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

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

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.

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 :)

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!

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.

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.

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.

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

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.

Loved your collab with Flammable Maths

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

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

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

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

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

Merry Christmas Mathologer!

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

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

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.

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.

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.

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.

WOW! Thank you! Merry Christmas doc.

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

You really teased pi with the thumbnail

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

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..

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.

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

Merry Christmas. Thanx for the vid.

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

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

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

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.

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

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.

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.

So cool quick and clean - I love it!

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

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

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

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.

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

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

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

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

Excited for your next video!

I did learn it in school! Im from iraq xD

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

Thank you for this Christmas present!

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'. :)

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!

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

This one was excellent - nice job guys!

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

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.

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

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!

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!

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

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.

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

Verryniice! Merry Xmas to you!

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

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...

Great video and super entertaining!!!

amazing way of explaining and an extraordinary video I ve faced telling you what in the last year while I was studying discrete math , a very sophisticated equation faced me so I went to you youtube and searched a lot , until I saw you cute video and the days went on today the same problem faced me so immediately and without thinking I headed towards your video it is really valuable and worth to see all the best and my God help you :)

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

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

Amazing as usual.

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

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

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

Great video, as usual

Marry Christmas to you what a nice surprise :D

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!

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

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

Loved it! Thank you

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

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.

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.

Yay ! another amazing video :-D I learned things again !!!!

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

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

Fröhliche Weihnachten und einen guten Rutsch ins neue Jahr.

Hi from Switzerland. I didn't know about this technik that you presented to try to hunt the solution of the equation. But I was teached a formula using complex numbers to solve 3rd degrees equation such as this one. Happy X-Mas

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

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!

Awesome. This was a ray of sunshine. Indeed.

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 🇮🇳

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.

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.

learnt the trick in first year high school in Switzerland... Happy christmas btw!

Great and insightful video, as always! Good to know there are at least 3,524 (as of now) happy boys and girls who got a very nice present for Christmas :) I'm confident that I've learnt both versions of the integral root theorem around 9th grade in school, but it must be noted that I was studying in a specialized math class in a very good school in Moscow, Russia. We were trained for math competitions and olympiads, so they couldn't have overlooked such a cool theorem. Not so sure about Gauss's lemma, but it's possible that we covered that one, too. *Solutions* to the video's puzzles and challenges: 1) sqrt(2)+sqrt(3) is irrational and can be proved by double-squaring and applying the theorem. Note that x = sqrt(2)+sqrt(3) x^2 = 5 + 2*sqrt(6) (x^2-5)^2 = 24 x^4 - 10*x^2 + 1 = 0, and x=1, x=-1 are obviously not solutions to this polynomial equation, meaning that x must be irrational. 2) If the constant term in the equation is zero, then we can note that x=0 is automatically a solution to the equation. Assuming that it is not the value we are looking for, we can divide the equation by x (to the minimal power present), and then apply the integral/rational root theorem to the resulting equation. 3) Massaging sqrt(4+2*sqrt(3))-sqrt(3) into 1 is not that hard. Note that we have sqrt(4+2*sqrt(3)) - sqrt(3) = sqrt( (1+sqrt(3))^2 ) - sqrt(3) = (1+sqrt(3)) - sqrt(3) = 1. 4) Coming up with a polynomial that has the number in challenge (3) as a root: Let x = sqrt(4+2*sqrt(3)) - sqrt(3), then note that x+sqrt(3) = sqrt(4+2*sqrt(3)) => (x+sqrt(3))^2 = 4+2*sqrt(3) => x^2 + 3 + 2*sqrt(3)*x = 4+2*sqrt(3) => x^2 - 1 = 2*sqrt(3)*(1-x) => (x^2-1)^2 = 12*(x-1)^2 => x^4 - 2*x^2 + 1 = 12*x^2 - 24*x + 12 => x^4 - 14*x^2 + 24*x - 11 = 0 This finishes our search.

That's a beautiful proof for the rational root theorem!

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.

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.

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.