Visualising irrationality with triangular squares

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Welcome to another Mathologer video. It's something that only a few experts among you will be aware of but pretty much every single Mathologer video features fresh maths, made up just for you. Every once in a while I even sneak in some of my own little theorems and proofs. Today's video is one of those videos. What I want to show you today are some new beautifully visual ways to prove the irrationality of some small integer roots like root 2, as well as some really cool and closely related mathematical gems. And it's all based on what I usually refer to as triangular squares. What the hell is a triangular square? Well, it's definitely not clickbait. Okay what I mean by triangular squares are equilateral triangles that are made up of mini equilateral triangles like that one over there. Why triangular squares? Because the number of mini triangles is always a square, namely the total width in terms of mini triangles squared. In this case the width of the big triangle is 5-mini triangles and I claim there are exactly 5 squared equals 25 of these mini triangles. Okay, I've prepared a really nice animated proof that this is the case in general. Ready to be dazzled? Here we go. Okay, yeah and it's getting squarish there. And done! There 5 squared. This is an incredibly beautiful proof and if you don't agree then there's something really, really really wrong with you and it's probably time to switch to a non maths channel. Also hiding just around the corner there's some more beautiful stuff, too good to pass over and so let's just savour that too before we go all irrational. Note that the layers of a triangular square are consecutive odd numbers. So one yellow triangle in the first layer, three orange in the second layer, then 5, 7 and 9. And so 1 plus 3 plus 5 plus 7 plus 9, the sum of the first five odd numbers is five squared. Doing the same for general triangular squares proves that the sum of the first N odd numbers is N squared. Also super beautiful isn't it? Anyway for what follows you only need to remember that a triangular square of width N contains N squared mini triangles. I now want to use triangular squares to prove that the square root of 3 is irrational, that is, that the number root 3 cannot be written as a fraction, as a ratio of integers. To prove this let's assume that it's in fact possible to express root 3 as such a fraction. Okay and let's also assume that A and B are as small as possible. Then squaring both sides and simplifying gives this here. Okay but then of course 3B squared equals A squared. That's the same as saying that B squared plus B squared plus B squared equals A squared. So what does this mean? This means that if root 3 was rational, then the equation X squared plus X squared plus X squared equals Y squared would have positive integer solutions and that B squared plus B squared plus B squared equals A squared would be the smallest such solution. But this would also mean that combined the three green triangular squares of width B would contain exactly the same number of mini triangles as the triangular square grid of width A on the right. And this would mean that all the mini green triangles on the left would fit exactly into a large grid on the right. And here's a super stylish way to begin an attempt to fit them. Pretty :) Anyway this attempt results in three dark green triangle overlaps and an empty triangle patch in the middle. This patch and the overlaps would also be triangular squares, right? However since they are just enough mini green triangles, the three dark green overlap triangles together must exactly fill the white triangular grid. But that's absolutely impossible! Why? Because we supposedly started with the absolutely smallest way to express a square as the sum of three other squares, that one up there. And, of course, no solution can be smaller than the smallest one. Well we've played this game a couple of times already in the last couple of videos so you should be okay with that. Now the only way to resolve this contradiction is to conclude that the assumption we started with, namely that root 3 is a fraction is false. In other words, root 3 is irrational, or equivalently, the equation X squared plus X squared plus X squared equals Y squared has no solutions in positive integers. How slick is that? :) Now, using similarly stunning triangle choreography we can also show that root 2, root 5 and root 6 are irrational. Now before I do that let me hammer the crux of our proof by contradiction just one more time. Okay, so root 3 being a fraction is equivalent to there being three identical triangular squares adding to another triangular square, right? And the contradiction hinges on us showing that if this was really possible for some triangular squares, then the same would be possible using even smaller triangular squares. So, now let's see how to prove that root 2 is irrational. Root 2 being a fraction is equivalent to there being TWO identical triangular squares adding to another triangular square, like that. And to make the contradiction work we then simply have to show that this equality implies another equality involving even smaller triangular squares. And here's how we can capture those smaller triangular squares.Llet's chase them down. Okay... pretty, pretty, pretty, pretty, very nice. At this point the dark green overlap would be exactly as large as the white empty area. Now let's shift the overlap down. This would fill some of the empty area and create new overlaps. There two smaller triangular squares adding to another triangular square, which then unleashes the contradiction and proves that root 2 is irrational as well. Here's a little puzzle for you. We have to have a puzzle! And there will be more. Have another look at this picture here. Can you see an even easier way to arrive at a contradiction, that is, can you see two other small triangular squares that add to one large triangular square. It's really jumping out at me but see whether you can see it, too. Let me now show you the root five and six triangular choreographies followed by some choreographies using actual squares and pentagons. First root 6 with 6 triangular squares adding to another triangular square. Here we go. (Music playing) Beautiful you must agree. Two things before I move on. When using the root 5 and root 6 choreographies to show the irrationality of these numbers one actually also has to make sure that the various dark overlaps in the animations always exist and are of the same size. I leave filling in the details as another puzzle for you. The root 5 choreographies are particularly tricky to pin down. For example, in the pentagon root 5 choreography it's not even clear where the squares are. Again consider filling in the details as a puzzle, or check out the write-up in the description of the video. Longtime Mathologer fans may remember that in the early days of the channel I did a video on proving that root 2 is irrational using the simple square choreography. In fact, it was this beautiful proof by mathematician Stanley Tennenbaum and popularised by the great John Conway that started this whole line of investigation. A paper expanding on this idea was authored by Stephen Miller and David Montague who discovered the root 3 triangle and the root 5 pentagon choreographies. My contributions are the triangular square root 2, 5 and 6 choreographies and using triangular squares to illustrate things. Anyway, all this is just the start of much more really, really beautiful mathematics, such as: 1) nearest miss solutions to our impossible equations; 2) best rational approximations to our small integer roots; and 3) a really nice paradoxical insight about triangular numbers to finish things off. Well let's go. Remember, by proving that root 3 is irrational we also proved that the equation X squared plus X squared plus X squared equals = Y squared has no positive integer solutions. This means that the three green triangular squares together and the white triangular square that I used to illustrate the proof actually cannot have the same number of mini triangles. In fact I chose the numbers on the left and right to be as close as possible differing by just 1, And so 15 and 26 form what I want to call an nearest miss solution to the equation X squared plus X squared plus X squared equals Y squared, the next best thing to an integer solution. Among other things this means that 26 over 15 is an extremely good approximation of root 3, sort of the root 3 counterpart of PI's 22 over 7. Have a look. Pretty good, right? When we run our choreography, we start with this propeller formation here and after our triangles finish their dance we end up with another propeller formation. Even better the new triangular squares correspond to another nearest miss solution, that one down there. Let's unleash the choreography on this small propeller. Okay, there, yet another propeller and another nearest miss solution one, there. Now clearly something goes wrong if we try this one more time and I'll leave it as yet another puzzle for you to figure out what but if we can't go on forever in one direction, let's go the other way. Let's run the choreography in reverse. Here we go. So everything going in reverse. Larger, even larger. Now we're back to where we started from. But why stop here? Let's just keep on going, right? Okay and now we are one step up. It's probably not so surprising and it's not hard to show that this new propeller also corresponds to a nearest-miss, like this. Okay, so the numbers pan out like this here and you can check that this is really true. In fact, it's not too hard to prove that if we keep on going we'll always get nearest miss solutions and in fact, we get all nearest miss solutions to our equation in this way. Anyway here's a list of the smallest of these nearest miss solutions. Okay now if we take the ratios from top to bottom this will give better and better approximations of root 3. Let's check this. So 2 over 1, 7 over 4, 26 over 15 and there I've highlighted how many digits we get correct. In fact the sequence of these fractions converges to root 3, so if we combine all our propellers containing all of these ratios into one large picture, in some sense this overall picture captures root 3, IS root 3. So this is root 3 have you ever seen root 3, this is it. Very pretty, right? And as you've probably already guessed everything I just said about root 3 can also be shown to work for the root 2, root 5, and root 6 choreographies. Now just for fun, here's a picture of root 2 corresponding to the original Tenenbaum choreography. Also super pretty right? Now there are some more footnotes that I really should add: 1) about a second type of nearest miss solution; 2) about irrationality of integer roots in general; 3) Pell's equation; and 4) some interesting connections with infinite continued fractions. For those of you interested I'll put some footnotes in the description of this video. But for the finale of this video let me tell you about a freaky alternate number reality to which our triangle choreographies can be applied, as well as a very paradoxical puzzle for you that pops up in this context. Have a look at this. I call this a triangular triangle. Well nobody in the universe does that except for me, but who's going to stop me, right, I'm the Mathologer :) Puzzle for you: Figure out the motivation behind this strange name. It's not hard. Anyway the number of hexagons in a triangular triangle of width N is called the Nth triangular number (hint :) T_n. So T_1 is just 1, T_2 is 1 plus 2 equals 3, T_3 is 1 plus 2 plus 3 equals 6, and so on. And so the nth triangular number T_n is just the sum of the first n positive integers which, as many of you will know, is equal to n times n plus 1 over 2 so for example T_5 the number of hexagons in this triangular triangle is equal to 15. Now, just like the integer squares, the triangular numbers also feature prominently in number theory and a lot of famous mathematicians have proved theorems about them. For example, up there is one of the most famous entries in Gauss's mathematical notebook. It says that every positive integer is the sum of at most three triangular numbers. You can tell that Gauss was really excited about having found a proof for this. Just have a look at the Eureka in really bold letters preceding the statement. Now important for us is the related fact that, unlike squares, there are instances where three times a triangular number equals another triangular number. Here's an example so T_20 plus T_20 plus T_20 is equal to T_35. Let's unleash our root 3 choreography on this set-up and see what happens. So, obviously, this propellor corresponds to a smaller triangular number sum that also works, this one here. T_5 plus T_5 plus T_5 is equal toT_9. In fact, we can run the choreography forwards and backwards to generate all instances of three identical triangular numbers adding to another triangular number. Now here's my main puzzle for you. Forget about the other ones. If you just want one, this is it: the smaller equation follows from the larger one, right, so why doesn't this prove, just as in the case of triangular squares, that three identical triangular numbers cannot add to another triangular number? That's a tricky one, let's see whether anybody can figure it out. Anyway, that's it for today.
Info
Channel: Mathologer
Views: 324,696
Rating: 4.8699794 out of 5
Keywords: irrational number, proof, root 2, root 3, root 5, root 6, Tennenbaum, John Conway, proof by contradiction, infinite descent, David Montague, Stanley Tennenbaum
Id: yk6wbvNPZW0
Channel Id: undefined
Length: 18min 9sec (1089 seconds)
Published: Fri Apr 13 2018
Reddit Comments

Mathologer says that some of the proofs he talks about in this vidoe are original and due to him! Somebody publishing math on YouTube is that a first?

👍︎︎ 56 👤︎︎ u/Romanmaths 📅︎︎ Apr 14 2018 🗫︎ replies

Triangular squares, I'll remember that one, .... and triangular triangles.

👍︎︎ 11 👤︎︎ u/11colours 📅︎︎ Apr 14 2018 🗫︎ replies

The abstract algebraic proof of the nth root of p such that n>1 is irrational is so beautiful and simple already. I guess I don't see the purpose :/

👍︎︎ 22 👤︎︎ u/[deleted] 📅︎︎ Apr 14 2018 🗫︎ replies

The youtube comments say you can do this in 3d with 4 tetrahedra to show the cuberoot of 4 is irrational. Is this true? Do the details work out appropriately?

👍︎︎ 1 👤︎︎ u/velcrorex 📅︎︎ Apr 14 2018 🗫︎ replies
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