 Hi, I'm Zor, welcome to Unisor Education. We continue solving non-trivial, unusual problems, unusual in the sense that it's not like the problem which is supposed to check your theoretical knowledge, it's just something which trains your brain, basically. So the purpose of the whole course, which I called Mass Plus and Problems, is basically a training exercise for your logic, for your analytical abilities, for your mind, for your creativity. All the problems presented in this course are not really related in any kind of hierarchical way, you just solve them one after another in any order. So we assume the theoretical knowledge sufficient to solve these problems you do have and well if you don't you can always take the course Mass for Teens which precedes this one and that gives you the foundation for all the mathematics needed to solve these problems. Okay, today's problem, we will have only one problem today and I kind of like it because it's really, it involves, that's why I have only one problem for today, not like three, four, five, sometimes I have one problem but we will discuss it in detail. So here is a problem, consider you have a sum of fractions. So starting from one half, all the natural numbers are in denominators and we have some kind of fraction as a result, I mean if we will sum them up together, get to the lowest common denominator or everything. My question is, can it be an integer number as a result of all the manipulation, all the addition of all these fractions? Well, no it's not, it cannot be an integer number and we have to prove it today. So this is basically the problem, prove that this is not an integer number no matter what n actually is for any n, for any n. By the way, a different way to write it would be sum from n equal to 2 to n, 1 over n. Okay, that's just another recording of the same thing, more compact I would say. So it's not really important right now but it's important is how can we prove this particular, well theorem if you wish, that this is not an integer number. It's not really kind of a usual problem which you have in regular course of mathematics at school. So we need something unusual, some creative approach and that's basically the purpose of solving all these problems which I'm offering. You need some creative approach, you have to think about how to approach this problem and exactly the process of thinking and finding the solution which is not the usual kind of solution is what is supposed to help you in practical life to solve the practical problems which always exist in front of you. So this has absolutely nothing to do with real life. It's purely artificial problem but it trains your brain to think creatively. So I will suggest two solutions. Okay so let's start, I mean first obviously you have to pause the video if you found this on the video somewhere on YouTube or whatever and think about this problem yourself and then continue listening to my solutions and you might actually come up with your own and I will gladly put it on a web in Unizor.com with your authorship so just send it to me as a solution. You have my email on every screen. Okay so here is my first solution. Among these numbers are numbers which have the form 1 over 2 to the power of K. Like one half, one fourth, one eighths, one sixteenths etc. So this is 1 over 2, 1 over 2 square, 1 over 2 to the cube, 1 over 2 to the fourth power etc. So let's choose the one which is the right most here somewhere here you have this. Now this is, well since this is the right most that's the minimum among all numbers, all fractions of this type this is the smallest one because it's to the right and our denominator is increasing. So the biggest K when 1 over 2K is still less, it's still greater than 1 over N. Okay now let's think about how do we add all these numbers together. Well we all remember something which is called least common denominator, right? So what is least common denominator? Well one of the ways basically to explain it is the following. Prime number has certain prime components because every number can be represented as a combination of prime numbers. Let's say number 18 can be represented as 2, 3 and 3, right? These are all prime numbers. Something like 65 is what, 5 times 13 again, prime numbers. Now then we take all the prime numbers from every number of these and compact them together without duplicating. So if you have already prime number here 2 then you don't have to really, one of these 2. Now there are 2 prime numbers too, right? 4 times 2. So we need at least 2, we have only 1. So we get 1 from here, then 1 from here, now from 4 we already have 2. So we take only 1, 2. 5 is a prime, 6 has 2 and 3 so we don't really need primes from 3, from 6 etc. So we combine them together and that would be our least common denominator. So this is the number which is divisible by each one of these and the smallest number which can be divided by any of these. Alright, let's consider a simple case. Simple case would be, let's say 10. Now for 10 the least common denominator, what is it? So we need 2, we need 3, 4 has 2 twos, right? So we need an additional 2, so it's 2 square, 5 we need 5 because we don't have 5 as a prime number, 6 we need 2 and 3, we already have 2 and 3, 7 okay, 8, 8 has 2, 2 and 2, right? So we need the power of 3 here. Then 9 has 3 and 3, we have only 1, 3 so we need 2, 3 and then 10 is 2 and 5, we already have 2 and 5. So this is the common denominator, least common denominator and it's 25, 20. Now how do we do the addition? Well, we assume that every one of these should be actually 25, 20 in denominator. So what would be numerator in this case? Well, we divide this by 2, so it would be 12, 60, 25, 20, 1, 3, we divide by 3, it's 20, 40, etc. And the last one would be divide by 10, it's 252 by 25, 20. I would like you to pay attention to one particular number here, the one which is 1, 8. One 8th would be here, 25, 20 and one 8th would be 315. So this is one 8th, this is one 10th, this is one half, this is one third and everything in between. Now I would like you to pay attention to one important factor that 315 is odd number. All others are even and here is why. So one 8th is the biggest, the 8th is the biggest denominator which is the power of 2. Now this is power, the k is equal to 3 in this case, one 8th is 1 over 2 cube and we have 3 twos in our list common denominator to satisfy one 8th. Now all other numbers have less number of twos in them, 2, 3, 4, 5, 6, they have 1 or 2, this guy has 2 twos but nobody except one 8th has 3. We specifically chose one 8th as the number with the greatest number of twos which means when we will divide this list common denominator by any one of these we will still have at least one 2 left. If we divide by 2 we will have 2 left so it's still even number as a result, that's why it's even. If we divide by 3 this, 2 is not involved so 2 will still be there and this is still an even number. By any other number we will divide except one 8th. We will have still some number of twos left but only if we divide it by the least number of this type we will have all twos exhausted and we will have everything else, all other prime numbers and obviously all other prime numbers are all odd so their product is odd and that's why we will have odd numerator in this case. So only in this case we will have odd numerator. But we have some other for example what if we will have to a hundred, right? What happens? If we will go to a hundred, obviously after one 8th we will have one 16th, we will have one 32nd and one 64th, right? So here we need 3 twos, here we need 4. So the one with the greatest number of twos will be 64 which is 2 to the 6th degree and they will have 2 to the 6th here and only this one will have all these twos basically cancel out when we divide by 64. Everything else would still have one or two or three or whatever number of twos left and will be even and only this one will be odd. So obviously this is a general rule that if we will take 1 over 2 to the power of k which is the right most to this boundary to the 1 over n we will have only one particular odd numerator after applying least common denominator and all others will be even. Now if you have all even numbers and only one odd, their sum will be odd, okay? So numerator of the entire thing when you add these numerators together will be odd but denominator obviously is even because there are twos in it. Now odd number cannot be divided by even number to get an integer number, right? So if you have odd number x divided by even number it cannot be some kind of integer number because integer multiplied by even will get even not odd. So that's the proof that this is not an integer number. But I would like to suggest another proof while I still have some time, okay, this is the wrong marker. All right, so another thing is instead of looking for one over 2 to the power of n or k whatever I will look for denominator which is a prime number. So there is a prime number which is the largest not exceeding n. So 1 over, so pn is the largest prime number which is less than equal to our n which is the end of our series. And this basically solution is very much analogous to the previous one. So again this is the right most fraction with the prime number in the denominator, okay? So now, now I would like to tell you that this prime number does not participate in any other denominator. Well, it does not participate in all those fractions which precede it because this is the largest prime number, right? And this is prime so it cannot be, it cannot divide any of these numbers so it's greater than these ones. So none of these is divisible by pn. How about those which are to the right of it? Well, that's very simple actually. The problem is what is to the right of pn? 1 over pn plus 1, 1 over pn plus 2, etc., right? Now what is the next number which is prime actually? Well the next prime number should be between this one and the next, right? So what I would like to say is that if this is the largest prime number there is no more numbers after that which are divisible by pn. Why? Very simple case. If you have 1 over pn what is the next number with denominator divisible by pn? Obviously 1 over pn times 2. That's the smallest number which is closest to this one with denominator still divisible by pn, right? Now, if this is the greatest prime number which does not exceed n, I'm saying that this is impossible, why? And here I would like to refer you to a so-called Bernhardt postulate. The Bernhardt postulate is as follows. If you have some prime number, if you have some any number, let's say m and the number 2m minus 2, there is always some prime number in between. Now this was not really proven by Bernhardt. He just proposed this and checked this for numbers from 2 to 5 million, whatever. But it was proven later on and it's not a trivial proof so that's why I would like to use this Berthrandt postulate as a given. We might address it in some other case but if this is true then it is impossible for pn to be the largest number, prime number less than n and still have this one. Why? Because between pn and 2pn, well between 2pn and 2pn minus 2, there is always another prime number. So that prime number will be bigger than this one and still less than this one. So if pn is the largest prime number, there is no other denominator divisible by pn. That's what's important. Obviously, there are no divisible by pn on the left but now I'm telling you that on the right from this number to 1 over n, there is also no numbers, no denominators divisible by pn because the smallest one, if it's still less than 1n, I mean this is actually greater than, the denominator is greater, then if this is the case then we will have another prime number and this will not be the largest prime number, not exceeding n. Okay, so being done with this, everything else is simple. If pn is the only, if 1 over pn is the only one which has prime number pn in its denominator, the least common denominator contains only one pn. So this least common denominator will be like 2 times 3 times blah blah times this pn times something else. Pn will be represented only once to cover this particular denominator. Now when we will multiply this to numerators and cancel out whatever is necessary, pn would cancel out and you will have all prime numbers here and we will have the pn for this one, we will have pn present for this one because it will not cancel out with anything else or even after 1 over pn. So all these will be divisible by pn except 1 and if all the numerators divisible by pn except 1 then we will not be able to divide, the result of division of some of the numerators divided by the common denominator will not be integer because in the numerator as a sum we will have everything divided by pn, so it will be like a times pn times something which is not divisible by pn divided by something which contains least common denominator which has many prime numbers including this one. So this one cannot be integer number because this is not divisible by pn, if this is integer number let's say k then you will have on the right you will have k times something which is divisible by pn and this is not divisible by pn. So again we have proven in some way it's similar to the first proof this one but it actually involves prime numbers and very important theorem which I was just talking about the Bert Rans postulate. I do suggest you to read the notes for this lecture they are more detailed and also there are maybe some other more precise information which I for instance I have given very brief kind of example of how to do the least common denominator I have elaborated this much more in the text for this particular lecture and to get to this lecture it's mass for mass plus and problems you choose arithmetic and this is a arithmetic O2 lecture. Okay that's it for today thank you very much and good luck.