 I'd like to bring up Neil Sloan. He is the person who is the creator of the online encyclopedia of integer sequences. And if you haven't checked that out, you'll have an opportunity to learn a little more about it now. And I encourage you to check it out on your own at home. So thank you. So thank you, Cindy, for the invitation, the chance to introduce my old friend, Mark Chamberlain, who is talking about single digit numbers today. We share a common interest in numbers. The numbers that I'm mostly interested in are bigger than single digit numbers, but still. I have a website called the OAIS, the Online Encyclopedia of Integer Sequences. And although if you read this carefully, they're now called Integers Squeensies. So the way this works is if you have a squint that you want to identify, somebody says to you, what's the next term after this? You type it in, and it will hopefully tell you what it is. So I just put a random example here. 71, 42, 12, 83, 54, 24, 95. You can probably all immediately guess what that is. It's a very simple sequence. So that's the kind of thing. So I'm going to mention two other sequences that are a little more mathematical than that one. And one is connected with a number that probably will be in your second volume. It's a two-digit number, namely 10, and then a sequence connected with the number 5. The number 10 is interesting because if you ever need a large prime number, of course 10 isn't a prime, and you may know that 2 to the 31 minus 1, if I remember correctly, is a prime. But suppose you want the actual digits of a large prime. So what you can do is you can say, let's build up slowly and look for primes. So look at the sequence. It starts 1, and then the next term is 1, 2, 1. And the next term is 1, 2, 3, 2, 1. And then 1, 2, 3, 4, 3, 2, 1, and so on. And keep doing this until you get to a prime. Well, 1, 2, 1 is 121, which is 11 squared. So that's not a prime, right? And 1, 2, 3, 2, 1 is 111 squared. So that's not a prime either. And the next one is 4 ones squared, all the way up to 1, 2, 3, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 4, 3, 2, 1, which is 1, 1, 1, 9 times squared. None of those are primes, obviously, since they're squares, except for 1. Well, 1 is not a prime, although in California, I'm told it's taught in schools, it's taught that 1 is a prime, although sometimes they say that this is not, in the rest of the world, it's not regarded as a prime, which is correct. But anyway, when you get to 10, you get the number. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1. And that is a prime. So it's a convenient 20-digit prime if you ever need one. And then, well, you can keep going. What if we go 7, 8, 9, 10, 11, 10, 9, 8, 7, down to 1? Do you get a prime? No. You don't get another prime. Last year, someone I don't know whose last name is Gupta found the next time you get a prime is if you go all the way out to 2, 4, 4, 6, and then down to 1, you get a prime. That's about 17,000 digits. So we've got the beginning of a sequence, 10, 2, 4, 4, 6. What's the next term? We don't know. And I would like to know if somebody would like to do a bit of computation and figure out what's the next time that the number's 1 up to n and then down to 1. When is that a prime? And when I was thinking about that, I realized there's an even simpler question that we don't know the answer to. Suppose you just write down the number 1, 2, 3, 4, 5, 6, blah, blah, blah, up to n and stop. When is that a prime? So you might go, well, what about if you stop at 2, obviously that's an even number, it's not a prime. 1, 2, 3 is not a prime, it's 3 times 41. If you stop at 7, you get 127 times 9721, not a prime. If you stop at 13, again, you don't get a prime. In fact, you don't get a prime for a long time. In fact, we don't know when you get the first prime. There are computers running night and day searching for the first prime. They've gotten up to about 300,000. So we know on probabilistic reasons that there should be infinitely many primes in this sequence. So the question is, write down the numbers 1 up to n, squash them all together and get a big number. We want that to be a prime. No one knows when it first happens. So that's my story about the number 10. And then I want to mention a story about the number 5, which I was involved with personally. And I think it's a pretty good story. But I have to tell you what a curling number is. So if you look at the sequence, squints 2222, that's got four copies of 2, and it has curling number 4. The curling number is the number of times you get an identical string at the end. So if I took the example 1, 6, 7, 2, 7, 2, 7, 2, that's got curling number 3 because it ends with three copies of 7, 2. So that's what the curling number is in general. You write your string of numbers as some prefix, which might be missing, and then the maximum number of copies of some tail. I have a slide with a pig, a drawing of a pig. The body of the pig is the prefix, which might be missing, and then the tail is the curl. And the number of curls in the tail is the curling number of the string, which in this case is 3. So years ago, a graduate student in Amsterdam, called Dianne Heistveit, said, let's start with 1 and work out the curling. He didn't use this language, but this is what he meant. Start with 1, work out the curling number. That's the next term. Recompute the curling number. That's the next term. Keep going. We start with 1. This is Heistveit's sequence. Start with 1. Its curling number is 1. So that's the next term. That's always the rule. Work out the curling number. That's the next term. Now we have 1, 1, 2. That has curling number 1, which I've written underneath. Now we have 1, 1, 2, 1. That's got curling number 1. This has curling number 2. Now we have 1, 1, 2, 1, 1, 2, 2. Now we have 2 twos, so we get a 2. Now we have 3 twos, and we get a 3. Now we go back to 1. And that repeats until we get to here, and we have 2 copies. So that's Heistveit's sequence. And the question is, what happens? Well, Dion noticed that when you get to 220 terms, you get a 4. And he said, is there a 5? He had worked out some 300,000 terms, and there was no 5. So I talked to some of my colleagues, and had them work out. John Lindemann worked out the first 2 million terms, and there was no 5. And then, well, we got clever. And over a long weekend, independently, I and Fokko Vanderbilt, a colleague in Amsterdam of Deons, we came to the conclusion that if you go out to about 10 to the 10 to the 23 terms, there is a 5. And if you go out, there is a 6 after a while. But to get to the 6, you have to go out to the number of terms, 2 to the power, 3 to the power of 4, to the power of 5. And if you want n, you'd have a tower of exponents. So those are a couple of sequences that I offered to Mark for later editions. So it's a great pleasure to introduce Mark. He's a professor of mathematics, and has been for nearly 20 years at Grinnell College in Iowa. He has many publications. He has two publications I'm particularly interested in, one on Duchy Sequences, and one on his continuous version of the 3x plus 1 map. But he's most famous, I guess these days, because he has his own YouTube channel called The Tipping Point Math, which looks, here's what you will see if you go there today. And I have been told. I have asked some friends to look at this. And one of my friends said that she finally understood Pascal's Triangle for the first time after seeing your videos. So let's welcome Mark. Thank you. All right, so mathematicians get excited about all kinds of different numbers, exciting sexy numbers like pi and e. But they miss the allure that just comes with the simple numbers from one to nine. So what I wanna do this evening is tell you one interesting fact that you may not, probably don't know about each of the numbers from one to nine. So there they are. They look simple enough. They wiggle. They turn around. Okay, but we'll start with the number one. And here and there I will offer a little fun fact about each number if possible. I couldn't think of something simple to say for one except it's the loneliest number. So here we go. All right, so what do I wanna start with? Can you cut out the square with only one cut? So of course, you can take your scissors. And most of you, if this was a paper, you would make four cuts going around but I don't wanna do that. I wanna do it with just one cut. How am I supposed to do this with only one cut? I need to fold. So I've done some pre-folding so I don't mess this up. So here's our square. Let me describe it on the video first. So let's do a fold across this diagonal. And that takes us here. And then let's do another fold. And that takes us here. And then we will do our cut like that. So let's match this up with some practice. So I have that one fold. I have this fold and you can see I just wanna do one cut straight across. Let's see if I can do this correctly. So what's left is our square. All right, nothing too magical about that. Contain your enthusiasm, please. Okay. So let's make this one step harder. Now I want this red region. I wanna get this with only one cut. So can you think of a way that we can do this with one cut piggybacking off of what we did before? Can you try to imagine it in your head? Does anybody have an idea? Tough crowd, hmm. Yes. Well, I don't, but I wanna do the cut around the outside and the inside all with one shot. Okay, just one cut. Let's see. So let's take this again. And I'm going to do like I did before. One fold along a diagonal, fold along another diagonal. So you probably can't see it from here. Imagine the red is just this little triangle here. Yeah, so what I'm gonna do is just fold it over a little bit so that the fold line is parallel to where that red line was. I did fail sandbox and kindergarten, so this might not work. So I cut that out and open it up. And there we go. All right, thank you very much. You're too generous. Okay, well, that's still too easy. So what if I tilt the square like that? Not only is this hard for you, this is hard for me as well. But let's push this even farther still. Suppose I wanna cut out this swan with one cut. It looks difficult, but if you do your folds the right way, you can do it. There's two types of lines here. You have the straight, you have the dash lines and you have dash dot lines. And one of them you make mountains and one of them you make valleys. And if you do it just right, and it's not simple because I've tried it, but if you do it just right, you can cut the swan out with one straight cut. So it did work, it did work. So in fact, there's a young mathematician, Eric DeMaine, who has spoken here in math encounters. And he has a theorem which he came up with. And it says if you have any polygonal region, there's always a way to fold it up so with one cut you will get your result come out. So that's something special about the number one that you probably didn't know about. All right, let's move on to the number two. So here's a bunch of numbers. Somebody tell me what these numbers are. They are prime numbers, numbers that only one and the number itself divide into. But if I'm gonna just look at these red ones in this list, what are these red numbers? Anybody know? Twin primes, okay? Twin primes, these are primes that are just two apart. So three and five are two apart, five and seven, 11 and 13, 281, 283, et cetera. So a nice classical result, which mathematicians just get really excited about, is that there are infinitely many prime numbers. That's not obvious, but there's a very beautiful simple proof of that. I'm not gonna go into it and it's an old proof and any math majors should know this in their sleep. But what's not known is, are there infinitely many twin prime pairs? Those red numbers I had up a moment ago. And it ends up the answer is, we don't know. And this question has been around for hundreds of years and we still don't know if there are infinitely many of these primes that are just two apart. I'll just let you in on an insight thing. I have a somewhat secret passion for this problem because I too am an identical twin. And I told this to my students the other day and they just wanted to completely monopolize the lesson with questions and I had to shut them down. So don't get started on me here. So these twin primes, we don't know if there are infinitely many of them or not, but primes are very important. They form the basis of modern cryptography that is the science of making and breaking secret codes. And you might think, well, do I need cryptography? Cryptography is used every day when you buy something on the internet, you need that internet security so that somebody doesn't walk off with your credit card information. And it's all based on large prime numbers. There is a website where people have been able to offer up their computer time and distributed computer project that looks for twin primes. And here's a colossal twin prime which was found. Take this huge number multiplied by this large power of two and if you either add one or subtract one, you get a prime, huge twin prime pair. Over 200,000 digits long. And here's a curious instance where twin primes came out to be useful. So in 1994, somebody was doing research on twin primes and they were trying to generate them and they found a discrepancy between what the computer was generating and some theoretical results. And what eventually turned up is there was a bug in this Intel chip. So Intel just swallowed their pride and they coughed up a lot of money to get all these bad chips replaced. Don't tell anybody I told you about this because I teach at Grinnell College and probably the most well-known alum is Robert Noyce who is one of the co-founders of Intel. So they wouldn't like it if I'm kind of giving a bad name here. So don't tell them I'm bad mouthing Intel. All right, let's move on to the number three. And here's a fun fact. Can anybody see something special about these three rings here? No two of them interlock. No two of, so if you take any pair of them, they don't lock up together, that's true. Can anybody go even farther? So that's a good first observation. Anything beyond that. No two of them interlock. But well, the three do, okay? The three you cannot pull apart. But here's something that looks strange. So the red one is over the green and the green is over the blue, but the blue is over the red. So you could not actually make these with three solid rings. They would have to be bent somehow. Not three traditional rings. And this is what we call Boromian rings and they have lots of nice structures and have appeared in lots of places in mathematical culture. All right, that was just an aside. I wanna ask you a question here. How many guards are needed to observe a museum? And I'll get to your papers in a second. So here's what I wanna do. I wanna think of this as the layout for a museum and these guards are gonna have two special characteristics. One is that they have to stand still. But the other thing is they can look in any direction. Maybe it's for some of you, it's clear if you see those thinner lines on the screens back there. So the question is what's the fewest number of guards you need to protect this museum? Are they what? Are they in a union? Oh, I think I will just refrain from answering that question. Here's some examples, okay? That was a good question. You caught me off guard there, no pun intended. All right, so here we have a convex set and no matter where you're standing in there one guard would be enough, right? And here's a slightly more sophisticated looking space and you see where I've marked those two points. Two guards is enough for there. And even for this pretty kind of complex space with these long narrow hallways, if I place my guards just right, they can see everywhere and guard the museum. So let's come back to this one. What I'd like you to do is take your paper, take your little pencil, talk to your neighbor and I'm gonna give you two or three minutes to think about what's the fewest number of guards you need to protect this museum? So go and you can talk. Okay. All right, let's try this together. So let's see how people did. Raise your hand if you could do it with eight guards or fewer. I kind of expected that. All right, keep your hands up, keep your hands up and keep it up if you could do it with seven or fewer. Six or fewer. Ooh, smart people. Five or fewer. Four or fewer. Three or fewer. And there's some people I don't believe back there because they still have their hands up. Okay. And once I talked about this and then one smart aleck in the back said, just one and a lot of video cameras. So that doesn't count, okay. All right, so let's think about this. Now, well just with this one layout here, I'll just mention one way I would think about it is to be able to guard this, I need a guard down here somewhere to cover this region and I need a guard somewhere over here to cover this region and I need a guard in this stretch somewhere to cover that region. Man, I probably need another guard somewhere here because I got this little cul-de-sac and I got that thing and it's, depends on your interpretation, what's going on. But anyway, that's how I would start thinking about this problem. But what if your museum has lots of edges to it, okay. Then this can become a much more complicated problem. So, all right, let's think naively. So I asked before how many could do it with eight guards or fewer. Let's just go crazy, okay. The union will be very happy with us. Let's put a guard at every corner of the museum. And the reason I say that is this is going to be enough because wherever you are, that's you, you can see a corner and so a guard at that corner can see you. That's a lot of guards. So how can we start firing some of these guards, okay. Just union busting here. How can we gently excuse some of these guards so that we still have enough? So there's a result called the art gallery theorem. And it says if you have n corners, or you could think of n sides of this polygon, then you need at most n over three guards. This is why this is under the number three. So no matter how big that n is, one third of them is enough. And we're gonna go through the idea of the proof and see why this works. First though, let me say that in general, you can't do better than n over three. Yes, maybe for the one I gave you, which has 28 corners, you can do a lot better than 28 over three. But in general, you won't be able to do better than a third of your corners. So this region is called Schvato's Cone. And we've got, I've drawn it with six of these triangular type regions. And we have a total of 18 sides. And you need a guard for every one of these triangular regions. If I try to add another one of these triangular regions, I would add three more edges and I would need another guard for it. So for every three new edges, I'm adding one new guard. So it's always gonna be this ratio of n over three. So in general, I can't do better than this. All right, let's talk about a proof. And we'll use our familiar gallery as our working example. So first thing I'm going to do is triangulate the region. And again, this might be easier to see on your monitors back there. So I've just drawn a thin black line here. And I'm going to keep adding these lines and I'm carving the region up into triangles. So here I have triangulated the whole region and this is always possible. I won't get into the reasons why, but it's always possible to carve your region up into triangles. Next thing I'm going to do is color the corners. So I'll color that first corner blue and then green and then yellow. And I'm gonna keep coloring it until I'm done. And the way I do the coloring, I want it to work so that every triangle in our triangulation, the three corners are the three different colors, green, blue, and yellow. You look at any of the triangles that we've carved out here, their corners are green, yellow, and blue. All right, so why is this helpful? This is also always possible no matter what shape we started with. So why is this helpful? Well, suppose I just focus on the blue corners to start with. So I said that every triangle has a blue corner, a yellow corner, a green corner. Every triangle has a blue corner. So if I put a guard at the blue points, then every triangle would be viewed, which means the whole gallery would be viewed. So if I just put guards where these blue points are, that's enough. But the same argument works with the yellow corners and with the green corners. So if you take your total number, we had 10 blues, 11 yellows, seven greens for a total of 28. So I have to have at least one of my corners as no more than a third of 28. Because if they were all greater than a third of 28, when I add those up, I would have more than 28 total and that's impossible. So at least one of these three has to be less than or equal to a third of the corners I started with. And that's basically our proof of the Art Gallery theorem. All right, here's a question I'll leave you with if you like this. You can take your paper home and think about it. So I call this the lonely guard conjecture and I put five points on here where you can convince yourself that the whole gallery is being covered by these five points, but why do I call it lonely guards? Because no guard sees another guard. They're lonely, okay? Sorry, it's a crappy union, all right? So does this always have to happen? I don't know. And I have posed this to people who work in this area of combinatorics and they don't know either. So the lonely guard conjecture, can you make a diagram, a polygon, and can you put guards in locations so that the whole gallery is viewed but no guard sees another? That's something for you to think about. All right, the number four. So here's a familiar tennis ball and a familiar baseball. And besides the fact that they're both balls, what do a tennis ball and a baseball have in common? Yes. The seams. The seams, yeah. So if you think of the seams in both, here, it's more articulated because of the stitching here. We've got the seam going around. We have the seam in the tennis ball. In fact, with these two balls, it's not, the seam is the same. You can make it by taking kind of like a figure eight shape, take two of those and you put them together and you get the covering for the appropriate ball. This one's obviously stitched and I believe a tennis ball is thermally bonded. So because they're an identical region, they are, the seam is cutting the surface area in half. Now if you follow, pretend you were an aunt and you were following walking along one of these seams, let me give you a better picture here. If you're walking along the seam, you see here we are bending towards the left as we move along, but we get to some point where it's no longer bending to the left, but it's bending to the right. And when we change our bending, we'll call that an inflection point. There's a fancier way to define inflection point with calculus and talk about a zero derivative, but generally speaking, we will think of an inflection point as changing from bending from one way to the other way. And if we look on this curve, you will find that there are four inflection points and this is part of a much bigger result called the tennis ball theorem, funny name. So the tennis balls theorem says, the most you have a smooth curve on a sphere like we do here and this curve cuts the sphere into two pieces of equal area. They don't have to be exactly the same like with the tennis ball, but they can be. Then you have to have at least four of these switching points, these inflection points, and that always has to happen. And this takes a little bit of work to prove this mathematically, but for any shape of this type, this will happen. So now if we look at a picture of the globe to get some feel for this, all points on the equator are actually inflection points. This has to do with this calculus idea I talked to. You're actually not bending to the left or the right at all. You're just going in a straight direction around the globe. And so here we have infinitely many inflection points, a lot more than four. But notice if you take this curve, the tropic of cancer or the arctic circle, look at the arctic circle and you're going around, it seems you just keep bending to the left and you can go around, there's no inflection points. Does that violate our theorem, our tennis ball theorem that I showed you a moment ago? Why is the tennis ball theorem still okay? Why is this not a problem? Doesn't divide it into equal parts, okay? So one thing that mathematicians will always emphasize when you have some claim, some theorem, you have the conditions of the theorem and then you have the conclusion. And one of our conditions is that our curve divides the sphere into two equal parts. And so this example with the arctic circle, it's not a problem, it doesn't cut the surface area in half. All right, so next time you're playing tennis, think about the tennis ball theorem. All right, the number five. And fun fact, does anybody know the name of this fancy shape I have up here? Any guesses? Pedersen graph. Say it again. Pedersen graph. That is correct, it is a Pedersen graph. So you have a pentagon on the outside, you have a pentagram on the inside and this is called a Pedersen graph. So people who work in graph theory, some of them, they look at huge graphs of interconnected components. For example, if you thought of the whole internet and you've got web pages and you've got links from one page to another, you can abstract that whole concept as a bunch of nodes of points and then the links are these connections from one node to another. That's a monstrously huge graph. So graph theorists have a very rich structure that they work with and the Pedersen graph actually fits in very nicely and you can see its connections to the number five. But that's not what I wanna talk to you about with the number five. So how many of you are familiar with the Fibonacci sequence? Lots of you, okay. So you know you start with a one and a one and the idea is let's add those two to get two then add one and two to get three, add two and three to get five, et cetera. So this is an example of a recurrence relationship and it's a recurrence relationship where you start with one term and you get that term by adding the previous two. Well I wanna build up a sequence of numbers like Fibonacci numbers but instead of just adding the previous two I wanna do something a little more fancy here. So the idea is how do I get the nth term in the sequence? I take the previous term in this case which is a seven, I add one to make eight and I divide by the term before that which is four, eight over four is two. So seven plus one divided by four is two. Two plus one is three divided by seven is three sevens. Three sevens plus one is 10 sevens divided by two is five sevens and you could keep going. And do you see something interesting in that sequence of numbers that comes out? It repeats, right? And how often does it repeat? After five steps, it repeats. We started with the four and five steps later we are back to a four and then a seven and since we have that four and seven and our sequence is defined by taking the previous two and combining them like that we will keep going in this manner. Okay, so this is what I want you to do. Flip your paper over and I want you to do a little arithmetic. If you get stuck on your fractions swallow your pride and talk to your neighbor, okay? So first question is suppose we start with a one and a two and we follow this rule I described what is the sequence that we will get? And if you think you've understood the pattern there then I want you to try a two and a five and follow this rule and see what pattern you generate. So give you a couple of minutes and again you can talk to your neighbor. What do you think? All right, let's talk about this together. So who can tell me what they found for that first one? If we start with a one and a two what happens? Who's a brave soul? Who's gonna tell the rest of us what they found? Yeah? It recycles, it repeats? After how long? Well, just count the first one is a one, one, two, three, four, five that's called the sixth step, the one that's... So we start with a one and when do you get one again? In position what? One, two, three, four, five, position six. So we start with a one and two and then you're saying in position six we get a one and in position seven we get a two so it repeats every five times. Good, that's what you should have gotten. Somebody else for starting with two and five what did you get? Yeah, same thing. It's a miracle, right? So what happens is almost no matter what you start with you will get cycling after five steps which is kind of bizarre, why should this be happening? So it's actually not too hard to convince yourself and I have one possibly scary slide because it has algebra and I know that's the A word and you don't say algebra to some people. So here suppose your first number I will represent it with a variable A the second one is a B when you combine those your next thing will be B plus one over A. After some simplification this is the fourth term after some simplification there's the fifth term and then when you use these two your next one will come back to A and then it will come back to B. Now I didn't say it will for any A and B I said almost all A and B where might you have a problem with this? Zero, okay? We are doing division in our process here and you can see here I've got the B or the A or in this case even both of them on the bottom so I cannot have A or B equal to zero. Are there any other numbers that are a problem? Is there any other way we can get a problem here for some other A's or B's? Yeah? It gives you zero down the line so like negative one would give you a zero. Okay so it's not so if A or B is zero we can see just from the algebra I've written here you would get zeros on the bottom and that would cause trouble but it's really if any of my terms are zero so I not only don't want either of these zero I don't want this to be zero which means I don't want B equal to negative one or I don't want A equal to negative one or I don't want the sum of them plus one equal to zero either and it ends up those are all the bad cases everything else works out fine so almost all starting cases work out okay so try to have some fun with a friend say pick two numbers try this process see what happens and you say and then say now pick another two and I bet with my mental powers that they will repeat after five as well just see what happens and just hope they don't pick one that gives a zero all right all right let's talk about the number six so the number six is one of my favorite numbers and so I want to illustrate a principle starting with six kind of pictures of random people so I have six random people that I've chosen here and all of these people are going to have some kind of relationship so if they know each other we will say they are friends if they're not friends by default we will say they are strangers okay so you're either friends or strangers if you're friends we will put a green line connecting the two of you and if you're strangers we will put a red line connecting the two of you and of course presumably with sufficient knowledge we could fill in the relationship of all of them does anybody notice anything kind of unusual about any of these people in their relationships? Yeah poor Kim Jong-un he's got no friends all right I didn't tell you anything you didn't know already all right so but if you look a little closer there's something else that's interesting here I can take out three of them and the three remaining people they are all pair-wise friends together so I want to investigate this idea so I need six volunteers who are willing to come up here put your hand up if you're willing to be chosen don't be so shy all right how about the two of you guys here and yes and yes and yes okay I think I chose six right come on up did I choose only five okay I need one more way in the back yes but like yes and I want you to make a circle up here so face each other in a circle number six has to get out okay right here all right so let's see do you know anybody here are you friends with anyone you just met okay well so I will call you friends because you're not strangers okay your friends is there anybody else that the two of you know I don't know anybody else no no okay do you know anybody else here no do you know anybody else do you know anybody else someone who's someone that's good enough your friends all right okay so your friends do the two of you know anybody else here no so it doesn't look like I've got three people like we've gone on the screen that are each pair wise friends all right so you don't know anybody else and let's see so you don't know anybody else and you don't know anybody else all right so if I pick you two and let's say this guy that means the three of you are each mutual strangers so instead of having green lines connecting you you would have red lines connecting you so I didn't get three green lines but I got three red lines here that's good enough for now thank you let's give these folks a hand okay but now they know each other so I hope they're gonna make it green lines all right so this leads me to what I will call the friends and strangers theorem actually that's not my name other people call it this so it says that in a group of six people either three people are all friends or three people are all strangers you have to have one of those two situations happen that's not so obvious is it so let's convince ourselves that are true what does a mathematician do we try to construct a proof so why does this work let's let's pick one of these people I had a little fun at with Kim Jong-un earlier so let's think about him for a moment so I could pick any one of these people but let's say we think about him with any one of these people think of their relationships with the five others so you're gonna have green lines and red lines and I claim there has to be either at least three red lines or three green lines why does that make sense why does there have to be either from any one person either at least three red lines coming up poor Kim Jong-un has five red lines coming out but at least three red or at least three green you have a total of five lines yeah so why do I have to have at least three either three all three red or three green in the back you can't that's true you cannot split five and half easily let me push this a little farther you have an idea there you go okay so if you didn't have at least three of green or three of red that means you have at most two greens and at most two reds but that only makes four so you have to have at least three of one so Kim Jong-un he's got he's drowning in reds but let's just pick three of them so he's got three here now what I want to do is try I'm going to make an effort to show that this Friends and Strangers theorem is false I'm going to try to make a counter example and if I reach a contradiction I'm going to try to make a contradiction I'm going to try to make a counter example to the theorem and if I reach a contradiction it means the theorem has to be true so I'm going to take these three red lines and if I don't want the theorem to be true that means this can't be a red line because then I would have a red triangle up here and this can't be a red line because then I have a red triangle down here and this can't be a red line because then I'd have a red triangle there so I don't want any of those things to be red lines or else I have trouble so if none of those is a red line then they must all be green lines but why is this a problem? If I'm trying to get a contradiction to my theorem a counter example rather why is this a problem? Yeah, that was a head scratch, sorry, not a hand Yeah? It has to be three green lines or three red lines Yeah, the theorem says that you're going to have either three reds or three greens and for the moment I'm trying to see if I can avoid either of those scenarios so we avoided having a red triangle but look what's happened in the process in avoiding that now we have created a green triangle which I don't want either so in trying to avoid stepping into this junk over here I stepped into a different problem over here and this gives us a contradiction so this proves why this proves the friends and strangers theorem so I'll just say the friends and strangers theorem is a special case it's a special theorem in a broader class of problems called Ramsey theory and Ramsey theory very abstractly put says if I have a lot of objects of a certain form if I have enough of them then I will have certain kind of structure I mean have you ever wondered why people for thousands of years have looked at star constellations and they see various patterns there it's not just that the stars happen to be aligned in a certain way it's because there are enough stars that there have to be certain configurations some geometrical configurations have to come up and it's the same so this is an example with enough people we have to have certain phenomena in this case either three friends or three strangers let's move on the number seven you recognize this from the seven samurai so here's another we've talked about fractions earlier here I have the fraction one seventh it has a funny decimal representation when you calculate it if you do a half or a third or a fourth up to one sixth sometimes you have repeating decimals like in one third or one sixth but they start repeating and just one digit is being repeated whereas with one seventh now I have six things which will just keep repeating indefinitely in fact it's not just for one seventh that this happens but if I take any of these numbers one seventh up to six sevenths I get repetition with six terms as well and does anybody see a special pattern relating those different digital representations yes yeah so the order is the same it's just shifted over so if you look at two sevenths here we have we're starting at two there's two eight five seven continuing to one four it's the same order just shifted over this is just shifted over all of these are just shifted over kind of surprising now but there's more with this result with one seventh so here I have an ellipse and I've got six points six pretty simple looking points on this ellipse now in general if you pick five points in the plane there is a unique ellipse which goes through them almost never will you have five points and you get more than one ellipse going through it it's usually one fixed ellipse that goes through them but here I've got six points and six nice points and I have an ellipse that goes through this this is surprising this shouldn't happen but moreover if you look at the coordinates of these points look at our representation for one seventh so I have one four one comma four and then four comma two and two comma eight and eight comma five and five comma seven and then with the wrap around seven comma one so these points come out nicely because of the digital representation of one seventh and there's another one like this these are called one seventh ellipses and again I've got six pretty nice looking points I didn't pick my ellipse and then just choose six points on it willy nilly these are nice points shouldn't be happening like this and again if you look at the digital representation fourteen twenty eight then forty two eighty five is right here and so on you get all the other possibilities so this is something really surprising that happens with one seventh what's a natural question you might ask here what if you do three three numbers in a row not just one digit at a time or two digits at a time but if you do three digits at a time I think it's not going to work because you're going to have some you're going to have some nice symmetry in the picture that probably shouldn't be there so I will leave it at that for now but even just one of these ellipses is a bit shocking to have two of them is to a mathematician rather remarkable okay the number eight and here's a fun fact who can tell me what kind of shape this is I thought I heard something a fractal it's a fractal so a fractal well it's kind of hard to define what a fractal is exactly but fractals exhibit what is called self-similarity there's two ways you can look at this picture one way is think of taking this huge square on the outside carving it up into nine sub-squares and then throwing away the middle one and then take each of those smaller sub-squares divide those into nine smaller squares still and throw out their middle ones and then just keep doing this process iteratively and you construct the shape another way and this reflects on what's called the self-similarity you can take this whole shape which I have drawn in here not without the blue numbers written in and I squeeze it down by a factor of three in both directions and I can place one copy in each of these eight slots that's the self-similarity that's going on here does anybody know the name of the shape by the way your mathematical lore yes is there Serpinsky carpet that's right Serpinsky gasket is a triangle that does something like this and this is the Serpinsky carpet okay but this is not what I want to talk about with the number eight so I need three volunteers to come up three new people let's spread the love around yes that hand there come on up who else here's a girl coming up I need one more person don't make me big darling one more yes in the back good come on up where was my other person my first person I called on alright so I have I have eight playing cards here okay try playing with these normally alright so we're going to try some shuffling here so you're going to help me out you're going to stand in the middle of the audience and you two stand on the sides so this is what I want you to do I have eight cards here I want you to take the top four and give it to your neighbor on the right and your bottom four to your neighbor on the left I'll do it the first time alright alright and now I want each of these people what you're going to do is hold out your hands and he's going to take his top card and put it face down and then you will take your top card and face down and then alternate like that put the first one face down second and keep going okay so now let's start over so give your neighbor on the right the top four count the first four and give your other neighbor the other four hold out your hands and let's do the same thing okay and if we look at them so far they are a bit mixed up so we're going to do it again do you know the rules without me telling you top four good okay and at this point we look at our cards and they are back in order again woo hoo okay so let's give these folks a round of applause you can take your seat alright so with eight cards what we did was three perfect shuffles or what is sometimes called three pharaoh shuffles so the idea is you have an even number of cards and you cut your deck in half here's a standard deck of cards so suppose I cannot but suppose I could cut it perfectly in half and then I do a perfect shuffle that's one pharaoh shuffle so what we did up here with the three volunteers was show after three pharaoh shuffles you will take whatever your original ordering is it comes back to that so how many pharaoh shuffles with the deck of 52 cards do I need to do to get it back to the original ordering give you a hint it's on the screen behind me okay eight pharaoh shuffles does the job let's see it on the screen so there's 52 cards so let's cut it in half and for visualization I'm going to spread these out and then I'm going to put them together that was one shuffle there's two three four five six seven I can't wait and eight brings us back to where we started so well some mathematicians find this exciting but as an aside I will say if you do a lot of practice and you can cut your deck in half which takes some work and you can do a perfect pharaoh shuffle and you can do it eight times in a row and you can repeat this as often as you like I would encourage you to catch a flight to Las Vegas okay eight shuffles is enough and that's surprising you would probably think it would take many many pharaoh shuffles to come back but it ends up for 52 cards eight is enough alright there's one other thing I want to show you connected to the number eight and Neil was generous enough to say something about my YouTube channel so I would like to show you a video from there and this was one of my early videos so David and Lucas are starving and want to share a pizza it's uncut and Lucas tells David we want four slices each make four cuts through the center vertical, horizontal and at the 245 degree angles while you're doing that I'll go get the pop or soda or coke that last phrase was used because depending on where you live or visit people call hyper sugar fizzy water different things anyway you expected the pizza slices to look like this but David made a fatal blunder he seriously misidentified the center so after the cuts there are troll and hobbit size slices but since David and Lucas are neither hobbits nor trolls they want to split the pizza evenly how is this possible there's an easy solution called the pizza theorem if each person takes alternating slices they each end up with exactly half of the pizza it doesn't matter where the center point for cutting is specified this strategy gives each person the same amount one can use calculus to prove the pizza but for an easier way look at this picture the pizza has been further divided into many regions each containing a number each number appears in two congruent regions once in violet and once in orange this implies that the orange area equals the violet area so each person gets the same amount of pizza this proof by picture works no matter where the center is placed the pizza theorem strategy also gives each person exactly the same amount of edge crust this may be good or bad depending on whether you don't like the edge or you gorge on stuffed crust this claim should we call it the crust theorem follows from two applications of the pizza theorem the areas on the slices of the outer pizza match but so do the areas of the inner pizza that is the part without crust the difference of the inner and the outer produce equal amounts of crust for each person the crust theorem implies that the perimeter of the circle is divided into two sets of equal length the pizza theorem does not work with only four slices the yellow area and the red area however the pizza theorem does work for more than eight slices of pizza if there are 12 16 20 or any greater multiple of four slices and the angles of each slice are the same taking alternating slices still splits the pizza into two equal amounts so the world may not be fair but you don't need to draw daggers over who gets more pizza ok so that was our fun with number eight and one more is number nine yeah oh what happens if you point your put your so what if you're really bad at guessing where the center is and you put your point on the edge of the pizza itself actually will still work you won't get now you won't get eight oh crust well alright alright alright you have think of it like this think of the circle think of the crust as infinitely thin so it's just the circle you're still will be cutting your circle equally okay you thought you got me on that one alright okay so one more number to go the number nine and this is not as glamorous as the others because things get harder with nine there's just not as many fun facts that come up here so let's see what comes up here suppose I have a bunch of cans that are all the same size and I want to pack them tightly in a case so suppose I have let's say cans of spaghetti sauce and I want to pack them tightly so that means tightly means there's no jiggling around of the cans and this is the standard what's called a honeycomb packing and so you see the centers of these form a hexagon so hexagonal or honeycomb kind of packing but suppose we want to do more than just bachelor cooking we want to make our spaghetti with spaghetti sauce and tomato paste as well so we have two different size cans and you can see in this packing we've got all these white places where there's empty space maybe if we can pack two different size cans in the same case we would have less wastage so is it possible to pack two different types of circles in a tight way no wiggling around like this and the answer is yes so here's an example where I've got two different size cans you could say and it's tight because nobody is wiggling around inside here so now the natural question is how many different ways can you do packing with two different size circles I'm glad you asked the number nine so here's the nine different packings of two different circles all the way from the one we first saw and things can be similar size but it's actually slightly different and then relative to the red circles the blues are getting smaller and smaller and smaller till we have hardly any tomato paste here at all but there are exactly nine different ways to do this okay so that's our number nine so our numbers from one to nine they have lots of interesting properties one last question I have for you here is there anything special about the arrangement of the numbers from one to nine as I've written them on the screen right here add up to 15 yes this is a magic square somebody said in the back so every row column and the two main diagonals add up to 15 you can check that and this is not a new result this goes back to 650 BCE Lao Xu did this and lots of people have had fun constructing magic squares since then but this goes back quite a ways it seems okay so do you want more from the numbers one to nine the shameless plug is by my book which came out with Princeton last year and all the topics I've talked about today have a section under the appropriate chapter in the book Princeton's been pretty happy with it and some people wrote some nice things to me a YouTube buddy so thank you for your attention alright we have some questions for our speaker tonight are there any generalizations that would work for the the friends friends and strangers that work for for nodes of n size I cannot give you a specific answer to that but as I said this is a special example of a theorem that fits into the area of Ramsey theory so if you wanted to guarantee at least four friends or four strangers then the question is how big does the number of people in the group have to be and I cannot tell you what that number is exactly but there is some finite number if you have enough you're guaranteed four friends or four strangers and if you go high enough still you get to five and you go high enough so if you push that number high enough you can guarantee as many friends or strangers as you want I can tell you how quickly that number is growing but it exists in your example of the four inflection points cutting a sphere and a half you explained that the equator cut has an infinite number of inflections what about the longitudinal cut well that would just be a different great circle so if you take any sphere and you take a plane that slices through that sphere and goes through the center where that plane cuts through the sphere forms what we call a great circle and the equator is an example of a great circle this horizontal slice but if I go like this just tilt it it's still a great circle it's going through the center of the earth but to me it has zero so this depends is it because you're thinking you're not bending at all? so somebody asked this at the four o'clock talk good the simple way to try to define an inflection point is it's a point where you were bending left and you switched to bending right or vice versa there's a more technical calculus definition that talks about derivative equaling zero and it ends up on the equator you're not doing any bending either way and so you have this derivative property holding often you usually that's not when you have an inflection point it's usually what I said the bending left changes to bending right the equator is somewhat of an anomaly but you just have to take my word on that one for now sorry back to the number two did Euclid or any of the Greeks conjecture that there were infinitely many pairs of twin primes? I don't know how old it goes I suspect it's a more hmm who first wrote about the question? I cannot give you an exact answer on that I suspect it's a more modern question and the progress on understanding twin primes is fairly modern really only in the last 100, 150 years there are postulated ideas of how not only are there conjectures that there are infinitely many twin primes but even how frequently they will appear has been conjectured and that's only about 100 years old that's fairly modern but as far as who was the first one who posed it that I'm afraid has been lost to antiquity sorry we have time for one more question so for the packing problem for number nine how would you go about finding the different ways of packing cans and how would you convince yourself that there aren't other ways? that's a hard question so this would be in what's called enumerative combinatorics and the idea of let's see and I don't go into these details in the book by the way just like I've tried to keep the talk non-technical to a large extent I've tried to keep the book non-technical as well so you get the ideas without being too burdened by a lot of details so this is tricky because I mean you would represent circles by identifying their centers and as I understand it the people who work in this area it's changing the packing problem into something like a graph problem so now you're thinking of points connecting centers and you've got edges and they somebody was able to show it boils down to a finite number of cases and then looking through every case and see what happens and a lot of problems in this type of combinatorial analysis you are breaking things down and showing there's just a finite number of cases and sometimes it's a huge number of cases and then you have to employ a computer to go through and check each of those but I'm sorry again because the particulars of this I can't give you an exact answer alright if anyone has any more questions Mark will be available where the books are so let's give Mark Chamberlain our speaker one last thank you