Mathematical Games Hosted by Ed Pegg Jr. [Episode 2: Fun with Fractions]

Hello, everyone. I am Ed Peg Jr. And this is episode two of Mathematical Games. Martin Gardner in the 60s started Mathematical Games, and it was a very popular column for 25 years in scientific American, and it was very influential on both me and many other mathematicians. So this series is kind of in the, the, the realm of those columns that Martin Gardner used to do. And today I'm going to be talking about fractions. But first, a little bit intro on the demonstrations project, in particular this demonstration, which I wrote a number of years ago on pentagon tiling, which is based on Martin Gardner's column on pentagon tilings. It turns out 100 years ago today, Marjorie Rice was born. She was a housewife with no mathematical training who saw Martin Gardner's column on pentagon tilings, which I'll go into here. Here's a bit of code from my demonstration. And Martin Gardner started out his column with the five types of pentagons cataloged by Carl Reinhart in the early part of the last century. And he found these five infinite families and basically said that that's all the possible ways you can tile with pentagons. But in the 70s, Richard Kirchner found three more, and he said Reinhardt was wrong about that being all the pentagons. Here's these three more, but I'm pretty sure this is now all the possible ways of tiling with pentagons. And Marjorie Rice loved this column. And for some reason I can't scroll down anymore, She said, I became fascinated by the subject and she found I need to go out of. For some reason I can't scroll down here. Ah, let's go back here. Sorry about that. I thought I'd put something further down. So Marjorie Rass found these four ways of tiling the plane with pentagons more than anyone else. She found more than one quarter of all possible ways of tiling with pentagons. So both Hershner and Reinhardt were wrong. Marjorie Rice found a whole bunch more, and it turned out there was two others that were found, the last one just eight years ago by Casey and Mann. So, bit of historical 100 years ago, Marjory Rice was born. For fractions, there's basically these four common ways of indicating fractions with horizontal bar virgo diagonals. And some fractions are actually built right into Unicode. These are usually called either vulgar or common fractions, and the vulgar means that it's commonplace. And here are the various fraction formulas that we all learn in grade school. You can add them, you can subtract, you can multiply and divide. But there's another function that isn't taught. And that is the mediant, where basically you add the numerators and add the denominators. This is also a very useful thing to do with fractions, as we shall see. A sequence of fractions of reduced fractions is often called the Ferry sequence. So for example, here, order 6. Here are all the fractions with a denominator 6 or less, where everything is reduced. So you don't see two over six here, because that's already one third. So these are the, the fairy fractions from zero to one. It turns out if you take the medians of these, you get numbers that are in or that are higher up in the Ferry sequence. But if you do the mediant again, you get back the original fractions. And that's a property of the Ferry sequence. When you just list fractions in order. One thing often done involving fractions is the greatest common denominator. This is one of the oldest algorithms, and a simple way of doing it is just to take a rectangle. So for you, if you want the greatest Common divisor of 7 and 4, just make a 7 by 4 rectangle, put in the largest possible square over and over again, and the size of the last square is the greatest common in the denominator. So for example, for 8 and 6, the largest square you get is of size 2. So the greatest common denominator is 2. And this algorithm is very fast. And this is basically how it works, just keeping, just tossing things out until there's just a small part left and then you're done. Also, at the time of Euclid, the Egyptians were also working with fractions, but they only allowed fractions if they had a 1 on top. So you could have the numbers 1, 2, 3, 4, 5, 6, 7, and you could have 1 over 2, 1 over 3, 1 over 4, 1 over 5, 1 over 6, 1 over 7. And with these fractions, they could sum them up to get any number. Though I don't know if Egyptians ever, ever did this particular one. They get pretty hairy quickly. So 77 over 1090, you can represent with these three Egyptian fractions. And one question you might ask is how many fractions are needed for a given number? Well, it turns out that's unsolved. Erdos noticed that the numbers with four on top could always be represented with three fractions. And Sierra Penske noticed three fractions were always noticed for anything with a five on top. Well, it turns out that whether this is always possible is still unsolved. For example, 4 over 74 has 32 solutions 5 over 74 has 60 solutions. But whether there's always a solution with three Egyptian fractions is still unsolved to this day. So even though this is a really old problem, it still has portions which are unsolved. So not everything about fractions is known. A common thing for, for messy fractions is to split them up with decomposition. And that is done by factoring the lower number. And from that you can get simpler fractions. So that's partial fraction decomposition. The slope of a line which goes through a grid point is a fraction. So here we have the, this the slopes of various lines. We can also take a series of fractions. For example the furry sequence with order 4. These are the furry fractions for order 4. If you square the denominator and multiply by two and move up the center to get this point, you can draw a circle with the radius of 1/2 times the denominator squared and you'll get this nested series of circles. So for the furry sequence you have 1/3 and 1/2 comes later at order 5, you have 2/5 which is tangent to both 1/3 and 1/2 is the median of 1/3 and 1/2, 3/7 here is the median of 2/5 and 1/2. Each of these infinite circles down here are part of a lower rank in the ferry sequence. And the tangent circles always are, are always found by taking the medians of the two higher fractions. So there's, there's the start of the first sequence at order for the median triangles are the median circles have these as the driving positions for, for the, where they would go. Fractions are, are so useful with, with circles that they usually will if it's, if it's an Egyptian fraction one over something, they will usually take the reciprocal and call it the bend. So for example, the circle with radius 1/3 is said to have a bend of 3 and that's measuring from the outside. If you're measuring from the inside of a circle, you consider that curvature of one over the radius as negative. One thing we can take a look at is what are the possible bins for the forward circles? Well, we're always squaring the denominator and then multiplying by two. So basically it's two times the squares and mod two. There's only four possible values that you can get. And that leads into the Descartes circle theorem with four mutually tangent circles. If you have the curvatures that are all bins, integer bins. So, so for example, this is 1/2 1/3, 1/6, 1:11, 1:18, 1:27 and so on. These big numbers here are the bends. 2 corresponds to a radius of 1, half 3 corresponds to a radius of 1 3rd, 6 corresponds to a radius of 1, 6. If you add up, if you, if you square those values. So this, this circle on the outside has a curvature of minus one. And if you square those and take twice the total, you get 100. If you just sum them up first and square, you get 100. The total of these is 2 plus 3 plus 6 minus 111 minus 1 is 10, 10 squared is 100. And similarly 1 plus 4 plus 9 plus 36 is 50 times 2 is 100. And by the discard circle circle theorem, if the, if the square of the curvatures times 2 is equal to the some of the curvature squared, those three, those four circles are tangent to each other on the complex plane. You can take the centers and just multiply them together as as so, and you get basically the same thing. So once you know where three circles are that are tangent, you can find the position and radius of the next circle. And that's what this, this function does is it just takes a starter seed, for example 3, 6, 7, and then everything else is generated automatically using Rene Descartes formula. And all of these circles mod 12, there's only four values, modulus modulus 4. Only four values can occur in with each of these. Once the initial four values are laid out, for example minus 6, 11, 14, 5, that gives a number of modulus values, mod 12. And that's, you're stuck with just those for all the infinite circles at that follow. And so these are colored by the modulus mod 12. If you have a interior circle, for example 198 here, the circles around it form a, what's called a Poppas chain. And the, the outer circle in each case has a negative value for the Ford circle. Ford circles that line at the bottom is a, is a circle of curvature 0 or a straight line. We can also enumerate fractions and basically the simple way is just to make a grid of them with 1 at the bottom 1 through 12, with 2 at the bottom 1 through 12, and so on. And then you just zig zag back and forth. And this basically shows that there's a one to one correspondence between all the fractions and the counting numbers. So index 49 corresponds to rational fraction 9, 4. However, this contains all of the fractions here are. Most of the fractions here are improper. You have 1 over 1, which is valid, but 2 over 2, 3 over 3, 4 over 4. Those are all equal to 11 over 2, 2 over 4, 3 over 6. Those are all improper fractions because you can reduce them. In fact, I don't know what this button does. So this, with this enumeration scheme we can also show reduced fractions. But I haven't played with this, so I shouldn't talk about it too much. One way of enumerating, enumerating them so that nothing is repeated is to use the Calcan WILF method. So once you have a fraction A over B, it has sub fractions A over A plus B and A plus B over B. So for example three halves. Here you have three over three plus two and three plus two over two. One of the oddities of the Calc and Wolf method is that each numerator here or each denominator here is the next numerator, 554-47733. And that also carries over to the between rows. So this is actually three over one. So one is the denominator, one is the next numerator and this tree continues on forever. Another way is the Stern Broca method which is based on medians. So for example, to get three four, you're basically between two thirds and one over one, two plus one is three, three plus one is four. And similarly between two thirds and three four you have two plus three and three plus four are five sevens. With both methods, the going from front to back is you. You can take the reciprocals and you get the sequence in reverse order. So for example, 1 over 5, the reciprocal is 5, 5 4, the reciprocal is 4 fifths. So this, the reciprocal is the reverse. This way, the Stern Broca sequence, you have the same thing. And once you build up these trees where you take each layer in turn, you can get the enumerator enumeration of the radicals by Calcan wilf. So for example, the hundredth fraction is by calcium Wilf is 7:19. And if you plug in 7:19, it's the 100th Stern Brock Hot work. Similarly, but since it's based on a method based on binary numbers, you can just put in any number and get that corresponding fraction. And when you plug in that fraction, you get back the index number. Whoever thing is nicely indexed and it's very fast. The enumeration by Calc and Wolf starts like that. The numerators give what's called the FUSC sequence F U S C because it's a confiscated method for generating numbers. And the Stern Broca sequence begins like that. And we can basically put in any integer and any, any numerator or denominator and pull up the indices. We can also use fractions within geometry methods. So, for example, we can take a triangle and find a point and then calculate the percentage of, of the triangle by each of the new generated by each of the new triangles generated by that point. So this point generates three triangles with areas 1, 6, 1/2, and 1/3 of the original triangle. These areas are considered to be signed areas. The point is outside. It turns out that this value here for this, this triangle is actually one. It actually has a spine area of minus one, six and minus one six plus one half plus two thirds is one. You're always wanting to get a sum of one when you're working with Ariel or barycentric coordinates. Ariel and barycentric both basically mean the same thing. One way of representing all of these fractions is to multiply by the greatest common denominator. And that turns everything into integers nicely. So this point here, which is basically 1/2, 1/3, 1/6, that corresponds to integers 3, 2, 1, and that's this point here. This, this point is the same as the point above. Divides the triangle into 1/2, 1/3, 1, 6. And each of these points divides up the triangle in different ways. For example, this point here doesn't divide up the triangle at all really. It's just the original triangle. And there's 0, 2, 0 area triangles that are degenerate. Once you have this series of integers representing the point, you can get the original fractions by dividing by their total. We could also do something like changing all of these fractions to generate a, an RGB color. So here's a start at a color triangle. And here's a color tetrahedron. Barycentric coordinates can work in any dimension. Here are the various fractions representing values in a tetrahedron. So, for example, this one is basically 1/2, 14014 is one of the one of the fractions where you're dividing a tetrahedron into four parts. And you could basically skew either the triangles or the tetrahedra in any way. The areal coordinates don't care how skewed your triangle is. It's basically working based on the way it divides the triangle into three areas or the tetrahedron into four areas. I showed this a few slides ago, and this is basically 18 lines of five points. And this was found using barycentric coordinates. We can extend it with this code and you get a 3D version of this which has a whole bunch of lines in three dimensions. Every single line here has five points. And this is all basically found with applying fractions. When you're working with baryol or barycenter coordinates, you can also look at lines. For example, this here is the line 1/2, 1/4, 1/4 the point here minus 1, 2, 0. If you multiply by the line, the dot product is zero. So a barycentric point such as that is on a barycentric line, if the dot product is zero, and that makes these very easy to work with, Every point corresponds to a line. Every line corresponds to a point. But the canonical versions have three forms. They can either a line can either have a sum totaling 1, or it can have a sum totaling 0, which means it goes to the center center point. And usually you want the last value to be 1 and then A and minus 1 minus A with a total 0, and that means it goes to the center point. And the toughest one to deal with is the vertical line through the center. And canonically that has this value. But basically you have 27 points and 27 lines here. With this configuration, which actually is a Delf Duel 4 configuration, every point first goes through, every point has four lines on it, and every line has four points. And also the points are the same as the lines. For example, this line here, 2, 1, 1, that means all blue here means all the values are positive. That corresponds to this point here. This is the barycentric point. This is the very centric line for 2010 is always neutral, so it's colored green. This line here re the red corresponds to negative. So this is the Barycentric line 3 minus 2 1. And the point is over here. So this is a set of 27 points and lines where every point is a line and every line is a point. Whether there's another four configuration like this is unsolved. So how else can we represent fractions? We can use repeating decimals. So for any fraction we can get a decimal version. And for any decimal we can also get a fraction. So, for example, this repeating decimal corresponds to this fraction. Another thing we can do is use continued fractions. So, for example, not sure why this isn't showing up. Well, we can. Represent this series of repeating fractions with this periodic sequence. When it's a repeating series and a continued fraction, that means that a square root is involved or a third. Not all, most rational numbers, our most irrational numbers, don't have a continued repeating continued fraction. Usually it's more varied and random. For example, here's the continued fraction for PI. It starts out three and then one seventh. 22 seventh is a pretty good approximation for PI, and so is 3. 3. 3 over 106 is the next good approximation for phi. And quite a few people probably know this one as well, which is a excellent fraction for representing a pie. And each fraction here is better. But usually most, most people who want to remember these will stop at this because it's a nice Easy to remember fraction 113355. You just 2 of each of the first odd numbers and then you put the bigger set over the smaller set. And that's very close to PI. One odd thing about these random sequences generate giving continue fractions is for almost any value, the geometric mean of those terms goes to what's called the Kinshin constant. So for example, if we look at the first of 200,000 terms is continuing fraction of PI, the geometric mean is about 2.685. The kitchen constant is 2.685, and so on. It's, it's very weird how this works, and it's still not quite understood. An exact value for the Kinshin constant is currently unknown. Some, some numbers have large values early on. For example, PI had 292. If you look at the simple value 9.1 to the fourth, you get this large value in the continued fraction. So it turns out that 33 over 19 is an excellent approximation for the fourth root of 9.1. It's good for eight decimal places. Usually that large value means that something weird is going on. So 91 over 10 is a to the 1 fourth is approximately 33 over 19, which means if you take the the fourth power of each side, it means that 91 over 10 is approximately 33 over 19 to the fourth. And if you multiply those out with the cross product, which is one of the one thing I thought was really neat when working with fractions when I was in grade school. If you look at the cross product 91 times 19 to the 4th minus 10 times 33 to the 4th equals 1. And if you multiply these out, you get these two large numbers 11,859,210 and one more than that. It turns out these are the largest two consecutive 19 smooth numbers, which means that they are all their factors are less or all their prime factors are 19 or less. And if you know what's known as the ABC conjecture, this is one of the best known examples for the ABC conjecture. And that's why the fourth root of 9.1 is weird because it's one of the most extremal values in what's known as the ABC conjecture. If you're playing with a continued fractions and you find something goes big, that means you're onto something. This is something. I found this playing around with Mathematica and it was so strange that I submitted it to the online encyclopedia of integer sequences and a few months later somebody found out this stuff. And this particular example is well known, has been known for over 100 years. But the reason why this acted so weird wasn't little interesting discovery. If this hadn't been found, basically I would have found one of the best known examples for this famous unsolved problem. But it was already known. I just found a weird way of getting to it. The Minkowski question mark function. Minkowski also did Egyptian fractions early on that five over n. Minkowski did a lot of work with container fractions. The question mark function is a way of representing a value based on its continued fractions. So for example, the Minkowski question mark value for PI is this thing. And you'll notice that there's a whole bunch of nines here. Well, that corresponds to the 292 in the continued fraction. A large term in a continued fraction means that the question mark function will have a lot of repeated nines or zeros. So if we look at, we can do the split, it turns out there's 72 nines in a row for PI. And if you divide 292 by 72, you get about 4. If we look at the, the fourth root of 9.1, we get 75,000. Well, if you do the Minkowski question mark sequence, you'll get a number which has 22,770 zeros in a row before it starts continuing on past that. And if you divide those big numbers, you get there's about 3.3 zeros per value in the large term of the continued fraction. This is an unsolved problem. How many zeros are nines are in the Minkowski question mark function based on the continued fraction? It's still not understood. And with all these things we can finally do bad fraction simplification. So this number here, if we, if we just cross out the repeating part, we get four fifths. And it turns out that this fraction actually is four fifths. And it turns out that if you look at the value we're crossing out here, if we look at that as a repeating decimal, it's equivalent to the fraction 4 7. So 4/5 corresponds to 4/7 as a bad fraction. And basically most, most fractions and numbers have a bad representation with, with simplification. And often it can be a very large number. So for, let's see if I can find a, a more reasonable size one. Oops, Here we go. That's 1/7th again. Okay, for example, 716 corresponds to this fraction 28 over 171. And here's another one here. 417 has a matching fraction of 68 over 23 83. But you need to use this number to get the bad simplification we've got. This is a demonstration on the in the demonstrations project. And I find it, it fascinated that there's this method for doing things badly involving continued fractions, repeating decimals and all this other nice stuff where you can set these things up so that things work out nicely when they really shouldn't. We can, we can also look at what integer values multiplied by the sum of their reciprocals gives a given value. If we take say the value. Oops. If we take the number 59, what are the options for possible integers and fractions? So for example, with 59, two 1585 works. If you do 2 plus 15 plus 85 and multiply by 1/2 over 115 plus 1 over 85, you get 59. So you can rewrite this as that which can multiply out to get this long polynomial. And solving for X, you get this elliptic curve which for the value 59 is solvable. It turns out that not all values are solvable. For example, 58 is not solvable. 52, 51 are. So 49, 50 and 51 are not solvable, but 49 and 52 both are. Once you have one solution, there's usually an infinite number of solutions. And I mentioned the elliptic curves here. A property of elliptic curves is usually any line will go through three points or it'll be tangent to the curve at one point and the that point of tangency is considered as two values. So if, if two points are given that are both rational. The property of elliptic curves is that the third point of intersection is also rational. And this is used in cryptography and other applications. Once you can set things up as an elliptic curve and you find some rational points, you can generate a whole bunch more rational points just by drawing lines. And each time you add a point, you'll usually be adding many more lines. And once you, once you start that process off, usually the points will fall into different ranks. And whether there's a infinite number of, whether there's A limit on the number of ranks for a given elliptic curve is still an unsolved problem. I, I believe Noam Elkes has the record for a curve with 21 different rational values for the possible ranks. Another place that fractions come up strangely is in probabilities. For example, the Sylvester four point problem. What are the odds of all points in a given convex hole? So this is for, for a, a bounded rectangle, if you're given five points, the odds that a convex hull has all five points is 25 over 36. If you're bounded by a, a triangle though instead. The, the chance of, of all the points being on the, on the convex hole is 2/3 instead. So every time I click there's a, there's a 1/3 chance that there's going to be a red point. So that's Sylvester's problem. And if you do five points, the chance that all five will be green is only 11 out of 36. And if we go to the full then the chance of all five is 49 over 144. There's many things dealing with average lengths and whether they a particular type of triangle will turn up. If I remember right, if you take a, a 3D ball and just take two points inside, the chance that the, the average length of that line I believe is if you take the points as being distributed normally is 33 over 32. If I remember correctly. There's all sorts of strange probabilities that boil down to simple fraction. If you're given a number, is it possible to make a right triangle which has that area where all the sides are rationed? It turns out that the smallest area you can get is five. And here's a triangle that has area five. Here's the next one up is area six which is a simple three four five triangle. And here's the smallest for seven. If that area corresponds to a rational triangle, then it's called a congruent number. And the, whether a number is congruent or not is still an unsolved problem. So here's they can get pretty, pretty nasty quickly. So for example, let's see here, here's. Here's the solution for 4006 and as we get to larger values. You can get very difficult fractions. They, they don't even show up on my screen any anymore. And let's take a look at the code here. Let's see, we'll un iconize this. Hope that doesn't crash anything. For 157. For example, the hypotenuse has this big fraction. These are solved up to 10,000, I believe. And these are very difficult to solve. The general method for finding these is unknown. Here's another one. You have a unit square, and you want to find a point so that the distances to the four corners is always rational. And this demonstration here has a whole bunch of solutions where three values are rational. This is an unsolved problem. So, for example, for this one here, this particular point at this location has these three rational distances and this irrational distance, If you put points into a bound region, for example here, the locations of these points, in this instance, all of these points here turn out to be rational values. It's not always the case. For example, for this case, everything is a strange root object. And here they're not even. They're not even well defined root objects. But for 16 points, everything is an exact fraction. And with the Heilbronn problem, you're basically looking for the largest possible smallest triangle. So all these red triangles here have the same area. Those are the smallest triangles. And these bigger ones, those are the bigger triangles. But the Heilbronn problem is still mostly unsolved for anything above. In fact, I don't even think 13 points has been solved yet. And that's the material I had prepared. I'll be glad to answer any questions that people have about fractions and similar items, and I can also take requests for future talks. Does anybody have any questions? Let's see here. And if anybody has any solutions to any of these edpolformat.com I'd be glad to take a look at anything you find. And also do feel free to send in new demonstrations. I'm always looking for nice new material, especially if it, if it is good at explaining something in mathematics in a new way or in an old way. I think we can wrap this up. Many thanks for attending and I'll see you in one month with another in the, in the, in the series Mathematical Games. Many thanks.