 Welcome back to our lecture series, Math 4230 Abstract Algebra 2 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. In lecture 28, we are going to talk about a very cool application of abstract algebra, particularly application of field theory towards the concept of actual geometric constructions. As you're watching these videos, you might think that we're actually talking about modern geometries Math 3130, but no, this is actually meant to be an algebraic lecture. Let's see what some of the connections we can find between algebra and geometry. In order to do so, we need to review a little bit of history. The ancient Greek mathematicians have been renowned throughout history for their skill in geometry, particularly in the topic of geometric constructions. In Euclid's very famous work entitled The Elements, 26 of his many propositions are actually geometric constructions. The Greeks placed a great importance on these type of constructions, and that is constructions using a straight edge and a compass. This is perhaps why so much effort was given to solve the so-called four notorious construction problems, which I have here listed on the screen. The first one was known as trisecting the angle. What this means is can every angle that's constructable be trisected? If you have an angle, can you construct an angle which is one-third the measure of the angle you start with? Now, bisecting an angle is no problem whatsoever using a straight edge and a compass, but trisecting gave a little bit of a problem. So can every angle be trisected? Can every constructable angle be trisect? That was the first question. The second problem I should say is referred to as squaring the circle. If you're given a circle, a constructable circle, whose area is given as A, can you construct a square whose area is the same as that circle you constructed? Duplicating the cube, what that problem was is if you can construct a cube, so something like this, your cube, if you can construct a cube whose volume say is V, can you construct a cube, larger cube obviously, whose volume is twice that of the original cube. So can you double the cube, duplicate the cube there? And then the last problem here, it's not as well known as the first three, but the last one is can you construct a regular in-gone using just a compass and a straight edge? So some of them are quite obvious, like we can construct an equilateral triangle, we can construct a square, you can construct a regular hexagon. Oh boy, my picture definitely does not look regular there. Even a regular pentagon can also be constructed using just a straight edge and a compass. Some of the other shapes a little bit more problematic. Some people struggled to build it to construct all of the regular polygons, but some of them were able to be done, some were not. And so these are the four notorious construction problems of geometry that were left unsolved for such a long time. Now why are these problems so notorious? Well, I kind of alluded to it just a moment ago. They're so notorious because for centuries, geometers tried and try as hard as they might, they were unable to find solutions to these problems. They couldn't find a construction that trisected every angle or that could square the circles and draw every regular in-gone. Right? We could make a hexagon, but why not arbitrary constructions, arbitrary in-gones in that situation? We had similar constructions of these four, but we couldn't master all four of them. They went unsolved for literally centuries. Now, why couldn't they do it? Why did they struggle with it so long? And the answer is actually pretty simple. These four problems are in fact unsolvable. Now, what does one mean by unsolvable? Now, this doesn't mean that we don't know how to do it. What this means is this actually is impossible. There is no possible construction. There's no possible construction using just a straight edge and a compass that solves these four problems. It cannot be done. Now, I have to be very careful with the language here, because people often say things like, oh, we did the impossible. No, no you didn't. You did not do the impossible. By definition, it's not possible. So if you did it, it means it was possible. Was it hard? Maybe. What did people think it was impossible? More likely, but it wasn't actually impossible. People often use the word, oh, this is impossible as in hyperbole. But I'm talking about in a literal sense. Solving these problems are impossible. It cannot be done. And one of the beauties of what we're gonna see in this lecture is the mathematics behind it to show that these problems are in fact impossible. Now, the first three problems, trisecting the angle, squaring the circle, duplicating the cube. It was first, it was first conjectured by Descartes, or probably Descartes at least. At least Descartes was the first we know about to conjecture that these problems were in fact insolvable, that they couldn't be solved. Kepler, he claimed that the problem of constructing the regular in-gon was unknowable, like we couldn't know if it's possible. And so there were, you know, hundreds of years ago, people who started to conjecture that these things were impossible, because after a millennia of trying, then it starts to beg the question, well, maybe it's not possible. So some of the brightest minds of the time were beginning to conjecture that these things could not be happened whatsoever. Now, to give some hysterical context to this thing, I wanna offer some quotes due to Dixon here. So Dixon explains that why do geometries fail to prove that these constructions are impossible? Okay, even after people started to realize that these problems were probably impossible to solve, they still struggled to actually prove it was impossible. Proving that you can't do something is a very hard thing. So getting back to Dixon's quote here, why do geometries fail to prove that these constructions are impossible? The answer is that the proof is beyond the scope of elementary geometry, since it requires the use of theory of abstract algebra. That's what we're gonna tackle in this lecture, lecture 20 in our series right now. The theory of equations and the theory of fields is the key to proving the insolvability of these four geometric constructions. Certainly for this reason today, several mathematical amateurs submit to universities solutions to these so-called problems, right? Another quote due to Dixon here. Why for centuries have there been an annual crop of angled trisectors? The answer is not easy. Some angled trisectors have not heard of the fact that there are proofs which show that it is impossible, it is absolutely impossible to trisect all angles by ruler and compasses. Others have heard of such proofs but ignore them. Often such a person regards impossible as meaning merely that mathematicians have not to date succeeded in finding a construction whereas he may have more luck. And I'll give these angled trisectors the benefit of the doubt here. It wasn't until the 1990s when Andrew Wiles finally proved Fermat's last theorem, which of course was a well-known theorem dating back to Fermat of course for hundreds of hundreds of years. And so stories like this where it took hundreds of years to solve Fermat's last theorem. It gives fruitless hope to many of these angled trisectors that Dixon is describing to here. I in fact had a professor when I was an undergraduate, he was teaching an abstract algebra class. I think we actually were talking about this very topic right here. And he had mentioned that when he was a graduate student their mathematics department would receive these proofs of angled trisection and the such in which case the professors would send a letter back to these people and be like, we appreciate you for submitting your proof but we unfortunately have to inform you that it is false. If you wanna find the holes, the problems with your proofs please be willing to pay our graduate students a stipend and they will find the error in your proof for you. And so that I think I was a little generous of them. But anyways, some people are convinced that you can prove these things because it's just cause they're hard, right? No, no, no, no, no, no, they are in fact impossible. This is not hyperbolic language. This is in fact an impossibility that we're gonna prove in this lecture, all right? So before continuing on, it ought to be defined what it means to construct with a straight edge encompass. For example, many of these angled trisectors, the problem is they don't even know what they're doing. Intuitively they think they know what they're doing but intuition can lead to lots of mathematical problems of circular reasoning paradoxes to name a few things. So in order to prove something about straight edges encompass we actually have to define what does it mean to use a straight edge and to use a compass. So going back to the elements where Euclid, one of the first geometers, at least in the modern sense, the axiomatic geometry began with Euclid. Let's go back to what Euclid considered a straight edge in a compass. So as first defined by Euclid himself, he said that a straight edge can draw a straight line from any constructible point to any constructible point and can extend a finite straight line continuously in a straight line. So what that means is something like the following. So let's say you have two points that you've already constructed in the plane. So these definitions are recursive. A straight edge allows you to draw the line segment between these two points. You can also use your line, your straight edge to extend the line in definitely one direction or in the other direction. So my line segment is hideous there. But the idea is we can use the straight edge, connect the dots, and then like I said, we can continue the line onward in any of these directions. So lines are essentially constructible through two points. Two points determine a line. We can construct that line segment and we can also construct the very important subsets of rays and segments. So that's what a straight edge does. It allows us to draw lines, draw segments. Now, what I should mention when it comes to a straight edge, a straight edge is not graded. That is, it's not graduated in a degree. There's no little marks on there that tell us how distance is. So we don't have a ruler when we talk about the ancient Greek notion of a straight edge, it's not a ruler. It is just a straight line. So we often use, like if we do these constructions by hand, we often use things like a meter stick or a yard stick, what have you, maybe a ruler. But we don't, some people even use a protractor because protractors generally have a straight edge on it. But we don't have any graduation on the straight edge. It's not a ruler or anything like that. In fact, it can be proven that if you have a ruler and you have a compass, you can actually trisect the angle. So the fact that a straight edge, it can be graduated gives you something different than just the straight edge that Euclid had in mind. So no marks are on that straight edge. The straight edge can't be used to measure distance. Well, what's the notion of a compass? So a compass, as we often think about it, right? You'll have like, it'll often look like this little V shaped thing where you have like a needle on one side and a marker on the other side. And so the needle acts as a pivot point and then you turn it around and then you can trace a circle using that. My drawing's not the worst, but it's not the best either. Anyways, so to Euclid, what was a compass? A compass can describe a circle with any constructable center and any constructable distance. So the idea is you have the center of your circle, you then place the needle point of your compass on that. That acts as your pivot. And then if you have any constructable point in the plane, well, then of course, you can construct the line segment using a straight edge, but in particular, you can then place the pen or chalk or pencil part of your compass on that other point. And then you can spin your compass around the center starting at that constructable point. This then gives you a constructable circle. And so circles can be constructed using a compass like this and lines can be constructed using a straight edge. But to build a straight edge, you have to have two points that determine the line. To construct a circle, you have to have one point which is constructable at the center and then a second point which is constructable on the circumference. So it's even with a straight edge and a compass, being able to construct a line or a circle is a recursive process. You have to construct it using initial parameters that were already constructed. So with that precise definition of a compass and straight edge now out of the way, we can define what it means for a number to be constructable. So we say that a real number alpha is constructable if a line segment of length alpha can be constructed using only a straight edge and a compass. So we're gonna let the set K denote the set of constructable numbers. And we claim that K is in fact an infinite algebraic extension of the rational numbers. So in particularly it's a field. The collection of constructable numbers is a field that the sum, sub-difference, product and quotient of constructable numbers is also constructable. We will show that the rational numbers can be embedded inside of this. And so we then get it's a field extension. It'll be infinite degree. And we claim that everything you can construct is algebraic. So these are some important observations. And this statement right here that the set of constructable numbers is an infinite algebraic field extension of the rational numbers. That is basically where the proof that those previous four problems were impossible will come from. There's a little bit more details to it but that's basically what it comes down to. The constructable numbers is an algebraic field extension. And that's how we utilize abstract downs where to prove the impossibility of those problems. We have to argue that to solve those problems we have to construct certain numbers and those numbers are not constructable because they don't belong to this field. All right, so let's then prove that claim. That's the main theorem for this video. In the other videos for election 28 they're gonna be much shorter and they actually provide the proofs of the impossibility of those four problems. In this video, we then establish the claim we just had that the constructable field is in fact this field extension. All right, so to get started for any recursion you have to have a seed of some kind. You have some places to get started. And so in order to form a plane we do have two points. So we're gonna have two points in the plane. It doesn't matter which two points you have. One of them we're gonna call it the origin and we're gonna give it the coordinate system zero zero. The other point we're gonna give it the label one zero and because we have a straight edge we can connect these two dots and we say that this is the unit length. The distance between these two points whatever these two points are is considered to be one. So that's how we get started. So that's just given to us. We have two distinct points we call their distance one. Now, if you pick different points the actual distance between them could be different but this is like a change of unit. Is it one meter, one foot, one yard, one mile, one kilometer doesn't make much of a difference in that situation. So this distance is relative and we call that distance one, okay? Now to make life more consistent with us I'm gonna think of it in the following way. We have our point the origin zero zero. We have our other point one one excuse me one zero and these are the two points here. Like I said, the distance between the origin and the point one zero is one. So the number one belongs to the constructable numbers. This is not an empty set but also you can construct the distance from zero to itself that is a line segment sort of a degenerate case but a point is a line segment where the end points are the same thing. That distance would be length zero. So zero also belongs there. So that's good. We have zero, we have one. So we have two numbers. What other numbers can we construct? Now, as you already see here on the screen we can use in a straight edge we can construct the line that passes through these two points. And we're gonna call that line because it can continue on indefinite in both directions. We're gonna call that line the x-axis. Now I've oriented it on the screen so that it looks horizontal as we usually think about the x-axis but it might not have been whatever but by up to rotation we could think of this as a horizontal line. Now, what we're gonna do next is we're gonna use our compass. So to, we're using our compass we're gonna place the center of our compass at the origin. Nope, we're not gonna do that one yet. We'll do that one in a second. We're gonna place the center of our compass at the point one zero and we're gonna point the marker of our compass here at zero zero. We're then gonna draw a circle of radius one centered at one. This circle will intersect the line at two different points. After all in the constructable number field we do have what's called circular the circular continuity principle. We've guaranteed, well circular continuity gives circles intersect circles. This is what we call elementary continuity principle. We have the lines intersect lines and circles intersect each other at two distinct points here. So we're gonna get another point of intersection. Let me erase that for a second. We're gonna get another point of intersection here, right? So we're gonna call this point two comma zero. Why is that? Because the distance between zero zero and this new point is two units. Cause after all you had a circle who's centered at one zero it's radius is one and this segment is a diameter of the circle. So that's gonna give you two, okay? Now if we play that same game where now the center of the circle is two zero and the radius is one zero we draw a circle like so that's gonna give us another point of intersection. We get this point here three zero. We play this game again. We center our circle at three. We take a radius of two. That's gonna give us a mark here. It's gonna give us a mark here. That second mark that second mark is gonna be the point four zero. And notice the distance between the origin and three zero is three. The distance between the origin and this point four zero is four which is why we're given at the coordinate systems that we are. And so this means the number three is constructible. I don't know if I said it, but two is constructible, four is constructible. And by induction we can repeat this process for every positive integer, okay? So in particular what we have now seen is that the set of natural numbers is a subset of K, all right? We can construct every natural number. We had zero, one, two, three by induction we get all of them. Now if we continue this process now center your thing at zero and take as your radius one if you draw the circle that's gonna give you a point of intersection over here. This gives us the point negative one zero centering at that point now and then you do a circle whose radius is one that'll give you another point over here. Whoops, this point right here that's gonna give you the point negative two zero. Again, centering at negative two zero take a circle of radius one you're gonna get a point right here at negative three zero. And then by induction you can do this for each and every negative integer. And then the distance we'll consider the length here as a signed operation going in the positive direction this direction here that is you have your two initial points if you call those points P and Q going towards, going from P towards Q is considered the positive direction and therefore going backwards is considered the negative direction. And so therefore we can get negative distances as well. We're thinking distance here as a vector. So we get negative one, negative two, negative three, negative four, et cetera, et cetera. By induction we can get all of the negative numbers as well. So it turns out we can improve upon this. We can visualize the integers as a subset of this set K as is illustrated right here. All right, I'm gonna move down a little bit and start my picture all over again. So we have the x-axis that we've now constructed. And now we can graduate the x-axis based upon integer markers. One, two, three, four, et cetera. And so I'm gonna put those markers here on the screen. So we get zero, one, two, three, four, like so. So this was our zero here. We can get negative one, negative two, negative three, negative four, et cetera here. One, two, three, four. Now what we can also do is we can take, well, we can construct the perpendicular bisector to the segment, consider the segment negative one to one. We can construct the perpendicular bisector. So that is this is the line that's going to pass through the midpoint of this segment, which is the origin here. And it's gonna form a right angle. Euclid in the elements prove that this construction is possible. I'm not gonna go through the details of every construction in this video, but some of these, of course, I can talk about here. So what we can do is we can take the circle centered at one whose radius is negative one. That circle's gonna have, part of it's gonna be here, part of it's gonna be down here. Now you'll have to apologize that my picture, I'm not actually using a compass, I'm just doing this by free with my hand. So you'll forgive me if it's not perfect. With an actual compass and straight edges, these graphics look a lot better. Next, take the circle centered at negative one, whose radius passes through one. Then using that circle, if you spin around, you're gonna get a mark here and you're gonna get a mark here, like so. If we connect these dots, because these two circles intersect each other, if we connect these dots, we get something, let me use the straight edge here. We get something like the following, like so. It connects those dots and passes through the origin. I confess my picture's not exactly perfect. It doesn't look like a perfect right angle, but Euclid's construction shows that this line will pass through the origin and it does form a perpendicular line. This was constructable using a straight edge and a compass. Now with this new line that's perpendicular to the x-axis passes through the origin, we're gonna call this new line the y-axis and using the same induction strategy before, we can graduate this line. So we get y equals one, y equals two, which admittedly with the location of my marker, that should have been the, since we have a radius of one, that's gonna, well, I won't worry about what that is. We're gonna get y equals three, y equals four, et cetera. Don't scold me or post in the comments if this thing is not drawn to scale. I don't claim it is. This is only meant to give us the intuition of what's going on here, but we can graduate the y-axis using the same thing here and much like we did before, like we were able to erect a perpendicular line out of the x-axis through the point x equals zero. We could do that for any point on the x-axis. We can erect a perpendicular line out of x equals one, out of x equals two, out of x equals three. We can do that through any point on the x-axis. We can also, looking at the y-axis, we can erect a perpendicular line out of y equals one, y equals two, y equals three, et cetera, et cetera. And so if you look at the points of intersections of these points, these are points that are gonna have integer coordinates. Like this point right here is two comma one. So therefore, what we've now argued is that every lattice point, that is every point whose coordinates is an integer and an integer can be constructed using a compass and a straight edge. Now, because we can construct integer points, we can also use this to measure distances. We can construct distances between them. For example, if you look at the line segment between the origin and the point one one, so we wanna find the distance between the point zero zero and the point one one, by the usual distance formula, you're gonna end up with the square root of one minus zero squared plus one minus zero squared, which is gonna give you the square root of one plus one, which gives you the square root of two. So for example, the square root of two is a constructable number. And there's a lot of possibilities we can do at this moment. So we know that all of these lattice points are at least constructable. But remember, our goal is to show that the set of constructable numbers K is in fact a field. At the moment, we do know it contains the integers. Why does it contain the rationals? What I actually wanna do now is show that this is a field, that it's closed under the field operations of addition, subtraction, multiplication, and division. So let's start with the most elementary of these. Let's start off with the idea that it's closed under addition, okay? So take two constructable numbers, alpha and beta. And so because they're constructable numbers, that means there do exist points in the plane so that if you connect the line segment here, this line segment has a length of alpha. And there's two other points when connected together, its line segment is beta. Now, in the elements, Euclid showed that if you have a line segment using a straight edge and a compass, you can translate the endpoint of a line segment to any other constructable point in the plane. So in particular, I can without the loss of generality, suppose that the line segment that gave us beta, that one of its endpoints coincides with the line segment that gave us alpha, okay? And also without the loss of generality, I can suppose that this point is the origin, zero, zero. So we can think of these line segments as line segments emanating from the origin, okay? Now Euclid, and this takes place in the elements, I dot 42 through 44, Euclid also showed that angles can be translated in the plane as well using a straight edge and a compass. So this angle right here, theta can also be translated. And so we'll talk about that one in just a second. So in particular, what we can do is we can take the line segment beta, so it comes out of alpha's endpoint, so we can also do it above here as well. So that's a beta, you can put this over here. And so you can try to make the claim that this is alpha, but that might be getting a little ahead of ourselves. We want that to be alpha. What we then try to do is basically the following type argument. Again, I don't wanna go through all of the details of this thing, but using combination of segment translation and angle translation, you can argue that this thing is in fact a parallelogram. So you have these angles phi and theta, like so. The sides of this parallelogram, the opposite sides are gonna be congruent, so you get alpha, alpha, one side, beta, beta on the other side. So because you have these two line segments, alpha and beta, going through various geometric constructions of translations of segments and angles, you can construct a parallelogram so that one side length is alpha and one side length is beta. Again, I'm just gonna skip the details of many of these constructions, but you could prove that you can construct a parallelogram with any two constructable side lengths, okay? Now I'm gonna go linear algebra here for a moment. If we think of this line segment as a vector, it's a vector whose magnitude is alpha. And if you think of this line segment that's associated to beta also as a vector, then this parallelogram situation might look familiar. In linear algebra, you often talk about the so-called parallelogram rule that when you add together vectors, the idea is geometrically you put the vectors head to tail so that as you can connect alpha and beta together like this, then the sum of the two vectors will go from the starting point, I should say, of the tail of alpha towards the head of beta. So in particular, the diagonal of this, the diagonal by the parallelogram rule from linear algebra, the diagonal of this parallelogram will have a length equal to alpha plus beta. And so this point was an endpoint of a segment so it's constructable. This is also the endpoint of a line segment so it is constructable. So because there's now a line segment, you can use a straight edge to connect this point and this point, we then have constructed a line segment whose distance is alpha minus beta, like so alpha plus beta, excuse me. And so that means that the sum of two constructable numbers is constructable. Thinking of linear algebra as well, if you take the segment in the reverse direction, that would give us negative beta. And then also you can see right here that the other diagonal of this parallelogram will have length alpha minus beta, like so up to absolute value, of course, it depends on the direction and things you're going here. But in particular, this shows us that by the parallelogram rule, since parallelograms can be constructed for any two side lengths using a straight edge and compass, since parallelograms are constructable, this shows us that addition and subtraction of constructable numbers is constructable. So with respect to addition, we then have that K is an abelian group. It's an abelian group that does contain the integers, remember. And so, okay, we are gonna go from there. What about the ring structure? What can we say about multiplication? Multiplication is a little bit trickier to explain, but it's not so bad. Think of the following type of construction, because that's what we wanna do next. We wanna show that multiplication between two constructable numbers is possible. So if alpha and beta are constructable numbers, alpha times beta will also be a constructable number. So come back to the x and y-axis, which we had constructed earlier. So imagine them as these lines here on the screen, the y-axis and the x-axis, like so. And consider the points, alpha comma zero. So this is a point on the x-axis, remember. Since alpha is a constructable number, there is some segment in the plane whose length is alpha. And we can, without the loss of generality, associate to it, we can pretend it's one of its end points is the origin. And then there's a line that passes through those points, right? We can identify that point with the x-axis up to rotation. Makes no bit of a difference whatsoever. And then construct this perpendicular bisector, we call it the y-intercept, and consider the point zero one on that line there, okay? So what we're then gonna do is we're gonna consider the line segment that connects together these two points. Let me draw a little bit better there with my straight edge tool. So consider the line between alpha zero and zero comma one. Like so. So then what we're gonna do is we're gonna take a different point, zero comma beta this time. So let's assume beta is bigger just for the sake of drawing it here than one. So take the point zero comma beta by Euclid's property i.31. We can construct a line parallel to this line right here that passes through zero beta right here. Honestly, what we're using here is the Euclidean Parallel Postulate. And what Euclid proved is in the Constructible Plane, the Constructible Plane satisfies the Euclidean Parallel Postulate. There exists a unique line parallel to a given line passing through a given point, like so. So by the Euclidean Parallel Postulate, there exists, let me make it blue, there exists a line parallel to the first one we drew. Maybe look something like this. Again, it's not perfect, but it looks okay right there. It's gonna be parallel in that situation. And yeah, Euclid proved this thing exists. This point will have a point of intersection. My drawing's a little bit crude. So for the sake of it, let me just draw it so I can see it on the screen right here. And so call this point, this point of intersection, we'll call it x comma zero, like so. And so with that point then in hand, look at the following points right here. So since these lines are parallel, they have the same slope. Please don't criticize the drawing right here. Since these lines are parallel, they have the same slope. And we know the x and y intercepts are both these lines. So as we compute them, you look at this one right here, one minus zero over zero minus alpha, that gives you the slope of this line right here. And then when you look at the second line, you get beta minus zero over zero minus x, that's this line right here. If you simplify this expression, it then turns out that you're gonna have one over alpha equals beta over negative x right here, like so, oops, that's because we have a negative alpha, one over negative alpha. So if you clear the denominators, you're gonna end up with negative x equals negative alpha beta, so therefore x equals alpha beta. We then constructed a distance whose length is the product of alpha beta. Now, mimicking this construction, we can also do alpha over beta, that is we can do the quotient. So in particular, you can use the points one, zero, zero beta, zero one. You draw a parallel line through zero one there, then you can make the argument that we've constructed the point zero comma one over beta. Same type of thing. I'll leave that exercise up to the viewer right here. So we get that alpha times beta and alpha divided by beta are inside of this set. So this shows that the set of constructable numbers is in fact a field. Since it contains the integers, then this field necessarily contains the integers field of fractions, which is the rational numbers. And so that then proves the first main statement here that the set of constructable numbers is a field that contains the rational numbers. Why is it algebraic? Well, we're gonna get into that in just a second. We've now shown that the set of constructable numbers is in fact a field, a subfield of the real numbers that contains q, of course. And so we then need to show that it is an algebraic extension. And the fact that it's algebraic is really where we're gonna derive the impossibility of the previous problems that we've talked about. Because in particular, in addition to being closed under addition, subtraction, multiplication, division, the constructable field is closed under square roots. And in fact, it's the smallest field that contains q that is closed under all square roots. And so how does one do that? Well, the idea is to take a line segment whose width is alpha plus one. So we have this distance here is alpha. This distance here is one. And we take this line segment. So the total distance is alpha plus one. We're gonna take a line segment, this line segment and make it be the diameter of a semicircle. I apologize, the crudeness here. My graphing tools don't have the compass here. So we're gonna take this and we're gonna inscribe inside of this semicircle two triangles. One of the triangles, let me use the straight edge, I do have that one. One of the legs is gonna go here and then we have goes here and then goes here. So this is, let me try that again. This is one of our triangles. And then the other triangle would look something like this. Like so. So we have these two triangles inscribed inside of our circle here. Now some properties to note is that the green triangle that involves length alpha, that is a right triangle. There's the blue triangle that involves one right here is likewise a right triangle. And then properties of inscribed triangles inside of a circle also gives that this angle right here is a right angle. So when you start looking at angles here, I want you to note here that these two angles right here are supplement, they're complementary angles to each other, excuse me. Now since you have a right triangle, the two acute angles are supplements of each other as well. So once, excuse me, they're complements of each other. So the complements of this angle is this angle, but it's also this angle. So those two angles are congruent to each other. And then likewise, when you look now at this angle, here's one compliment, here's another compliment. So they have to be congruent to each other as well. So we actually have three triangles in play here. There's this triangle, the blue one. There's the green triangle right here. And then there's their combination, this triangle right here. These three right triangles are all similar to each other. And so their sides will be proportional to one another. I want you to consider their shared leg right here. And consider that that segment, say that its distance is X. I wanna argue to you that the length of X is in fact the square root of alpha. So how do we do that? So because these triangles are proportional to one another, all right, notice the following situation. You have the green triangle. So ignoring the hypotenuse here. You have alpha and X. And then looking at the blue triangle, I'm going to arrange them so that they look similar to each other. So these angles were congruent and these angles were congruent. With the blue triangle, we have this side is one and this side is X. So because these triangles, let me first of course mention that this diagram we talked about earlier here, this is in fact a constructible diagram because we can construct the segment alpha. We can construct the number one here. We can put these two segments together, okay? Then since we can construct a unit of one, we can then direct the perpendicular line that comes out of the segment at that point. And then we can construct the midpoint of this segment. Then we take a circle for which is centered at the midpoint and whose radius is half of this segment. So that gives us this semicircle right here. This line will then intersect the semicircle at this specific point. We can then connect all the dots and we're good to go from there, all right? So this is in fact a constructible diagram here. So coming back to our proportionality statements here. So X coincides with one, alpha coincides with X. So this gives us the equation, one over X is equal to X over A. Click cross multiply, you get one times alpha times X times X. You're gonna get that X squared equals alpha, therefore alpha, excuse me, X is a square root of alpha. Or I guess you could say it's the negative square root of alpha, but if you have one, you have the other since we have a field. So using compass and straight edge, we can construct any square root. So if a number is constructible, it's a positive number of course, then its square root will also be constructible. And so the field of constructible numbers is closed under square roots. Now it's important to note that when you take the square root of a number, so you have some field, say alpha belongs to a field F right here. This is not necessarily the rational numbers. Let me fix that like so. Then you look at the field, the square root of alpha over F. Like so, one of two things happens because I want you to consider the polynomial X squared minus alpha. One of two things happens since this is a quadratic polynomial. Either it's irreducible or it factors as X minus the square root of alpha times X plus the square root of alpha, okay? Now, if this polynomial is irreducible, then that means that the square root of alpha does not belong to F, and this would be a degree two extension. If the polynomial does factor, that means the square root of alpha already belong to F, and so F adjoined the square root of alpha is the same thing as just F, and that situation, the fields are the same, and so you get a degree extension of just one. So every time you take the square root of a number in a field extension, you either are going to have a degree two extension or a degree one extension. So if you do this recursively, that is if alpha is some element which is constructed from rational numbers using some sequence of square roots, then Q adjoined alpha will have as its degree two to the N, where N is some natural number. Now that could be zero, right? Like if we adjoined the square root of four, well, that's a rational number, that's two, no big deal there, nothing new, but that is a possibility. So in particular adjoining a constructable number to the rational field will only extend you by two to the N, okay? Now I need to make mention that, okay, this is if you construct a number constructed, that is if you adjoined a number constructed by square roots, but why is that the only thing constructable? We're going to do a little bit of some geometric geometry here. If you look at a line, the equation of a line in the plane is basically AX plus BY equals C, okay? The equation of a circle, this is gonna look like X minus H squared plus Y minus K squared is equal to R squared, like so. For which the numbers A, B and C are gonna have to be constructable numbers, and HK and R are likewise gonna have to be constructable numbers for these equations as you build them. And so then we start looking at systems of equations, right? If you take two lines to intersect each other, there's only a couple possibilities. The lines are parallel, maybe the two lines coincide or they intersect at unique points, right? For which if you have two rational lines, the intersection will be a rational point. In particular, if you have two lines with constructable coefficients, the intersection will likewise be constructable, okay? And the important thing is that this is a linear equation. Even though there's two variables, it's a linear equation. So adjoining roots, when you intersect them, you can take one equation and substitute it into the other. So when you intersect two lines, their solution will be the solution essentially to a linear equation that doesn't offer you anything new. When you come to a circle, how can a circle intersect with a line? Well, they might not intersect at all. They could be tangent if you get one point of intersection or you get two points of intersection because it's a secant line. Now, think of this as a nonlinear system of equations. You can take the linear equation and substitute it into the circular equation. That'll rule one of the variables. And so then you have a quadratic equation with respect to one of the variables, like so. In which case, because of that, because of that, when you look at the, when you look at the, you're solving a quadratic equation, that would lead to a simple extension, kind of like this, a square root. Basically by the quadratic formula, you're only including square roots. The worst case scenario is you have circles intersecting circles, for which if the circles don't intersect, you get nothing. They could be tangent to each other, something like that, but most likely the circles are gonna intersect at two different places, like so. And again, as you work through these nonlinear systems of equations recursively, basically since you put a quadratic inside of a quadratic, you get in degree four polynomial. And so the degree of that extension could be four and all of those situations possible. And so because of we can only construct things using lines and circles and analytically speaking, lines give you degree one polynomials and circles give you degree two polynomials. When you join a, when you join a constructible number, recursively you're always gonna get a power of two for the degree. And you can always do this in a sequence. So you can take step by step by step by step, like for example, the number of the square root of one plus the square root of one plus the square root of two. This is a constructible number and you can build a sequence of fields where you have like the rational numbers. This is inside the square root, excuse me, Q would join the square root of two, which sits inside the field where you join the square root of one plus the square root of two. And then this sits inside the field Q would join the square root of one plus the square root of one. You get the idea there. One plus the square root of two, like so. So every constructible number can be fit into a sequence of field extensions where each extension along the way is 2, and that's what verifies this formula right here. So this is the very important observation when it comes to the constructible number field. Now you can just take square root after square root after square root after square root add infinitum, and that makes this an infinite extension because I can get more square roots, more square roots, more square roots, more square roots going on and on and on and on. So there is no finite basis that takes care of it because but at each step you're taking a square root that makes every element algebraic and so k is in fact an infinite algebraic extension of the rational numbers and it satisfies this relationship right here. If you join one constructible number to the rational field the degree of that extension is always a power of 2 and this fundamental observation is how we're going to prove the impossibility of the four geometric constructions we mentioned earlier. Now we're going to do those in short videos by themselves just to make it easier for people to study them but before we do that I do want to conclude with the following observation here because some might consider the solutions to these classic problems somewhat disappointing because the solution is actually that there's no solution but these impossible these impossible results should really be considered as triumphs of geometry and triumphs of abstract algebra. So a quote due to Oscar Morgenstern is the following. Some of the profoundest insights the human mind has achieved are stated in negative form such insights are that there are no perpetual mobile that the speed of light cannot be exceeded that the circle cannot be squared using ruler and compass now technically that should say straight edge and compass there I'm breaking breaking the quote right now sorry continuing on here that similarly any angle cannot be trisected or so on and so on each of these statements is the culmination of great intellectual efforts all are based upon centuries of work those stated negatively these and other discoveries are positive achievements and great contributions to human knowledge the difficulty that the ancient Greeks had and later geometers had in solving these problems was not from their lack of geometric skill it actually came from their lack of algebraic skills and this is actually a huge irony here because the oldest school of jim of Greek Geo Greek mathematics I should say was the school of Pythagoras as as the legend goes the Pythagoreans were in fact pupils of algebra but they discovered that there were numbers which could not be written down exactly what nowadays we call an irrational number can't be written as a ratio of integers and they were disgusted by this in abandoned algebra entirely they couldn't accept that the square root of two was not a rational number this was disgusting to them so they converted instead to geometry since a continuous segment could always be drawn to express any real numbers including irrational numbers so if we take for example in a Saucili's right triangle let's say the lengths are one and one right here then the hypotenuse will its side length would be the square root of two and now while we can't write the number we can't draw the line and this apparently gave gave comfort to the Pythagoreans here now the irony of this in hindsight is that the mathematics the ancient Greeks loved could only be solved using the mathematics the Greeks loathed they loved geometry hated algebra but algebra was the secret to solving these geometric problems they studied so hence the history of these classic geometric problems comes full circle which I know is a horrible pun but I couldn't help myself there um and by the way a circle of course is constructible using a compass and a straight edge