As a studied graphics designer and 'hobby mathematician' I'm always astonished by Burkard's animations and appreciate the work being put into it in order to make the mathematics look this beautiful.
Fun fact. A calculative way of getting sqrt(3) via the same numbers you got visually from the four triangles is this: a(n) = 4*a(n-1) - a(n-2); for a(0)=1 and a(1)=2 b(n)=4*b(n-1) - b(n-2); for b(0)=0 and b(1)=1 n 0 1 2 3 4 5 ... a(n) 1 2 4 7 26 97 ... b(n) 0 1 4 15 56 209 ... sqrt(3)=a(inf)/b(inf-1) source: oeis.org/A001075 and oeis.org/A001353 I wrote a Bash program to check the results for higher n and it works like a charm.
The geometrical difference between the triangles (square numbers) and hexagons (triangular numbers) is that with the triangles the center point falls within a triangle in all cases, whereas with the hexagons, the center point can fall on a vertex between hexagons. So you can have symmetry around that point and still have all the hexagons only used once, while the triangle in the center can only be used 0 or 3 times.
As for the last puzzle, we didn’t start from the assumption that the solution is the smallest one, hence, finding a smaller solution doesn’t contradict with any assumptions. I have to thank you and your colleagues for sharing these beautifully created math ideas to us. It’s amazing!
Actually, The supplied answer would not have been shown to be the smallest solution if we had started with that premise. For all we know, the absolute smallest solution might be the next smallest or the one less than that. Despite the fact that root 3 is irrational, this renders the proof invalid. We might also demonstrate that the sum of three triangular numbers could never be a triangular number using this same proof.
@@pokemonjourneysfan5925it doesn’t need to be shown to be the smallest solution, just a smaller solution than the original, which then completes the contradiction, and so the proof is perfectly valid
@@amansparekh So by your logic, I'll try to prove there are no pythagorean triples in pos. integers. Let's assume there are pythagorean triples. The smallest one I found yet is 5^2+12^2=13^2, Using analysis, I can deduce there is a smaller solution 3^2+4^2=5^2. Thus, there are no pythagorean triples.
@@pokemonjourneysfan5925 no, the reason this proof works and yours doesn’t is it shows that given any starting solution, there is a smaller one, whereas if you pick 345, there is no smaller solution
"Why doesn't this prove that 3 triangular numbers cannot add to another triangular number?" - Because you have a base case where there are no leftover hexagons: 3*tri(1) = tri(2). Then you just say that the minimal case is that case.
Because the question was 3x=y where x and y are triangular numbers, and "T_2*2 being equal to T_3" is 2x=y, not 3x=y =P And of course, this generator is for the strict case of 3x=y, not any arbitrary x+y+z=n (where x, y, z, and n are all triangular)
How do you know that there is not a similar base case solution for 3*n^2=m^2, but where n and m are very large numbers that you cannot easily find/check as you did for the triangle numbers?
subh1, That's not what Burkhardt asked. But anyways, the answer has already been given on this board: because for triangles, the center always falls within a triangle, so you can't divide the big triangle into 3 symmetric parts because you can't subdivide the center triangle.
The answer to the last puzzle is that in case of aa+aa+aa=bb you can prove that a+a > b and triangles will overlap. It follows from the fact that sqrt(3) < 2 because sqrt(4) == 2. In the last example you can't prove that they will overlap. In fact 1+1+1=3 is the simplest counterexample.
velcrorex No. Each of the four smaller tetrahedra touches the bigger one with three of its faces and the leftover inside with the fourth face. Since there are four tetrahedra touching the leftover with one face, it must too have four faces. It's just that the small tetrahedra must cover all the surface of the bigger one (so that the inside touches only the small tetrahedra)
I love those animations where you see an equation that you've used a thousand times visualized so intuitively that it simply 'clicks' and your mind is just blown.
At 9:00 I was afraid you were going to try to tile the plane with pentagons. That would have been scary! Also, I think the first half of the video would have benefited from a "triangle choreography" showing that the square root of 4 is rational, since you can perfectly pack 4 identical equilateral triangles into one big equilateral triangle with no overlap. (This would look like step 1 of a Sierpinski triangle.) This is kind of the trivial case, but I think showing this complementary case of the proof going the other way would provide a great contrast to the way the proof demonstrates the irrationality of the square roots of 2, 3, 5, 6, etc. But still, fascinating! Thank you!
Number theory is another fun way to show the square-root of any non-square is irrational: Suppose sqrt(n) is rational. Let sqrt(n) = a/b, for relatively prime a and b. Then n = a^2 / b^2, then n * b^2 = a^2. Since n divides the lhs, then n divides the rhs. Since n is not a perfect-square, then let p equal the product of the primes with odd powers in n's factorization, and p != 1. Now, p divides a. So divide both sides by p. (n/p) * b^2 = a * (a/p). Since p still divides the rhs (p divides a), it must divide the lhs. But a and b are relatively prime, so p cannot divide both a and b. Contraction, so sqrt(n) must not be rational. QED
17:17 Three years late but everyone here seems to be be proving it using a counterexample, when the question really was "why doesn't this same line of reasoning apply to the case shown?". All the proof was based on the fact that we were taking the smallest two triangles X and Y such as 3X = Y, considering X and Y to be the number of tiles in each triangle. By finding a smaller solution, it contradicts the fact that X and Y were the smallest possible in the first place. The point here is that in the example shown, where he took 3 triangular triangles of side 20 and one with side 35 and created a new smaller solution, he didn't start with the smallest triangles possible, so there's no premise to be contradicticted by the smaller solution to begin with. As many people stated, the smallest case here would be with three triangular triangles of side 1 fitting in one with side 2 (2*(2+1)/2 = 3 hexagons), which is trivially correct and doesn't create a smaller solution, meaning the contradiction from the square triangles doesn't happen here.
7:24 The white _”Salmiakki”_ (rhombus) shape can be cut into 2 identical triangular squares, which must exactly add up to the area of the dark green triangular square.
And I must tell that when sometime in primary school our teacher draw the numbers 1, 4, 9, 16, 25 and 36 on the blackboard and asked us what they are. I got to answer and said that they are the sum of consecutive odd numbers. I really can't remember why I did not see them as squares, but I do remember that my teacher said that I was correct and that it was well spotted and then asked again what those numbers are.
Yeah, might be what someone would guess if they haven't learned much about exponents, and that person has a good chance to understand squares better than anyone else in the room
I JUST HAPPEN to be giving a presentation this week involving continued fractions, as well as their connection to a certain graph in the hyperbolic half-plane called the Farey Graph, which is built off vertices from the Farey Sequence. Edges connect vertices if they are neighbors in any level of the Farey Sequence. What is interesting is that paths in the Farey Graph to any irrational hit all the vertices that are continued-fraction convergents of that irrational. It's interesting to see that, in this video, there are EVEN MORE ways to visualize approximations to irrationals! Math is full of connections between its sub-disciplines!!
Thanks for your response! I haven't heard of the Real Projective Line before, but I just looked it up. It looks like it's the same line on which the Farey Graph is built, in that it consists of all real numbers PLUS a point at infinity. EDIT: Ok, I just read that the Projective Real Line is, in fact, the boundary of the Poincare-disk model of the hyperbolic plane. Nice!
About the easier proof, it also jumped to me when you showed it the first time. The white rhombus divides to two equal triangular squares that have to equal the overlapping area.
nope, the upper triangle has base of 7 triangles so there are 49 in there. the white diamond is made of 2 triangles with base of 5 triangles, so 5²+5² = 50.
Exactly. Actually this is a very interesting alternative because the two triangular squares are now empty whereas the ones we started with are filled (green). The same only happens with the original square choreography. This also has an interesting effect on the nearest miss solutions in that they alternate in overshooting and undershooting by 1 (so plus and minus 1 and not just minus 1 :)
I remember the day I discovered that squares are the sums of odd numbers. I was looking at a black and white tile pattern and could see that the squares were made by adding layers of odd numbered tiles. 1 + 3 makes a square + 5 makes a larger square + 7 etc. For a non mathematician it gave me a good feeling that I had discovered something my math teachers never taught me. I even figured out the summation formula for it. I have always enjoyed math “tricks”. Interesting formulas that solved complex problems in simple to understand steps.
I also remember my math teacher trying to egg us on for extra credit if we could find a way to trisect an angle. Everyone started trying to use their compasses and angles and I started trying to fold the paper. I figured there had to be some way you could fold the paper to divide the angle into three parts. And years later someone did figure out how to use origami to trisect an angle.
The assumption that our supposed solution was the smallest "solution" with integer values, simply led to a contradiction that meant our smallest "solution" wasn't the smallest in either case, but in the case of the triangle numbers there is no statement that necessitates that the animated image is the smallest solution, however there was such a statement in the squares argument that there has to exist some solution 3A^2= B^2 that is non reducible, however this always leads to a smaller solution. In the case of the triangular numbers there is no such statement that forces any one solution to be the smallest solution so no contradiction is reached by reducing the triangle
Mathologer oh I will. I find myself working them out and I use the internet to find answers. I find the more I watch the more I "see" it. If you know what I mean. I've always had a hard time seeing math. I've got to write it down and then sometimes it comes to me. Sometimes not. But I won't give up. Thanks for your videos.
'If you don't agree you should really switch to a non-maths channel' This is exactly why I like math so much! When the "truth" of a subject has been proven and settled there's no use arguing it. You either understand or you don't.
I love when someone explains why it works and why it doesn't work but doesn't at any point differentiate between the two, making the assumption that you can see what is going on in their head.
The reason the "make smaller construction" proof doesn't generalize to triangular numbers is because the construction does not always produce a smaller triangular triangle *on the positive integers*. In particular, applying the construction to 3 T_1 = T_2 produces 3 T_0 = T_0 which is not a smaller solution on positive integers. If you try to fix this by allowing T_0, then the proof would instead fail because applying the construction to 3 T_0 = T_0 produces 3 T_0 = T_0 (which is not smaller). Basically, it is crucial that you prove your construction produces a smaller solution *within the allowed range*. When demonstrating sqrt(3) is irrational, you must show that your construction does not produce the trivial 3 0^2 = 0 solution. You might do so by pointing out that it would require the three smaller triangles to exactly meet in the middle and yet not overlap elsewhere, which is not possible.
Well said. Your response is the first one I found that gave some insight into how and when exactly a line of reasoning parallel to the triangle case would break down for the hexagon case.
(3*Tm = Tn implies 3*Th = Tk for some h < m) does not imply 3*Ts != Tt for any s and t, because the latter implies that the triangular root of 3 is irrational. But the triangular root of 3 is 2, a rational number. Even if the triangle root of 3 were unknown, (3*Tm = Tn implies 3*Th = Tk for some h < m) would not imply 3*Ts != Tt for any s and t because the right hand side is a non-sequitur.
i love you Burkhardt! You're teaching my post-Covid Grade 9 IB class for me!! Yesterday, I did the infinitely shrinking hexagons on a grid proof. Fantastic production value! Proofs by contradiction and Rational v. Irrational Numbers exquisitely done! It's actually wonderful, under the Hybrid model here in Ontario, I see ONE class from 8:15 to 12:45 with a break for lunch....so lots of time to delve deeper when desired!
Just some feedback. I think it would've been great if you could've shown the 1 + 3 + 5 ... is square through filling a square from a corner and adding to a larger square. So close! Also, if you could've shown that 4 is square because 4 smaller equilateral triangles can make one bigger one.
Yes, both the facts you mention (and quite a few more) would have been nice to include. It's always a judgement call where one stops in this respect. For example, I felt that including the standard 1 + 3 + 5 ... would have led us too far off course and it's also something that a lot of people watching this video will be familiar with. On the other hand, the little triangular square proof that I showed was probably new to most people, even those who know a lot of maths :)
Thanks for that video, the triangular proofs were definitely beautiful. I thought I'd add, as I did in another video for you, that in modular arithmetic irrational numbers can be rational mod a p*q modulus. For instance, while 3*x^2=y^2 is impossible in continuous numbers, this equation is certainly possible in modular arithmetic for certain p*q modulus. For instance 3*8^2=192 mod 61*73 and 192^(1/2) mod 61*73 ===241. Then the square root of 3 is then found by 3*8*241^(-1) mod 61*73 === 1700, and 1700^2 mod 61*73 === 3
I just recently discovered the way to create repeating continued fractions for any sqrt(n), and using this as a way to calculate the minimal solution for the Pell's equation a^2-nb^2 = +/- 1 - this is really awesome both as a way to see how continued fractions relate to extensions of rationals, and to generate optimal rational approximations to square roots! The method is as follows: if you want to generate the continued fraction for sqrt(18) (for example), we start by calculating the floor of this (4), and rewriting sqrt(18) = 4+ (sqrt(18)-4) - we can now multiply the term in parentheses by (sqrt(18)+4)/(sqrt(18)+4) to give sqrt(18) = 4+1/((sqrt(18)+4)/2) - and repeating the process by taking the floor of (sqrt(18)+4)/2 we get sqrt(18) = 4+1/(4+ 1/(8+(sqrt(18)-4))) - and we're back where we started, showing that sqrt(18) = [4;4,8,4,8,4,8,...] as a continued fraction, giving rational approximations of 4, 17/4, 140/33, 577/136 - and 17^2-18*4^2=1, 577^2-18*136^2=1 - and you can take every second convergent to generate a new and more accurate rational approximation of sqrt(18). Amazing!
Someone probably already said this but in your original proof, you assumed the smallest solution and showed any solution either admits a smaller solution or isn't a solution (ie: no base case). Here, a smallest solution exists, ie: 3 = (1+2)/1 so contradiction avoided. This video gave me some inspiration to try some weird stuff. Thanks!
At 7:30, the white diamond is made of two identical triangular squares that by definition must be the same area as the dark green. Sorry I know that's super obvious but I never get these challenge questions right
I could and I would definitely enjoy it. The problem is that there are so many great things waiting to be explained properly and there is so little time. Having said that the plan is to branch out to topics other than purely math later this year once I've survived my teaching at uni. So, we'll see :)
Before creation, God did just pure mathematics. Then He thought it would be a pleasant change to do some applied. - J E Littlewood. It's about time for Mathologer aswell. µπΣ
In Galileo's experiments with balls rolling down an inclined plane, he noticed that the increasing distances rolled during equal time intervals was in the ratios of the odd integers and that sums of the distance rolled was equal to the square of the number of intervals. In one experiment he used a straight ramp that had a curve from side to side that was lightly coated with flour. He started the ball out high on one side. Rolling from side to side was like the pendulum of a clock taking equal time intervals leaving a serpentine track that gradually stretched out demonstrating the above relationships.
You probably could, but figuring out how to snip up the triangles to make the right number of double covers and gaps while ensuring everything matches up in size would be an utter mess.
This is beautiful! It reminds me of the ratios in music somehow, but maybe that's just me. Your animations really are on fire lately, visual proofs are so strong if made right.
1:37 "This is an incredibly beautiful proof, and if you don't agree, there is something really, really, really wrong with you". You shouldn't have said that. Not because it is not a really beautiful proof, but because: 1. It's ambiguous. The quote could be understood to refer either to "agreement with the proof" or "agreement with the beauty of the proof". Someone who doesn't understand the proof at first glance could feel offended (rightfully so). 2. It's elitist. There are many people who despite being curious about mathematics, have not yet grown to appreciate a proof's elegance the way we do. It requires some sort of familiarity with mathematics that many people who watch these TH-cam videos are still in the process of acquiring. I don't think it was ever your intention to be offensive or elitist, and I'm not even saying that you are/were regardless of your intention. But that it's reasonable to assume that this sentence could be understood that way by a reasonable person. I know all this is very nitpicky, but I wouldn't even bother commenting were I not such a huge fan of your work. I just don't want people to misunderstand what you actually mean and dislike a channel I like so much. That aside, this was a great video, even for Mathloger standards. Keep the great work up!
> They might then feel compelled to grab a coat; offense isn't always a bad feeling to have. And the closest thing to grabbing a coat when feeling offended is saying "the way you said that made me feel bad". After all, sometimes jokes don't land, and there is nothing wrong with improving them by listening to feedback. It did rub me the wrong way too. "Something must be wrong with you if you don't agree with me." sounds exactly like a clumsy appeal to consensus/authority, even if it was not meant that way. I happen to prefer proofs that accentuate and formalize the underlying "common sense" assumptions, and I think slick animations hide the axiomatic basis of the proof too much in the name of being elegant. When I saw the triangle proof, my first thought was not "how beautiful" but "why *exactly* is this true?" Does this mean "something is wrong with me"?
True scotsman fallacy: Person A: "No Scotsman puts sugar on his porridge." Person B: "But my uncle Angus is a Scotsman and he puts sugar on his porridge." Person A: "Ah yes, but no true Scotsman puts sugar on his porridge." No true mathematician / truly sane person wouldn't find this proof beautiful...
This is kind of weird. What if you just chose the wrong triangles to disprove X²+X²+X²=Y². Maybe you just needed a different set of triangles for X² and Y². What if there is a set of triangles where this equation works? It looks like you chose a wrong example and then say "It does not work. So it's proof by contradiction." 😹
Jkl oe actually the set of triangles chosen is just to show that the sides are integers. The main thing to note is that the second set also has integer sides. Why? Because it is created by subtracting integers from integer sides. Additionally this set is smaller than the starting set. Now here is the contradiction, we had assumed at the start that x and y are smallest integers, but we found another set as if x and y had a common factor which got cancelled.
There are only two ways for it to be "wrong": 1) There is leftover space 2) There are integers in your solutions. The "integers in your solutions" thing is why the triangular triangles work. You can go from A=0, B=0 to A'=1, B'=3, and you can't accomplish that if A' is just a sum of As and Bs. And if there are leftover spaces, then if A/B is sqrt(2), then A'/B' is not sqrt(2). The thing is, if a solution is wrong due to leftover space, it's useless. If it's wrong because integers, and you can reverse your operations (so you can go from small solution to bigger solution), then you can find what is the next step up from A=0, B=0, and get your smallest possible solution, which is now a strong proof (It is possible to get A and B that do this, and here it is, a weak proof would be "It is possible to get A and B that do this, but I have no idea how to get them). And if you get a "right" solution, then it is not possible to get A and B.
Think about the types of triangles you are allowed to use (where you have 3 small "X" and 1 large "Y"). The X triangles can't be too small, otherwise you could snuggly fit 4 of them in, but we're trying to fit exactly 3. X obviously must be smaller than Y, otherwise we could fit 1 of them. This means that the triangles we use to fit 3 X's into Y must overlap with each other. This overlap will always produce the pattern he showed, where the unfilled empty space becomes our new Y, and the 3 tiny overlaps become our new X's. Mathologer chose "near misses" for demonstration purposes, but the actual size of the X's doesn't matter so long as it's smaller than Y and greater than 1/4th the area.
To answer the final question, the contradiction comes from assuming that we have the *smallest* solution, but then showing that such a solution generates a smaller one (which, as you pointed out, was not actually rigorously shown in this video). No such assumption was made for the 3T(m) = T(n) question. In fact, you can take the smallest solution -- 3 ⋅ T(0) = T(0), or the one just above, 3 ⋅ T(1) = T(2) -- and show that it *doesn't* generate a smaller solution, but fits perfectly. To really show the difference, you have to prove that, given the set of solutions S = { (x, y) ∈ ℕ² | 3x² = y² }, ∀s ∈ S . ∃s' ∈ S . fst s' < fst s I may come back in a bit with an Idris program that proves this (in a type-theory way, not a set-theory way).
Recurrence : For a new row of width n , 2n - 1 mini triangles are added to the precedent triangle. For example Triangle_6 = Triangle_5 + 2*6 -1 = 25 + 11 = 36 If we start with 1 we have the arithmetic serie : 1,3,5,7,9,11... A well know result : the sum of n odd numbers (1,3,5,7....n), is : n**2 (square of n)
For the last question, there is an obvious smallest solution (1,3). But as to why the original proof does work, whereas this one doesn't, is because in the original you can guarantee overlap between the smaller triangles as they needed to be specifically fitted in the corners. This is not the case in the (1,3) solution, as a result of how the hexagons fill up space.
On each row (bar the first one), there was a center triangle. The hexagons had that feature only every other row. With triangular squares, that means there was a center triangle left over. There is no center hexagon on T35. Hence none will be left from the final, smallest case of your choreography.
For the last question, it’s because: in the case of triangular squares, you assumed root three to be rational, therefore would have a smallest integer representation, and from there you proceeded to show that this assumption led to a contradiction, namely, an even smaller integer representation; whereas in the case of triangular numbers there’s no such requirement because you’re not trying to prove rationality or anything else requiring a smallest set of two integers.
That was my first guess too, but if there are any solutions in the positive integers, then there is a smallest solution. You can reframe that into an attempted irrationality proof. First let's define a function f and its inverse. Let f(a/b) = tri(a)/tri(b) where a/b is a rational number in reduced form f^-1(tri(a)/tri(b)) = a/b Then we use some algebra on the puzzle's equation. 3 * tri(b) = tri(a) where a and b are the smallest integers that fulfill this equation 3 = tri(a)/tri(b) where "..." f^-1(3) = f^-1( tri(a)/tri(b) ) where "..." f^-1(3) = a/b where "..." Which implies that f^-1(3) is a rational number.
I really appreciate you taking out some time to explain this in such detail. I've been away from math for years, but now with resources such as this channel and its comment section, there's pretty much endless content that's insightful and elucidating. Requiring nothing else but an hour free of distractions and a willingness to think, I can take a dip in a math universe with such ease that I almost feel spoiled. Thank you again for the help.
Here is a way to generate "nearest miss solutions" in their fraction form, for any square root. Given may be u/v as a (good) approximation to sqrt(a). Then, (u^2+av^2)/(2uv) is an equally good or better approximation to sqrt(a). You can use this rule recursively. Here it is with sqrt(3): u' = u^2+3v^2 v' = 2u*v Guess u = 2, v = 1; 2 is about sqrt(3). Error of about 0.25 Applying the rule gives 7/4 as a better approximation, with an error of about 0.018 Next iteration: 97/56. Error of about 10^-4 Next iteration: 18817/10864. Error of about 2.5*10^-9 So we have an exponential convergence. Even better: as soon as |u^2-av^2| = 1, the resulting fractions will be as good as possible. If you are lucky with the initial pick, you will not have to cancel common factors from u and v. You can use this method for the nth root of a using this rule: u' = (n-1)*u^n + a*v^n v' = n*v*u^(n-1) EDIT: Might not be exponential, but only quadratic convergence, have not yet had the time to sit down and investigate further. Found this by using newton's method for finding zeros of x^n-a
"For example, in the pentagon root 5 coreography, it's not even clear where the squares are. Again, consider filling in the details as a puzzle." This comes from a more general theorem that, if you have some shape, and scale it up by a factor of n, the measure of that shape will scale by n^D, where D is the dimension of the shape. Since pentagons are 2 dimensional shapes, the measure (area) of a pentagon scales up by n^2 when you scale the length of its line segments by a factor of n.
You can see the contradiction at 7:50 because the three gray triangles plus the three small triangles are equal to the white triangle which means that B²+B²+B²+C²+C²+C² = A² when C < B
Stemming from the original equation 3Tn=Tx, Tx/Tn=3 which does not require the smallest possible values of Tx and Tn to approach the correct value “3” (like is required to approach root 3). Therefore Tx and Tn can be of higher values that continue with the choreography until the triangular triangles are equal.
As my other comment may imply, I already made the connection to the square based proof of root two before you proposed the puzzle and was momentarily shocked that you didn't do that equivalent in triangles first
Hi Mathologer, Love your contents!! Spontaneous Answer without paper, just a suspicion: Hexagons pack differently to triangles. The most efficient way to pack area into a shape is by choosing a circle. A hexagon is closer to a circle than a triangle. So i suspect you can pack the same area closer by choosing hexagons than by choosing squares or triangles. If you would pack the same area in triangles as close to each other as the hexagons, they would overlap (i suspect). So in the choreography of hexagons as opposed to the choreography of triangles, the three little overlapping dark green groups of hexagons do not touch the bigger area of white triangles in the middle, while in the choreography of triangles the dark green overlapping triangles touch the middle white triangle. I guess that is where the difference in area per object center distance manifests itself. Love your t-shirts! Can I get one if i am right? Njoy! Cheers...
since the amount of tiles in a triangular triangle is (n^2+n)/2 instead of n^2 for normal triangles the equation of 3 triangular triangles adding up to another one corresponds no longer to 3*X^2 = Y^2, but to 3*(n^2+n)/2=(m^2+m)/2 3*(n^2+n)=(m^2+m) 3n(n+1)=m(m+1) 3=m(m+1)/(n(n+1)). the final animation therefore proves the impossibility of expressing 3 as to quotient of the product of consecutive integers instead of the irrationality o sqrt(3)
2:43 I think that’s a more rigorous proof for the number of mini-triangles in an order-n triangle being n²: Just start with the difference of 2 consecutive squares: (n+1)² - n² = n²+2n+1-n² = 2n+1 Then, using the fact that the numbers of mini-triangles, in each layer, are consecutive odd numbers, you automatically prove the ”A(T(n)) = n²” -statement. 🙂
Fantastic video as always :) Regarding the final puzzle: I think that we can derive correct statements from a correct statement. So if we can prove that any of equations related to one another is true then all of them must be true as they "emerge" from one another. On the other hand side of we prove that one out of the "emerging" equations is false then all of them must be false. To me this is what Logical consequence stands for. But I don't know how to motivate the better.
(re: final puzzle) At a guess, it's something to do with step 1 of the sqrt(2) proof: the notion of a smallest possible fraction. That step doesn't apply to the triangular numbers. 1 is a triangular number and the choreography stops there. Or possibly - if *all* positive numbers are the sum of 3 triangular numbers, then we must treat zero as a triangular number, and so again the choreography stops.
The "trick" is the use of induction: in the first case, when you go down to step #1 (when you can't go further), it doesn't work, so every step by which you've been through is wrong, while in the second case, when your dancing leads you to 3*T_1 = T_2, that is true, so are every step your choreography has shown you. By the way, it works with your starting step, if you tried T_4 and T_5, it doesn't work. If you looked for sqrt(4), it would work (y=2 and x=1) sometimes (y=3 and x=1) One Step to prove them all [...] and by the induction demonstrate them
The reason the makes smaller things is because your smallest instant is T0+T0+T0=T0. Even if you tried to go into negetive number of hexagons (somewhat confusing) it still works T(-1)*3=T(-1).
I know you showed 5^2 in the beginning, but I'd like to see sqrt(4) or sqrt(9), just to follow the system of counting upwards and showing that it actually works for some.
17:00 in this case it's not an infinite reduction (3*T_1 = T_2, but when you try to do the propeller thing there is no overlap to construct the next step) so no contradiction arises. With the roots we picked the smallest solution and found a smaller one, here we just picked a solution and constructed another smaller one. In order for the irrationality proof to work you need to show that your smallest propeller image actually does have the overlap needed to construct the next step. For root 2 and 3 that works no problem since 2^2>3 thus the edges of the two smaller triangles must be longer than half the large triangle. As stated above, for the triangular numbers that doesn't work as 1+1+1 = 3 is a counter example
I remember reading the sum of odd numbers property in a book, and went about proving it algebraically, getting really close to it and messing it up twice or thrice before I got it, without realizing that there was the exact same proof on the next page of the book.
Mathologer needs a second channel with answers to the homework questions he gives
What do you think of the idea of publishing original mathematics on TH-cam instead of mathematics journals?
Mathologer, it's a fantastic idea
I will likely watch whatever you will publish
Nice idea!
Its a good idea, keep doing it :)
Much less formal for one.
As a studied graphics designer and 'hobby mathematician' I'm always astonished by Burkard's animations and appreciate the work being put into it in order to make the mathematics look this beautiful.
:)
Fun fact. A calculative way of getting sqrt(3) via the same numbers you got visually from the four triangles is this:
a(n) = 4*a(n-1) - a(n-2); for a(0)=1 and a(1)=2
b(n)=4*b(n-1) - b(n-2); for b(0)=0 and b(1)=1
n 0 1 2 3 4 5 ...
a(n) 1 2 4 7 26 97 ...
b(n) 0 1 4 15 56 209 ...
sqrt(3)=a(inf)/b(inf-1)
source: oeis.org/A001075 and oeis.org/A001353
I wrote a Bash program to check the results for higher n and it works like a charm.
The geometrical difference between the triangles (square numbers) and hexagons (triangular numbers) is that with the triangles the center point falls within a triangle in all cases, whereas with the hexagons, the center point can fall on a vertex between hexagons. So you can have symmetry around that point and still have all the hexagons only used once, while the triangle in the center can only be used 0 or 3 times.
As for the last puzzle, we didn’t start from the assumption that the solution is the smallest one, hence, finding a smaller solution doesn’t contradict with any assumptions. I have to thank you and your colleagues for sharing these beautifully created math ideas to us. It’s amazing!
Actually, The supplied answer would not have been shown to be the smallest solution if we had started with that premise. For all we know, the absolute smallest solution might be the next smallest or the one less than that. Despite the fact that root 3 is irrational, this renders the proof invalid. We might also demonstrate that the sum of three triangular numbers could never be a triangular number using this same proof.
I had this same concern. The proof mainly depends on our initial assumption.
@@pokemonjourneysfan5925it doesn’t need to be shown to be the smallest solution, just a smaller solution than the original, which then completes the contradiction, and so the proof is perfectly valid
@@amansparekh So by your logic, I'll try to prove there are no pythagorean triples in pos. integers. Let's assume there are pythagorean triples. The smallest one I found yet is 5^2+12^2=13^2, Using analysis, I can deduce there is a smaller solution 3^2+4^2=5^2. Thus, there are no pythagorean triples.
@@pokemonjourneysfan5925 no, the reason this proof works and yours doesn’t is it shows that given any starting solution, there is a smaller one, whereas if you pick 345, there is no smaller solution
Took me 15 minutes (while watching square triangles dance around) to figure out the solution to your shirt. The tree has square roots!
Yes, AND the tree is a square root sign ! :)
Here was me thinking he was going to demonstrate irrationality by showing us the infinite orchard problem.
True, the tree has square roots, but did you also notice it was a "root tree" (root three) and the video was mostly about the root of three?
@@MathologerYour shirt reminds me of this: i.snag.gy/SUZHku.jpg
I need this shirt.
11:30 "15 and 26 form...a nearest miss solution" You know that there's a name for this, right? It's a Parker Square.
a Parker Triangular Square.
At first you describe things, then eventually you name them
@@washizukanorico /r/whoosh
@@michalnemecek3575
With a PARKER SQUARE you really have to watch those diagonals.
aaaaaaaaayyyyyy
Matt's gonna be mad tho.. XD
"Why doesn't this prove that 3 triangular numbers cannot add to another triangular number?" - Because you have a base case where there are no leftover hexagons: 3*tri(1) = tri(2). Then you just say that the minimal case is that case.
Tehom Yeah, exactly. Not really sure how T_1*3 being equal to T_2 and T_2*2 being equal to T_3 can 'generate' all equations for integers, proof-wise.
Because the question was 3x=y where x and y are triangular numbers, and "T_2*2 being equal to T_3" is 2x=y, not 3x=y =P And of course, this generator is for the strict case of 3x=y, not any arbitrary x+y+z=n (where x, y, z, and n are all triangular)
Because they beehive better!
How do you know that there is not a similar base case solution for 3*n^2=m^2, but where n and m are very large numbers that you cannot easily find/check as you did for the triangle numbers?
subh1, That's not what Burkhardt asked. But anyways, the answer has already been given on this board: because for triangles, the center always falls within a triangle, so you can't divide the big triangle into 3 symmetric parts because you can't subdivide the center triangle.
The answer to the last puzzle is that in case of aa+aa+aa=bb you can prove that a+a > b and triangles will overlap. It follows from the fact that sqrt(3) < 2 because sqrt(4) == 2. In the last example you can't prove that they will overlap. In fact 1+1+1=3 is the simplest counterexample.
I've never really learn a single thing from these videos but the visual representations of mathematical concepts are nice.
Man, it's nice to imagine how a bunch of tetrahedrons demonstrate the irrationality of the cube root of 4 just as easily.
Casual Graphman We could keep going into extra dimensions this way... Very fun to think about it.
Aw, you've just made my bleeding day. So satisfying.
Does that work? Don't you get an octahedron left over when you cut tetrahedral corners off of a tetrahedron?
velcrorex No. Each of the four smaller tetrahedra touches the bigger one with three of its faces and the leftover inside with the fourth face. Since there are four tetrahedra touching the leftover with one face, it must too have four faces.
It's just that the small tetrahedra must cover all the surface of the bigger one (so that the inside touches only the small tetrahedra)
It's the Desmos man himself! Keep it up, your work is wonderful and even inspiring.
I love those animations where you see an equation that you've used a thousand times visualized so intuitively that it simply 'clicks' and your mind is just blown.
At 9:00 I was afraid you were going to try to tile the plane with pentagons. That would have been scary!
Also, I think the first half of the video would have benefited from a "triangle choreography" showing that the square root of 4 is rational, since you can perfectly pack 4 identical equilateral triangles into one big equilateral triangle with no overlap. (This would look like step 1 of a Sierpinski triangle.) This is kind of the trivial case, but I think showing this complementary case of the proof going the other way would provide a great contrast to the way the proof demonstrates the irrationality of the square roots of 2, 3, 5, 6, etc. But still, fascinating! Thank you!
Number theory is another fun way to show the square-root of any non-square is irrational: Suppose sqrt(n) is rational. Let sqrt(n) = a/b, for relatively prime a and b. Then n = a^2 / b^2, then n * b^2 = a^2. Since n divides the lhs, then n divides the rhs. Since n is not a perfect-square, then let p equal the product of the primes with odd powers in n's factorization, and p != 1. Now, p divides a. So divide both sides by p. (n/p) * b^2 = a * (a/p). Since p still divides the rhs (p divides a), it must divide the lhs. But a and b are relatively prime, so p cannot divide both a and b. Contraction, so sqrt(n) must not be rational. QED
В школе давали готовые формулы, не поясняя откуда они взялись. Этот канал раскрывает всю первичную суть. Просто супер!!!
"Wth is a triangular square? Well it's definitely not click bait, OK" 😂😂 earned that like
17:17 Three years late but everyone here seems to be be proving it using a counterexample, when the question really was "why doesn't this same line of reasoning apply to the case shown?". All the proof was based on the fact that we were taking the smallest two triangles X and Y such as 3X = Y, considering X and Y to be the number of tiles in each triangle. By finding a smaller solution, it contradicts the fact that X and Y were the smallest possible in the first place. The point here is that in the example shown, where he took 3 triangular triangles of side 20 and one with side 35 and created a new smaller solution, he didn't start with the smallest triangles possible, so there's no premise to be contradicticted by the smaller solution to begin with. As many people stated, the smallest case here would be with three triangular triangles of side 1 fitting in one with side 2 (2*(2+1)/2 = 3 hexagons), which is trivially correct and doesn't create a smaller solution, meaning the contradiction from the square triangles doesn't happen here.
7:25 Of course! The rombus there can be divided into two triangular squares by just slicing it by the middle.
7:24 The white _”Salmiakki”_ (rhombus) shape can be cut into 2 identical triangular squares, which must exactly add up to the area of the dark green triangular square.
And I must tell that when sometime in primary school our teacher draw the numbers 1, 4, 9, 16, 25 and 36 on the blackboard and asked us what they are. I got to answer and said that they are the sum of consecutive odd numbers. I really can't remember why I did not see them as squares, but I do remember that my teacher said that I was correct and that it was well spotted and then asked again what those numbers are.
Yeah, might be what someone would guess if they haven't learned much about exponents, and that person has a good chance to understand squares better than anyone else in the room
I now understand how the decimals places of irrational square roots may go on forever! This is so cool.
0:33 he used his shirt as the radical, well done
I had to pause the video and think. It takes time to really appreciate the beauty of mathematics.
I JUST HAPPEN to be giving a presentation this week involving continued fractions, as well as their connection to a certain graph in the hyperbolic half-plane called the Farey Graph, which is built off vertices from the Farey Sequence. Edges connect vertices if they are neighbors in any level of the Farey Sequence. What is interesting is that paths in the Farey Graph to any irrational hit all the vertices that are continued-fraction convergents of that irrational.
It's interesting to see that, in this video, there are EVEN MORE ways to visualize approximations to irrationals! Math is full of connections between its sub-disciplines!!
Woah. Just looked up the Farey Graph. Have you heard of the "Real Projective Line"?
Thanks for your response! I haven't heard of the Real Projective Line before, but I just looked it up. It looks like it's the same line on which the Farey Graph is built, in that it consists of all real numbers PLUS a point at infinity.
EDIT: Ok, I just read that the Projective Real Line is, in fact, the boundary of the Poincare-disk model of the hyperbolic plane. Nice!
About the easier proof, it also jumped to me when you showed it the first time. The white rhombus divides to two equal triangular squares that have to equal the overlapping area.
this video is precisely why Mathologer is one of favorite channels :)
My brain became bigger and able to store a lot more information than before just because of your videos and some other videos from other math channels
7:20 the white diamond has an upper half and a lower half that add together to the darkgreen triangle.
nope, the upper triangle has base of 7 triangles so there are 49 in there. the white diamond is made of 2 triangles with base of 5 triangles, so 5²+5² = 50.
Uhm... well of course it doesn't actually add together. Did you watch the video?
Supremebubble you just said they add together
Exactly. Actually this is a very interesting alternative because the two triangular squares are now empty whereas the ones we started with are filled (green). The same only happens with the original square choreography. This also has an interesting effect on the nearest miss solutions in that they alternate in overshooting and undershooting by 1 (so plus and minus 1 and not just minus 1 :)
Sigh, of course I meant the triangles that Mathologer is talking about and not the ones that are only there for the sake of visualization.
I remember the day I discovered that squares are the sums of odd numbers. I was looking at a black and white tile pattern and could see that the squares were made by adding layers of odd numbered tiles. 1 + 3 makes a square + 5 makes a larger square + 7 etc. For a non mathematician it gave me a good feeling that I had discovered something my math teachers never taught me. I even figured out the summation formula for it. I have always enjoyed math “tricks”. Interesting formulas that solved complex problems in simple to understand steps.
I also remember my math teacher trying to egg us on for extra credit if we could find a way to trisect an angle. Everyone started trying to use their compasses and angles and I started trying to fold the paper. I figured there had to be some way you could fold the paper to divide the angle into three parts. And years later someone did figure out how to use origami to trisect an angle.
The assumption that our supposed solution was the smallest "solution" with integer values, simply led to a contradiction that meant our smallest "solution" wasn't the smallest in either case, but in the case of the triangle numbers there is no statement that necessitates that the animated image is the smallest solution, however there was such a statement in the squares argument that there has to exist some solution 3A^2= B^2 that is non reducible, however this always leads to a smaller solution. In the case of the triangular numbers there is no such statement that forces any one solution to be the smallest solution so no contradiction is reached by reducing the triangle
No, there must be the smallest solutions because no natural number are smaller than 1
@@themobiusfunction tbh. I don't remember what I was on about
Happy to see that the compass and straightedge lives strong on TH-cam.
I don't quite understand it but I love it! Thanks for the great video.
I hope you love it enough to eventually watch it again to really get all of it. Well worth the effort :)
Mathologer oh I will. I find myself working them out and I use the internet to find answers. I find the more I watch the more I "see" it. If you know what I mean. I've always had a hard time seeing math. I've got to write it down and then sometimes it comes to me. Sometimes not. But I won't give up. Thanks for your videos.
'If you don't agree you should really switch to a non-maths channel'
This is exactly why I like math so much! When the "truth" of a subject has been proven and settled there's no use arguing it. You either understand or you don't.
The total no of hexagons is a triangular number. That's why its called a triangular triangle
Yes, at least that is why I call it a triangular triangle (and because I think it sounds funny :)
I love when someone explains why it works and why it doesn't work but doesn't at any point differentiate between the two, making the assumption that you can see what is going on in their head.
The reason the "make smaller construction" proof doesn't generalize to triangular numbers is because the construction does not always produce a smaller triangular triangle *on the positive integers*. In particular, applying the construction to 3 T_1 = T_2 produces 3 T_0 = T_0 which is not a smaller solution on positive integers. If you try to fix this by allowing T_0, then the proof would instead fail because applying the construction to 3 T_0 = T_0 produces 3 T_0 = T_0 (which is not smaller).
Basically, it is crucial that you prove your construction produces a smaller solution *within the allowed range*. When demonstrating sqrt(3) is irrational, you must show that your construction does not produce the trivial 3 0^2 = 0 solution. You might do so by pointing out that it would require the three smaller triangles to exactly meet in the middle and yet not overlap elsewhere, which is not possible.
Well said. Your response is the first one I found that gave some insight into how and when exactly a line of reasoning parallel to the triangle case would break down for the hexagon case.
He does mention this at 9:23, w.r.t. root 5 and 6. Maybe he thought the root 3 one was obvious.
(3*Tm = Tn implies 3*Th = Tk for some h < m) does not imply 3*Ts != Tt for any s and t, because the latter implies that the triangular root of 3 is irrational. But the triangular root of 3 is 2, a rational number.
Even if the triangle root of 3 were unknown, (3*Tm = Tn implies 3*Th = Tk for some h < m) would not imply 3*Ts != Tt for any s and t because the right hand side is a non-sequitur.
Actually it would imply that.
14:05 nice little quasi-example of the Four Colour Theorem you got there.
This guy is truely amazing! Yes ! have watch all and continue to watch every video he will make! I am mesmerised
holy cow, i was not expecting my jaw to drop in the first 2 minutes...
i love you Burkhardt! You're teaching my post-Covid Grade 9 IB class for me!! Yesterday, I did the infinitely shrinking hexagons on a grid proof. Fantastic production value! Proofs by contradiction and Rational v. Irrational Numbers exquisitely done! It's actually wonderful, under the Hybrid model here in Ontario, I see ONE class from 8:15 to 12:45 with a break for lunch....so lots of time to delve deeper when desired!
Just some feedback. I think it would've been great if you could've shown the 1 + 3 + 5 ... is square through filling a square from a corner and adding to a larger square. So close! Also, if you could've shown that 4 is square because 4 smaller equilateral triangles can make one bigger one.
Yes, both the facts you mention (and quite a few more) would have been nice to include. It's always a judgement call where one stops in this respect. For example, I felt that including the standard 1 + 3 + 5 ... would have led us too far off course and it's also something that a lot of people watching this video will be familiar with. On the other hand, the little triangular square proof that I showed was probably new to most people, even those who know a lot of maths :)
brayton goodall just what i thought
Yes, the 3x3 + 3 + 4 = 4x4; 4x4 + 4 +5 = 5x5 etc.
Thanks for that video, the triangular proofs were definitely beautiful. I thought I'd add, as I did in another video for you, that in modular arithmetic irrational numbers can be rational mod a p*q modulus. For instance, while 3*x^2=y^2 is impossible in continuous numbers, this equation is certainly possible in modular arithmetic for certain p*q modulus. For instance 3*8^2=192 mod 61*73 and 192^(1/2) mod 61*73 ===241. Then the square root of 3 is then found by 3*8*241^(-1) mod 61*73 === 1700, and 1700^2 mod 61*73 === 3
"Now is ze time on Sprockets vhen ve MATH!"
I love your channel. It has put my mind in to math comas, filled algorithms, crystalline and quasicrystalline polymorphism.
so can you use this to make new fractals?
I just recently discovered the way to create repeating continued fractions for any sqrt(n), and using this as a way to calculate the minimal solution for the Pell's equation a^2-nb^2 = +/- 1 - this is really awesome both as a way to see how continued fractions relate to extensions of rationals, and to generate optimal rational approximations to square roots! The method is as follows: if you want to generate the continued fraction for sqrt(18) (for example), we start by calculating the floor of this (4), and rewriting sqrt(18) = 4+ (sqrt(18)-4) - we can now multiply the term in parentheses by (sqrt(18)+4)/(sqrt(18)+4) to give sqrt(18) = 4+1/((sqrt(18)+4)/2) - and repeating the process by taking the floor of (sqrt(18)+4)/2 we get sqrt(18) = 4+1/(4+ 1/(8+(sqrt(18)-4))) - and we're back where we started, showing that sqrt(18) = [4;4,8,4,8,4,8,...] as a continued fraction, giving rational approximations of 4, 17/4, 140/33, 577/136 - and 17^2-18*4^2=1, 577^2-18*136^2=1 - and you can take every second convergent to generate a new and more accurate rational approximation of sqrt(18). Amazing!
@Danil Dmitriev, я не знаю кто ты, но я тебя обожаю за то что ты переводишь эти ролики(ну и этот канал тоже)
Someone probably already said this but in your original proof, you assumed the smallest solution and showed any solution either admits a smaller solution or isn't a solution (ie: no base case). Here, a smallest solution exists, ie: 3 = (1+2)/1 so contradiction avoided. This video gave me some inspiration to try some weird stuff. Thanks!
At 7:30, the white diamond is made of two identical triangular squares that by definition must be the same area as the dark green. Sorry I know that's super obvious but I never get these challenge questions right
Well you got this one right :)
Those 18 minutes has so much knowledge and home work , some one could probalably write a text book of it...
Wat a video #mathologer purely rocks.
Can you explain the Bitcoin selfish miner problem?
I could and I would definitely enjoy it. The problem is that there are so many great things waiting to be explained properly and there is so little time. Having said that the plan is to branch out to topics other than purely math later this year once I've survived my teaching at uni. So, we'll see :)
Before creation, God did just pure mathematics. Then He thought it would be a pleasant change to do some applied.
- J E Littlewood.
It's about time for Mathologer aswell. µπΣ
Please‽
In 2013 I thought Bitcoin was cool...😫
I have found out about this triangle square thing, shown here within the example with 25, many years ago by myself. This is so cool!
"if you don't agree then there is something really really really wrong with you" lol
Personally, I thought that wisecrack was beneath him. He’s a great teacher. He really should know better than to math-shame anyone.
@@jkershenbaum you are right
math shaming is not very encouraging for learning
word up!
In Galileo's experiments with balls rolling down an inclined plane, he noticed that the increasing distances rolled during equal time intervals was in the ratios of the odd integers and that sums of the distance rolled was equal to the square of the number of intervals. In one experiment he used a straight ramp that had a curve from side to side that was lightly coated with flour. He started the ball out high on one side. Rolling from side to side was like the pendulum of a clock taking equal time intervals leaving a serpentine track that gradually stretched out demonstrating the above relationships.
I REALLY CARDIOID THIS CHANNEL !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 😍😍😍😍😍😍😍😍😍😍😍😍😍
7:28 the empty triangles at the bottom can be separated into two large triangles that have to add up to the large overlap triangle.
Why can't this method be used to prove that sqrt7 is irrational? Or any other number higher than 6 for that matter
It probably can. But as you saw in the 5 and 6 cases, the animation got really complicated. More triangles -> more complicated to show.
You probably could, but figuring out how to snip up the triangles to make the right number of double covers and gaps while ensuring everything matches up in size would be an utter mess.
After watching several of your videos, I am starting to understand your methodology. it really opens up a new way of thinking.
THANKS!!!!!
Cool :)
Friday the 13th.
Who cares
It's already Saturday the 14th here in Melbourne :)
Ryan Truong Mumbiker
In Hispanic countries we consider Tuesday the 13th an unlucky day
Dude is your pic convergent?...wait....nvm
This is beautiful! It reminds me of the ratios in music somehow, but maybe that's just me. Your animations really are on fire lately, visual proofs are so strong if made right.
1:37 "This is an incredibly beautiful proof, and if you don't agree, there is something really, really, really wrong with you".
You shouldn't have said that. Not because it is not a really beautiful proof, but because:
1. It's ambiguous. The quote could be understood to refer either to "agreement with the proof" or "agreement with the beauty of the proof". Someone who doesn't understand the proof at first glance could feel offended (rightfully so).
2. It's elitist. There are many people who despite being curious about mathematics, have not yet grown to appreciate a proof's elegance the way we do. It requires some sort of familiarity with mathematics that many people who watch these TH-cam videos are still in the process of acquiring.
I don't think it was ever your intention to be offensive or elitist, and I'm not even saying that you are/were regardless of your intention. But that it's reasonable to assume that this sentence could be understood that way by a reasonable person.
I know all this is very nitpicky, but I wouldn't even bother commenting were I not such a huge fan of your work. I just don't want people to misunderstand what you actually mean and dislike a channel I like so much.
That aside, this was a great video, even for Mathloger standards. Keep the great work up!
Felipe Hindi i agree. I thought the same.
seriously, does a linear transform preserves area? how did he even said it?
He does this *a lot* in his videos. I think he's just insecure.
> They might then feel compelled to grab a coat; offense isn't always a bad feeling to have.
And the closest thing to grabbing a coat when feeling offended is saying "the way you said that made me feel bad". After all, sometimes jokes don't land, and there is nothing wrong with improving them by listening to feedback. It did rub me the wrong way too. "Something must be wrong with you if you don't agree with me." sounds exactly like a clumsy appeal to consensus/authority, even if it was not meant that way.
I happen to prefer proofs that accentuate and formalize the underlying "common sense" assumptions, and I think slick animations hide the axiomatic basis of the proof too much in the name of being elegant. When I saw the triangle proof, my first thought was not "how beautiful" but "why *exactly* is this true?" Does this mean "something is wrong with me"?
True scotsman fallacy:
Person A: "No Scotsman puts sugar on his porridge."
Person B: "But my uncle Angus is a Scotsman and he puts sugar on his porridge."
Person A: "Ah yes, but no true Scotsman puts sugar on his porridge."
No true mathematician / truly sane person wouldn't find this proof beautiful...
This is beautifully elegant. It strains my smooth brain to follow the reasoning, but I managed to get the gist of it (barely).
This is kind of weird. What if you just chose the wrong triangles to disprove X²+X²+X²=Y². Maybe you just needed a different set of triangles for X² and Y². What if there is a set of triangles where this equation works?
It looks like you chose a wrong example and then say "It does not work. So it's proof by contradiction." 😹
+
Jkl oe actually the set of triangles chosen is just to show that the sides are integers. The main thing to note is that the second set also has integer sides. Why? Because it is created by subtracting integers from integer sides. Additionally this set is smaller than the starting set. Now here is the contradiction, we had assumed at the start that x and y are smallest integers, but we found another set as if x and y had a common factor which got cancelled.
There are only two ways for it to be "wrong":
1) There is leftover space
2) There are integers in your solutions.
The "integers in your solutions" thing is why the triangular triangles work. You can go from A=0, B=0 to A'=1, B'=3, and you can't accomplish that if A' is just a sum of As and Bs.
And if there are leftover spaces, then if A/B is sqrt(2), then A'/B' is not sqrt(2).
The thing is, if a solution is wrong due to leftover space, it's useless. If it's wrong because integers, and you can reverse your operations (so you can go from small solution to bigger solution), then you can find what is the next step up from A=0, B=0, and get your smallest possible solution, which is now a strong proof (It is possible to get A and B that do this, and here it is, a weak proof would be "It is possible to get A and B that do this, but I have no idea how to get them). And if you get a "right" solution, then it is not possible to get A and B.
Think about the types of triangles you are allowed to use (where you have 3 small "X" and 1 large "Y"). The X triangles can't be too small, otherwise you could snuggly fit 4 of them in, but we're trying to fit exactly 3. X obviously must be smaller than Y, otherwise we could fit 1 of them. This means that the triangles we use to fit 3 X's into Y must overlap with each other. This overlap will always produce the pattern he showed, where the unfilled empty space becomes our new Y, and the 3 tiny overlaps become our new X's. Mathologer chose "near misses" for demonstration purposes, but the actual size of the X's doesn't matter so long as it's smaller than Y and greater than 1/4th the area.
Thanks for the explanations, all. I understand it now. I missed the part at 3:13 where Mathologger said "smallest possible" A and B.
I have no idea what you’re talking about but I love it
One of the prettiest math videos on YT, the 3Blue1Brown level of prettiness.
So pretty! Lovely proof with no numbers and no words. Just the shapes moving around are enough! Love it!
To answer the final question, the contradiction comes from assuming that we have the *smallest* solution, but then showing that such a solution generates a smaller one (which, as you pointed out, was not actually rigorously shown in this video). No such assumption was made for the 3T(m) = T(n) question. In fact, you can take the smallest solution -- 3 ⋅ T(0) = T(0), or the one just above, 3 ⋅ T(1) = T(2) -- and show that it *doesn't* generate a smaller solution, but fits perfectly.
To really show the difference, you have to prove that, given the set of solutions S = { (x, y) ∈ ℕ² | 3x² = y² }, ∀s ∈ S . ∃s' ∈ S . fst s' < fst s
I may come back in a bit with an Idris program that proves this (in a type-theory way, not a set-theory way).
I wish I'd had access to these videos back in high school. The explanations and animations make sensical fun.
Recurrence :
For a new row of width n , 2n - 1 mini triangles are added to the precedent triangle. For example
Triangle_6 = Triangle_5 + 2*6 -1 = 25 + 11 = 36
If we start with 1 we have the arithmetic serie :
1,3,5,7,9,11...
A well know result : the sum of n odd numbers (1,3,5,7....n), is : n**2 (square of n)
Keep up the great work on these. Brilliantly presented, challenging, and yet still accessible and entertaining to anyone. Truly inspiring.
Awesome! Finding new ways to look at the simplest numbers. That's what resonates with me.
The representation when you said “like root 2” at the start (I.e., what was shown on screen) was super nice and subtle!
For the last question, there is an obvious smallest solution (1,3). But as to why the original proof does work, whereas this one doesn't, is because in the original you can guarantee overlap between the smaller triangles as they needed to be specifically fitted in the corners. This is not the case in the (1,3) solution, as a result of how the hexagons fill up space.
Still trying to wrap my brain around the Euler video… love the content my brother
I love your work! You really put effort in making them. Keep up the good work 👍🏼👌🏼 and thanks for the great content
On each row (bar the first one), there was a center triangle. The hexagons had that feature only every other row.
With triangular squares, that means there was a center triangle left over. There is no center hexagon on T35. Hence none will be left from the final, smallest case of your choreography.
Burkhard, I must thank you for your beautiful reductiones, and I must purchase your square roots shirt.
Exceptionnal way of thinking about square roots.
For the last question, it’s because: in the case of triangular squares, you assumed root three to be rational, therefore would have a smallest integer representation, and from there you proceeded to show that this assumption led to a contradiction, namely, an even smaller integer representation; whereas in the case of triangular numbers there’s no such requirement because you’re not trying to prove rationality or anything else requiring a smallest set of two integers.
That was my first guess too, but if there are any solutions in the positive integers, then there is a smallest solution. You can reframe that into an attempted irrationality proof.
First let's define a function f and its inverse.
Let f(a/b) = tri(a)/tri(b) where a/b is a rational number in reduced form
f^-1(tri(a)/tri(b)) = a/b
Then we use some algebra on the puzzle's equation.
3 * tri(b) = tri(a) where a and b are the smallest integers that fulfill this equation
3 = tri(a)/tri(b) where "..."
f^-1(3) = f^-1( tri(a)/tri(b) ) where "..."
f^-1(3) = a/b where "..."
Which implies that f^-1(3) is a rational number.
I really appreciate you taking out some time to explain this in such detail. I've been away from math for years, but now with resources such as this channel and its comment section, there's pretty much endless content that's insightful and elucidating. Requiring nothing else but an hour free of distractions and a willingness to think, I can take a dip in a math universe with such ease that I almost feel spoiled.
Thank you again for the help.
Here is a way to generate "nearest miss solutions" in their fraction form, for any square root.
Given may be u/v as a (good) approximation to sqrt(a).
Then, (u^2+av^2)/(2uv) is an equally good or better approximation to sqrt(a).
You can use this rule recursively. Here it is with sqrt(3):
u' = u^2+3v^2
v' = 2u*v
Guess u = 2, v = 1; 2 is about sqrt(3). Error of about 0.25
Applying the rule gives 7/4 as a better approximation, with an error of about 0.018
Next iteration: 97/56. Error of about 10^-4
Next iteration: 18817/10864. Error of about 2.5*10^-9
So we have an exponential convergence.
Even better: as soon as |u^2-av^2| = 1, the resulting fractions will be as good as possible. If you are lucky with the initial pick, you will not have to cancel common factors from u and v.
You can use this method for the nth root of a using this rule:
u' = (n-1)*u^n + a*v^n
v' = n*v*u^(n-1)
EDIT: Might not be exponential, but only quadratic convergence, have not yet had the time to sit down and investigate further.
Found this by using newton's method for finding zeros of x^n-a
"For example, in the pentagon root 5 coreography, it's not even clear where the squares are. Again, consider filling in the details as a puzzle." This comes from a more general theorem that, if you have some shape, and scale it up by a factor of n, the measure of that shape will scale by n^D, where D is the dimension of the shape. Since pentagons are 2 dimensional shapes, the measure (area) of a pentagon scales up by n^2 when you scale the length of its line segments by a factor of n.
I figured out the Tn by myself in class using sigma, while trying to find a solution of the shortest formula for adding integers, glad I was correct!
You can see the contradiction at 7:50 because the three gray triangles plus the three small triangles are equal to the white triangle which means that B²+B²+B²+C²+C²+C² = A² when C < B
Stemming from the original equation 3Tn=Tx, Tx/Tn=3 which does not require the smallest possible values of Tx and Tn to approach the correct value “3” (like is required to approach root 3). Therefore Tx and Tn can be of higher values that continue with the choreography until the triangular triangles are equal.
The sqrt(5) coreography for the triangle is quite cumbersome...all of them are really cool!
As my other comment may imply, I already made the connection to the square based proof of root two before you proposed the puzzle and was momentarily shocked that you didn't do that equivalent in triangles first
Great animations. It really makes these proofs quite intuitive.
oh man, this might be my favorite mathologer video. absolutely perfect
Hi Mathologer,
Love your contents!!
Spontaneous Answer without paper, just a suspicion:
Hexagons pack differently to triangles. The most efficient way to pack area into a shape is by choosing a circle. A hexagon is closer to a circle than a triangle. So i suspect you can pack the same area closer by choosing hexagons than by choosing squares or triangles.
If you would pack the same area in triangles as close to each other as the hexagons, they would overlap (i suspect). So in the choreography of hexagons as opposed to the choreography of triangles, the three little overlapping dark green groups of hexagons do not touch the bigger area of white triangles in the middle, while in the choreography of triangles the dark green overlapping triangles touch the middle white triangle.
I guess that is where the difference in area per object center distance manifests itself.
Love your t-shirts!
Can I get one if i am right?
Njoy!
Cheers...
It's the exact opposite, right?
since the amount of tiles in a triangular triangle is (n^2+n)/2 instead of n^2 for normal triangles the equation of 3 triangular triangles adding up to another one corresponds no longer to 3*X^2 = Y^2, but to 3*(n^2+n)/2=(m^2+m)/2 3*(n^2+n)=(m^2+m) 3n(n+1)=m(m+1) 3=m(m+1)/(n(n+1)).
the final animation therefore proves the impossibility of expressing 3 as to quotient of the product of consecutive integers instead of the irrationality o sqrt(3)
2:43 I think that’s a more rigorous proof for the number of mini-triangles in an order-n triangle being n²: Just start with the difference of 2 consecutive squares:
(n+1)² - n² = n²+2n+1-n² = 2n+1
Then, using the fact that the numbers of mini-triangles, in each layer, are consecutive odd numbers, you automatically prove the ”A(T(n)) = n²” -statement. 🙂
Fantastic video as always :)
Regarding the final puzzle: I think that we can derive correct statements from a correct statement. So if we can prove that any of equations related to one another is true then all of them must be true as they "emerge" from one another. On the other hand side of we prove that one out of the "emerging" equations is false then all of them must be false. To me this is what Logical consequence stands for. But I don't know how to motivate the better.
Wow! This is fantastic! There is enough material here to puzzle for a year. Maybe more.Great job!
(re: final puzzle) At a guess, it's something to do with step 1 of the sqrt(2) proof: the notion of a smallest possible fraction. That step doesn't apply to the triangular numbers. 1 is a triangular number and the choreography stops there. Or possibly - if *all* positive numbers are the sum of 3 triangular numbers, then we must treat zero as a triangular number, and so again the choreography stops.
The "trick" is the use of induction: in the first case, when you go down to step #1 (when you can't go further), it doesn't work, so every step by which you've been through is wrong, while in the second case, when your dancing leads you to 3*T_1 = T_2, that is true, so are every step your choreography has shown you.
By the way, it works with your starting step, if you tried T_4 and T_5, it doesn't work.
If you looked for sqrt(4), it would work (y=2 and x=1) sometimes (y=3 and x=1)
One Step to prove them all [...] and by the induction demonstrate them
The reason the makes smaller things is because your smallest instant is T0+T0+T0=T0. Even if you tried to go into negetive number of hexagons (somewhat confusing) it still works T(-1)*3=T(-1).
I know you showed 5^2 in the beginning, but I'd like to see sqrt(4) or sqrt(9), just to follow the system of counting upwards and showing that it actually works for some.
17:00 in this case it's not an infinite reduction (3*T_1 = T_2, but when you try to do the propeller thing there is no overlap to construct the next step) so no contradiction arises. With the roots we picked the smallest solution and found a smaller one, here we just picked a solution and constructed another smaller one. In order for the irrationality proof to work you need to show that your smallest propeller image actually does have the overlap needed to construct the next step. For root 2 and 3 that works no problem since 2^2>3 thus the edges of the two smaller triangles must be longer than half the large triangle. As stated above, for the triangular numbers that doesn't work as 1+1+1 = 3 is a counter example
I remember reading the sum of odd numbers property in a book, and went about proving it algebraically, getting really close to it and messing it up twice or thrice before I got it, without realizing that there was the exact same proof on the next page of the book.
17:27
because in the smallest case, the three ‘triangles’ (at that point they are hexagons) fill the larger without any overlap.