Gresham College Lectures

Mathematical Puzzles and Paradoxes - Sarah Hart

February 12, 2024 Gresham College
Gresham College Lectures
Mathematical Puzzles and Paradoxes - Sarah Hart
Show Notes Transcript

Many puzzles have a long history, such as water pouring puzzles, where you need to measure (for example) one pint of water equipped only with an eight-pint and a five-pint jug. The mathematics behind the solution has many useful applications.

Meanwhile, paradoxes such as: “some men shave themselves; those that do not shave themselves are shaved by the barber: who shaves the barber?” lead us to deep questions about set theory.

We will discuss several examples and the related mathematics.


This lecture was recorded by Sarah Hart on 30th January 2024 at Barnard's Inn Hall, London

The transcript and downloadable versions of the lecture are available from the Gresham College website:
https://www.gresham.ac.uk/watch-now/maths-puzzles

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So today we are going to look at puzzles, riddles, conundrums and paradoxes, and how they can lead to fascinating mathematical ideas and discoveries. We as humans are inherently curious and playful, so it's no surprise that puzzles date back as far as recorded history can take us. And probably further, we just love entertaining our minds with a little conundrum. So some of these puzzles I'm gonna talk about today, date back hundreds, even thousands of years, and we'll see that they can morph and change as they cross from different languages to one another. They're moving over, uh, uh, through cultures across time. We can see little changes creeping into some of these puzzles. It's quite interesting to observe. Let's start with a puzzle that dates back quite a while, and it's called the 17 Camels Problem. The Gresham College lecture that you're listening to right now is giving you knowledge and insight from one of the world's leading academic experts, making it takes a lot of time, but because we want to encourage a love of learning, we think it's well worth it. We never make you pay for lectures, although donations are needed. All we ask in return is this. Send a link to this lecture to someone you think would benefit. And if you haven't already, click the follow or subscribe button from wherever you are listening right now. Now, let's get back to the lecture. So the story is, there's lots of variants of this. It's not always camels, but the, this is the basic idea. So a father in his will leaves provision, um, for his, his herd of camels to be left to his sons. So a bit unfairly, half the herd goes to the eldest son. There are three sons. One third of the herd goes to the next son, and the youngest gets one ninth of the herd. Okay? So after the father dies, the sons gather together to divide up this herd of camels. And there's a problem. There are only 17 camels in the herd, and if you try and divide 17, uh, into half even, um, you hit a problem and half a camel isn't much good for anyone. So what are they going to do? They're standing around perplexed, and then a traveler passes by and says, I can help you with this. Uh, I'm gonna lend you my camel. Right? So now there are 18 camels and everything's easy 'cause 18 is divisible by all of these things. So now the eldest son, half of 18 is nine, so that's his bit. And the next one, he gets a third of the herd. So a third of 18 is six, so he's happy, he's got his six camels. And then the youngest son, he gets one ninth of the herd, and one ninth of 18 is two. So he gets two camels, they're all happy. That just leaves the traveler's camel leftover, and he gets back on it and off he goes. Problem solved. But of course, the, this is a bit confusing because we moneyed we couldn't solve the puzzle. Uh, we couldn't divide up the herd with 17 camels. Then the 18th camel allowed us to do it, but we didn't use the 18th camel. So what on earth is going on? And, and this kind of problem, problems like this where you are trying to divide up, um, some amount into, into a, a collection of fractions, um, in various guises go back a long, long time. Uh, the expert on recreational mathematics, David Singh master, who's most famous actually for bringing the Rubik's cube to Britain, uh, 1980s. Uh, he has analyzed this, this problem and, and tracked it back through history, and it dates back in, in, with these fractions about 600 years, but with other fractions much, much further, even almost 4,000 years. So what's going on behind this, behind this puzzle? Well, you are hoodwinked right at the beginning because if you actually look at those fractions and add them up, you find that they don't add up to one. You can't divide, um, an amount into a a half and a third and a ninth. Uh, exactly. There's some leftover, the sum of these fractions is in fact 17 eighteenths. So that's why we could divide up in this way using 17 camels out of 18. But each brother actually gets a little bit more than the amount to which he's entitled, right? So the first brother should really get eight and a half camels, but he gets nine. So of course they're happy. They've all got a bit more than than the will provided for, and the traveler can go away on, on his way with his camel. So this kind of question about how do we divide things up into fractions that don't add up to one? I mean, that potentially is a, is a challenging thing to deal with. And if you go back far enough, this is really, you know, a problem that has to be addressed if you've got to divide things up in certain ways. So seeing the answer to this problem, and even the fact that it is being asked as a puzzle or a little conundrum, tells you something about the mathematical knowledge that is, that is there in that society at that time. And that's one of the great things about, uh, about mathematical puzzles, just like fairytales or folk tales or, you know, creation myths. We can see stories like this moving across different cultures and getting changed and developing, but if it's a mathematical puzzle, we get the extra thing. We are not only seeing knowledge being transferred, we're also seeing what the mathematics is at the time and how the, and how the mathematics of the time deals with the questions like these. Um, so for example, if you had a, a puzzle whose solution requires you to do something involving, you know, a quadratic expression, something with X squares in it, then that tells you that, um, at that time and in that place there was some knowledge of something like quadratic expressions. And that goes back quite a long time. Uh, for example, there's a, there's a puzzle on this Babylonian, I was gonna say tablet, I don't know if that's the right word for it, prism or something. Um, that dates back as you can see, nearly 4,000 years. And the puzzle is about finding the dimensions of a mysterious rectangle. And so you are told some things, you are told that to the area of this rectangle, um, was added the excess of length over width. So in other words, that's area plus length minus width equals 183. And you're also told that length plus width is 27, and you have to then find the length, the width, and the area of that rectangle. Now you can solve this puzzle if you like, on your way home. Um, if you do, you are gonna find that you end up probably with a quadratic equation to solve, uh, at some point. Babylonian mathematics did not have quadratic equations in the way we think of them now, but they had an algorithm that would solve problems like this. So in some sense, they could solve certain kinds of quadratic equation. That's very interesting to know. Sometimes we find that when puzzles move between different, uh, different cultures and different traditions, they can change in subtle ways that actually affect the solution to the problem. So I'm gonna tell you a, a puzzle. This is a modern version of it. Uh, it's the one that my dad told me when I was eight years old and I got it wrong. So, you know, this is now my rehabilitation finally complete. Um, the snail in the well might have seen a puzzle a bit like this. So a snail, uh, Paul thinks at the bottom of a, well, two meters deep, and starting first thing Monday morning, sunrise on Monday, it starts to climb up very slowly and it climbs up half a meter during each day. But then at night it kind of slides back down again, one fifth of a meter. And the question is, um, what day is it gonna escape from this? Well, so, okay, first thing, Monday, um, it's at the bottom and then it is going up half a meter, but then it slides back down a fifth of a meter. So kind of the net progress each day from sunrise one day to sunrise. The next is a half minus a fifth, which is three tenths. Okay? So, so on Tuesday morning, first thing it'll be n 0.3 meters above the bottom, and then it's gonna make n 0.3 meters progress in every 24 hour period. So we can mark off those point threes and Wednesday, Thursday, Friday, Saturday, Sunday, uh, but look by Sunday, 1.8 meters up. And so it's only got 20 centimeters left to climb. So it's gonna do that at some point on Sunday, right? So it escapes sort of Sunday teeter, okay? No, it doesn't. That is, that is the mistake little Sarah made, and that's okay because I learned something that day.<laugh>, I learned that 30 or 40 years, hence I would be able to get it right in front of an audience. Um, so rewind, rewind, Saturday morning, 1.5 meters, that's where it's, but during the day on Saturday it goes up, it can climb half a meter. So it actually escapes right at the end of the day on Saturday. It doesn't get to the point of sliding back at night because it's already out by then. So actually the answer is that the snail will escape from this well, last thing on Saturday. But the reason I'm telling you this puzzle is that it started off or puzzles a bit like this began, uh, life probably around about the sixth century in India. And then they traveled, you know, along trade routes with other knowledge that was being transmitted. They got translated into different languages and they made their way to Italy, uh, in maybe the 13th century ish. And in the Italian books that had this puzzle in, or puzzles like it, the solution was for the first few decades wrong, it was the Sunday tea time solution. But there, there's a reason for this. It's not that they were stupid, there's the, the reason is that something's lost in translation. The original puzzles of this form had this a half minus of fifth or something like that. But that's because it was a common way to express more complicated fractions. You write them in terms of simpler unit fractions with a one on the top. So actually originally the puzzle puzzles like this would say something like the snail or whatever animal climbs a half minus a fifth meters per day, not meters, but whatever unit was it at the time, right? And then this is all happening in one go, so you don't have this progressing and then separately falling back. So actually that it's simply, it climbs three tenths of a meter every day. And so you don't have this sort of end special case at the end. So those puzzles, as they originally were, the solution was the Sunday tea time solution when these puzzles were translated. Um, and the pe people who were doing the translations and copying out the, the puzzles, they saw this a half minus a fifth type expression and kind of thought, oh, it must be because first it goes and then it comes back. And therefore they started to say, that doesn't seem like much of a change, but it totally changes the puzzle. So it's very interesting to see, yeah, actually it took, you know, decades for the solutions to catch up with the puzzle. So maybe if you're doing, if you find an old puzzle book and you dunno why the solution given is the correct one, maybe it isn't <laugh>, it's always, always worth like keep it, keep your hope alive. Uh, okay, so we can see knowledge moving and transmitting and sometimes changing. Let's look at another old puzzle. Uh, this diagram is a very misleading in terms of how you actually shove to solve this puzzle. Um, so this is one of a, a, a whole collection of puzzles around this kind of theme that are often called decanting problems, where you have, uh, one or more jugs, uh, of fixed volumes and you have to create a different volume of, of water or sometimes wine. So in this case, you have a, a jug that measures exactly five liters and another one that measures exactly three liters, and there's no markings or anything on them. Uh, but you have to use those jugs to create a volume of exactly one liter. Okay? So, so you've gotta somehow work out how to do this now, solving this problem and problems a bit like it, um, turn out that to have important applications in kind of the real world, not in puzzle world. Um, obviously the most important one, uh, was discovered by noted mathematician. John McClain in the film Die Hard with vengeance <laugh>, which I'm sure you remember being highly mathematical. So in, I'm being a bit silly, but in this film, uh, Bruce Willis and Samuel L. Jackson are somehow pitted against this mean, uh, bad guy, Jeremy Irons, who sets them these problems to do against the clock or, you know, and if they don't solve them, then everything explodes or something. Um, and Bruce Willis noted, mathematician manages to solve this puzzle, <laugh> and you know, save the day. So I wonder how he solved it. Maybe he, he did what I'm about to show you, um, just a bit of terminology. I'll call this the five three, uh, Walter Jug problem. Uh, but actually as mathematicians of course, we immediately want to know can we solve this? But can we solve all the problems like it, can we solve the XY jug problem where the volumes of the, the jugs that we have are X and y, we're still trying to make one liter. So how do we approach a question like this? Something we could trial and error will get us some sort of distance. Um, we've got our five liter jug and our three liter jug. We've gotta fill and empty them in some way to end up with one liter. So the crucial observation is it, this is the same as finding some combination of fives and threes that gives you one, just the numbers five and three. So for instance, uh, two times three take away one times five is one six minus five is one. Okay? Now, why should that help us with the problem? Well, you can turn an expression like this into a, a rule for filling and emptying the jugs. You've got a positive multiple of three. So that's going to mean filling the three liter jug two times and you've got minus, uh, one outta five. That means we're gonna empty the five liter jug once. So minus sign means empty, plus sign means fill. So let's do this. So we're gonna fill the three liter jug for the first time. What are we gonna do with that water? We'll just pour it into the five liter jug. Um, then we fill three to jug again and pour as much as we can into the five liter jug. Um, there's room for two more liters, so we'll pour those in. Now the five liter jug is full, we'll empty it, aha, wonderfully. We have exactly one liter left in the three liter jug. Okay? And you see we've used plus two, lots of three and minus one lots of five. So you can translate this, you know, expression just in terms of numbers into a solution to the problem. Uh, we can do it from a slightly bigger problem version. So this is the seven five jug problem. So first step we have to try and find an expression, a combination of sevens and fives that make one. If you look around for a bit, you might notice that three lots of five take away two lots of seven is one. Great. So let's use our little process. So we are adding five, so that means we are filling the five liter jug. So we fill it, pour it into the seven liter jug, fill the five liter jug. Again, there's room for two liters in the top there so that they'll fit in there. Now the seven liter jug is full, let's empty it. And now we can keep going and fill the five eth jug one more time. Now look, there's room for four liters left in that seven liter jug. So we'll pour those in. Now it's full empty it and we're done hurrah. So solve the problem. So any, any bad guys who want to make us solve these things at speed, we now have a process that will allow us to do it if we can find expressions like this. So how do we find expression like this? Do we, how can we be sure there even is one? Um, I mean there might be many. It will work for any expression you like. So three times seven minus four times five is one as well. That would give us a solution involving filling the seven liter jug and emptying the five liter jug. But how do we know there's even one of these? Well, what we've said so far is that if we want to be able to do these problems in general, we, it's the same as finding an expression like this. So au minus bv, it's one, uh, U and V are the, the volumes of the jugs in some order. I'm not going to care which one's bigger or small at this point. I just wanna find some expression like this. And, and A and B are are whole numbers. So if we can do that, we are happy. There's one instance where we definitely are never going to be able to do that. This, we are never gonna be able to find an expression like this. And that's when the jug volumes, um, have a common factor bigger than one. So if for example, they're both U and V, if they're both uh, divisible by three say then the whole of the left hand side of that expression that a U minus bv, it'll also be divisible by three 'cause U as a factor of three and so does V. So that combination will, but that's bad news if you want the answer to be one, 'cause one definitely does not divisible by three. So we know if, if UNV have a common factor, common divide are bigger than one, we are in trouble. So that means if their greatest common advisor is anything other than one, we can't do this, we absolutely, definitely cannot solve this problem. So it's impossible in that situation. Well, okay, what about if they are co prime, if they greatest common divides really is one, can we do it there? So then the question is, is it enough if they're co prime these numbers? So like five and seven or three and five, can we then find an expression like this? And the answer is brilliantly that we can, we always can. We're guaranteed to be able to do it and there's a way that we can do it, a simple process that allows us to find these expressions. And it comes from something called the Euclidean algorithm, as in Euclid who uh, devised it a couple of thousand years ago, but it's still very useful. So actually this algorithm is really for finding the greatest common divisor of numbers. Now, the numbers we are looking at, you can instantly spot what the greatest common DI advisor is.'cause you can just factorized them and know what the factors are. Um, so we'll just do it. An example with, uh, 35 and 11, obviously they're co prime, so their greatest common DI advisor is one, but we're still gonna go through the little calculation. It's very quick because then we're going to do something with that process to allow us to get this, this equation that solves the water jug problem. So what you do, you do the same thing repeatedly just yeah, um, you take the bigger number, divide it by the smaller one. So we've got 35 and 11 and you'll get a quotient and you'll get a remainder. So 35, 3 times 11 plus two, fine. Now there's something interesting we can say. Now this 35, we've written it in terms of 11 and two, any number that divides any factor of 11 and two. So a common factor of those two things divides the whole right hand side of this and therefore divides 35. So it's also a common factor of 35 and 11. And the other thing is true, the other way round is true as well. Anything that divides both 35 and 11 also must divide two. So what that tells you is, um, after a bit of thought, it tells you that these pairs 35 and 11 and 11 and two have the same greatest common divider. Anything that divides both 35 and 11 must also divide two. Anything that divides both 11 and two must also divide 35. And that gives you the solution. So what do we do? Well, we, we are gonna carry on. So we take the 11 and two, we've made ourselves a smaller problem, and that's something we often do in mathematics. We take hard problem and reduce it to a simpler one. So we now have a simpler challenge. Can we find the greatest common advisor of 11 and two? Well, yes, of course we can, but let's do this little process. We do the same thing. Divide the bigger number by the smaller 11 is five times two plus one. And now we can say the same, uh, thing about these three numbers here, that the greatest common DI advisor of 11 and two is equal to the greatest common DI advisor of two and one. And at this point, we really can be absolutely sure that that is one because one, the o the only divisors of one. Uh, well it's just one if you're talking about positive divisors. And so any divisors of the greatest common di advisor of one with anything is always gonna be one. So, okay, well we already knew the greatest common DI advisor of 35 and 11. That wasn't, that isn't why we're doing this. But we've obtained now two uh, equations that are gonna be very useful to us because what we do is we're gonna rearrange them. So those two equations there, 35 equals something 11 equals something, we're gonna rearrange them. So 11 equals five times two plus one. We are rearranging it to be one is now 11 minus five times two. And then the top one, we've rearranged that to make two the subject. So two is 35 minus three times 11. And the reason we are doing that because we want to find an expression that looks like something times 35 and something times 11 equals one. So we take that first one there, one is equal to something and it's equal to 11 times 11 minus five times two. We don't really want a two. We want instead to involve 30 fives and elevens. So we've got elevens and twos in that expression. I'm gonna just replace that two with this expression for it that I've got on the right hand side. Okay? So if you do that, you've now got an expression for one, which is just in terms of 30 fives and elevens. That's what we're after. And if you just gather up your terms, there you go, we've got that now 16 times 11 minus five times 35 equals one. And that's right because 16 times 11 is 176 and five times 35 is a hundred thirty five, a hundred and seventy five. So we are happy, uh, we've got this expression that allows us to solve the, I'm sure we all wanted to solve it, the 35 11 jug problem. Okay? So now hopefully that little example shows you there is this technique, um, which allows you to solve these problems. So we can solve any XY water jug problem as long as X and Y are Cori, that their greatest common advisor is one. Now this is all very nice if we are presented suddenly with emergency drug, uh, solving <laugh> conundrums. But this algorithm, this process to find combinations of numbers that equal one turns out to be absolutely vital in something many of us do almost every day, if not every day, which is online transactions because it's a crucial component of the so-called RSA encryption algorithm. And this is used a lot online for kind of, uh, banking securities that prevent your details being stolen. So if you wanna send, uh, some message, you know, please transfer X amount of money from my account to your bank. You don't want anyone, anyone intercepting your bank details. So of course that information is encrypted, and I won't go into the details of the algorithm of how this is done, but to encode, encrypt your message, um, you use a number called the public key to do something to your message that encrypts it mathematically. And then you don't want just anyone, anyone ought to be able to encrypt and send messages, but only the person to whom the message is really sent, uh, should be able to decrypt and, and work out what the message actually says. So at a crucial moment in the process, in order to do this decryption, what needs to happen is this Euclidean algorithm process has to be applied, um, for two numbers, K and x. Now, K is this public encryption key that you know, and everyone can know it's public, but this number X is not, that is a private key. Only the bank or whoever it is, will will have that. And so only they will be able to apply this. You plead in algorithm process to decre the message. So this little cute little algorithm that helps you solve the water jog problem actually also helps, uh, online banking stay secure. But this is very nice. It links to a nice little puzzle. However, the, this algorithm was not developed in order to solve the puzzle. It helps to solve it, but wasn't, it was not developed in order to solve it. So you might say, well, okay then why should we be interested in what's called recreational mathematics? Um, it's just a bit of fun. Well, that's actually my answer. It's a bit of fun, <laugh>. What's wrong with that? Right? Actually, for a mathematician, the word recreational here is completely redundant.'cause what do, what's recreational? It's something that's fun. We do 'cause we enjoy it, and that's what mathematics is. So, okay, it is mathematics and we don't need that word recreational, but if you insist, if you insist, there are plenty of examples of puzzles that have given rise to genuinely new and exciting mathematics. Um, the most famous of these, which I feel sort of, there's a bylaw of mathematics that you cannot give a lecture involving recreational mathematics without mentioning it, is the famous, uh, bridges of burg problem. So I'll just briefly mention this, um, the situation was there's a real life, uh, town burg, uh, and there's a river through the center kind of splits and joins back together again. And, uh, there were, there aren't anymore, but there were seven bridges, uh, at various places in this river. And the townsfolk had this little pastime on the weekend. They'd go for a walk around the town and they wanted an interesting fun walk. So the question was, is it possible to design a route that will cross each of these bridges but not go back on itself? So you don't wanna go over the same bridge twice, so you want to cross every bridge once and only once is this possible to do. And it seemed somehow it didn't seem possible, but they couldn't quite work out really. Is it impossible or are we just not clever enough to think of the answer? Um, or, or, you know, can it be done? So enter the great mathematician, Leonard Oiler, who we, we have mentioned a few times in in Gresham lectures, uh, because he was, uh, just an amazing genius and he was involved in many, many developments in mathematics. If you came to my lecture on Sudoku puzzles, we talked about him briefly then. Uh, so he came along and he did what mathematics, um, does very well, which is to get to actually the real structure hidden underneath this problem about, you know, towns and rivers. He said, what's really going on here, and this is, uh, his diagram is you've got four regions of land, A, B, C, and D. There they are. And connecting them are these bridges. So each bridge gives you a path between two regions. So we can draw those in, uh, that's where those bridges are. And now in fact, that's the information we need to deal, deal with this problem. That's all the information that is required to solve the problem. Um, so we can throw away everything else. We get that and we can make it look a bit pretty. Uh, so now it looks like this and that gives, that is the essence of the problem right there. We are now asking, translate that, that the problem into a problem involving this, this network. Here we are asking, is there a way you can walk around this, this network or graph where you go along each edge exactly once those edges correspond to the bridges. And once you've got it in this format, um, it's becomes easier to see how we might proceed. And all it came up with this brilliant argument to prove that it is impossible to do this. Uh, so what you have to do, if you're walking around this, this graph from the nodes traveling along, along the edges to each node or vertex, every time you come into a node and go out of it, you are using up two edges, an in edge and an out edge. So you can sort of cross those off. And except for possibly the starting point where you're just leaving and the finishing point where you're just arriving, that must mean that for all the other points, the edges can be paired off with each other, an in edge, in an outage, an in edge, in an outage. So you've taken 'em away in pairs. And that means that apart from possibly the start and the end points, all the other vertices or nodes have to have an even number of edges join to them.'cause half of them are in edges and half of them are out edges. But in this picture, unfortunately, you can see that none of them have an even number of edges. Uh, so most of 'em have three edges, and one of them, that one up there has five edges. So you cannot do it. It's impossible to traverse all the bridges of kosberg exactly once. And converting the problem into this really simple graph actually was the way to see the solution. But graphs like this are incredibly useful and important in many, many areas of life. The map of the London underground or, um, the, the connections between webpages online, you know, Google's search engine relies on this, uh, kind of concept of a graph like this where things are joined together, uh, pages are joined together, if there's a link from one to the other. So this oiler oiler idea actually kickstarted a whole new area of mathematics called graph theory, uh, which is incredibly important and useful today, but it came from trying to solve a puzzle. Uh, I'll mention another little puzzle, which, because I like it. Uh, this is, uh, called the Puzzle of the Birds. And the example I'm giving is from a book by Fibonacci, but actually it dates back probably a thousand years before, uh, this, um, puzzles of this form probably originated in China, um, in maybe 500 ad this is about constraints in a problem. So you've got 30 birds, you buy 30 birds for 30 pennies in the market, and there are three kinds of birds, and you're given the prices. And then the question is, how many of each bird did I buy? So in order to solve this, you have to take that information and use the constraints that you've got, 30 birds, 30 pennies. And to do that, what you end up doing is something that we now call linear programming or linear optimization that again, has loads and loads of applications in terms of, you know, manufacturing or transport links. Um, and it's, you know, a whole area of mathematics on its own, but you can also solve, these are other charming, uh, bird problems with it. Okay, the final example of problems or puzzles that have very interesting mathematical ramifications, uh, and kind of lead to new mathematics is our segue into paradoxes. So what's the most famous paradox ever? Um, well, I don't know, but <laugh>, but zeno's paradoxes are probably up there. The list. And the most famous of those is probably Achilles and the tortoise. Actually, I'm, I'm not sure this is Achilles. I think it might be Mercury this in this, but let's pretend it's Achilles. No one will tell me. Uh, so somebody who can move very fast is in a race against a tortoise. Of course we know they can beat the tortoise, okay, we'll give the tortoise a head start. But even if we give the tortoise a head start, you know, Achilles is so fast, he's gonna catch up to the tortoise only there's a problem. And the problem is that there's achilles, there's a tortoise. So he runs along very fast where the tortoise has been, but in that time, the tortoise will have moved on a little bit, okay, fine. So he's just gonna now run to that point, but then in that little time, the tortoise has moved on again and then so he catches up, but then the tortoise still moved, however small the distance is that he has to cover to catch the tortoise in the time it takes him to cover it, the tortoise has moved on still a little tiny bit so he can never catch the tortoise, which is sort of paradoxical because we all know he, he can catch the tortoise. So what is going on? And a, a similar problem. So I think there were nine paradoxes of Xeno, um, similar flavor. So a similar one, uh, is about lanter. Um, so she, again, very, very fast runner, uh, but she can't ever get between two points. Why? Because to get from A to B first she has to get to the midpoint, uh, and then she has to get to, to the midpoint of those two, and then half again and half again and half again. Um, but each one of those, we know she can do it very quickly, but it's still a thing to do. She still has to cover that distance. So there's an infinite number of small distances that she has to cover. So anything you do infinitely, an infinite number of however small time it takes you, it's surely gonna take infinitely long. So she can't get anywhere. Um, both of these problems really boil down to the, the question of how do we add an infinite number of things together? How do we sum that infinite sequence? Uh, now we can sort of think, well, a half plus a quarter plus an eighth, plus 16th, we.dot. That will eventually the limiting case of that, if we did do it to infinity, we kind of get one, um, and we can convince ourselves that might be true, but it's still a bit, maybe feels a bit dodgy, right? We really, how do we really, really know? We're adding these infinitely many things and you know, how can we be sure that it really is going to stop at some point? For example, you could say, well, these numbers are getting smaller and smaller, so it's gonna be fine. They'll stop at some finite place, but these numbers are also getting smaller and smaller, a half a third, a quarter, a fifth, and so on. Uh, if you try to add all of those up, it can actually be shown that however big of a number you can think of, eventually the sum of this will overtake it. So this goes to infinity, this second one, and the top one doesn't, it only stops at one. It's sort of limited above by one, but discerning which is which and how we can really be sure that took a long, long, long time. It took until really the, uh, 18th, 19th century to, to get that sorted out. Um, well beyond Zeno's time, let's mention just one more paradox of Xeno, the arrow paradox. So this is an arrow flying through the air very fast, but if you actually try and look at its motion and how it can move, um, if we just freeze frame at any point, it's still, it's motionless, just fix time at an instant. At that instant, the arrow isn't moving. So how do we then put all the, what is time? It's an infinite number of instance, but at each instant there's no motion. So how on earth can there be movement? And then, you know, your brain sort of melts and dribbles out of your ears.<laugh>, I think, how do we solve this? Uh, if we think, how would we really think work out how far say the arrow had gone? Um, if it's going at a constant speed, well, you could draw, you could draw a graph of speed against time. Um, constant speed there. I've just randomly said it's going at six meters per second. How far has it traveled in 10 seconds? Well, speed is distance over time. So distance is speed, time is time. So you take that really what you want is the area under the graph then in that, in that, uh, ten second time interval and the distance travel will be six times 10, 60 meters fine. But what do we do if the speed is not constant? What if the speed is like this? Then how are we gonna find the area under that graph? Well, you can kind of, you can make approximations, you can say I'll divide it up into, into rectangles and that, you know, we'll say, we'll sort of pretend the speed is constant over each of these little rectangles and then add up all of those areas over the width times the height. Fine, that's not very accurate. So let's make the rectangles a bit thinner. Um, and we can do that. We'll get a better estimate. And as we make the rectangles thinner and thinner, the estimate gets better and better. But in the limit, the width of those rectangles is zero. And, and so the area of each rectangle will be zero. So then they're somehow supposed to add up to make something that isn't zero, that's a bit confusing. And dealing with these infinite decimals, right? These, these tiny quantities that close, closer and closer to zero is exactly what, uh, Newton and liveness still at the calculus to cope with. But that again, took many, many centuries after Z know. Uh, and really it wasn't until then that we could say these paradoxes, we sort of understood what was going on. It's the difference between the discreet and the continuous and the mathematics of the continuous is what calculus is so good at. Okay? So for the rest of the time today, we are gonna think about some more paradoxes, but they are logic paradoxes. Now, this is in honor of all those times I sat an exam when I was at school and I got very cross because there would be a page and it would say on it, this page is left intentionally blank. Remember that? I used to think I should write a very sniffy letter to the examiner saying that due to extreme emotional distress caused by malfunctioning logic in your exam, I was unable to complete it and you should give me an a star immediately. Um, you know, this kind of sentence, it, it, it, it has an antecedent that go back a long way. Um, it refers to itself and that's, that's really the essence of what's going on. Um, the classical version of, of something like this is the, so-called liar paradox. So epi amenities says all credences are liars. Um, okay, well it's not very nice of him, but you know, no paradox seemingly until you know that epi amenities came from knossos and knossos is on Crete. So epi amenities, the Creon is saying that all Creon are liars. So if he's telling the truth, so then it would be right that true that all cretins are liars, but then he himself as a creon would be lying. So if this statement is true, then it's also false. So that's a bit confusing. However, this isn't really a paradox because the statement could just be false. He could lie being a cretin. He could lie and say that all cretins are liars. That makes him a liar. But it doesn't mean that all cretins are liars. Now, the, the opposite of all cretins are liars is, you know, at least one cretin tells the truth at least some of the time, something like this. So it's not absolutely a paradox, but we can say this, we are pretty sure this is a false statement.'cause if it is true, that would imply its own falsehood. But try this one. Is this sentence true or false? It asserts that it is false, right? Is that a true statement? Suppose it's true. Um, then it would be true that it's false, so it'd be false. Okay? On the other hand, suppose it's false. Um, so it's false to say this sentence is false. Um, but then the sentence is actually true because it's saying that it's false, which it, it's, so, if it's true, then it's false. And if it's false, then it's true. Um, the third way out of this is to notice, I forgot to put full stop at the end. So it isn't even a sentence <laugh>, but you know, there, that's, that's cheating. Um, there are lots of ways to think about this and people have had various theories or, or ideas almost philosophically about how to deal with a statement like this. Um, I quite, so you could say it's both, it's true and it's false, maybe I'm not sure. I like that because I think especially if we want to do mathematics, we don't want to have statements that are ambiguous in that way. I mean, it's one thing not to be able to prove something, but if you can prove it's true and then the next day your friend comes along with their equally valid proof that it's false, then you know, it's very difficult to imagine mathematics being put at a firm foundation. We'd like, we'd like phagocyte either to be true or to be false. Um, not both, and not neither. So things being both true and false at the same time, we perhaps want to avoid, uh, there was, uh, one idea put forward by the logician, uh, Arthur Pryor. And he thinks you can get round it by saying or noticing that every sentence, you know, if we're just talking normally, every sentence that we say implicitly asserts its own truth. So we are saying something and that kind of implies that we're saying, and it's true. So if you say the sky is blue, that is logically equivalent to says Arthur Pryor saying this sentence is true and the sky is blue. Uh, if we agree with that, then the sentence that we have here, this sentence is false, is logically equivalent to this sentence is true and it is false, and that is isn't possible because we believe things can't be both true and false at the same time. So if you look at it that way, then you can say what it, it's false because something can't be both true and false at the same time. So I think this is one we don't have to absolutely decide on. We can argue about it and discuss it. Um, what we think about this sentence, as long as we don't try and have things like this at the heart of, you know, at the mathematics probably to be avoided, um, since I enjoy finding mathematics everywhere, and especially I enjoy finding it in, uh, literature, I wanted to just mention a paradox a bit like this called of liar paradox or self-referential paradox that crops up in, uh, donte. So, uh, here they are, there's donte and there's his loyal faithful Sancho Panza, uh, beside him. And at one point in the, in the story, uh, some, some not very nice people are kind of teasing Sancho panza and they're saying, we're gonna make you the governor of this province and you are gonna have to make all these decisions. So just to make sure you can cope with the responsibility, what should we do in this situation? So there's a town where, you know, there's a bridge and you have to, to get into the town, uh, you are questioned by the, by the guards, and you have to say what you're going to do in the town that day. So if you speak the truth, fine, off you go, it's good, you can get enter. But if you lie, then you'll be hanged, you know, in the town square. So one day a traveler arrives and says, and they say, what are you gonna do in the town? And he says, I will be hanged. So should you hang him? Um, if you do hang him, then actually he was telling the truth, so you should have let him go. But if you don't hang him, then his assertion that he would be hanged well's false. So what he actually, he should be hanged. So what are you gonna do? Um, well, if you want to know what Sancho Panza decided, you can either read the book or you can ask me at the end, <laugh>. Alright, onto another, another sentence. So we've dealt with this sentence is false, sort of what about this? This is not, this is at the moment, this is fine, right? This sentence has seven words. Well, it doesn't, okay, fine. So it's a false statement, nothing to see here. Um, well, but we know if something is false or we believe if, if, if a statement is false, then it kind of, it's opposite statement. The negation ought to be true, right? If something's false, the opposite is true. If something's true, the opposite is false. Okay? So, um, this sentence does not have seven words, right? Only unfortunately, uh, it does. So that is also a false statement, <laugh>, right? So how, what we've got a a statement and its negation both being false, this ought to not be possible. And so this kind of problem is really good for helping our brains think very clearly about what do we mean by negation and true and false. What's really going on, I think here is that we haven't got the correct negation. This sentence at the top, this sentence has seven words. It's talking about its self. If we say that that's sentence A for example, uh, let's call that sentence a then the top sentence A is asserting that itself. IE sentence A has seven words. So the opposite of that, the negation of that is not, this sentence does not have seven words. That's a different sentence. The negation is that sentence. Sentence A has seven words, right? That deals with it because it's true that sentence A does not have seven words. So we have to be really careful with self-referential statements. They're talking about themselves. But then the negation is a different statement that it should be talking about that over there. So it's quite a fun little thing. Takes a while to reason out what's gone wrong in that situation. Um, so this can be resolved by being really careful about the meanings of words, kind of semantics, um, thinking not just about the statements, but what the statements are saying and being sure we can distinguish between those two things. But sometimes, uh, these kind of paradoxical self-referential statements can really have a, a genuine deep effect on, on mathematical thought. And one instance of this, it's not quite this, but it's a, a statement very like this one. And actually, um, mathematician Bertrand Russell and philosopher, um, used this example to illustrate a mathematical argument that he, he had made, which we'll mention in a moment. So it's the barber paradox and it says in a certain town there's just one barber and he shaves only those men who don't shave themselves. So, um, you shave yourself, okay? If you don't, then the barber will shave you question, who shaves the barber? Okay? So if the barber shaves himself, um, then he's one of the men who do shave themselves. But the barber only shaves the men who don't shave themselves. So the barber definitely does not shave the barber. Okay? Oh dear, all, let's suppose the barber doesn't shave himself. Well then we know that the barber precisely shaves the men who don't shave themselves. So then the barber does shave the barber. So do you have a way you start from, you get into this kind of, um, doom spiral of logic from which there is no escape. If he does shave himself, then he doesn't, and he doesn't, then he does. So we dunno what to do with this quandary and actually what you do, I mean, there's, so there's ways around it to like try and sort of wriggle out of it semantically and say, well, perhaps the barber, you know, is bald or something, but <laugh>, let's not do that. Let's take it in spirit in which it's intended. The only conclusion really is to say there can be no such talent. This cannot happen.'cause if it did, we'd get into this logical, um, vicious cycle. However, unfortunately, well, or interestingly, there was a serious bit of mathematical research, um, that came up against something very like the Bible paradox in the late 19th, early 20th century. Mathematicians were really trying very hard to put mathematics on a very, very firm foundation, a rigorous logical footing, and do a bit like what Euclid had had done with geometry. You know, you start off with ground rules. You say what points and lines are you say, um, some, some basic facts about them, the axioms that, uh, you know, I dunno, you can, you can draw a circle from any point of any radius, for instance. Um, these basic rules can then be combined and you can make deductions from them. And then you create the whole beautiful edifice of geometry. Well, it, wouldn't it be nice if we could do that for all of mathematics? And in particular, um, there was a guy called Got Love frager who tried to do this and come up with a theory that would cover all of, um, everything you could do with sets. So if you take, you know, sets of numbers or sets of anything you like, you could combine them. You can make the union of two sets. You can make the intersection. Uh, you can look at subsets, you can define subsets according to various rules. So you might think, you know, the prime numbers are a subset of all numbers. You can have sets, offsets. Uh, you might look at all the possible subsets of a particular set, all the possible subsets of numbers or, or all the possible potential friend groups in a <laugh> in a class or something like this. So he frager, um, wrote a huge book actually of two volumes about his idea of how to do this really rigorous, uh, theory of sets. Um, but unfortunately the kind of sets he allowed, made way for a problem, and this is called now Russell's Paradox. It's basically the barber paradox. Um, but in the form of sets. So if you define this set, but a strange set, let's call it s it's the set of all sets that are not elements of themselves. So you can have a set of sets. So then you can think of the concept of a set that's actually a member of itself. This set S is all of those sets that are not members of themselves. Now is s an element of s and this is just like the Barbara paradox. If S is inside s then the entry criterion for being in s is that you are not a member of yourself. So that doesn't work because if it's in, if it's a member of itself, that would imply it's not a member of itself. Bad. Okay? What about if s is not inside s Well then it fits the bill to be a member of S because s is all the things that are not inside themselves. So either way we get this. If it's in itself, then it isn't. And if it isn't in itself, then it is round and round and round and round forever till we fall over dizzy. But unfortunately, uh, Rieger's lovely setup allowed this for this set. It, it said this set exists, not in so many words, but it sort of implied by the work he'd done. And so Ho and Russell wrote to Frago about this and the poor guy, he had to horribly put an appendix in his book. Um, it was just kinda at the printers and he wrote hardly anything. I feel so bad for him. More unfortunate can fall a scientific writer than to have one of the foundations of his edifice shaken after the work is finished. This was the position I was placed in by a letter of Mr. Bert and Russell, just when the printing of this volume was nearing its completion. So yeah, it like it all just fell apart because there was a paradox. Something that couldn't work was at the heart of, you know, it it implied by his setup. So that was obviously he was devastated, of course. Um, but the problems for this kind of great idea of putting mathematics on a, on a, on a ideal footing, where there'd be one wonderful system that within which you could express every mathematical idea, and that if you just worked long enough, you could prove anything that was true and there wouldn't be any contradictions. That itself, and that was something that Bert and Russell was involved in as well, was ultimately doomed. Um, that worry, there is good news at the end, <laugh> doom is, is featured a bit too much. So the question was, can we come up with some really kind of great system of, of mathematical logic within which in theory, I mean they didn't have computers at this time, but in theory you could start off with the basic rules, the axioms, and you could say, what are the rules of inference? What kind of rules of logic are you allowed to use? And you plug it all in and you kind of crank the handle. And eventually, um, every kind of statement that you could make about numbers, you get either a proof that it's true or proof that the opposite is true. So you know that it was false. So that's what would be great to do. Um, so what we want from this, the three Cs, what we want from this, this putative wonderful system that's gonna solve all our problems and we want it to be consistent. So we don't want anything to be both true and false. We don't want any contradictions. Um, the system should not be able to prove a false statement. In other words, you shouldn't be able to start from the axioms and use the, the allowed rules and end up with something that's false. So only true things ought to be provable. I think we could all agree that's a good thing. Um, it also should be complete. So not only, uh, are the only things that you can prove true things, but actually if a thing is true, if a statement that you can make in that system is true, then it's got to be provable within that system. So in principle, every kind of statement you could make about numbers, uh, whether that's, I don't know, the gold back conjecture, the Raymond hypothesis or any one of these things, um, if it's true, there should be, there will be a proof of it within this, within this system you're setting up. And so this is the final C that it, this big edifice we're making has to be complex and sophisticated enough to be able to make any mathematical statement about numbers. And then with those three things, it really would be, you could kind of start it going and, and eventually outward come a proof of everything that was true and nothing that was false could be proved. So wouldn't that be lovely? Only there's a problem. And the problem is in the form of a, uh, a young man called Kurt Girdle who did this amazing thing. He managed to come up with a proof that said, actually this is impossible. Because if you've got a system like this that is sufficiently complex enough to be able to make any statement that you might want about numbers and it's consistent and complete, as I've said, then it would be possible to make a statement within this system that says, essentially this statement cannot be proved within this system. Now the the hard, this is really hard thing to do.'cause you have to prove that such a thing is possible, always gonna be possible. Um, but if you have got a statement like this that asserts that, it cannot be proved. So this statement cannot be proved within the system. Now let's think about this. Uh, if, if this statement can be proved right, either can be or it can't be proved. Because in this system, it's a complete system. Every statement is either true or false. And every statement therefore can either be proved or can not be proved. If it's false, it can't be proved. If it's true, it can be proved. So this statement, suppose it can be proved. So you can prove that the statement cannot be proved, but that then we can only prove true things, right? This is supposed to be consistent. We can't prove a false thing. So if you can prove it, then it's true. Uh, but that would mean it cannot be proved 'cause it's asserting that it can't be proved. So if you can prove it, then it's true. And that means you can't prove it. So that doesn't work. If you can prove it, then you can't prove it. That's, that, that is a contradiction that doesn't work. Okay? So that means you can't prove it within the system. But hang on a minute, that's exactly what it's telling you. It's saying I cannot be proved within the system, and we've just deduced that it can't be proven in the system. So it's a true statement, but it can't be proven in the system. So what this tells you is, and this is called girdles incompleteness theorem, it tells you that there will always be however brilliant you, the system that you set up, is there are always gonna be things that are true that you can't prove within that system. You're gonna have to think of other ways and other mathematical arguments to prove them. So that's where, this is the good news we're finishing on. Mathematicians are not going to be out of a job anytime soon, <laugh>. Okay, well, I will stop there next time, coincidences, but thank you for listening today. How do you think mathematic mathematicians are getting on board with this idea of things like superposition or things being simultaneously happening and yet contradictory? How is mathematics coping with that? Yeah, so we can address things like that with, with ideas of probability really, and statistics. So we can, you know, if you roll a dice or something, um, you don't know exactly what's going to happen, <laugh>, but you could say there's, there's a probability of each different event and you might not know until you look, which it's going to be. So I suppose my feeling on that is, yeah, something has happened, but we don't know what it is yet. So I'm still clinging to things are either true or they're false <laugh>, but we may not know and that's okay. But I of course, in the quantum world, yeah, things can be they, you know, until you collapse the wave packet, you don't know. And, and things are sort of how in this state of unknowability until, until you look though. So I suppose at the point at which you observe, then things become yes or no. But, you know, we have to allow our math, mathematics does allow for uncertainty. Yeah. Very clever. I I thought that was a horrible thing to chuck you, but that was a fantastic answer. Do infinitesimal exist? Uh, <laugh>? Well, it depends what you mean by infinite decimal. I, I mean, I would say not, I don't think so, right?'cause ultimately, kind of calculus lets you take the limit all the way to zero. And I'm, I'm happy with that. But there are alternative viewpoints of how it should all work that do allow for these infinite SALs. Um, so it's different schools of thought, but kind of ultimately when you're done, you end up differentiating in the same way. So maybe it's not <laugh>, maybe it's not terribly problem, but problematic for real life. But I don't think so. Like, not in that sense, not, not infinitely small numbers that are nevertheless bigger than zero. Worked On this for 200. Well, yes, yes, yes. We should, we should have a lecture just about, just about calculus, I think. There you go. You heard it here first. Infant decimals, possibly not too worrying for everyday life.<laugh> Yes, we heard. I'm very sorry if you didn't get a chance to ask a question possibly Sarah might be willing to, to, um, have a bit of a chat at the, at the end. But in the meantime, please join me in thanking professor Sarah. Ha.