 Hi, I'm Zor. Welcome to Unisor Education. We continue talking about solving problems. Now, as I was actually saying many times, solving problems is probably one of the most important goals of this course. Now, there is a prerequisite course called Math for Teens. It's presented on Unisor.com, and this course is called Math Plus and Problems, which is based on that theoretical course, Math for Teens, but presents certain unorthodox problems, not exactly to verify how well you know the theory, but really kind of train your mind to think creatively. So, most of these problems are not of a kind you probably can be presented in school. But again, in school usually you're presented with problems just to verify whether you know certain theoretical concepts. For example, if they teach you how to solve quadratic equations, they will give you a lot of quadratic equations, which you have to really solve using the same formula, not very creative process. Now, in this course, I'm trying to present certain problems where you don't really know how to solve it. You have to come up with certain new idea, new methodology. Obviously, after solving rather large number of these problems, they will start basically kind of looking like one another, which means some thoughts which you came up with before during the working with previous problems will definitely give you a clue how to solve the next one. So that's why it's very important to solve as many problems as possible, because every new problem of this unorthodox type presents you certain challenge, and if you overcome this challenge, it will be much easier to do the next one. Okay, enough about philosophy. Let's just solve problems. So the first problem is how to prove that this sum of square roots is irrational number. Now, what does it mean it's irrational number? It means that if I will suggest that it's represented as a fraction between two integer numbers, and let's assume that if there is some kind of a reduction, like both numbers are divisible by three, so I will divide it by three. So this is a minimalist, so to speak, representation of this rational number. There are no common divisors between t and q. Okay, so if we assume this, we have to come up with some kind of a contradiction. If we will, it means our initial suggestion that this is irrational number is wrong, which means it's irrational. So now, before approaching this problem, I would like to solve something much, much simpler, which I think, I'm not sure, but maybe I have presented this problem in the main theoretical course called Mass for Teens. What if I would like to prove that square root of two is irrational? Same thing. I assume that this is p over q, where p and q are integer numbers without common divisors, and I will just change it into square root of two times q equals p. I will square both sides. I will have two q squared equals to p squared, which means that p is even number. It must be divisible by two, because p squared is divisible by two. Now, if p squared is divisible by two, it means that p must be divisible by two. Why? Well, you can always represent any number as a product of prime number, p1, p2, and p3 are prime. So this is the representation of p. Now, one of them, if there are no two's among these, then there are no two's among these. So p and p squared must be even or not even at the same time. So p squared is even, which means I can represent it as p is equal to two r, let's say, which means two q squared is equal to four r squared, which means q squared is equal to two r squared, which means q is even number. So it appears that both q and p are even numbers, which means we can cancel two and we can reduce this fraction by two, which I assume there are no common divisors between p and q. The type of already reduced it as much as possible. So that's a contradiction. Now, why did I present this simpler case? Because this is exactly the same kind. However, it involves much more calculations, so to speak. So what are we doing? Basically, the same thing. So I will put square root of five to the right and multiply everything by q. So I will have q square root of two plus q square root of three equals to p minus q square root of five, right? So I transfer five to the right and multiply everything by q. Now I will square both sides. So I will have q square two plus two q square root of two square root of three, so it's square root of six plus q square times three equals p square minus two pq square root of five plus q five. That's a square. Now q square three q square, well, two plus q square three q square five so I can just reduce it. They are equal. So why is it better? Because there are three square roots and now only two square roots. Well, I will square it again. So I will have four q fours times six so I can put it 24 equals p to the fourth minus four p cube q square root of five plus four p square q square and five. So it's 20. Now I will leave this on the right everything else on the left plus 24 q to the fourth minus p four minus 20 p square q square equals minus four p cube q square root of five. Square root again. And now I'm not going to continue anymore because after this you will see that everything, if you will square it, all the members except p to the eighth when we will square this one you will have p to the eighths. Everything else will have some even number 24, 24. So if you will square both sides you will have eight here, you will have something here definitely all even numbers which means p to the eighth must be even number which means p must be even and if you will put it here same thing will be with q. So basically I don't want to spend time it's obvious how to proceed but basically that's the way how you will get everything. So eventually for q you will also need to have it as an even number which means p over q can be reduced. Well that's basically it. I'm just quite frankly don't want to spend much time lecturing time when you will do all this. I'm not sure but maybe I will put some more detail explanation in the written part. You see every lecture on unizord.com including this one obviously will have a textual part so you have the video representation and the textual part and the textual part sometimes I might actually do all these calculations and put it there but I would encourage you to do it quite frankly because it's a good exercise you have to do it accurately square of these things is a big one it's three of them so in square it will be nine members it's a big deal but in any case it's a good exercise so I suggest you to do it this way. Alright so that's the first problem basically it's like almost finished I just don't want to spend any more time on the trivial calculations but idea is the same so p over q must be p and q both must be even number because of all this and reducible by two which we assumed is not the case okay another one is kind of interesting it's extremely simple however there is an interesting theory interesting story actually which happened with this particular problem and artificial intelligence which is right now in bulk kind of thing so the problem is here is a function five minus four x to power one thousand times three x minus four to the power one thousand and one now p of x is basically a polynomial open the parentheses if you will really use five minus four x you will raise it in the power of one thousand you will have a polynomial and this thing also will have a polynomial then you multiply one by another you will have another polynomial so basically it's kind of can be represented as a zero times x to the power what's the maximum power two thousand and one plus a one times x to the power two thousand plus etc plus a two thousand times x to the first plus a two thousand one times x to the power of zero which is one so this is polynomial in some kind of form which is not really a canonical form this is a canonical form of the same polynomial so what's the problem the problem is I would like to find some of all coefficients from zero to two thousand one so what I did I went to a artificial intelligent website and just ask this question what is the sum of the of these multipliers, these coefficients in the canonical representation of the polynomial which is given as a formula like that well guess what artificial intelligence program well it's a computer program it's a good search engine actually it found that this is supposed to be like a Newton's binomial so they used the formula for Newton's binomial got sigma of members whatever it is according to the polynomial binomial sorry same thing with this you have two sigmas then multiplied gave the whole page of functions and came up with the right answer now what is the right answer but let's just think about it the smart person if it's asked if this is a canonical representation of the polynomial p of x what is the sum of all these all these coefficients well that's actually p of 1 if you will put 1 instead of x all these x to the power will be 1 which means you will have the sum of these coefficients but now let's calculate p of 1 in this representation which is much easier because 4x for x is equal to 1, 5 minus 4x is 1 1 to the power of 1000 is 1 3x minus 4 when x is equal to 1 is minus 1 minus 1 to the power of 1001 it's odd number 1001 is odd which means 1 minus 1 will be multiplied by itself odd number of times and the result will be minus 1 so it's 1 times minus 1 which is minus 1 and again as I was saying this program which I was talking about did it I would say force method without thinking and that's the difference that might point to it that's the difference between artificial intelligence and the human being smart human being because in some cases you need some creative moment I mean it's kind of smart way of doing this and in most cases I'm not saying all of them in most cases all these programs they are like robots they are doing something mechanically whatever they have taught how to solve the problems they are solving exactly the same method they do not really invent anything new at least at this particular moment they are just using whatever already known and that's the problem there are some students if you ask them something which they were not really explained how to do beforehand they will say well I don't know how to do it teach me and I will do it you teach him and then they will do it this is one kind of way one way one approach with one type of students other students if you will give them the problem which was not explained how to solve it before they will start thinking about it creatively and that's the most important part creative thinking the purpose of the whole course whatever I'm just presenting is to kind of force you to think creatively to think about certain methods which might not be something which you have already been explained in school or in textbook or somewhere else and your own way how to approach this particular problem and solve it this way and that's exactly the good example how smart people would do it thinking about some of these coefficients they will say hey if you put x is equal to 1 that's what you will get so let's just put x equal to 1 in the original formula and that's easy so that's the difference and that's why I think this particular problem gives you the idea about well what's the purpose of the whole course actually you have to think creatively and think about this you look at this you don't know how to solve it obviously I do not expect anybody in a reasonable mind to do these newtons binomial and multiply them etc it's just not the way people really should do in this particular case computers can do it but people don't people do it smart in a smart way and the smart way is this one maybe there is no smart way and that's why we have computers and let them do it whatever whatever is necessary in brute force way but still I mean it's very important to find something which computers cannot find to computer program which basically stands behind the artificial intelligence that if you would like to find a sum of coefficients of polynomial you have to really find it the value of the polynomial with x is equal to 1 once I put it in there it's already kind of a known thing and the next time maybe computer program will do it the right way but I have to teach it first the programmers should put it into the artificial intelligence program which basically does all these search things etc and they will find ok there was such a problem it was solved that particular way so that's what I will do it next time only things which completely new require of this creativity but again it means that only people who are creative can make certain progress anywhere computers don't do this type of thing computers just repeat whatever they have been told already about ok next problem ok the reason equation with two unknown x plus y equals 3 x square minus xy plus y so what I need is I cannot solve this equation in just as it is because I have to really understand that one equation with two unknowns obviously will give infinite number of solutions but question is where I am looking for solutions traditionally all these xy equations are solved in the realm of real numbers sometimes complex numbers but in this case I would like to solve it in the different domain of numbers only among prime numbers integer prime numbers I would like to find solutions among integer prime numbers so we are narrowing down the possible solutions to only integer prime numbers ok now in one case since the numbers integer prime prime numbers is much smaller set than all real numbers it is supposed to be easier but still it is two variables and one equation so there is no obvious methodology to solve this equation it is not like a quadratic equation with one variable ok as usually it presents certain challenge I did not ask by the way this solution artificial intelligence program I wonder what it will give me maybe it will start presenting all the different checking all the numbers but the number of prime numbers is infinite so I am not sure anyway maybe I should try it or you should try it so you have to solve this when X are and Y are prime numbers prime ok here is what I suggest first of all let's express this as equation for X where Y is hidden in the coefficients so it is X square now all the coefficients with X minus X 3 Y and 13 would be with a minus sign yes but minus we have so it is plus 13 ok minus 13 X it would be here everything else is Y Y is 3 Y square minus 13 Y equals to zero so this I can consider to be an equation with X as unknown and Y is hidden in the coefficients now if I am looking for a solution for X and X is supposed to be prime number my discriminant of this quadratic equation must be non-negative so discriminant which is B square minus 4 AC so it is X square plus B X plus C equals to zero this is my equation well A in this case is 3 B is minus 3 Y plus 13 and C is 3 Y square minus so this is discriminant it is supposed to be greater or equal to zero to have real solutions right? we need prime solutions primes are real so obviously remember this formula B plus minus square root of B square minus 4 AC divided by 4 by 2A these are solutions so we need this non-negative under the root ok fine so that is discriminant ok in this case what is the discriminant it is B square now B is minus 3 Y plus 13 so it would be 9 Y square 2 39 plus 78 Y plus 169 that is B square minus 4 AC A is 13 so times 4 is 12 times this so minus 12 times this is 36 Y square plus 12 times 13 144 156 Y and this is supposed to be greater or equal to zero right? so let's just simplify it a little bit so it's minus 27 Y square 9 and minus 36 is minus 27 78 and 156 it's 234 234 Y and minus 169 that's supposed to be greater or equal to zero ok that's interesting now you see this is basically a parabola with horns horns down because it's a minus sign so if I will draw the graph of this parabola I will have something like this since the horns are down it's minus so it's open this way now what are the roots of this? the positive when it's positive between the roots between the places where it's equal to zero actually calculate at least approximately the roots of this equation I did and one of them was something like minus one and zero and another was between 9 and 10 if I'm not mistaken but I think I'm not which means that if we are looking for prime Y and X well in this case Y there are how many prime numbers are between minus one and zero goes out so it's two three five seven that's it there are only four numbers of Y which can be solutions of this equation if we are looking for solutions among prime numbers ok great you just do it very very simply you put it two into this equation if you can for prime number or three and solve the equation for X if there is a prime number as a solution or five or seven and as a result you will see that as soon as you put two into this solution you will have into this equation sorry what you will have is thirteen this is two so it's six nineteen minus nineteen X two square is four twelve minus twenty six minus fourteen right equals to zero so the solutions are six nineteen plus minus square root nineteen square so it's twenty minus one square is four hundred three sixty one right three sixty one and plus four times this times this that's twelve by fourteen twelve by fourteen that's one forty four and twenty four one sixty eight one sixty eight now three sixty one plus one sixty eight is five twenty nine am I right five twenty nine five twenty nine unless I mistake twenty three it's sixty nine six four yeah five twenty nine is twenty three square so that's twenty three okay twenty three and nineteen is forty two divided by six is seven okay we've got it so two for Y and seven for X is a solution so we found the pair two seven by the way this is symmetrical to X and Y so if two seven is pair of solutions seven and two will also be pair of solutions so X is equal to seven Y is equal to two or X is equal to two and Y is equal to seven so we have already found two of these now if you will try three and five for Y you will see that there are no other solutions I mean I did try it so basically that's it we found all the solutions it's a pair two and seven in this and that order well that's it so all these problems are presented on the textual part of this lecture go to Unisor.com courses mass plus plus problems Algebra and Algebra zero six okay thanks very much and good luck