Irrational Roots

Поділитися
Вставка
  • Опубліковано 23 гру 2018
  • iNEW (Christmas 2019). Two ways to support Mathologer
    Mathologer Patreon: / mathologer
    Mathologer PayPal: paypal.me/mathologer
    (see the Patreon page for details)
    For the final video for 2018 we return to obsessing about irrational numbers. Everybody knows that root 2 is irrational but how do you figure out whether or not a scary expression involving several nested roots is irrational or not? Meet two very simple yet incredibly powerful tools that they ALMOST told you about in school. Featuring the Integral and Rational Root Theorems, pi Santa, e(lf), and a really cringy mathematical Christmas carol.
    As usual thank you very much to Marty and Danil for their help with this video.
    Merry Christmas :)

КОМЕНТАРІ • 1,1 тис.

  • @evaristegalois6282
    @evaristegalois6282 5 років тому +1158

    *_All I want for Christmas is a complete proof of the Riemann Hypothesis_*

    • @joryjones6808
      @joryjones6808 5 років тому +7

      Me too

    • @samueldevulder
      @samueldevulder 5 років тому +5

      Ask to Michael Atiyah ;-)

    • @AttilaAsztalos
      @AttilaAsztalos 5 років тому +79

      I have discovered a truly remarkable proof of this theorem which this text box is too small to contain... ;)

    • @helloiamenergyman
      @helloiamenergyman 5 років тому +4

      @@AttilaAsztalos where could you fit it?

    • @gautamdiwan5952
      @gautamdiwan5952 5 років тому +22

      @@AttilaAsztalos no shit Fermat

  • @nemo2803
    @nemo2803 4 роки тому +356

    saying something equals 3.14... is the clickbait of the math community

    • @rubenvela44
      @rubenvela44 4 роки тому +21

      There are a lot of things that equal to 3.14.

    • @contasecreta1234
      @contasecreta1234 4 роки тому +66

      @@rubenvela44 like 3 + 0.14

    • @realbignoob1886
      @realbignoob1886 3 роки тому +7

      Lmfao

    • @uRealReels
      @uRealReels 3 роки тому +20

      whats funny is that the cube root of 3 + cube root of 2 is about e

    • @joshuascholar3220
      @joshuascholar3220 3 роки тому +8

      Not so much. Few would be naive enough to think that pi could be algebraic.

  • @unalcachofa
    @unalcachofa 5 років тому +253

    it's a square of a binomial inside the first square root :)
    √(4 + 2√3) - √3
    √(1 + 2√3 + 3) - √3
    √( (1 + √3)^2 ) - √3
    1 + √3 - √3
    1

    • @tashigawayuno7503
      @tashigawayuno7503 5 років тому +14

      thanks, u saved my life

    • @MathNerd1729
      @MathNerd1729 5 років тому +4

      Astute observation, Jose. Bravo!

    • @rosebuster
      @rosebuster 5 років тому +5

      I didn't see this comment before writing my own with the same solution. I would have just replied here.

    • @jsutinbibber9508
      @jsutinbibber9508 5 років тому

      How does the second one transform into the third one?

    • @SgtSupaman
      @SgtSupaman 5 років тому +6

      @@jsutinbibber9508, (1 + √3) is the square root of (1 + 2√3 + 3). So, he got the square root and squared it. Basically the steps were this: √4 = √2^2 = 2

  • @sashabell9997
    @sashabell9997 5 років тому +169

    6:50
    sqrt(2) + sqrt(3) = x
    x^2 = (sqrt(2) + sqrt(3))^2
    x^2 = 5 + sqrt(24)
    x^2 - 5 = sqrt(24)
    (x^2 - 5)^2 = 24
    x^4 - 10x^2 + 25 = 24
    x^4 - 10x^2 + 1 = 0
    Only possible integer solutions are +1 or -1. Subbing them in gives -8 for both, not 0. Therefore solution must be irrational. Therefore sqrt(2) + sqrt(3) must be irrational.
    8:04
    If the constant is equal to 0 then all other terms are divisible by x, so simply factorise x out of the equation and solve this new equation. The constant being 0 simply means one of the factors of x is 0.

    • @Mathologer
      @Mathologer  5 років тому +33

      Very good :)

    • @Jo_Es_Chess_Channel
      @Jo_Es_Chess_Channel 5 років тому +2

      I think you made a mistake in the third line:
      x^2=5+2sqrt(6) not 5+sqrt(24)

    • @TheKingartur007
      @TheKingartur007 5 років тому +18

      @@Jo_Es_Chess_Channel 2 * sqrt(6) = sqrt(24)
      I think it is right.

    • @Sordorack
      @Sordorack 5 років тому +9

      @@Jo_Es_Chess_Channel well, but it is:
      2*sqrt(6) = sqrt(2²) * sqrt(6) = sqrt(4*6) = sqrt(24)
      is it not? ^^

    • @Jo_Es_Chess_Channel
      @Jo_Es_Chess_Channel 5 років тому +3

      Yeah I'm sorry I thought she did 2x the rest and got 24

  • @Mathologer
    @Mathologer  5 років тому +265

    I am spending Christmas in Germany this year and we've just exchanged Christmas presents (this happens on the 24th in Germany and not on the 25th). Now it's your turn :)

    • @gr1nder07
      @gr1nder07 5 років тому +6

      We do the same, always the day before :)

    • @andreaaristokrates9516
      @andreaaristokrates9516 5 років тому +12

      In what part of Germany are you, because you said "fröhliche Weihnachten", this sounds weird to me, where I live we/most say "frohe..", but I know there people out there saying it the other way.

    • @miniwizard
      @miniwizard 5 років тому

      Wishing you a Merry Christmas - and looking forward to more brain stimulation in the new year!

    • @maxsch.6555
      @maxsch.6555 5 років тому +6

      Frohe Weihnachten 😄

    • @John77Doe
      @John77Doe 5 років тому +1

      Mathologer X =.sqr(2) + sqr(3) is the same X*X -2 * sqr(2) - 1 = 0 and +1. and -1 are not roots of this quadratic equation. Happy Holidays!!! 🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻🎄🎅🏻

  • @Lurco8
    @Lurco8 5 років тому +53

    I'm from Poland and we definitely had both the Integer Root Theorem and the Rational Root Theorem in their proper, full statements in High School. In fact it is still taught, since I tutor a few teenagers and have seen their textbooks. We used them only to find roots of polynomials of order higher then 2 however and not the other way around as you do. So this is indeed a very cool trick. Thank you for the video and Merry Christmas!

    • @PC_Simo
      @PC_Simo Рік тому +2

      I’m guessing you took Long Maths, then? I’m from Finland, and don’t remember either from high school (I took Intermediate Maths, though); and we do pretty well, in the Pisa comparisons. 🤔
      Also: *_*Than._*

    • @morgoth4486
      @morgoth4486 Рік тому +3

      yeah, schemat Hornera

  • @cybisz2883
    @cybisz2883 5 років тому +171

    3:05 Holy crap. I'm from the United States and have a university degree in mathematics. I mean, it's totally obvious now that I see it, but I don't recall any of my school teachers or professors ever actually pointing out this trick.

    • @stephenbeck7222
      @stephenbeck7222 5 років тому +16

      Cybis Z every high school algebra 2 and precalculus curriculum I’ve seen has the rational root theorem as the primary method of “guessing” roots of large polynomials to attempt to break them down into at least quadratics (at which point the quadratic formula can be used).

    • @e11eohe11e
      @e11eohe11e 5 років тому +4

      @Cybis Z Same here, senior getting a BS in Math in NC USA, and haven't seen this even once.

    • @ProfessorPolitics
      @ProfessorPolitics 5 років тому +5

      USA also. It was never explained explicitly to my math classes either. I vaguely noticed a pattern when I did algebra homework, but this video was like a lightbulb turning on in my brain.

    • @DavidSharpMSc
      @DavidSharpMSc 5 років тому +1

      went to school in Scotland and pretty sure this integral root theorem was never taught to us

    • @ThePharphis
      @ThePharphis 5 років тому +8

      it is taught in canada in high school

  • @endolp2624
    @endolp2624 5 років тому +9

    I'm going to a German school and so far nobody showed us such a useful way to solve our tasks. Thank you and Merry Christmas.

  • @xyz.ijk.
    @xyz.ijk. 5 років тому +52

    NY, USA ... never heard of these until university. Sad.
    However, I was recently visited by the Mathologer Ghost of Christmas Yet to Come ... it warbled at me that .999... = 1. I said, "wait, I've already learned my lesson. I agree ... I agree." "No," it said. "There is more yet to come."
    Merry Christmas to you and Marty and your staff and families. You have brought much joy and learning over the year.

    • @Mathologer
      @Mathologer  5 років тому +6

      :)

    • @jessehammer123
      @jessehammer123 5 років тому +1

      J G Really? At Stuy, I learned this sophomore year.

    • @willgd6666
      @willgd6666 5 років тому +1

      In MA, I learned this in high school.

  • @Fustincho
    @Fustincho 5 років тому +93

    It's 00:00 in Germany and Santhologer arrived with this cool present :)

    • @AlexVasiluta
      @AlexVasiluta 5 років тому

      @Destro so 60 toffee? I envy you.

    • @shadowbane7401
      @shadowbane7401 5 років тому

      How can it be 00:00?

    • @AlexVasiluta
      @AlexVasiluta 5 років тому

      @@shadowbane7401 he means midnight

    • @DDKKAY
      @DDKKAY 4 роки тому

      11:59 time for Grand Finale......this is german.😁

    • @ffggddss
      @ffggddss 3 роки тому

      @Agustin Vargas Toro: And he arrived to the tune of Tannenbaum.
      Fred

  • @alicetroise5785
    @alicetroise5785 5 років тому +16

    In high school in Italy we teach rational root theorem for the factorization of polynomials with "Teorema di Ruffini" and "Regola di Ruffini"

    • @electric7487
      @electric7487 4 роки тому

      They teach Ruffini's theorems in secondary schools?!

    • @3mon3y94
      @3mon3y94 3 роки тому

      I d

    • @raffaeledivora9517
      @raffaeledivora9517 3 роки тому +2

      @@electric7487 Yes.

    • @dlevi67
      @dlevi67 3 роки тому +2

      @@electric7487 Yep. First year of high school.

  • @mehditabti2658
    @mehditabti2658 5 років тому +469

    My mother gave me 1 present, my father gave me 2, my grandfather 3, my grandmother 4 and at the end im crying bc i only got -1/12 present

    • @ajfiaewoipaweipefawejpfwoe1049
      @ajfiaewoipaweipefawejpfwoe1049 5 років тому +18

      Cristalboy lol well played

    • @stevethecatcouch6532
      @stevethecatcouch6532 5 років тому +112

      It's worse that you're telling us. If you got -1/12 presents, you must have infinitely many relatives you had to buy presents for.

    • @looney1023
      @looney1023 5 років тому +9

      If you continue in that fashion, does that mean your great great great . . . grandparents are still alive???

    • @Chubtato
      @Chubtato 5 років тому +24

      Go back to numberphile please

    • @nischalmr5978
      @nischalmr5978 5 років тому +2

      Steve the Cat Couch epic!

  • @franciskatende1566
    @franciskatende1566 4 роки тому +9

    Learnt the integral root theorem in high school in Botswana. We then used the solution to divide the polynomial and finally use the quadratic equation

    • @PC_Simo
      @PC_Simo Рік тому

      You must’ve taken Long Maths, then? I don’t remember being taught that in high school, in Finland. Though, I took Intermediate Maths. Maybe I’ve just forgotten about it. Wouldn’t put it past me, to forget a single topic, in 12-14 years. 🤷🏼‍♂️

  • @TheZMDX
    @TheZMDX 5 років тому +66

    3:05 I was taught this method in 2nd year of high school, but note I was attending mathematics-physics profiled class. I’m from Poland.

    • @TheZMDX
      @TheZMDX 5 років тому +2

      Moreover we were taught “real solution theorem” but we didn’t really named it and didn’t consider irrational solutions since polynomial equations that we considered always had a rational solution.

    • @maciejmanna9246
      @maciejmanna9246 5 років тому +3

      Same here (also in Poland).

    • @pszm5085
      @pszm5085 5 років тому +1

      I'm in 2nd year of maths-physics in high school, and I'm also from Poland! :)

    • @adimoon9
      @adimoon9 5 років тому +1

      Yep, exactly the same.

    • @teodorkowalski3873
      @teodorkowalski3873 5 років тому +1

      Same. Except for we learned it in 1st class.

  • @OlavRH
    @OlavRH 5 років тому +8

    Done three years of maths at the uni in Norway. Never heard of the integral root theorem until now. Neat

  • @Gelo2000origami
    @Gelo2000origami 5 років тому +22

    Loved your collab with Flammable Maths

  • @BigNWide
    @BigNWide 5 років тому +3

    I've taught the integral and rational root theorems in high school math classes in the United States. We then reduced polynomials (using discovered rational factors) to find complex roots as well.

  • @Henrix1998
    @Henrix1998 5 років тому +44

    You really teased pi with the thumbnail

    • @MathNerd1729
      @MathNerd1729 5 років тому +4

      I know, right?
      By the way, pi is transcendental so it cannot equal sqrt(2)+sqrt(3) because sqrt(2)+sqrt(3) is algebraic [polynomial: x^4-10x^2+1=0].

    • @mitchmarq428
      @mitchmarq428 5 років тому +4

      it really DOES equal 3.14...

  • @thomaskirsch2209
    @thomaskirsch2209 5 років тому +1

    As always, for me at least, another wonderful video. What a great Christmas present. May you and yours and the crew that supports you have the most wonderful of Holidays.

  • @user-fp6pi6wi5l
    @user-fp6pi6wi5l 5 років тому

    Thanks for the video and all other videos of this year. Merry Christmas and Happy New Year!

  • @aaryannakhat1842
    @aaryannakhat1842 5 років тому +5

    Merry Christmas Mathologer!

  • @quackers.
    @quackers. 5 років тому +37

    In school in northern England, we were just told to guess random integers

    • @olbluelips
      @olbluelips 5 років тому

      ryan2698 - Minecraft Not if the leading coefficient divides the constant term

    • @ryan2-518
      @ryan2-518 5 років тому

      Ol' Bluelips if it divides it to an integer it’s fine. It’s it’s 3/4 you’re screwed

    • @brianrobinson4825
      @brianrobinson4825 5 років тому

      Also from Northern England. Also told to try sensible possible integer values. Even the current A Level exam questions give a root to divide by first. Why was I not told?

  • @billrussell3955
    @billrussell3955 5 років тому

    WOW! Thank you!
    Merry Christmas doc.

  • @lourdjonsalen8465
    @lourdjonsalen8465 2 роки тому +1

    I really appreciate that I found this video. I was looking for some lessons and videos which tackle finding irrational root because I failed to find real irrational roots on four items on our activity so I was worrying too much.

  • @evanyao8475
    @evanyao8475 5 років тому +3

    Love your videos as always! Curious, what animation software you use for these equations? They're quite neat!

  • @mimithehotdog7836
    @mimithehotdog7836 5 років тому +4

    Is √(2) + √(3) a complex number?
    √(2) + √(3) = x
    x^2 = [√(2) +√(3)]^2
    x^2 = 5 + √(24)
    x^2 - 5 = √(24)
    (x^2 - 5)^2 = 24
    x^4 - 10x^2 + 25 = 24
    x^4 - 10x^2 + 1 = 0
    Possible integer solutions: x = +1 or x = -1.
    Let f(x) = x^4 - 10x^2 + 1
    f(1)= f(-1) = -8, not 0.
    Therefore √(2) + √(3) must be irrational.
    What if the polynomial has an ending constant = 0?
    At least 1 of the solutions is x = 0.
    Algebraically show that √(4 + 2√3) - √3 = 1
    Let x=√(4 + 2√3) - √3
    x=√(1 + 2√3 + 3) - √3
    x=√( (1 + √3)^2 ) - √3
    x=1 + √3 - √3
    x=1
    Find the equation that has √(4 + 2√3) - √3 as a root?
    Let x=√(4 + 2√3) - √3
    √(4 + 2√3) = x + √3
    4 + 2√3 = (x+√3)^2
    4 + 2√3 = x^2 + x(2√3) + 3
    2√3 = x^2 + x(2√3) - 1
    2√3 - x(2√3) = x^2 - 1
    2√3 (1 - x) = x^2 - 1
    12 (1 - x)^2 = (x^2 - 1)^2
    12 (x^2 - 2x + 1) = x^4 - 2x^2 + 1
    12x^2 - 24x + 12 = x^4 - 2x^2 + 1
    0 = x^4 - 14x^2 + 24x - 11
    let f(x) = x^4 - 14x^2 + 24x - 11
    f(x) has a root of 1

  • @stevenpreissler4205
    @stevenpreissler4205 4 роки тому

    I very much enjoyed this video. I've had differential calculus and never came across ether of these theorems. Thank you for your charming presentation.

  • @NikhilKumar-im8ls
    @NikhilKumar-im8ls 5 років тому

    You are the best maths channel I have watched till date :)

  • @benjaminbrady2385
    @benjaminbrady2385 5 років тому +8

    I learned the polynomial guessing trick in school, they also said to start with 1 then -1 then the other factors as they were most likely to be picked by examinators. I'm from Ireland

  • @miap2605
    @miap2605 4 роки тому +3

    Thank you for your videos! My University degree is Arts orientated, so I haven’t been focusing on maths as much as beforehand. My enthusiasm for mathematics has been rekindled due to your videos. Again, thank you.

  • @JS-rt7kp
    @JS-rt7kp 5 років тому +2

    You’re awesome! Thanks for the maths and the laughs!!
    I’m from the US and this was my first time learning both theorems 😀

  • @F.E.Terman
    @F.E.Terman 5 років тому +2

    I'm from The Netherlands. I have no complaints about the high school math I had, but we did not hear about these integral and rational root theorems. So I know now! Thank you.
    From all the answers we could build a 'root theorem proliferation map'. :)

  • @menukasharma4890
    @menukasharma4890 5 років тому +7

    sqrt(4+2sqrt(3))-sqrt(3) then this can be written as
    sqrt((sqrt(3))^2+1^2+2(1)(sqrt(3))-sqrt(3)
    using identity a^2+b^2+2ab=(a+b)^2
    so we get
    sqrt(((sqrt(3)+1)^2)-sqrt(3)
    cancelling the square roots we get
    sqrt(3)+1-sqrt(3) hence
    1

    • @royhedine5153
      @royhedine5153 5 років тому

      You are missing critical brackets ...
      sqrt((sqrt(3))^2+1^2+2(1)(sqrt(3))-sqrt(3)
      let A=sqrt(3)
      sqrt((A)^2+1^2+2(1)(A)-A
      sqrt((A)^2+1+A ????

    • @AymanSussy
      @AymanSussy 5 років тому

      nice men

  • @STD282
    @STD282 5 років тому +7

    3:03
    I was taught this trick at the 9th year of education in school in Russia.

    • @zhasulanaset6514
      @zhasulanaset6514 3 роки тому

      We taught this trick at 10th grade in Kazakhstan

  • @Dradan659
    @Dradan659 3 роки тому +2

    I'm from California, USA and I learned about the Integral Root Theorem. Though we learned that in order to find all possible solutions we would need to find our "ps and qs". Where p was the set of factors of the constant, and q was the set of factors of the leading coefficient. Then all possible solutions were made by matching all ps and qs together in the form of p/q.

    • @ffggddss
      @ffggddss 3 роки тому +1

      That's the Rational Roots Theorem; the Integral Roots Theorem is a special case of it.
      Fred

  • @jean-micheltorres6925
    @jean-micheltorres6925 5 років тому

    Thank you for the present ! once again : very much appreciated ! In France, 35 years ago, for polynomial equations of degree > 2, I was told to try small integers : 1, -1, and 2, -2..

  • @jessehammer123
    @jessehammer123 5 років тому +4

    NYC, Stuyvesant High School, Sophomore Honors Algebra 2. Though I learned it as the Rational Roots Theorem.

  • @KnakuanaRka
    @KnakuanaRka 5 років тому +3

    As for getting that root 3 mess into an integer, here: (note I use rx for the square root of x)
    (1+r3)^2=
    1^2+r3^2+2*1*r3=
    1+3+2r3=
    4+2r3
    Thus, r(4+2r3)-r3=
    (1+r3)-r3=
    1.

    • @KSignalEingang
      @KSignalEingang 3 роки тому

      Thanks for this! I kept getting stuck with an equation that either spat out nonsense or something useless like "0x = 0".

  • @johnwelter7081
    @johnwelter7081 5 років тому

    Merry Christmas. Thanx for the vid.

  • @letheodobertdetrelorneux8149
    @letheodobertdetrelorneux8149 5 років тому

    I learned this in secondary school, merry Christmas from Belgium and keep going with the good work !

  • @yaseralaa9902
    @yaseralaa9902 5 років тому +73

    I did learn it in school! Im from iraq xD

    • @ekuchsbdyshxdhzu7dhdhx949
      @ekuchsbdyshxdhzu7dhdhx949 5 років тому +2

      I want to say an edgy joke so badly

    • @yaseralaa9902
      @yaseralaa9902 5 років тому +9

      @@ekuchsbdyshxdhzu7dhdhx949 My Christmas present for you is letting you say that joke, go ahead!

    • @nevokrien95
      @nevokrien95 5 років тому

      so i guess thats why in Israel we dont do cubics

    • @johnmc1011
      @johnmc1011 4 роки тому

      yaser alaa I’m from Iraq and I learn everything from UA-cam 😂😂😂😂

    • @hybmnzz2658
      @hybmnzz2658 3 роки тому

      @@ekuchsbdyshxdhzu7dhdhx949 puss

  • @professorryze3739
    @professorryze3739 5 років тому +4

    Just tell me where do you buy those T shirts! Please! I want to buy all of them!

  • @DMSG1981
    @DMSG1981 5 років тому

    What a wonderful Christmas present!
    Fröhliche Weihnachten to you and your family as well.

  • @tansoon8257
    @tansoon8257 3 роки тому

    My favourite Mathologer video of all time! (I know there r many other great ones but I simply love that proof at the end)

  • @brendanward2991
    @brendanward2991 5 років тому +3

    Dublin, Ireland. I never heard of the Integral Root Theorem.

  • @sergeikalashnikov556
    @sergeikalashnikov556 5 років тому +10

    3:03 We were taught the Ruffini method in high school, that was in Spain.

    • @frabol02
      @frabol02 5 років тому +3

      I was taught the same in Italy

    • @willa4you
      @willa4you 5 років тому +1

      Me too. Italy. :)

    • @overfield18
      @overfield18 5 років тому

      i just was looking for this comentary, im from spain too

  • @christopherkingsland4604
    @christopherkingsland4604 5 років тому

    Thank you for your videos. As my family and I are cloudy-minded due to so much Christmas Cheer today, your easy to understand video is very much appreciated.

  • @MathNerd1729
    @MathNerd1729 5 років тому +2

    In a mathematical training session in the Philippines, I learned how to simplify certain nested radicals. So, I actually knew that the second one wasn't irrational and was equal to one before you told me it was. By the way, great video!

  • @kurtilein3
    @kurtilein3 5 років тому +3

    I am from Germany. Never heared about the trick, because it would not work, because while equations are checked to see that getting the solution does not get too overly complicated, the solutions will generally not be nice. We do get fractions, roots and stuff.

    • @narutosaga12
      @narutosaga12 5 років тому

      kurtilein3 a

    • @xario2007
      @xario2007 5 років тому

      @kurtlein3: Wut? I'm also from Germany and for this kind of problem, where you had to guess the first root to then be able to use pq or abc formula, the first root would usually be an integer or 1/2.

  • @looney1023
    @looney1023 5 років тому +3

    From New Jersey; we learned this in the u/v form, but not really the integral form.
    Also wanted to say that in the beginning, you briefly brought up root(2) being irrational. A while back, a video went viral where Micheal from Vsauce walked through a proof of root(2) being irrational, but he made a super subtle mistake near the end that pissed me off. He used the fact that an even number squared is even to imply that the square root of an even perfect square was even, which is a converse error. It made me wish you were as big as Vsauce, as you're so thorough in your presentation and explanation I doubt you'd have made such a mistake.

    • @Laufissa
      @Laufissa 5 років тому +1

      So you think that there exists an even perfect square with a root that isn't even?

    • @rema0126
      @rema0126 5 років тому

      Lmao

    • @MathNerd1729
      @MathNerd1729 5 років тому

      Look, odd x odd is odd so it cannot be odd. If it is not even either then what is it?

    • @looney1023
      @looney1023 5 років тому +2

      @@Laufissa Of course not, but the method of proof was wrong and misleading, especially for the mathematics level of people the video was for.
      A=>B does not mean B=>A

    • @Laufissa
      @Laufissa 5 років тому

      @@looney1023 Ah, I see. It's hard to spot these kinds of mistakes when the statements are very simple and you already know the answer.

  • @lucasamorim8699
    @lucasamorim8699 5 років тому +2

    Here in Brazil we don't learn these root theorems at school. I always thought I should guess integer roots until reducing my polynomial to a second degree polynomial by using long division haha. Thanks for sharing!

  • @Jelle_NL
    @Jelle_NL 5 років тому

    Thank you for this Christmas present!

  • @johnchessant3012
    @johnchessant3012 5 років тому +3

    8:35 "So while the root theorems are fantastic tools for determining the irrationality of an algebraic number, they are of no use for transcendental numbers."
    But every transcendental number is irrational!

    • @effuah
      @effuah 5 років тому

      But proving that a number is transcendental may be really hard, if you want to determine for example if π²-e is rational you have some work to do.

    • @dlevi67
      @dlevi67 5 років тому

      @@robbraman5869 That is (almost) the definition of transcendental...

  • @AlejandroBravo0
    @AlejandroBravo0 5 років тому +3

    I learnt this as the Ruffini method, or something like that, I wasn't paying too much attention tbh. I'm from Spain.

  • @alexanderli5987
    @alexanderli5987 5 років тому

    Whoaa the ending was neat.
    These videos always put a giant childish grin on my face
    Merry christmas!

  • @jofox8066
    @jofox8066 5 років тому

    So cool quick and clean - I love it!

  • @supercobra1746
    @supercobra1746 5 років тому +3

    Soviet school taught me to go through integer solutions first.
    Vieta formula and stuff.
    Damn you bloody Bolsheviks!

  • @TrimutiusToo
    @TrimutiusToo 5 років тому +4

    I learned not exactly this trick for solving polynomial equations, but I cooked up something like that myself without help from textbooks or teacher when I was in high school back in Russia, and was a lot faster with eliminating potential roots once I got a hang of it... And I wasn't even always searching for intereger solution, because if coefficient before the largest power isn't 1, then I would use it as a potential denominator for possible rational fraction solutions too... But I was always good at spotting patterns, I came up with a recursive solution for Tower of Hanoi when I was 6 years old, all on my own, it was a bit cruel of my dad to ask me find the smallest number of moves for 8 tiles high tower of Hanoi just so that I don't bother him, but hey spending 3 hours as a pre-school kid on such a puzzle taught me a lot, and it was a perfect solution too, I still use same thinking I did back then whenever I solve this puzzle...

    • @trueriver1950
      @trueriver1950 Рік тому

      Just think how neat it would've been if you'd already worked out the general result for the Hanoi puzzle and replied instantly "255; and by the way for a stack of ten it's 1023".

    • @TrimutiusToo
      @TrimutiusToo Рік тому

      @@trueriver1950 well i mean I was 6 I just learned how to solve it back then, forget general result i didn't even know the multiplication table yet back then...

  • @KipIngram
    @KipIngram 8 місяців тому

    This one was excellent - nice job guys!

  • @jpalreis
    @jpalreis 5 років тому +2

    Firstly, thank you for the video und frohe Weihnachten! I'm from Brazil. I was first presented to the rational root theorem (not it's proof) in my late high school years. Nowadays, it's hard to see it being taught. I would like to suggest a topic related to this one (has to be done after minimal polynomials, if recall it right): unnesting nested radicals. If I'm not mistaken, there's an article about it.

  • @SergiuszOlszewski
    @SergiuszOlszewski 5 років тому +9

    There is a mistake at 8:37 - transcendental numbers do not have to be real, can be complex as well.

    • @nathanisbored
      @nathanisbored 5 років тому +1

      the reals are part of the complex numbers, so just imagine another both around the reals that says 'complex'

    • @SergiuszOlszewski
      @SergiuszOlszewski 5 років тому +1

      Still, the transcendental numbers box should float over the frame of real numbers. Especially given that they can be imaginary only, for example πi is transcendental.

    • @nathanisbored
      @nathanisbored 5 років тому

      Sergiusz Olszewski that wouldn’t work either, not every real number is transcendental

    • @SergiuszOlszewski
      @SergiuszOlszewski 5 років тому +1

      I've put up a visual for you. i.ibb.co/7nFs72g/transcendental.png Tho I'm not sure if every real number has to be either algebraic or transcendental.
      Edit: the same story is with algebraic numbers - can be complex or imaginary as well.

    • @becomepostal
      @becomepostal 5 років тому

      Sergiusz Olszewski I think there are algebraic numbers that are not real.

  • @deniskhryashchev7245
    @deniskhryashchev7245 5 років тому +4

    Learned it in grade 7 (at the age of 12) along with Horner's method, polynomial division, factorization, etc. I am originally from Russia.

    • @RapGeneral11
      @RapGeneral11 5 років тому +1

      Hail slavs? I am form Bulgaria, got taught the same!

  • @Israel220500
    @Israel220500 5 років тому +1

    I didn't learn this trick explicitly, but I was taught something in High School that I think is related called the Girard Identities, one of them states that the product of all the roots of a polynomial is equal to the constant term divided by the leading coefficient. I'm from Brazil.

  • @MuffinsAPlenty
    @MuffinsAPlenty 5 років тому

    I'm in the US, and I was taught the rational root theorem in a high school algebra class.
    I teach the rational root theorem to my students if I'm ever teaching linear algebra (finding eigenvalues) or differential equations (characteristic polynomial for an nth order linear differential equation with constant coefficients).
    Great video by the way! Loved it!

  • @denelson83
    @denelson83 5 років тому +3

    3:05 - Canada here, and that wasn't ever taught to me.

    • @reinymichel
      @reinymichel 5 років тому

      I taught Math in Canada for 33 years and I always taught this. So it must be a regional thing, and depends on your teacher.

  • @mehtapuru
    @mehtapuru 5 років тому +12

    0:05 When she cheats on you with a guy just because he's richer

    • @rubico1894
      @rubico1894 5 років тому

      Well at least it was with a guy and not the neighbors dog

  • @stromboli183
    @stromboli183 5 років тому

    Burkard, your t-shirt is absolutely brilliant... yet again! 👏

  • @CastorQuinn
    @CastorQuinn 5 років тому

    We learned this in year twelve maths in NSW in the '90s. Our selective school had this peculiar way of teaching content, where we'd be taught to be canny to the way exams are put together. So we were taught this for the purpose of self checking, or digging ourselves out of a hole, but we were also taught that, since maths exams mark on working, not on solutions, we still needed to know how to work the question in the expected way. End result was a grade of students who were all hyper cynical about exams, but also had a lot of confidence that we understood the maths we were being tested on, and why we were learning it.

  • @aidanokeeffe7928
    @aidanokeeffe7928 5 років тому

    Excited for your next video!

  • @pronounjow
    @pronounjow 5 років тому

    Very nice video! I had no idea about this at all (from the US). Merry Christmas!

  • @karangupta1825
    @karangupta1825 3 роки тому +1

    There's an identity to add to irrationals:
    m^(1/2) + (m +1)^(1/2) = (4m+2)^(1/2)
    It was first derived by ramanujan
    I have generalized it a bit:
    M^(1/2) + (m+n)^(1/2) = (4m+2n)^(1/2)
    I am just 13 and I have been watching your videos for a year
    💘 Love from India 🇮🇳

  • @enricolucarelli816
    @enricolucarelli816 5 років тому +1

    👏👏👏👏What took you so long? I have been missing this piece of knowledge since ever! 😁 What a cool video! Thank you very very much!

    • @Mathologer
      @Mathologer  5 років тому +1

      Just wait, three, four videos from now I'll show how a closely related simple result can be used to solve some really ancient problems :)

  • @yjin378
    @yjin378 5 років тому

    Thanks for the great videos. Merry Christmas. Bon Noël.

  • @ponsi83
    @ponsi83 5 років тому +1

    Merry Christmas Burkard and everyone! I never heard about the integer or rational root theorem or a similar trick in school (neither did I during my math degree). Greetings from Austria, Thomas

  • @benjaminbrady2385
    @benjaminbrady2385 5 років тому +1

    I have returned after a six month pilgrimage to understand the generalised hypergeometric function... Worth it

  • @chickitychina100
    @chickitychina100 5 років тому

    Thank you so much for another fine video...keep up the work.

  • @MushookieMan
    @MushookieMan 5 років тому

    This truly is a gift, some mathematics on your channel that I can actually do!

  • @fmcore
    @fmcore 3 роки тому

    Amazing as usual.

  • @e.s.f.8185
    @e.s.f.8185 5 років тому

    3:05 I'm a highshchool student in Catalonia, and we learn this in math class at about fifteen y-o while learning Ruffini's method and we call it "Teorema del residu" (Remainder's Theorem in catalan). Love your videos, by the way.

  • @jmochogi
    @jmochogi 5 років тому

    I once asked my teacher how I would apply these stuff in life and he told me this: One day, you will be walking home from the market and find the equation blocking your way. It will ask you to solve it before proceeding. I never took my teacher seriously but now I know how mathematics is a gatekeeper to many juicy careers. It took me a while but thank God I am now a mathematics educator, trying to help people learn this stuff in the nicest way possible that makes sense to them.

  • @ginosantillo4655
    @ginosantillo4655 5 років тому +1

    It was taught as the “fundamental theorem of algebra” and we do some synthetic division by one of the zeros to simplify the equation to a quadratic and then just use quadratic formula to find the remaining zeros

  • @podemosurss8316
    @podemosurss8316 5 років тому +1

    2:59 I learned it in 9th grade (actually in Spain it's "3rd of ESO", ESO standing for Obligatory Secondary Education in Spanish). Our teacher called it the "Ruffinni method".

  • @BechirZalila
    @BechirZalila 5 років тому

    Loved it! Thank you

  • @PC_Simo
    @PC_Simo Рік тому +1

    I’m from Finland; and I don’t remember being taught the Integer Root Theorem, in middle school (or high school (Intermediate Maths), for that matter); and Finland’s schools are among the best in the world; but we only solved first-degree equations or partial higher-degree equations, like: x³ + 6 = 0, or: 2x² - 8 = 0, in middle school 🇫🇮🤷🏼‍♂️.

  • @thorscape3879
    @thorscape3879 3 роки тому

    I learned the Integral Root Theorem and Rational Root Theorem in my Sophomore year of high school in the USA in my Pre-Calculus class. It was essentially a class on Euler and a few other mathematicians. However most of the year was spent with polynomial equations, deriving them, and using those derivations so construct shapes with ruler/compass, and prove constants like Pi, e, and i.

  • @AlessandroDruetto
    @AlessandroDruetto 2 роки тому

    I'm from Italy and studied (in high school) about that method by the name of "Ruffini's Rule": for a polynomial of degree n, we know that an eventual rational solution r must have the form p/q; where p is a divisor of the term that multiplies x^0 and q is a divisor of the term that multiplies x^n.

  • @iammaxhailme
    @iammaxhailme 5 років тому +1

    I am from New York City and I was never shown anything like the integral root theorem (or rational root theorem) in school or uni, but I read about it on my own time 8) merry Christmas!

  • @aee220phmunirabad
    @aee220phmunirabad 5 років тому

    I don't know IRT before. Thank you very much. It is the basic and most important theory. Well explained

  • @thienngthe116
    @thienngthe116 5 років тому

    My own childhood hypothesis is proved true, thanks to this man :)

  • @QDWhite
    @QDWhite 5 років тому

    I was taught that integer root trick in Canada, but it was never explained as eloquently as this. I was just told to look for common factors between the coefficient of the highest power and the coefficient of the lowest power and then brute force my way throw them until I find (or didn’t find) a root. I didn’t know I was exploiting a fundamental property of polynomials until watching this video.

  • @picklevoncrunchnmunch7946
    @picklevoncrunchnmunch7946 5 років тому

    i love rooty combo numbers. Also I loved when the transcendental numbers floated out of the screen

  • @__-cx6lg
    @__-cx6lg 5 років тому +1

    3:00
    I learned this as the Rational Root Theorem (which is the generalization and easy consequence of the version you presented). It was basically identical to how you presented it; I don't remember the packaging being very different. I'm from the US East Coast.

  • @carolynk2
    @carolynk2 5 років тому

    Fröhliche Weihnachten und einen guten Rutsch ins neue Jahr.

  • @Bl4ckiii
    @Bl4ckiii 5 років тому

    Marry Christmas to you what a nice surprise :D

  • @mohameddib6069
    @mohameddib6069 5 років тому

    Great video, as usual

  • @frankharr9466
    @frankharr9466 5 років тому

    Und schoene Ruetsch ins neue Jahr!
    I had no idea. That is so cool.

  • @dominicstewart-guido7598
    @dominicstewart-guido7598 3 роки тому +1

    I learned the trick. They called it the “rational root theorem” and told us to divide all factors of the constant term by all factors of the leading coefficient. From New Orleans, USA

  • @madpad759
    @madpad759 5 років тому

    Your video is really good. I hope I found this channel as a student!

  • @bemusedindian8571
    @bemusedindian8571 5 років тому

    Awesome. This was a ray of sunshine. Indeed.

  • @jhawk2402
    @jhawk2402 5 років тому +2

    11:35
    Lets start with √(4 + 2√3) only
    lets say √(4 + 2√3) = √a + √b
    square both sides
    4 + 2√3 = a + b + 2√ab
    Because of the square roots, we can deduce this system of equations
    a + b = 4
    2√ab = 2√3
    Solving this we get a = 1 and b = 3 (no it doesn't matter which order you put them in)
    so then going back, √(4 + 2√3) = √1 + √3 = 1 + √3
    now adding in the -√3 at the end,
    1 + √3 - √3 = 1.