 Welcome back everyone for our continuing lecture series for Math 3130 Modern Geometries at Southern Utah University. This will comprise lecture 26 in our course about rectangles and is loosely based. This lecture is loosely based upon section 3.6 in the textbook Roads to Geometry Third Edition by Wallace and West. It is entitled The Search for the Rectangle. As usual, I am your professor, Dr. Andrew Missildine. So I just want to say hi to all my students and also other viewers of this video here today. Now this section 3.6 here, we're going to be talking about the existence of rectangles inside of neutral geometry. All of the theorems and definitions for this lecture will be comprised in neutral geometry. So that using Hilbert's axioms, the incidence axioms, the betweenness axioms, the congruent axioms, and axioms of continuity. We have at the moment not making any assumptions about any parallel postulates whatsoever. And in fact, I'm going to make one slight revision to the name of our section. Instead of calling it the search for the rectangle, we're going to change its name to be the quest for the rectangle. It feels much more epic that way. And so we're going to make that modification. So first of all, what do we mean by a rectangle? First of all, a rectangle is a quadrilateral, we'll call it a quadrilateral ABC. And so we say this as a rectangle, if all of the four angles associated to this quadrilateral are right angles. The word rectangle itself comes from the Latin for right angle. And so quadrilials with all four right angles, we call a rectangle. Now, in Euclidean geometry, you could define a rectangle to be an equal angular slash equilateral quadrilateral. Now a equilateral quadrilateral is what we would normally call a rhombus. One could likewise define the idea of an equal angular quadrilateral, that is all angles are congruent to each other. Now in neutral geometry, this doesn't necessarily give us what we mean by a rectangle. Because in hyperbolic geometry, you can have equal angular quadrilateral that are not rectangles, because the angle sum is strictly less than 360 degrees. And so each individual angle doesn't have to add up to be 90 degrees themselves. So we're going to see in this lecture and in the next couple of lectures that the existence of rectangles is actually equivalent to Euclidean parallel postulate. Now as many geometers such as Sikeri, Lambert, Lajandra and others were trying to prove Euclid's fifth postulate as a theorem of neutral geometry, they tried many different techniques to get the Euclidean parallel postulate. And one of those approaches was this discussion of rectangles, but try as they might over many decades. I think Lambert was working on these problems for a good number of 40 years or so, and maybe I'm confusing the mathematicians here, but I mean many of these mathematicians spend a lifetime working on these geometric problems and were never able to successfully prove Euclid's fifth postulate as a theorem of neutral geometry. This of course, as we hopefully know by now, Euclid's fifth is in fact logically independent of the neutral axioms. We know this because of the existence of consistent models of hyperbolic geometry for which Euclid's fifth postulate is negated. So in today's lecture, we're going to focus on this critical word, you can see at the beginning of this lemma, if. If a rectangle exists, what can we say? Now Sikeri's approach and some of the others who followed him, with Sikeri's approach, I mean Sikeri was a master of the reducto-audisertum that is proof by contradiction technique. For him, in his day and age, this is sort of like cutting edge logical technology. It's a pretty standard argument nowadays, but Sikeri was trying to create a inconsistent logical system to rule out the possibility of what we now call hyperbolic geometry. That is, he was sort of assuming what he believed to be contradictory statements to the neutral axioms, and he was hopefully trying to find a contradiction he never was able to do so. So we're going to sort of change pace a little bit. If we have a rectangle, what can we say about our geometry? And so if one rectangle exists, we're going to prove some equivalent statements to the existence of rectangles. So for this first lemma here, if one rectangle exists, then there exists a rectangle with two arbitrarily large side measures. So we're going to start off with a rectangle, and so the rectangle that we're guaranteed by assumption will call the rectangle ABC. And I'll draw for you here on the screen. So we have this rectangle A, B, C, and D. And as it is a rectangle, we're going to assume that each of the four angles here are right angles. All right. And so what we want to do is make rectangles that get bigger and bigger and bigger and bigger, and they can increase without bound. So these vertices B and C, I actually want to think of like step one of an induction argument we're going to do in a moment. So let's relabel them B1 and C1. Because the idea is by segment translation and the extension axioms, we are going to copy the segment A, B1. There's some new point B2 so that the segment B1, B2 is congruent to the segment AB1. And likewise, we're going to use segment translation to copy the segment DC1 with a new point C2 so that C1, C2 is congruent to DC1. And this then creates a new quadrilateral. And what can we say about this new quadrilateral AB2, C2, D? Well, first of all, the angle AB1, C1 is supplementary to the angle B2, B1, C1. And as the first angle is a right angle, its supplement will also be a right angle. Likewise, this angle above here will be a right angle as well. And so we have these two quadrilaterals AB1, C1, D and B2, B1, C1, C2. And I claim that these two, the red rectangle and the blue quadrilateral are actually congruent to each other. And this actually follows from a side angle side argument here. The basic idea here is the segments on the top top are congruent. The right angles are congruent. They share an edge which is congruent, right angles are congruent, and the bottom edges are congruent as well. So by side, angle, side, angle, side, we know that these quadrilaterals are congruent to each other. That is, this new quadrilateral B1, B2, C2, C1 is congruent to the original one. The significance of this, of course, is that the corresponding parts are congruent and we can guarantee that angle C2 and angle B2 are right angles. And in fact, we've actually created a larger rectangle. So the rectangle AB2, C2, D is now a, it's a rectangle as well. But the significance of this is that the length of this rectangle is now twice the measure of the original side length. So we got bigger. In fact, this is just the, basically the base case of an induction argument. We can replicate this argument again using some like B3 and C3 as well. The exact same argument guarantee that these are right angles here as well. And therefore we get a rectangle AB3, C3, D3, whose length is equal to three times the original rectangle. And we can just replicate this over and over and over again, guaranteeing that the rectangle AB and C and DN has a side length of n times the original length. And if we do this horizontally, we can extend the length, but we can also do this argument vertically and form a much larger rectangle whose width and length are arbitrarily large. So we can guarantee that the length is n times the original length, and we can also guarantee that the original width is m times the original one. And so we can get these things bigger and bigger and bigger. And invoking the Archimedean principle right here, we can choose to find a rectangle whose length is larger than any prescribed real number, positive real number x, and whose width is larger than any prescribed positive real number y. So these rectangles can get arbitrarily large in their size. Alright, let's actually improve the previous lemma this time around. And so again, under the assumption that if a rectangle exists, we can actually construct a rectangle with specific sizes x and y. X and y are going to be two positive real numbers in this context. Previously, we had shown that the rectangle can be arbitrarily large, but actually this time we can get a rectangle of any size that we want. So how's that going to work? Well, using the previous lemma, we have a rectangle a, b, c, d. So we know that rectangles exist, and the previous lemma then says that we can pick a rectangle a, b, c, and d. So this is a rectangle, so all four angles are right angles right here. We have a rectangle, but then we can, we can choose this rectangle so that the length and width of the rectangle are larger than the numbers x and y. So there's some rectangle whose length is larger than x and whose width is larger than y. And I don't know, I keep on calling the horizontal length and the vertical width. If you don't like that, feel free to switch the numbers around. Honestly, I never knew which one is which, which is length, which is width. All I know is width is not length, it's the other one. So make a choice. I'm sorry if I'm missing one up here. But we have this rectangle whose length and width are larger than x and y respectively. And so that means there has to be some point over here on the length side of this rectangle. So we'll call this point here b prime, so that the segment a to b prime, its measure is exactly x. And likewise, there's going to be a point along the width there, but we'll come back to that in a moment. So taking this segment a, b prime, what we're going to do is we're going to take the unique perpendicular line, that is the unique line perpendicular to the line dc, that intersects this point b prime. And that is construct the perpendicular line going through b prime and is perpendicular to the line segment dc. And let c prime be the foot of that perpendicular line. So we would love, we would love to say that angle right here at b prime is right, but we can't have our cake and eat it too. So what we can do is we can guarantee that the segment from a to b prime is exactly x distance. And we can guarantee that we actually have three right angles, one at c prime, one at d, one at a. But what about this fourth angle at b prime? What can we say about that angle? Well, I can at least say that this is the fourth angle of a Lambert quadrilateral. This guy right here is a Lambert quadrilateral, because it has at least three right angles. But also if we look at the other rectangle b, b prime, c prime and c, this would also be a Lambert quadrilateral. Lambert, the sheepish lion. Anyways, we have these two Lambert quadrilaterals that share the common edge of b prime, c prime. So we're going to try to manipulate that right now. We would love to say that these are both rectangles. We want right angle, right angle. But what if it's not a rectangle? What if we have a Lambert quadrilateral that's not a rectangle? Consider that possibility for a moment. Well, if they're not right angles, that means looking at these two supplementary angles, one of them is acute and one of them would be obtuse. Without the loss of generality, let's assume that the a side is the acute angle, you know, a for acute. And then the other side, the b side is the obtuse angle. Well, that's kind of a problem here, because as b, b prime, c prime, c is a Lambert quadrilateral, its fourth angle cannot be obtuse in neutral geometry. And so that goes as a contradiction here, which actually says that the other side can't be acute either, because if either one of those was acute, the other would have to be obtuse. And so because of that contradiction there, we go back in time and correct what we said before, turns out these are actually right angles, like I was hoping for before. So this would tell us that the two Lambert quadrilauts are rectangles, in particular a, b prime, c prime, and d forms a rectangle, whose side length is exactly this distance x. Well, if we replicate this argument with respect to the width, we can also get another rectangle whose dimensions are exactly x and y. So if a rectangle exists, we get a rectangle of any dimension that we want. And so we're going to use that for our forthcoming theorem here. So I want to now, for another lemma here, make a connection between the existence of rectangles with triangles. Alright, so let's again assume there is a rectangle and by the previous two lemmas, if we have a rectangle, we have every rectangle you could ever imagine, any dimension of rectangle we want. And so I would claim that if at least one rectangle exists, then every right triangle will have an angle sum of 180 degrees. And so we're going to talk about this a lot in this lecture and also in future lectures as well, the next lecture in particular. So if a triangle has an angle sum equal to 180 degrees, we'll say that is a 180 triangle. So this lemma will say that if rectangles exist, then every right triangle is a 180 triangle. And let me emphasize here that we are only working with right triangles for a moment. We won't say things about general triangles until a little bit later, but right now let's suppose we have a right triangle. And so the idea is the following, if we consider an arbitrary triangle, a right triangle I should say, and so let's call it P, Q and R. And let's assume that Q is the right angle. Suppose we have this right triangle. Well, that means that the side, the dimensions of this right triangle will be the segment P, Q and the segment Q, R. And let's assume that the angle sum of this thing is going to equal X. So X equals the measure of angle P plus the measure of angle Q plus the measure of angle R, like so. Let's say that equals X. We want to show that X is equal to 180. So what we're going to do is we're going to create a rectangle using this right, or using this right triangle here in the following manner. We're going to come up with a rectangle whose, so we're going to label the rectangle A, B, C and D. And we want that the, well let me finish drawing the picture here. So we have this rectangle, here we go. So we have a rectangle and this is going to a genuine rectangle. But we want the rectangle to have the following property that the side A, B has the same length as P, Q did before. And we're going to have that the width A, D has the same dimensions as Q, R did before. So this is going to be a rectangle which dimensions will coincide with the legs of the previous right triangle. So we have this right triangle which is congruent to the P, Q, R triangle we had before. And in this triangle, this A, B, D triangle, its angle sum is going to still be X because it's congruent to the original triangle as well. And we're going to say that this angle sum of the B, C, D triangle, its angle sum equals Y. Now we're not ready to say that the B, C, D triangle is congruent to A, B, D and we don't need that here. I mean that will be a consequence later to be shown but we don't care about that right now. So let X be the angle sum of A, B, D and let the Y be the angle sum of the triangle B, C, D. Notice that B, C, D is likewise a right triangle here. Now if we add up the angles of this triangle, so you have angle A, angle A, B, D, angle B, D, A, that will add up to be X. And then if you take angle C plus angle C, B, D and angle C, D, B, that will add up to be Y. And so if we take the sum of these, of the angle sum, if we take the sum of the angle sums of the two right triangles, this will add up to be the angle sum of the rectangle which is 360 degrees. All right? So X plus Y equals 360. But by the Secure Ligeon or Theorem which we talked about before, we have that X has to be greater than or equal to 180, so I'm sorry that's the wrong way. In neutral geometry the angle sum of triangle has to be less than 180 degrees, but this also applies to the triangle B, C, D, so X and Y themselves are going to be less than 160 degrees. And if you put these two properties together, a sum of two numbers adds up to 360 and the numbers themselves have to be positive numbers less than or equal to 180, we can actually infer from this that X equals 180 degrees. It's also true that Y equals 180 degrees, but we don't really care about Y right now. We're trying to show this for X because the only way that X and Y can come together to equal 360 is that X and Y, the thing is one of them can't be less than 180 because that would force the other one to be greater than 180. So if X was strictly less than 180 degrees, that would force Y to be greater than 180 degrees which would violate the Secure Ligeon or Theorem. We can conclude that our right triangles have an angle sum of exactly 180 degrees. This applies for every right triangle under the sum under the assumption that their rectangles exist. And so we'll actually follow this up with a Theorem, which it continues to the same assumptions here. So if a rectangle exists, then all triangles will have an angle sum of 180. That is all triangles would be 180. This is going to be left as a homework exercise and the basic idea of that exercise is going to be the following. If you choose an arbitrary triangle, you can basically cut this triangle into two pieces, two triangles, using a so-called altitude. An altitude is the perpendicular line to a side that goes through the opposite, the corresponding vertex of that side here. If we have our triangle, say A, B, and C, you see illustrated the altitude associated to the vertex C. It's perpendicular to the side, A, B, and it goes to the vertex C. And so if you dissect a triangle into two right triangles, you can apply the previous limit to try to argue why every triangle is 180 if one rectangle exists. Now there's a little bit of an issue here, though, is that when it comes to altitudes, it's very possible that an altitude actually might live outside of a triangle, like something like this. And so if one's constructed an argument using altitudes, you have to be careful that your altitude is interior to the triangle so that it actually dissects into two right triangles. Otherwise, the trigonometry is going to be a little bit more complicated here. The good news is, if you always focus on the largest angle, then the associated altitude will always be interior to the triangle. The next theorem will actually supply an argument for that, so just hold on just for one moment. Don't hold your breath. I mean, it'll be more than 30 seconds here. You might get a headache or something if you do that, but it's coming just a moment. So for your homework exercise, this is the hint I wanted to give you. Use altitudes in the previous limit to prove this theorem right here. Alright, let's get to the major theorem for this section here, what we'll call the all or nothing theorem. What this theorem tells us is that if one triangle has an angle sum of 180, then rectangles exist. What does that have to do with all or nothing? Well, let's actually go to some corollaries of this theorem. So the all or nothing theorem tells us if one triangle is 180, then a rectangle exists. Well, if we combine the previous theorem to the all or nothing theorem, then if one triangle has an angle sum of 180, then all triangles will have an angle sum of 180, because with the previous theorem that the students are going to prove in the homework, then we see that if one triangle is 180, then rectangles exist, and if rectangles exist, then all triangles will be 180 here. There's some details to supply there, but I'm also leaving this corollary as a homework question for students to do. And so this is why we call the previous theorem all or nothing, because if one triangle is 180, then all triangles will be 180. But let's say that if one triangle is 180, but another triangle has a sum less than 180 degrees, what happens there? Well, if one triangle's angle sum is 180, then rectangles exist, and then all triangles will be 180, so you get a contradiction. So you can't have sum equal to 180 and sum less than 180. If one is 180, then they're all 180. And so if one triangle is less than 180, then in order to be a consistent geometry, that would imply that if one triangle has an angle sum less than 180, then all triangles have an angle sum less than 180. And this is why we call it the all or nothing theorem, that if we combine these results, then either all triangles are 180 or no triangles are 180. And by the Sicari-Ligiannis theorem, the only other option is that all triangles are strictly less than 180 degrees. And this then provides, these two corollaries provide us the two paths one can take in neutral geometry. There's the first possibility, which you have a 180 triangle, which is going to be equivalent to Euclidean geometry. The second path is that triangles have angles sums less than 180 degrees, in which case that gives us hyperbolic geometry. For my students, their homework questions, I'll ask you to supply the details of these corollaries here, for which I've kind of told you what to do. So if you watch the lecture right now, then you might have a good head start in the homework. Shocker there, come to class and get help on the homework. All right, so let's get back to the major theorem here, the all or nothing theorem. So we're like a gambler and we're betting all on black, you know, because that's what Batman would do. Anyways, if one triangle exists with an angle sum of 180, then all of them do. So let's actually first get to that argument of altitudes right here. Why is it safe to assume the altitude is interior to the triangle? So let's consider a triangle for a moment where we have, we'll call it A, B, and C. And for the sake of argument, let's assume that the altitude is exterior to the triangle. All right, so maybe we get a picture like the following. We have a point D that lives on the line AB, B will be between A and D, and then the altitude associated to the vertex C has D at its foot. Why could something like this not happen? Well, what if angle C was the biggest vertex of the triangle? Well, what I can say is the following by the security of the genre theorem, because we have a right angle in the triangle right here. The only way, well, if we have a right angle, then there can't be any other right angles in this triangle, the triangle CBD. And there can't be any obtuse angles right here. So this guarantees that angle DBC will be an acute angle. Well, then it's supplement, which is angle ABC would have to be obtuse. And this is sort of the problem that, again, if you look at the angle sum of ABC, if you have an obtuse angle, that's guaranteed to be the biggest angle inside the triangle. And therefore that would contradict the CBD, the biggest angle in the triangle here. So what we can do is if you pick the biggest angle of the triangle, this will guarantee that the altitude is interior to the triangle. That's useful in that previous homework question. We're going to use that assumption right here. So we can safely assume that the altitude associated to C is an interior, and therefore D, the foot of that altitude is between A and B. All right, let's slide this thing up a little bit. So we've dissected our triangle into two pieces. We're going to let the angle sum of ADC be called X and the angle sum of BCD, we're going to call that Y. All right. So in this triangle right here, let's play around with some angle sums. If we look at just X for a moment, then that comes from the angle A, the angle ADC and the angle DCA. If we look at the angle Y, sorry, the angle sum Y, that comes from the angle B, the angle BDC and the angle DCB. Well, if we rearrange these things right here, we get angle A, angle DCA and DCB. That comes together to form angle C of the original triangle and angle B right here. These three angles add up together to be the angle sum of the original triangle, which by assumption we're going to take to be 180. And then the two other angles we haven't accounted for yet, ADC and BCD, these are both right angles because they're associated to the altitude there. So they're both 90 degrees, that adds up to be 180 between the two right angles. If you take those two right angles plus the original triangle, we get an angle sum of 360 degrees. So we get that X plus Y adds up to be exactly 360 degrees. But then this puts us in a situation very similar to what we had in the previous argument. Both X and Y, if they add up to be 360 and they themselves cannot exceed 180, we have to conclude that X and Y both equal 180 degrees. So these angle sums are both X and Y, these angle sums will add up to be exactly 180 degrees. So if we have a triangle who has an angle sum of 180, we can construct a right triangle whose angle sum is 180 as well. And so with this right triangle, I'm going to build a rectangle. So we're going to get a picture that looks something like the following. So of the above picture, take the triangle ACD, we showed this one was a right triangle by construction. Its angle sum is 180 degrees, this was a right angle right here. And we're going to take this triangle ACD and we're going to copy it along the line AC. And so there's going to be a point B prime, it's on the opposite side of the line AC from D. And this is going to be congruent to the triangle ADC. So we've constructed this right here as corresponding parts are congruent. This new triangle AB prime C is a right triangle as well. So angle B prime is right and then corresponding angles are also congruent. So angle CAB prime is congruent to angle ACD. Likewise, angle CAD will be congruent to angle ACB prime. So we have some correspondence of angles going on right there. And so if we play around with these angles we have right here, what do we get? Well, angle A of the rectangle can be broken up into two pieces. There's CAD and CAB prime associated to the two triangles. And likewise, angle C is going to be dissected into two parts ACD and ACB prime. The other two angles of the rectangle D and B prime we know to be right angles. So those are going to add up together to give us something that... They'll write angles. We'll come back to that in a second. So we can break up angle A and angle C into these two parts and if we rearrange them we can get the angles associated to the first triangle ADC, which you get right here. And then we can also get the angle sum for ACB prime right here. And so by our previous argument the angle sum of ADC is 180 degrees and because AB prime C is congruent to it, it's also going to be 180 degrees. The angle sum of this quadrilateral is 360 degrees. We don't know this quadrilateral is yet a rectangle. We do know it has two right angles, but because they're kitty corner from each other we don't actually know that these... We don't know this like a security quadrilateral thing, but we do know there's two right angles. But let's play around with these angles A and C a little bit more. We know the angle sum of the quadrilateral is 360. What else can we say about this quadrilateral? Now if we look at the angle sum in a slightly different decomposition we still have angle A, we still have angle C broken up. But if we put them together in a different way, because after all remember we know there's a correspondence of angles in this triangle, in this rectangle here corresponding angles from the different triangles. If you combine these together, the entire angle A is going to be congruent to the entire angle C. If we make that rearrangement, we're going to see that the two different parts of angle A will be congruent to the two different parts of angle C. These will come together and actually make two copies of angle A. The other two angles D and B prime are both right angles, so their angle measure is 180. And since the total of the quadrilateral is 360, we can subtract the 180 part to give us two times the measure of angle A is equal to 180, divide that by two. We get the measure of angle A is 90 degrees. And since C was congruent that also gives us 90 degrees and therefore we're going to have a B prime CD is a rectangle. So what we've seen today is that if a rectangle exists, then all triangles have an angle sum of 180 degrees. We've also seen that if a single triangle has an angle sum of 180 degrees, then rectangles exist. So logically, 180 triangles is equivalent to the existence of rectangles. And we're going to see next time that these are equivalent to the Euclidean Parallel Postulate. All right, thank you for watching. If you do have any questions or comments, please post those in the comments below. As usual, in the description of this video, you can actually see a complete transcript of this lecture we went through. You can download those either from the link provided or from my faculty website. And yeah, feel free to subscribe if you want to hear some more about these fun little videos. And if you are one of my students, please turn on your homework and I'm available to answer questions as usual. See you next time. Bye.