 Hi, I'm Zor. Welcome to Unisor Education. Last lecture was about similarity of angles, and it was a very simple theorem that scaling preserves the angles. Now we will go into a little bit more difficult area of scaling. We will talk about scaling of segments. Well, let's start with something simple. If you have center of scaling and some factor f, and we will consider any segment which has an end point in the center, exactly the center of scaling, and end point somewhere else. Well, just for example, factor is equal to 2. By definition of scaling, point m will be transformed into point n, which is f in this case 2 times greater than the distance from q to m. So, what is the image of segment qm? It's a segment qm, which is exactly f times greater. So, obviously we see that scaling is stretching these segments with one end point in the center in exactly the number of times which corresponds to the factor. Well, but this is a definition of scaling, so there is nothing new. What is more interesting is the following. Let's say you have another point, let's say point k, and we also scale it by the same factor into point l. Obviously, segment qk will be transformed into segment ql, length of which is exactly f times greater. But how about this segment in between? Well, it will be transformed into this segment, and k will be transformed into nl. Why? Because m will be transformed to n, k will be transformed to l as points, and straight line as we assumed, we didn't really prove it, but we assumed that the straight lines are transformed into straight lines. Now, we proved actually that the image will be parallel. So, these two straight lines are parallel. Now, consider this triangle qml. In case factor is equal to 2, it's really very easy to see that mk is a mid-segment of this triangle. It connects two midpoints of sides, and we know from the theorem about triangles that this particular segment mk mid-segment is half the size of the base. In this case, base is nl, which actually proves that in case our factor is equal to 2, any segment is stretched by the factor of 2, any segment mk. So, if we have started from this segment mk, then we would just stretch m into n, k into l, make this picture, and obviously nl is twice as big as mk. Okay, so for factor of 2, we have proven this particular theorem. How about factor of 3 or 4? How about the factor of three quarters? That's not as obvious, right? So, but let's do it one by one, step by step. When we do it step by step, it's easier, right? So, let's start with f equals to some natural number n. Now, what happens is the following. Let me just here. So, it looks like it's four times. I'm stretching, right? So, I have a segment mk, and I'm stretching by the factor of n. So, qm will be stretched to the qn, and qk will be stretched into ql. Now, we have this number n. So, we know that this is one nth of this. So, if I will put these dividing points along the segment qn, the first point will be fm, then it will be the second point and the third point and the fourth will be n itself. Now, if I will draw parallel lines through each of these points, you know that if equal segments are positioned on one leg of an angle, and there are parallel lines which cross another leg of the same angle, then I will have congruent segments of equal lengths cut on the other leg. Again, if you don't remember, I can refer you back to one of the theorems which we have proven before whenever we were talking about parallel lines and angles, etc. It's all proven. So, if equal segments are on one leg of the angle and there are parallel lines, it will be equal on this side as well, which means that this qk will be one nth of ql. So, the parallel lines will actually divide ql also in n points. And what actually follows from here that these lines are also increasing proportionally. How to do this? Well, we can always draw parallel lines this way, right? So, if these are equal, then these also would be equal in size. So, this is, and considering mklp is a parallelogram, obviously this segment mk is equal in lengths to pl, which is one nth of the total lengths of nl. So, from drawing parallel lines, we see that on each line we have equal number of segments, small segments, which is actually nth equal segments. So, if these are n for in our case, then these will be n for in our case, all equal in lengths. And one of them is equal exactly to the beginning to the original segment from which we started. So, we see that this total big segment nl, which is an image of mk, is exactly n times longer than this particular segment. So, for integer n, for natural actually, we talk about positive, negative will be exactly the same thing because it just goes into another direction. So, for all the natural numbers n, we have proven that scaling of any segment wherever it is would convert it, would transform it into another segment, which is n times longer. Now, next is, we will do one over n, where n is integer, like one-half, one-third, one-quarter. Now, it's actually quite similar. Let's consider that this is a segment which is given to us, and this is the image. So, from here, we transform to here. Now, applying exactly the same logic, we can see that if we will divide our given segment into n equal parts and draw parallel lines, we will obviously come to this segment mk. And in this case, it will be one n of this segment, but now this is the original segment, and this is this image. So, because this scaling is reversible in this case, so this multiplied by n will give you this, and this divided by n will give you this, and from this, it's easily following that multiplying by the factor which looks like this, one over some integer number, then the length would be shortened by this factor n, or multiplied by one n, so if you wish. So, again, our theorem which says that the factor is actually the multiplicative factor for the lengths of the segment in case of transformation of scaling, so we have proven it for natural numbers, we have proven it for one over natural numbers. Now, what if I have any rational number? Now, instead of going through some geometrical considerations, let's do it a little bit more, how should I say, it's not scientifically, but I like this approach which I'm going to offer you a little bit more. It's a little bit more mathematical. It just proves that you can think not only in terms of concrete pictures, but also in terms of abstractions, for instance. Now, you remember that factors can be multiplied if consecutive transformations of scaling are made. So, first we made a scaling of one particular factor, and then scaling of another particular factor, they are commutative and dissociative, if you remember that, right? So, basically, scaling by the factor m over n can be represented as a scaling of m, and then scaling of one over n. But both, we have already considered before, we know that the first scaling will multiply the lengths of every segment by the factor of m. The second scaling will multiply by the lengths of any segment by the factor of one over n. But the gain because multiplication of numbers is commutative and dissociative, etc., we can say that multiplication by two numbers is equivalent to multiplication by one number. So, for any rational number m over n, we basically have proven this theorem that the factor is just a multiplicative factor for the lengths of any segment by referring to certain properties of transformation of scaling like commutative and dissociative property. Not to, like, explicit proof in geometrical fashion. And they do like this particular proof a little better because it's more, well, at least to me, it appeals to my kind of longing for abstract thinking in terms of operations rather than in terms of geometry. Okay, so now we have proven that scaling by any factor which looks like this, which means any rational number as a factor, is multiplying the lengths of any segment, not only the one which has an end point at the center but also any other segment, it multiplies by this factor. Well, have we exhausted all the different factors? No. I mean, we are talking about all the real numbers right now, which means rational and irrational. Well, this is, again, something which I don't want to mislead you offering some quasi-proof, which is not really a rigorous proof. An honest point of view is that it's not easy to prove for irrational numbers and we have to really go through a proper definition of irrational numbers, etc., which is not easy. However, there are certain considerations which might help our intuition. Now, we know that irrational numbers can always be approximated with rational numbers to any degree of precision. Let's say a rational number is represented as an infinite, unperiodical decimal fraction. Well, we can always cut it at any point, however far towards the tail, and say, okay, this is a rational number, which is approximation of our irrational number. So, for any rational number, this theorem is true and that's why intuitively it is kind of obvious that there must exist some proof that the theorem is true for irrational numbers as well. And that's where I would like to leave. I don't want to go into details of irrational numbers and rigorous proof. It's related to some limits, etc., etc., which is a little bit different kind of category. Let's just assume that you intuitively understand that since the theorem is true for an irrational number, it must be true for any irrational, and that's why for all real numbers, the theorem is true. So, scaling or homophagy of any segment basically is converting, transforming it into another segment, which has the lengths equal to the original lengths multiplied by the factor of scaling. Any segment, no matter how it's positioned relative to the center of scaling. Okay, that's it for this particular lecture. Everything is on Unizord.com. I do encourage you to take a look at this particular website. It will help you not only just to learn some material, but also you can take exams. Your parents or supervisors or teachers can actually participate in this educational process by enrolling you in certain courses, like in the course of similarity, for instance. It's a course of lectures, which will have one particular exam at the very end, and they can check the score and they can say, okay, you have to go again through these lectures and try to solve these problems, etc. So it's a very good tool for not only for students to learn, but for parents and teachers to supervise the educational process. Thanks very much for listening.