Finite Fields & Return of The Parker Square - Numberphile

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  • Опубліковано 26 гру 2024

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  • @numberphile
    @numberphile  3 роки тому +222

    Extra footage & become a millionaire by winning The Parker Prize: ua-cam.com/video/hn8SwBhhDvU/v-deo.html
    The Original Parker Square video: ua-cam.com/video/aOT_bG-vWyg/v-deo.html
    Stand-Ups Maths on UA-cam: ua-cam.com/users/standupmaths
    Matt's Books (Amazon): amzn.to/3absFfV
    Matt's playlist on Numberphile: bit.ly/Matt_Videos
    Parker Square Merch: numberphile.creator-spring.com/listing/the-parker-square

    • @nowionlywantatriumph
      @nowionlywantatriumph 3 роки тому +8

      A millionaire, or a Parker Millionaire?

    • @felixlaroche8039
      @felixlaroche8039 3 роки тому +11

      Btw, Matt Parker got something wrong! Z mod powers of primes are *not* fields! For instance, in Z_4, 2x2 = 4 = 0, so that Z_4 has zero-divisors. Hence, since it has zero-divisors, it cannot be a field

    • @ZainAK283
      @ZainAK283 3 роки тому +4

      @@felixlaroche8039 Exactly - given a power of a prime, there is a finite field of that size, but it's NOT just modular arithmetic (it's a bit more complicated than that)

    • @zunaidparker
      @zunaidparker 3 роки тому +2

      Man I feel attacked...

    • @baronhannsz8900
      @baronhannsz8900 3 роки тому

      How do we get the article you referenced?

  • @wilkmarton
    @wilkmarton 3 роки тому +3661

    I don't mind Numberphile's filler episodes, but I love it when they seriously advance the main plot like this.

  • @JulietKneeled
    @JulietKneeled 3 роки тому +3388

    When I saw that "Parker" was a property of something in an actual, published research paper I legitimately doubled over laughing. The parker square is officially a real mathematical term!! I never thought I'd see the day.

    • @tsawy6
      @tsawy6 3 роки тому +247

      See, at first I was surprised, but after a certain point it's like... Damn Matt and Numberphile's fans have gotta include a significant fractions of budding mathematicians

    • @jamesonhardy2126
      @jamesonhardy2126 3 роки тому +6

      Same

    • @DomenBremecXCVI
      @DomenBremecXCVI 3 роки тому +189

      @@tsawy6 I feel like there are 10 types of mathematicians watching Numberphile; those that came here because they know maths, those that were brought into maths by Numberphile and those who forgot this comment isn't supposed to be a spin on the classic binary joke.

    • @AaronRotenberg
      @AaronRotenberg 3 роки тому +55

      @@DomenBremecXCVI That's a real Parker list, if I do say so myself.

    • @Deus_Almighty
      @Deus_Almighty 3 роки тому +4

      It's not published though

  • @dig_dus
    @dig_dus 3 роки тому +2559

    That P vs NP killed me

    • @RanEncounter
      @RanEncounter 3 роки тому +64

      That was golden :D

    • @Saka_Mulia
      @Saka_Mulia 3 роки тому +44

      Had to pause for my lols to come to a side-stiched stop

    • @sgttomas
      @sgttomas 3 роки тому +9

      Best part 😁

    • @beev
      @beev 3 роки тому +15

      surely, NP should be rebranded IP - Inverse Parker.... ;-)

    • @GreRe9
      @GreRe9 3 роки тому +1

      +

  • @SkywalterDBZ
    @SkywalterDBZ 3 роки тому +833

    In the Parker Square video, Matt said something like "In mathematics, fame is different. It's when someone looks you up once a century.". This must mean Matt is REALLY famous now.

    • @custodeon
      @custodeon 3 роки тому +77

      he is some hybrid of maths-famous and regular famous which is both more famous than maths-famous and less famous than celebrity-status

    • @thealkymyst
      @thealkymyst 2 роки тому +73

      Parker Famous.

    • @SG2048-meta
      @SG2048-meta 2 роки тому +10

      @@custodeon TL;DR a superposition of different famousnesses

    • @tinkut8960
      @tinkut8960 2 роки тому +10

      @@custodeon he’s a Parker square of a celebrity

    • @crisdunbar4753
      @crisdunbar4753 2 роки тому +3

      He's on a coffee mug fer gosh sake. Millennia from now, archaeologists (probably alien) will dig them up and he'll still be famous.

  • @QuantumHistorian
    @QuantumHistorian 3 роки тому +1222

    It's so rare, and incredibly delightful, to see a grown man beaming with joy at what is literally a consequence of being mocked in front of an audience of millions.

    • @landsgevaer
      @landsgevaer 3 роки тому +29

      He is a meme.
      I would be happy too...

    • @warasilawombat
      @warasilawombat 3 роки тому +81

      Honestly I think it’s quite sweet that they named it after him.

    • @almoglevin
      @almoglevin 3 роки тому +50

      But affectionally mocked.

    • @broadleyn
      @broadleyn 3 роки тому +5

      Well, self-mocked, but yep. Matt is awesome.

    • @josephjennings7932
      @josephjennings7932 2 роки тому +6

      All this mockery just earned him a place in mathematics for posterity.

  • @saraqael.
    @saraqael. 3 роки тому +1303

    8:33
    Kid: “Mom can I have P vs NP“
    Mom: “No, we have P vs NP at home“
    P vs NP at home: Parker vs Non-Parker

  • @sakkikoyumikishi
    @sakkikoyumikishi 3 роки тому +816

    Also:
    "They're all non-Parker - because they work." *dies inside*

  • @wecantry4393
    @wecantry4393 3 роки тому +724

    Parker square video was one of the most fun video I've ever watched. I never thought how a simple mathematical puzzle can be so enchanting.

    • @goldnutter412
      @goldnutter412 3 роки тому +2

      1/7 is a cool number with 6 recurring digits and the 0 is the FP function
      How many are there for 1/49 ? 😎

    • @veggiet2009
      @veggiet2009 3 роки тому +2

      @@goldnutter412 but not in integer fields

    • @goldnutter412
      @goldnutter412 3 роки тому +1

      14:48 there's 13 ? really ? 🤣
      This seems right to me for personal reasons hahaha also distribution wise you wouldn't expect, but possibly suddenly another group appears wayyyy up there in the giant numbers.. hm

  • @WMTeWu
    @WMTeWu 3 роки тому +810

    Everybody seem excited that "parker" has been mentioned in real, published research paper - but I think most of you underestimate how exited the authors of the paper are, that their paper has been featured in real, published Numberphile video.

    • @nomekop777
      @nomekop777 3 роки тому +71

      It's basically numberphile bait

    • @mati.benapezo
      @mati.benapezo 2 роки тому +13

      And we got tricked.

  • @olifantoliver
    @olifantoliver 3 роки тому +442

    Everytime he said "Non-Parker.. because.. it's working" you can see in his eyes, a part of him dies. :D

    • @simonmultiverse6349
      @simonmultiverse6349 3 роки тому +6

      It's fame... don't knock it!

    • @loturzelrestaurant
      @loturzelrestaurant 3 роки тому

      @@simonmultiverse6349 Anti-Science is on the Rise. Uneducation causes Muffled Logic to be be more and more accepted, so casual B.S. is getting more and more popular.
      People embarass themselves all the time now by claming NASA is faking the Sun,
      the moon is a hologram,
      the Earth is flat,
      Aura and Chakra are kinda Science, so trust me bro, i know we are all immortal - oh, and one last thing: Koalas are Fake; they are ALL CGI. All.

    • @SillyMakesVids
      @SillyMakesVids 3 роки тому +7

      A part of him becomes Parker.

    • @loturzelrestaurant
      @loturzelrestaurant 2 роки тому

      @Irony What a silly comment, Irony.

    • @Triantalex
      @Triantalex Рік тому

      false :D.

  • @alancash6420
    @alancash6420 3 роки тому +469

    I look forward to seeing Matt being awarded the Inverse Fields Medal

    • @LeonardChurch33
      @LeonardChurch33 3 роки тому +107

      Would that involve paying $15,000 for damages done to the field of mathematics?

    • @tobiaswilhelmi4819
      @tobiaswilhelmi4819 3 роки тому +70

      I would much more like to see a Parker Medal for mathematical innovations that almost work.

    • @MattMcIrvin
      @MattMcIrvin 3 роки тому +28

      The Parker Finite Fields Medal

    • @cookieninja2154
      @cookieninja2154 3 роки тому +19

      The medal for math that doesn't work but you gave it a go.

    • @simonmultiverse6349
      @simonmultiverse6349 3 роки тому +4

      @@MattMcIrvin Damn! You got there before me!

  • @TECHN01200
    @TECHN01200 3 роки тому +791

    I love how mathematicians use Parker as an adjective meaning "almost works"...

    • @dexter2392
      @dexter2392 3 роки тому +69

      If the large mathematical community finally caught it... Parker will be a legend.

    • @pvic6959
      @pvic6959 3 роки тому +40

      idk if I would be happy or sad if my name was given that definition. On one hand, my name has become an ACTUAL property in math. Like in a published paper - it will live on forever. but on the other hand, the property my name describes is "doesnt work" LOLOL

    • @Relkond
      @Relkond 3 роки тому +40

      Matt Parker is a comedian. Some of the best jokes in life are where things almost work.
      I’m sure he’s elated.

    • @TECHN01200
      @TECHN01200 3 роки тому +3

      @@pvic6959 At the very least, they have a sense of humor...

    • @brandonthesteele
      @brandonthesteele 3 роки тому +8

      I would be tremendously honored to have my name used in math in any capacity. Matt seems pretty jazzed about it.

  • @HopUpOutDaBed
    @HopUpOutDaBed 3 роки тому +181

    Finally someone explaining P vs. NP in a way everyone can easily understand.

  • @YosmHere
    @YosmHere Рік тому +23

    For those who might've missed a pun at 8:38:
    P v/s NP (Which in video is used as a short form for Parker v/s Non-Parker) is actually one of the seven millenium problems by the Clay University. Each problem worth a million dollars. That means if you solve it you'll get a million dollars.

  • @feudiable
    @feudiable 3 роки тому +573

    The 6x6 table says 3*2 = 1 mod 6, but I guess that is a parker-one.

    • @Minihood31770
      @Minihood31770 3 роки тому +126

      The Parker Times Table

    • @cybisz2883
      @cybisz2883 3 роки тому +66

      Lol, I caught that too. Seems closeups of that table were edited out due to the mistakes in it.

    • @laurihei
      @laurihei 3 роки тому +31

      Plus he also circled that one when circling all the ones in the table ':D

    • @simonmultiverse6349
      @simonmultiverse6349 3 роки тому +13

      Can we have a Parker Timetable, (not "Times Table") where the trains almost but not quite arrive at the times they're supposed to?

    • @laurihei
      @laurihei 3 роки тому +11

      @@simonmultiverse6349 I think we already have that :D

  • @karlwaugh30
    @karlwaugh30 3 роки тому +248

    For finite fields of prime power orders there was some confusion in this video. The integers mod 49 or 4 or 8 etc don't produce finite fields of those orders. It's just that there do exist other finite fields of those orders with different structure to them.
    Eg. In Z mod 4 the multiples of 2 are 2x1 = 2, 2x2=0, 2x3=2 and 2x0=0 and so there is no inverse for 2.

    • @samuelthecamel
      @samuelthecamel 3 роки тому +31

      The true Parker Finite Fields

    • @MrSamwise25
      @MrSamwise25 3 роки тому +1

      Thanks for pointing this out! :)

    • @probablyapproximatelyok8146
      @probablyapproximatelyok8146 3 роки тому +20

      And I think the way you can get finite fields of prime power order p^k is by adding zeros of particular polynomials to the finite field Z/pZ, much like you can add i (one of the zeros of x^2 + 1) to the real numbers to get a new, bigger field: The complex numbers

    • @djyotta
      @djyotta 3 роки тому +1

      I was thinking that finite fields of order of "powers of primes" could be things other than Z mod (p^r), but note that the paper says: Finite Fields and Rings - which implies to me that they're claiming that magic squares of squares don't just work in (most) finite fields of the form Z mod (p), but also some rings of the form Z mod (p^r) where p is prime...

    • @johanrichter2695
      @johanrichter2695 3 роки тому +8

      Yes, that is a very important point, hope they correct that.

  • @agar0285
    @agar0285 3 роки тому +378

    I love the fact that "Parker" defined as "not working" is an actual term in a math research. I just started laughing so much, this was awesome.

    • @ГеоргиГеоргиев-с3г
      @ГеоргиГеоргиев-с3г 3 роки тому +3

      6:50 ( left square 3;2 )(seen the meme, just was about to comment on the Parker square, and was informed that it's actually a Parker Parker square. The circle later really helped )

    • @RaiinWing
      @RaiinWing 3 роки тому +2

      lets gooo you watch numberphile too

    • @agar0285
      @agar0285 3 роки тому +1

      @@RaiinWing Hi rainwing 😀

    • @Triantalex
      @Triantalex Рік тому

      ??

  • @MrQwefty
    @MrQwefty 3 роки тому +67

    He gave it a go, he tried, and finally he's achieved infamy in actual mathematical research! Kudos to you Matt

  • @namduong8437
    @namduong8437 3 роки тому +5

    The fact that you still have the mug at 7:52 makes me super happy to follow math community

  • @illesizs
    @illesizs 3 роки тому +295

    "Every real number has a buddy real number, where if they multiply together, you get 1."
    1: "Am I a joke to you?"
    0: "Yes."

    • @pulsefel9210
      @pulsefel9210 3 роки тому +28

      1 is such a lonely number. so powerful they wont even let it have its proper title of prime of primes.

    • @Sibula
      @Sibula 3 роки тому +31

      @@pulsefel9210 You could even say that one is the loneliest number

    • @neopalm2050
      @neopalm2050 3 роки тому +18

      -1:

    • @BizVlogs
      @BizVlogs 3 роки тому +5

      1? One’s buddy number is 1.
      0? Zero is the same as n (limit as n goes to zero). So its buddy number in that case is 1/n (limit as n goes to 0).

    • @allanolley4874
      @allanolley4874 3 роки тому +3

      It is after all an ancient mathematical proposition that one is not a number but the unit first enunciated by Aristotle. If 1 is not a number then 0 is right out.

  • @KSJR1000
    @KSJR1000 3 роки тому +10

    This is the most clear explanation of N vs NP I've ever seen.

  • @JanxakaJX
    @JanxakaJX 3 роки тому +39

    Matt Parker is a great teacher and quite funny too. I love seeing him here.

  • @GivenFailure
    @GivenFailure 3 роки тому +418

    I think I'm going to start saying "don't go trivial" randomly to people.

    • @goldnutter412
      @goldnutter412 3 роки тому +5

      Just answer any complex question with relativity
      Meaning of life ? relativity (or 369)

    • @MrAlRats
      @MrAlRats 3 роки тому +1

      For String Theorists, every sequence of "Why" questions leads ultimately to the answer "String theory".

    • @goldnutter412
      @goldnutter412 3 роки тому

      @@MrAlRats but they have to be "strings" of physical matter, with 2 dimensions 😅

  • @ugu8963
    @ugu8963 3 роки тому +219

    I'm feeling the need to hear the word "Parkericity"
    "Hey how about the Parkericity of that field ?"

    • @prashantadhimal
      @prashantadhimal 3 роки тому +11

      Parkerness?

    • @annie4424
      @annie4424 3 роки тому +2

      This. This needs to become a thing.

    • @mond256
      @mond256 3 роки тому +12

      Why not have degrees of Parker for how far off from working it is

    • @Games_and_Music
      @Games_and_Music 3 роки тому +11

      Margin of error is now called "Parker approximation".

    • @NoNameAtAll2
      @NoNameAtAll2 3 роки тому +1

      @@Games_and_Music approximation is already Parker property (Parker action?)
      Parkerximation?

  • @Sam_on_YouTube
    @Sam_on_YouTube 3 роки тому +31

    I tip my hat to the author of this paper. Well done.

  • @MonzennCarloMallari
    @MonzennCarloMallari 5 місяців тому +3

    When Matt said "most finite fields are non Parker" and then he smirked, I died laughing

  • @floyo
    @floyo 3 роки тому +53

    5:13 The finite field with 49 elements is not actually the integers mod 49 (Z/49Z), because 7 has no inverse. The construction of this field is more complicated.

    • @jaredbitz
      @jaredbitz 3 роки тому +7

      For the curious - to actually construct that finite field, consider the set of polynomials with coefficients modulo 7. You can get a field with 49 elements by taking all polynomials of the form ax + b, and then doing arithmetic on them modulo x^2 - 3 (again all the coefficients are modulo 7). 7 choices for a and 7 choices for b make 49 elements, and you can never multiply two polynomials to get zero because x^2 - 3 doesn't factor modulo 7.
      You can get finite fields whose sizes are higher prime powers (i.e. 7^n) by doing arithmetic modulo some irreducible polynomial of degree n.

    • @FireSwordOfMagic
      @FireSwordOfMagic 3 роки тому +1

      Same with any number that isn't a prime.

    • @danielyuan9862
      @danielyuan9862 3 роки тому +2

      @@jaredbitz why modulo x^2-3 and not x^2?

    • @danielyuan9862
      @danielyuan9862 3 роки тому +2

      @@jaredbitz no wait, it's because you can imagine x=sqrt(3)

    • @TimHardcastle-i9g
      @TimHardcastle-i9g Рік тому

      @@jaredbitz , or for people who don't know how to do that with finite fields, but do know how complex numbers work, imagine that i is the square root of 3 mod 7, and consider things of the form a + bi where a and b are in Z7.

  • @shaftahoy
    @shaftahoy 3 роки тому +7

    14:46 'Parker' being in Comic Sans is the cherry on the top of this video.

  • @baguettegott3409
    @baguettegott3409 3 роки тому +2

    This made me so happy. I can't believe this is actually in the paper - what a wonderful thing the community has created here.

  • @iah7264
    @iah7264 3 роки тому +308

    "Return of the Parker square"
    This is probably the most clickbaity title possible, for numberfile fans ;)

    • @Neefew
      @Neefew 3 роки тому +26

      Is it clickbait if it's true?

    • @SheldonBird
      @SheldonBird 3 роки тому +2

      It's the only reason I clicked instantly

    • @simonmultiverse6349
      @simonmultiverse6349 3 роки тому

      Return of The Pink Parker?

    • @simonmultiverse6349
      @simonmultiverse6349 3 роки тому +1

      ...featuring Peter Parker? (different superhero, I know)

    • @alfieomega
      @alfieomega 3 роки тому

      it did reappear, not as main focus though
      more like a cameo old character in the new series

  • @ferraneb
    @ferraneb Рік тому +2

    5:04 Just to clarify the the integers mod a power of a prime do NOT form a field in general (for example, 7 does not have an inverse mod 49). It is only the case when the power is 1 (that is, the integers mod a prime). There exist finite fields of size p^k for p prime and k > 1, but they are constructed differently.

  • @gordonwiley2006
    @gordonwiley2006 3 роки тому +7

    We tease because we love you, Matt. Your enthusiasm is infectious. I consider myself, to be a Parker Person.

  • @TheSummoner
    @TheSummoner 3 роки тому +101

    5:09 - Is he implying that the integers mod 49 are equivalent to the finite field of order 49? Because as far as I know this only work for primes, for prime *powers* the multiplicative structure is actually different.

    • @Vodboi
      @Vodboi 3 роки тому +21

      Yea, just noticed that, in Z_49 you have 7*7=0, and a field doesn't have zero divisors, so its not a field. I guess he kinda confused it with the fields of order equal to that prime power.

    • @pianissimo7121
      @pianissimo7121 3 роки тому +2

      I am a bit confused, does a Z7 field for example, have 0 in it? Cause 0 doesn't have a multiplicative inverse does it?

    • @AGLubang
      @AGLubang 3 роки тому +12

      @@pianissimo7121 Yes. All fields must have a 0. The rule for multiplicative inverse doesn't include 0, as with usual real numbers, rationals, etc.

    • @dabluse3497
      @dabluse3497 3 роки тому +7

      @@pianissimo7121 In fields, zero is a special number that follows different rules. In every field, 0*a=0, for any a in the field, and 0 is the only number that doesn't have a multiplicative inverse, because a field needs 0 to work. That's true in the real numbers, complex numbers, and any other field. Hope that clears it all up.

    • @Vodboi
      @Vodboi 3 роки тому +5

      @@pianissimo7121 The statement of being a field is that: "Every nonzero element has a multiplicative inverse", where zero is defined as the element satisfying 0+x=x+0=x for all x in the field (in other words, 0 is the additive identity). So yes Z_7 has the elements {0,1,2,3,4,5,6}, where all but 0 have multiplicative inverses

  • @cereal_chick2515
    @cereal_chick2515 2 роки тому +3

    This is one of the greatest character arcs I've ever seen!

  • @shawon265
    @shawon265 3 роки тому +23

    Matt Parker: You cannot find a whole number inverse of an integer.
    1: I will pretend I didn't see that.

    • @purrplaysLE
      @purrplaysLE 3 роки тому

      1*1=1

    • @ragnkja
      @ragnkja 3 роки тому

      Unless it’s the identity. Just like the only non-negative number with a non-negative additive inverse is 0.

  • @chimiseanga9054
    @chimiseanga9054 3 роки тому +26

    Correction: only "integers mod a prime" is a field, not "integers mod a power of a prime". There are finite fields of size "power of a prime" but they are not a quotient of the integers.

    • @keineangabe8993
      @keineangabe8993 3 роки тому +1

      Thank you! I didn't think they would miss such an obvious mistake..

    • @mbartelsm
      @mbartelsm 3 роки тому +6

      It was a Parker-explanation

    • @kijkbuis8575
      @kijkbuis8575 3 роки тому +1

      These are the Parker finite "fields"

  • @danielbergman1984
    @danielbergman1984 3 роки тому +2

    This video made me happy! Not that any other Numberphile video makes me otherwise, but this one's special. Congratulations Matt!

  • @davidwilsch4668
    @davidwilsch4668 3 роки тому +28

    Z mod 49 and Z mod 25 are NOT fields. There exist fields with 49 or 25 elements but they aren't simply integers modulo some number.

    • @davidkalichman
      @davidkalichman 2 роки тому

      THANK YOU for pointing this out. An uncharacteristic error from Matt :(

  • @EdwardCree
    @EdwardCree 2 роки тому +14

    "What about _infinite_ rings?" Well, if a magic square of squares "works" in ℤ, then it must also work modulo n ∀ n∈ℕ. However, in some of those ℤₙ, the square may have repeated entries that weren't there in ℤ; in particular we know that this must be the case for all n for which ℤₙ is Parker. (As the paper points out, and as you mention in the extra footage, a solution in ℤ would imply there are only finitely many Parker rings.) Thus those rings give us constraints on any possible solution in ℤ; for instance, ℤ₆₇ being Parker implies that a magic square of squares in ℤ cannot have all nine numbers distinct modulo 67, because otherwise it would imply a solution in ℤ₆₇. It's the Parker rings, and _only_ those rings, which help us by cutting down the search space for ℤ; Parker rings are _useful_ because they help us identify what _won't_ work, and that can be valuable in itself :)
    Hope that helps Matt feel a little better about his eponymy.

  • @antezante
    @antezante 3 роки тому +7

    This was great, having in-depth math on a higher level than usual! Please do more of this!

  • @meeDamian
    @meeDamian 2 роки тому +1

    This is the greatest video I've watched this year by far 👏🏻👏🏻👏🏻.

  • @jan-pi-ala-suli
    @jan-pi-ala-suli 7 місяців тому +3

    parker is finally a true mathematician, he has a thing named after himself

  • @stardustpan
    @stardustpan 3 роки тому +18

    PARKER SQUARE LES GOOOO

  • @johnchessant3012
    @johnchessant3012 3 роки тому +17

    Important note (for anyone who, like me, is going to spend a few hours looking into this): The finite field F_(p^k) is NOT the integers mod p^k. For example, F_9 = {0, 1, 2, i, 1+i, 2+i, 2i, 1+2i, 2+2i} where i = sqrt(-1).

    • @redapplefour6223
      @redapplefour6223 3 роки тому +3

      well you know for pedantry that it's actually that i^2 = -1, thats the technical definition

    • @StoicTheGeek
      @StoicTheGeek 3 роки тому +1

      Darn I just went and typed all that out less clearly and then I saw your comment!

    • @StoicTheGeek
      @StoicTheGeek 3 роки тому +3

      Please also not that the field F_(p^k) has character p ie. np = 0 for all n in the field

    • @leftaroundabout
      @leftaroundabout 3 роки тому +1

      @@redapplefour6223 that's not the technical definition either. Or, well, it is part of the definition, but of the technical definition of the _multiplication operation_ in ℂ, not of i. The imaginary unit can't be defined like this. (Note that e.g. in the quaternions there are three distinct values that all fulfill this equation!) To make it a technical definition, you need to first define ℂ as a 2-dimensional vector space with unit vectors 1 and i, and only then equip it with the multiplication that has this property, in order to form a field.

    • @redapplefour6223
      @redapplefour6223 3 роки тому

      @@leftaroundabout right, thanks! makes sense that that's how that works. so are field extensions are just unit vectors in disguise?

  • @mathieudehouck9657
    @mathieudehouck9657 3 роки тому +2

    This amazing 1 in the column of 2.
    Made my day Mr Parker.
    Thank you.

  • @helpme6599
    @helpme6599 3 роки тому +412

    It's been 5 years, but Matt Parker is still Matt Parker.

    • @yourguard4
      @yourguard4 3 роки тому +22

      Matt Parker + 5 years = Matt Parker ? :D

    • @2D_SVD
      @2D_SVD 3 роки тому +2

      And that's great!

    • @Ravendragon52
      @Ravendragon52 3 роки тому +29

      Matt Parker is officially invariant wrt time

    • @proloycodes
      @proloycodes 3 роки тому

      88th like!

    • @idahogie
      @idahogie 3 роки тому +5

      And I'm still non-Parker.

  • @vindj2391
    @vindj2391 3 роки тому +5

    14:10 i don't know why but seeing those parkers pop up on the screen cracks me up

  • @rubenlarochelle1881
    @rubenlarochelle1881 3 роки тому +36

    "Technology has moved on since", showing a 3D-printed version of what he once wrote on brown paper.

  • @ImaginaryMdA
    @ImaginaryMdA 3 роки тому +44

    The Parker prize needs to become a reality, surely!

    • @nielskorpel8860
      @nielskorpel8860 3 роки тому +13

      For all maths research whose results do not accomplish what they aimed for,...
      ...but which do make some headway towards it, which gives an insight into the subject, which explores useful perspective on the subject, or which studies the hardship of proving what you are trying to prove,...
      ... so that maybe one day we can make more informed maths research that DOES achieve what it was trying to do.
      In other words, for all the disappointing, unglamorous near-misses which might eventually lead to actual results.
      Not a bad thing to have a prize for, actually.
      If this approach of near misses does at some point answer the question whether the integers are parker or not, then it actually becomes a serious proposal: the approach worked.

    • @camicus-3249
      @camicus-3249 3 роки тому +7

      @@nielskorpel8860 Basically, "Give it a go"

  • @terraqueo89
    @terraqueo89 3 роки тому +8

    This is one of the best gags of this channel lol

  • @nopetuber
    @nopetuber 3 роки тому +4

    I've been following these channels forever and I'm like, look at you Matt! Congrats!

  • @smallishkae
    @smallishkae 3 роки тому +59

    “If you’ve got a number, I dunno… a.”
    Can’t wait to see that one out of context

  • @shuetomtqasaab
    @shuetomtqasaab 3 роки тому +3

    That's absolutely insane! Parker is not only a scientist, but also a living meme - we know that for quite some time. But the fact, that he's not just an ordinary walking meme (albeit this in itself is something to be proud of), but a meme which is included in scientific papers. Incredibly amazing!

    • @jd9119
      @jd9119 Рік тому

      Are mathematicians scientists? And if so (or not so), what exactly are the criteria we're using to define what a scientist is?

    • @fregattenkapitan
      @fregattenkapitan Рік тому

      ​@@jd9119they do research in universities in a scientific field. Difficult to be more of a scientist....

    • @jd9119
      @jd9119 Рік тому

      @@fregattenkapitan Except scientists usualy apply the mathematics to another discipline.

  • @PopeLando
    @PopeLando 3 роки тому +25

    8:09 I am made up, and enormously proud of you, Matt! Edited: doubly proud of your joke at 8:33 🤣🤣

  • @Astromath
    @Astromath 3 роки тому +3

    A Numberphile video with Matt Parker AND a Stand-Up Maths video on the same day? Nice!

  • @asdfghyter
    @asdfghyter 3 роки тому

    15:26 I love that the previous video is in the citations for this paper!

  • @ModeDecay
    @ModeDecay 3 роки тому +50

    I wish there was a compilation of every time Matt says "big fan..."

  • @DemoniteBL
    @DemoniteBL 3 роки тому +2

    I just love the fact that "Parker" is a term accepted by most if not all mathematicians.

  • @WGSen
    @WGSen 3 роки тому +4

    I am in love with this whole saga

  • @codelerias
    @codelerias 3 роки тому +9

    I love it when they bring back season 1 characters!

  • @matheusspable
    @matheusspable 3 роки тому +4

    Ok. When you get named in a paper that actually delivers, and sets a new standard for maths... This is amazing.

  • @jasonburt8069
    @jasonburt8069 6 місяців тому +1

    Every step towards a better understanding, in every field of study, has the name of the person who discovered it, attached. Matt Parker should be proud to have his name linked to this little step.
    "True understanding is built upon a mountain of mistakes." Paraphrased from someone important I don't remember at this time 😅.

  • @MeTalkPrettyOneDay
    @MeTalkPrettyOneDay 3 роки тому +3

    Truly the most troll-y way to get something professionally named after you. I love it.

  • @iansragingbileduct
    @iansragingbileduct 3 місяці тому +1

    I shrivelled up into a small human bean when "P vs. NP" showed up on the screen. Amazing. Level 99 math-dad joke.

  • @argentvixen
    @argentvixen 3 роки тому +15

    This is about right. We had the "Mould effect" so now Matt is just catching up to Steve with the "Parker property". I assume this is the omen that Matt will catch up with a million subs soon. 😘

  • @bootesvoidband
    @bootesvoidband 3 роки тому +22

    I’m waiting for the OEIS entry for Parker Numbers

    • @babel_
      @babel_ 3 роки тому +6

      A308838, the Orders of Parker finite fields of odd characteristic, aka the list shown ignoring 2. The "state of the art" has improved and it was shown 243 is a Parker finite field.

  • @mohamedaminekoubaa5231
    @mohamedaminekoubaa5231 3 роки тому +3

    a small mistake at 5:04. It only works for prime numbers. If you take a power of prime numbers, it is not modular arithmetic anymore. So basically if you are working in the finite field with four elements, 1+1 is still 0 just like the field with two elements, but you have an extra element x which satisfies x^3=1.

  • @henrygreen2096
    @henrygreen2096 3 роки тому +8

    I actually find the the fact that Parker is rare a really cool thing. Sure they “don’t work” but they got people talking first, and there aren’t that many

  • @EmC_98
    @EmC_98 3 роки тому +3

    10:58 nice surprise seeing myself in a Numberphile video!!

  • @julesbrunton1728
    @julesbrunton1728 3 роки тому +2

    I've always enjoyed how the multiplication symbol is the addition symbol nudged over 45° and the division symbol is the minus symbol with some dots or recently also just pushed over at an angle /

  • @ИльяИльяшенко-р7т
    @ИльяИльяшенко-р7т 3 роки тому +7

    It's not true that integers mod 49 (or any non-trivial prime power) form a field. For example, 7 doesn't have an inverse mod 49.
    I think Matt got confused by the notation F_{49} for a finite field with 49 elements.

  • @expomath9348
    @expomath9348 3 роки тому +2

    Excellent comme d'habitude ! Un plaisir de regarder cette chaine.
    Translation for non french people : " Hi, it's sunny today but it depends where you live actually"

  • @cyaneya
    @cyaneya 3 роки тому +6

    This was soooo interesting, thank you Parker for being very knowledgeble and funny. I wish i was able to sit with you with a glass of beer and just ask basic questions about math, which i'm terrible at, and the answers would be probably unexpected. Yeah, thanks again!

  • @brunoalejandroandrades354
    @brunoalejandroandrades354 3 роки тому +2

    Just a heads up, mod p^k is not a field for k>1. It's just there are field with that amount of elements, but they're not Z/p^kZ. Z/49Z is not a field, since 7,14,21,...,42 do not have inverses

  • @IamBATMAN13
    @IamBATMAN13 3 роки тому +14

    The P vs NP reference killed me

  • @rachelsimons2040
    @rachelsimons2040 3 роки тому

    Thanks for the cameo 😅 (also congrats on becoming a fully-fledged mathematical term!)

  • @മിന്നൽമുരളി-ഠ1ച

    This guy has such comical facial expressions, he would probably do well in comedy movies if he did acting.

    • @EM-pb7lk
      @EM-pb7lk 3 роки тому +74

      He does math related stand-up

    • @elevown
      @elevown 3 роки тому +21

      Well he does do stand up about math lol

    • @yousorooo
      @yousorooo 3 роки тому +12

      Matt Parker is a comedian after all

    • @abhijiths5237
      @abhijiths5237 3 роки тому

      Mallu spotted 😂

    • @Triantalex
      @Triantalex Рік тому

      false.

  • @SquirrelASMR
    @SquirrelASMR 2 роки тому +2

    The people with Parker square shirts are probably your biggest fans haha

  • @rubenlarochelle1881
    @rubenlarochelle1881 3 роки тому +4

    Parker and non-Parker being used in an actual paper was an hilarious twist ahahahah

  • @davidharmeyer3093
    @davidharmeyer3093 3 роки тому +2

    I burst out laughing when you put "P vs. NP" as an overlay on the screen for Parker vs. Non Parker fields

  • @shirou9790
    @shirou9790 3 роки тому +26

    5:17 that's not exactly true, the integers mod 49 do not work as a finite field. However there is indeed a finite field of 49 elements, which can be constructed as 1st-degree polynomials over the integers mod 7.
    In fact [Theorem 1] the integers mod n are a field if and only if n is prime, and [Theorem 2] there exists a finite field with n elements if and only if n is the power of a prime p (constructed as polynomials over integers mod p)

    • @andrewharrison8436
      @andrewharrison8436 3 роки тому +2

      Ahhh - useful comment. Since 2 is a prime and powers of 2 crop up in computers this creates lots of possibilities once you realise the fields are more complex than just mod n. Now I need to look up polynomials over integers as fields - well that's this afternoon gone.

    • @mxpxorsist
      @mxpxorsist 3 роки тому +1

      It's a parker field.

    • @twohoos
      @twohoos 3 роки тому +1

      Correct, the powers of primes correspond to extension fields, i.e. ordered n-tuples of elements of the base prime field. It's analogous to how the complex numbers may be viewed as ordered pairs of real numbers.

    • @shirou9790
      @shirou9790 3 роки тому

      @@twohoos Yeah exactly. Now that I think of it complex numbers are essentially polynomials modulo x²+1, which is really similar to the way we construct finite fields of order p^n.

  • @calebcopeland6425
    @calebcopeland6425 3 роки тому +1

    It brings me joy that the Parker Square has left the numberphile bubble and ventured into general mathematics and is being used in published research papers

  • @CHIEF_420
    @CHIEF_420 3 роки тому

    Thanks!

  • @eathonhowell7414
    @eathonhowell7414 2 роки тому +3

    This is the equivalent of how Gary Larson is now credited as naming "the spiny bits on the end of a Stegosaurus" the Thagomizer because before him nobody had a name for it. It was done as a joke and then someone saw value outside of it being funny.

  • @ssdd28561
    @ssdd28561 3 роки тому

    Thanks

  • @DiracComb.7585
    @DiracComb.7585 3 роки тому +4

    Being diagnosed with Parker finite-fieldness is a truly heartbreaking event, my condolences.

  • @video99couk
    @video99couk 3 роки тому +2

    Many years from now when you're pushing up the daisies, at least you will be forever remembered having had a mathematical property (even a duff one) named after you. Quite an honour.

  • @bur2000
    @bur2000 3 роки тому +2

    8:42 is that this P-NP problem that's one of the great unsolved questions?

  • @arnauarnauarnau
    @arnauarnauarnau 3 роки тому +1

    Wow this is so cool! Awesome sequel to parker square. Can’t wait for part 3 in a few years

  • @Frownlandia
    @Frownlandia 3 роки тому +13

    Maybe a Mathematician gets to be upset when their name is associated with a kind of failure, but a Standup Mathematician is just happy to setup a punchline.

  • @bentationfunkiloglio
    @bentationfunkiloglio 3 роки тому +1

    Seriously, true genius. You actually made number theory funny and interesting.

  • @cgibbard
    @cgibbard 3 роки тому +18

    Polynomial rings typically aren't fields, but you can make fractions of them (rational functions) and those will be a field.

    • @JM-us3fr
      @JM-us3fr 2 роки тому +2

      True, but those aren’t _finite_ fields. You have to mod an irreducible polynomial to get a field.

  • @nokillnina
    @nokillnina Рік тому +1

    you made math history! congratulations!

  • @shanathered5910
    @shanathered5910 3 роки тому +6

    Finite field F₄ isn’t technically integers mod 4, it’s a bit more complicated than that. Example: 2² = 3, it’s not mod 4 because 2² = 0 mod 4. This is true for all non-prime order fields.

    • @shanathered5910
      @shanathered5910 Рік тому

      I also showed that integers mod 4 has zero divisors and therefore NOT a field

  • @tth-2507
    @tth-2507 3 роки тому +2

    5:15: No, the Integers mod 49 (or any other non prime) do not form a field.
    It is true, that finite fields exist exactly for the powers of primes, but the higher powers are of a different from.

  • @XtReMz98
    @XtReMz98 3 роки тому +3

    Well. I can only guess that Matt Parker’s ego went from finite to non-finite after being established as an (in)famous legend of mathematics! I love this guy!

  • @MrMaelstrom07
    @MrMaelstrom07 3 роки тому +1

    Since the conjecture says all fields of prime (or powers of prime) size above 67 are non-parker BUT they can't show that there exists a non-Parker magic square of squares, then ALL (if the conjecture is true), of the non-Parker fields above 67 must use a sum somewhere on the square that is at or above the field number.
    For example, the non-Parker square for say F121 MUST, at some point, have a sum that is equal to or greater than 121. If this wasn't true, then the Magic Square of Squares would exist using that sum

  • @falquicao8331
    @falquicao8331 3 роки тому +6

    The sequel we always knew we needed

  • @SigmaSixSoftware
    @SigmaSixSoftware 3 роки тому +1

    I haven’t started the video yet and this is the best explication of fields I’ve heard