 Hi, I'm Zor. Welcome to Unisor Education. We will talk today about algebra of segments. Basically, it's something which is in between algebra and geometry, or rather, it accumulates in itself, algebraic and geometrical properties of segments. Well, let's think about segments. The only important characteristic of a segment is, basically, its lengths. Well, there is no much shape in it, basically. So, all the different segments have the same shape in general, just different lengths. So, length is a numerical value, and that's why working with segments can be, in some way, replaced or supplemented by working with their lengths using purely algebraic approach. And we have already spoken about certain algebraic, if you wish, manipulations with segments. Like, for instance, we can add two segments, gauging another segment of the length, which is equal to some of the two lengths. At the same time, so we can add two segments, we can subtract from a smaller, from a bigger segment, we can subtract the smaller one. So, in this case, we just put them together. In this case, we have to go backwards by the length of the second segment, and whatever is left will be the difference. We also can multiply segments by any natural number, like, by ten, for instance, because multiplication by a natural number is basically addition with itself, a certain number of times. We also can divide the segment into equal parts, and the number of these parts can be basically any, for instance, how to divide the segment into three different parts. Well, you have this angle, you have another segment of any lengths, three times put on this side of the angle, then connect the ends and put the parallel lines, and these will be also equal to each other parts of the given segment. We have already learned that during our previous lecture. So, if we can add, we can subtract, we can multiply by a natural number, we can divide by a natural number, which means we can multiply by any rational number, where m and n are any natural numbers. Basically, first multiplying and then dividing by n. So, we have certain operations already covered. I will go into a little bit more complex operations. I have put them in a row, which we will go through during this lecture. And basically, during the explanation of how these manipulations are done, what you will be introduced to these algebraical concepts of manipulation with segments. Okay. Number one, how to multiply a segment by square root of two. Well, this is an easy part, and all you need to do is remember the Pythagorean theorem. Remember, if you have two, if you have a category with a and b, then c is equal to square root of a square plus b square. So, you remember this. Now, what if a and b are equal in lengths, and they are equal to that particular segment a, which we would like to multiply by square root of two. So, replace b with a. A square plus a square, which is equal to square root of two a square, which is equal to a square root of two. So, to multiply a segment by square root of two, we have to build this triangle. It's a right triangle with both categories equal to our segment in lengths. So, you have the right angle. You have your segment on one side of the angle, and then another, and then the diagonal, which is actually a hypotenuse of this particular triangle, is exactly the segment we're looking for. Its length is equal to the length of the kaito juice, which is a times square root of two. Now, how to divide by square root of two? Well, obviously, we have to use something similar, but in this case, it's the kaito juice, which is equal to a. And we have to build the right triangle with both categories equal to itself, equal among themselves, and kaito juice being a given segment. Well, how to do it very easily? You have to remember that locus of all the points from which given segment is viewed at 90 degree, this is 90 degree. So, locus of all the points from which this angle is 90 degree is a circle with a as a diameter. If you remember, inscribed angle is measured by half of the arc supported it, and the arc supported is basically half a circle, because this is the diameter, that's how we built our circle, which is 180 degree, so half of this is 90. So, any angle with vertex on this particular circle is 90 degrees, even if it doesn't look like on my drawing, although it actually a little bit almost like. So, we start from having our segment as a diameter of a circle. Now, out of all these right triangles, we have to choose the one which has equal quantity. How to do it? Well, draw another diameter perpendicular to this one, and wherever it crosses the circle is the vertex we are looking for, because obviously since these two diameters are perpendicular to other, to each other, then this angle will not only be 90 degree, but also these two corners will be equal to among themselves. So, this is a right triangle with equal quantity, and hypotenuse equal to 8. Now, if this is x and this is x, obviously again, x square plus x square is equal to a square, 2x square equals 2a square, and x is equal to a divided by square root of 2, and that's what's needed for the second problem. Alright, next. How to multiply by square root of any number, any natural number n, square root of 3, square root of 4, square root of 27, etc. Well, let me just give you the following process, if you wish. We start first with a and a, and this is a square root of 2, as we have already proven before, right? Let's wipe out these guys. Let's add a and calculate this. This is the hypotenuse. This is the right triangle. One side is a square root of 2, which we know how to do, and this is still a. So, this x would be a square plus a square plus a square root of 2, which is equal to a square plus 2a square, which is equal to 3a square, from which x is equal to a square root of 3. Okay, so we got a square root of 3. Next, well, I'm sure you have already guessed. If this is again a, then this would be x. But in this case, x square is equal to a square plus a square root of 3 square, right? Which is a square plus 3a square, which is 4a square, from which x is equal to a square root of 4. So, this is a square root of 4. Yes, square root of 4 is 2. I know that, but let's not just go into this. Next, again, as you probably have already figured out, again, right angle and a again. So, what would be the length of this guy? All I have to do is change 3 to 4. So, this is a square root of 5. Now, as you see, we can continue this process indefinitely, and eventually we will get a very nice shape, actually. Right angle, a square root of 6, right angle, a square root of 7, etc., etc. So, it's just unwinding, actually. We have bigger and bigger ranges, and we can continue this process indefinitely. Our ranges, well, it's not really a radius, because it's not a circle, but it looks like a circle, but its radius is actually increasing. So, there is a special name for this particular figure, and it's part of the higher level mathematics, but it doesn't really matter. So, what this figure is, we have the process, which you can actually compare with the induction. First, we have learned how to calculate a times square root of 2. So, this is like n equals 2, the beginning of the induction process. And then, if you already know how to build a square root of k, then it's very easy to build a square root of k plus 1 by having this is a square root of k, this is a, and this will be a square root of k plus 1. So, this is the process, the inductive process. So, you know how to begin this process, and you know how to make one step from k to k plus 1, which means you can make a step from 2 to 3, from 3 to 4, and that's how you can build any square root of n. That's this third problem. The algebra which I'm using right now is not much of an algebra. It's basically a Pythagorean theory so far, but you will get a little bit more complicated algebra. Now, how about this one, number 4? So, there are many different means. There is an arithmetic mean of two numbers, which is this. There is a quadratic mean, which is this square root of quadratic average. There is a geometric mean, which is square root of their multiplication. So, these are a little bit more complex formulas. So, if A and B represent lengths of two given segments, the question is how to build average segment between these two? How to build the segment which has certain mean, and there are many different means. All these means are in between A and B. It's just different to calculate it. Let's just consider if A is equal to 2 and B is equal to, I don't know, let's say 8. What will be? A plus B divided by 2. So, this is the arithmetic average. Arithmetic average would be equal to 2 plus 8 10 divided by 2, 5. Now, this quadratic average, excuse my voice, 2 square 4, 64, 68, 39, square root of 39, well, it's 6 something. So, quadratic average would be 6. And finally, the geometric average between 2 and 8 would be multiplication is 16, square root is 4. So, geometric average would be 4. So, these are numbers in between 2 and 8, but they are different. So, we can calculate differently the mean between two numbers. Many different means. Arithmetic average is more often occurred in the real life. This type of average is more applicable to some statistical calculations. This is rarely used, but in any case, it does exist. It's called geometric, quadratic, and arithmetic average. All right, so let's build all these three averages using A and B as the lengths of two given segments. Well, the first one, A plus B divided by 2 is simple, because if you have one and then you have another segment, you put them together, that's A plus B. How to divide this in half? You just draw a perpendicular bisector. So, that's easy. This one. A squared plus B squared divided by 2. Well, let's consider this as square root of A squared plus B squared divided by square root of 2. Now, we do know how to build this particular segment. The segment which has the lengths equal to the square root of sum of squares of given two segments. Well, very easy. It's basically a game, the figurant theory. You have one calculus A, another calculus B, then the diagonal would be square root of A squared plus B squared, right? That's the theory. That's the piece of argument theory. Now, after you have built this particular segment, next step is divided by square root of 2. And you already know how to do it. We did it as number 2. So, that's the solution. That's how to build quadratic average. Now, how to build geometric average? That's an interesting point. I would like to remind you an interesting theorem in geometry of right triangles, which we actually did cover before. When we were talking about similarity. So, if this is A, B, C, and this is an altitude, C, H. Now, obviously triangles A, C, H, which has this acute angle is similar to triangle A, B, C, which also has the same acute angle, because another angle is 9°. Same thing about B, C, H. It's also similar to the big triangle because they have common acute angle and another one is 9°. So, all these triangles, all these three triangles, I would say, two small inside triangles and the big one are all similar to each other. Angles are equal, these angles are equal, this is 9° and this is 9°. Now, from similarity, we can have the correspondence of the sides. So, let's consider this piece to be A, this piece to be B, and the altitude H. Now, since these guys are similar to each other, then the ratio of the corresponding sides should be the same. Now, in this small triangle A, C, H, A lies against double-arct angle and in this triangle B, C, H against the same angle you have the H. So, A over H is equal to... Now, let's talk about different angle. Against this single-arct angle, in this triangle A, C, H lies H and in this triangle B, C, H against the same angle you have B. So, that's the proportionality. A relates to H as H relates to B, from which we can do A, B is equal to H squared and H is equal to A, B squared root. So, that's the answer to this particular problem. The question is, given A and B how to build H? Well, elementary. We will do exactly the same thing as before by having A plus B as a diameter of a circle which is a locus of all the points like C where this particular segment AB can be viewed at 90 degrees. So, all the angles are 90 degrees from any point on this circle. Now, we have to find a point on the circle which has the altitude equal to H. Well, very simply. Let's just have this altitude from this point of the height H, of any height. I mean, let's just draw a perpendicular. We don't know the H. And whatever the point of intersection is this segment would be the one which we are looking for because, number one, this angle is 90 degrees because it's on the circle. And these two segments are equal to whatever we wanted, A and B, are given segments. So, this particular segment would be equal to in lengths to square root of multiplied lengths of each of those. So, again, we put them together, A plus B basically, have a circle on this as a diameter and from this point, we draw a perpendicular and whatever the lengths from this point to the circle is actually the answer to this. Well, obviously it crosses on both sides but these two are equal in lengths because the perpendicular to the diameter is always, chord is always divided in half. All right, so that's how you build a geometric average. That's the next one. And the last but not least and probably more interesting is related to so-called golden ratio. Now, what is golden ratio? The golden ratio in geometry is the following thing. If you have a particular segment, then let's say this is X and this is A minus X. So, the golden ratio is basically the point which divides our given segment of the lengths A in two halves where the smaller part relates to the bigger part as bigger part relates to the whole segment. So, A minus X divided by X is equal to X divided by A. So, that's basically the equation for X which we have to solve somehow but solve it geometrically. Now, just as a side note, this golden ratio is very much popular in the nature. Lots of different things, trees, human body, etc. They contain certain division points which are golden ratio. Like, for instance, elbow is dividing the arm in golden ratio on this level. Something with leaves, something with many different aspects. Golden ratio is really very much a natural thing, so to speak. So, how to divide a segment in golden ratio? So, this part is A minus X and this is X. The whole segment is A. So, the smaller part relates to the bigger part as bigger relates to the whole segment. Well, let's use algebra. Now, how can you use algebra? Well, this is basically a quadratic equation. Let's do it this way. From this, we go to A minus X times A is equal to X times X, which is X squared. So, this is an equation from which we can derive the value of X. Well, let's simplify. X squared minus AX goes to this AX A squared minus A squared is equal to zero. All right, fine. This is algebra so far. Now, what are the roots of this equation? X first is equal to minus A plus square root of A squared by four plus A squared. And the second value is minus A2 minus A squared by four plus A squared. So, two solutions. Obviously, it's a regular solution to a quadratic equation. Now, this one is definitely not good for us because this is negative. So, negative cannot be the length of the segment. So, let's forget about this one and this is a solution. All right, now the question is how we can construct a segment A at a segment X which has this particular length? Well, easy. First of all, let's construct A divided by two. So, if this is A, this is A divided by two. We know how to do this. Now, what is this? This is a Pythagorean theorem of a hypotenuse of a triangle where this is A divided by two and this is A, right? Because A divided by two square is A squared by four and A is A squared. So, this is our square root of A squared divided by four plus A squared. So, we get this point. Now, all we have to do is subtract from this this. Just put the compass here, subtract this and whatever is remaining would be the solution of our equation. So, we have geometrically solved the quadratic equation and well, basically we have found this particular piece which should be used here. This one is equal to this one to divide the given segment in golden ratio. Now, incidentally, for strictly algebraic purposes we can simplify this, obviously, because this is equal to minus A divided by two plus square root of five A squared divided by two. A goes out, so it's A divided by two square root of five minus one. That's what it is. So, this is an algebraic algebraic solution to a golden ratio problem. But this is a pure algebra that I did before was pure geometry. Well, that basically concludes my different problems which I wanted to address in some kind of synergy between algebra and geometry. So, you can call it algebraometry. But it's only about segments because segments have only one characteristic, which is lengths. So, that's why we are using algebra for segments. Number is always corresponding to some kind of segment and segment to a number. There is no extra shape kind of collages. The risk, however, the subject in mathematics which is called algebraic geometry, this is a much more complex subject and it actually is explaining certain geometric collages using algebraic methods or it's describing certain geometric properties of certain objects in multi-dimensional space using algebraic methodology. It's very interesting, but very, very complex piece of mathematics requires huge amount of imagination and creativity. Well, this is just a preparation for something maybe more difficult, but it's still interesting that using geometry you can actually solve certain algebraic problems or use algebra to solve certain geometric problems. That concludes the lecture for today. Thanks very much for your attention. Don't forget that Unisor.com contains this and many other lectures. It's supposed to be studied in sequence as a course of advanced level mathematics and it's open for everybody and parents are encouraged to take charge and control over this educational process. Thank you very much.