.g. on the Game of Life wiki.

[@uwepleban3784]

As of 12.10.2024, we know that at position 2^136279841-1, there is a 1 in the binary representation of the prime constant.

[@lopesdoria]

Okay, but why even encode it in base 2? Can't I just say that all primes are contained within 0.23571113171923... ?

[@MichaelDarrow-tr1mn]

why is 71 encoded twice

[@petrkdn8224]

@@MichaelDarrow-tr1mn its primes in order, 2 , 3, 5, 7, 11, 13, 17, 19, 23

[@theadamabrams]

@@MichaelDarrow-tr1mn It's not. The digits come from 0.[2][3][5][7][11][13][17][19][23]... with the brackets just added for clarity.

[@landsgevaer]

Interspersing zeros would, to make the decoding unique, in decimal. Like 0.20305070110130170..

[@vsm1456]

@@landsgevaer how do you know it's 13, 17 and not 13017?

[@BohonChina]

this prime constant representation is very close to the arithmetic coding in the coding theory, Matt Parker should make a video about this.

[@kurotoruk]

AAHH THE DIALUP HANDSHAKE SCREECH

[@MaGaO]

And Mom just picked up the phone to call someone. "Noooooooooooooo!"

[@kurotoruk]

@@MaGaO MOOOOOOOM I WAS GRINDING RARE DROPS IN RUNESCAPE!!!!!!!

[@wyboo2019]

my favorite way of encoding sequences as numbers is making them into a simple continued fraction. i.e, the primes would be 2+1/(3+1/(5+1/(7+...)))≈2.31304 it's base-neutral (also my favorite way to represent real number in general, because it is base-neutral)

[@konan4heather]

Fun fact: if you apply reverse technique to pi/4 (where we convert the fraction into base 2, and create series from the "1" indices: 2,5,8,13,14,15...), the difference-1 looks very random. I failed to find any patterns, it appears to be distributed by Negative Binomial mean=1 disperison=1.

[@skyscraperfan]

So the sequence of all natural numbers is encoded as just "1", because 0.11111111111 in binary equals 1.

[@alexpotts6520]

Yes, the thing about binary that is especially nice is that not only is every subset of the positive integers encoded by a real number between 0 and 1, but every number between 0 and 1 encodes a subset of the positive integers. Another nice property is that x and 1-x will be representations of complimentary sets (ie sets that contain all the integers between them - so for example one minus the prime constant picks out all the non-prime numbers). Binary is the natural base for this because there are two options for each number - it's either in the set or it isn't.

[@skyscraperfan]

@@alexpotts6520 In theory you could encode even encode all texts every written into a single binary number. Imagine all human knowledge can be represented by a single number. That is somehow even more mind-boggling, although would only be a finite amount of information.

[@alexpotts6520]

@@skyscraperfan I mean, this is essentially what computers do, isn't it? Everything is just converted into ones and zeros, and if you strung all those ones and zeros together you'd make a single very large integer. In fact, nature got there first! DNA works basically the same way, except it uses base 4 rather than binary.

[@UMosNyu]

What I like about the constants for each series is that is tells you how many numbers are "hit" by the sequence. The closer the number to 1, the more numbers appear in the sequence. So... kinda usefull

[@Tumbolisu]

being close to 1 only tells you that several small integers are included. for instance, the sequence {1, 2, 3, 4} (finite) becomes 0.9375 while the sequence {2, 3, 4, 5, 6, ...} (infinite) becomes only 0.5

[@UMosNyu]

@@Tumbolisu True. You are weighting small numbers bigger

[@vturiserra]

There's no need to complicate it so much. The number 0.235711131719232931... has all the primes in base 10.

[@CaoticoFanegasO_o]

I think the point is, the beauty of binary, is that it's the simplest encoding you can get. I mean, there's no term between being and not being. And that's simple and beautiful. You either have a sack of grain, or you don't. We agreed on the weight a sack of grain should have, and it doesn't qualify. But I want that grain, so I will separate a bag on as many parts we need to get a fair trade, and then when people were specialized enough, we established measures. But the start of it all, was the binary, even if we forgot of it.

[@metacob]

I can represent Pi very elegantly: 10 (That is in base Pi)

[@huawafabe]

how many different digits do we have in base pi? Because in base n we have n different digits 😅

[@Faroshkas]

Pi is 10 in base Pi

[@BryanLu0]

​@@huawafabeI think you round down. Irrational bases aren't really useful anyways, because normally rational numbers like 5 become "irrational"

[@metacob]

@@Faroshkas you're right, thanks

[@therealax6]

@@huawafabe In general, in base z your digits are all integers d so that 0 ≤ d < |z|. Note that this works for any z - it can even be complex. (Complex bases are fun!)

[@alecbader7433]

I love the way that mathematicians will say "I'm going to do something *infinitely*" and then pause, side-eye the interviewer, and clarify - "... in green, if tha'ts ok."

[@DeGuerre]

It is NOT true that any sequence of increasing integers has a real number, in the sense that they aren't uniquely "decodable". We all know that in decimal, 0.999... = 1, and 0.2999... = 0.3 for the same reason. Well, the same holds in binary. In binary, 0.0111... = 0.1, so 0.5 (in decimal) is both the "greater than 1 constant" and the "only 1 constant".

[@hoebare]

I haven't re-watched to check, but I don't think he said sequences had unique mappings, only that they had mappings. In your example, the mappings for ">1" and "1" are the same number in base 2, but if we used '1' to indicate "part of the sequence" then we could say "1/9 - 0.1 in base 10" is the mapping for ">1" and "0.1 in base 10" is the mapping for "1". There are probably (I haven't counted them) infinitely many ways to map sequences to numbers between zero and one, uniquely or otherwise. I think part of what's throwing some people off is thinking that these mappings are useful for anything other than giving Matt another chance to be wrong about something like the binary representation of 1/3.

[@RobinDSaunders]

I came here to say the same thing. But the sequences which are ambiguous are exactly the finite and cofinite ones, so you can exclude either one of those and then all the other sequences will have a unique representation.

[@WK-5775]

This criticism is not valid. The "only 1 sequence" is not a (strictly) increasing sequence - it's not even a sequence in the sense considered here.

[@RobinDSaunders]

@@WK-5775 This is a semantic matter. "Sequence" is widely used to refer to (sub)countable lists which may not be known to be infinite, or may even be known to be finite. The Online Encyclopedia of Integer Sequences has plenty examples of both.

[@Robinsonero]

This is a great argument for the absurdity of Real numbers. If "0.4146825098511..." encodes all the primes, then surely "0.414682509851..." and "0.41468250985..." do as well. Generalizing we find that "0.4..." encodes all the primes, and so does "0..." For that matter. The problem here is that when you put the "..." You have an undefined term. Does "0.414682508511..." equal "0.414..."? If it does, then does "0.414999..." also equal this number? And if there's not enough information to say, then how can we say "3.1415..." equals pi? The next digit could be anything. Whereas saying "3.1415 is a useful approximation of the circumference of a unit circle." Is totally rigorous.

[@Prinrin]

Half of this video is trying to figure out what the correct binary expansion of 1/3 is (and still getting it wrong). This is then not relevant to the point of the video, since it's just "the sum of 1/2^p for primes p". No usage (even in pure math) is given, just "you can make this construction". I didn't even see it mentioned that it's transcendental!

[@JohnDoe-ti2np]

I don't think that the prime constant has been proved to be transcendental.

[@lonestarr1490]

​@@JohnDoe-ti2np But it totally is, probably. Because if it weren't, that would be HUGE! Probably.

[@Poizon-]

10:28 - 10:50 "primes are always primes" This is not really true though. Another civilization might consist of a species with 4 fingers on each hand making base 8 their daily base. 11 is a prime number in base 10 yes, but in base 8 the number 11 can be represented by 3x3. If they had for example 3 arms they might even use a base easily divisible by 3 like base 9, 12 or 15 and that would make base 10 even more arbitrary. 3x5 in base 10 is 15 (obviously not a prime), but the number 15 has no divisors in base 8 and is therefor a prime in base 8. I did however now while writing this comment realize that the order sequence of the primes are not dependent on the base, as for example the 11th number is a prime in both bases just that we write that number as 11 in base 10 but as 13 in base 8. So the constant works in all bases and signaling it out actually works none the less! Damn, that's clever.

[@ghaydn]

That's an insane compressing method

[@StefanReich]

Except it isn't

[@JordanBiserkov]

@@StefanReich Care to elaborate? The way I see it, you take any finite sequence of monotonically increasing integers, say the primes below 64 (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61) and then you compress that into a single 64 bit number M. I can then be used to quickly answer the question "is K a prime?" (or "is K a member of the sequence?" in the more general case) simply by looking up the value of bit K. To "decompress" the sequence, loop over the bits of M and whenever the bit is equal to 1, add the index i to the accumulator.

[@theslay66]

That's just a flag system encoded into a binary value. Nothing new.

[@StefanReich]

​@@JordanBiserkov Primes are spaced out more and more as you go higher. There are roughly N/log(N) primes between 1 and N. Writing a number down in, say, decimal, takes (on the order of) log(N) bits. So we have N bits for your method, and roughly log(N)*N/log(N) = N bits for just writing down all the primes as numbers with commas in between. So no real compression except maybe a constant factor.

[@MichaelPohoreski]

It is _horribly inefficient._ Aside from 2 and 3 every pair of prime is always of the form 6n-1, 6n+1 so this wastes 4/6 or 66% encoding.

[@abracadabra6324]

The doggo living its best life there lol

[@davidcahan]

The dog sacked out on the couch is hysterical

[@dannyboy1350]

Just felt the need to point out that base 2 is actually better than base 10 for fractions. 1/n for any prime n that is one more or less than a power of the base is very easy to calculate (for 1/3 in base 2: 1/4+1/16+1/64... or 0.010101...) that is why 1/11 is so nice in base 10, but the same can't be said for base 10 1/7 which takes a 7 digit repeat whereas in base 2 it's 1/8+1/64... or 0.001001... basically a 1 digit decimal repeat. The whole binary is worse for 1/3 also breaks down when you remember that 2 binary digits is less than 2/3rds of a decimal digit. The first prime that b-2 does worse than b-10 is 11 because it's not a mersen prime but is 1 more then 10.

[@stevefrandsen7897]

Interesting and fun as always with Matt

[@Skeeve-Magick]

This is similar to an idea I read in a science fiction story. The idea is, to take all the literature of the world and put its numerical (ASCII, UTF-8… you name it) representation as a fraction and then take a 1 meter stick and mark the position of this fraction. So all the literatur of the world in one single mark on a 1m long stick.

[@Nmmoinn]

To summarize - Let S be any non repeating sequence of numbers, 0

[@Pehr81]

Different species of cicadas wake up on different prime number year intervals

[@johnbruhling8018]

Thats neat. There was a recent paper about a proof for calculating large Drielicht Primes (spellcheck) opertating largely by using a mod30 function. They are (all primes above 29), in effect, some small prime of 1,7,11,13,17,19,23,29 and some quantity of 30. See for yourself by applying a mod30 to any prime, you get one of those numbers.

[@therealax6]

The simple proof of the property you just mentioned is that all other numbers between 0 and 29 have some divisor in common with 30. So if a number can be written as 30n + r with r < 30 (and all positive integers can be), if 30 and r have some common divisor d, then both 30n and r are divisible by d, meaning that their sum is also divisible by d. The same is true for any other number, not just 30: a prime divided by any smaller positive integer has a remainder that is coprime (i.e., shares no divisors) with that positive integer.

[@fakjbf3129]

Another thing you can do is encode text. Simply chop the fractional expansion into eight bit chunks and use ASCII to give each symbol a binary representation and string them together one after another. Convert back to a decimal and you can use a single number to encode an entire book.

[@Verlisify]

"Astute viewers can try to predict this" Golden Ratio

[@oscaro.172]

It's basically rt(2)-1