 I'm Zor. Welcome to Unisore Education. This is the second part of lecture about decimals and relationship between decimal fractions and rational numbers. In the first part, I have proved that every finite or infinite periodic decimal fraction is a representation of a rational number. For instance, the periodic will be something like 0.142857 in a period, which means infinite number of times repeated, and this is actually one-seventh. Now, in this second part, I will try to prove the converse theorem that every rational number can be represented as a finite or infinite periodic decimal fraction. First of all, a couple of purely philosophical observations. Why am I proving this type of thing? I mean, this is definitely proved by many different people in many different ways, so the purpose of this proof is not really to prove. It's already proven. It's purely educational. More than that, I was trying to put together a proof which does not really depend on any kind of prior knowledge, any kind of a theorem which was proven by somebody else. It's a very, I would say, homegrown proof which anybody can do it himself, and that's why I think it's very important to really not to pay attention to whatever has been proven, but the way of the proof itself, because it might actually present certain logical benefits for you, and basically, again, the purpose is purely educational. That's number one. Number two, I would like to remind one very important piece of the previous lecture where I was talking about infinite geometrical progression. I will do it slightly more generally than in that lecture. If I have a sum of this type of members, so every next member of this progression is the previous multiplied by Q. Now, this is called the geometrical progression, and if Q is less than one, let's talk about the positive numbers only. Then, obviously, it's decreasing. Every member is smaller than the previous one by a factor of Q. Now, in this case, the sum of this infinite geometrical progression can be calculated very easily. I did it in the previous lecture in a slightly less general case, but anyway, the way how it is done is the following. I multiply by Q, all members of this thing. So, A times Q is AQ. AQ times Q would be AQ square, AQ, Q, etc. By the way, I do not remember the final formula. I always do this type of derivation and come with the formula, and that's basically something which I think is a better way. Well, unless you have an infinite memory, I don't know. It's easier to remember the way how the formula is derived than the formula itself, which might be quite complex. In this particular case, what I did was I multiplied by Q this sum, which means every particular member is multiplied by Q. As you see, this sum and this sum are very close to each other. You see, the difference is actually only A. Everything else is repeated, and this goes to infinity. So, the whole tale will actually disappear if I will subtract from one another. From S, I subtract SQ with the S minus SQ, and here, all these members will give the zero result. So, basically, that's the final equation which I have, which I can solve for S. S, obviously, is equal to A, A divided by 1 minus Q. This is the formula for a sum of infinite geometrical progression with the first member, A, and the factor Q, or multiplier Q, whatever you call it. Let's just remember it. I'll put it aside here because we will use it in the previous lecture. I also use this formula because if I have a geometrical progression, sorry, if I have a decimal fraction, periodic decimal fraction, actually it is the geometrical progression summed up because what this actually means is, well, let's illustrate it in a slightly more in a shorter example. Let's say you have 1 over 33, which is 0.03 in a period. Now, what is this? It means it's 0.03, 0.03, 0.03, etc. up to infinity, right? Now, these are periods which are repeated, infinite number of times. Now, the weight of this is 1 tenths, and the weight of this is 1 congrats. So, altogether, it represents 3 congrats. Or, if you wish, I'd rather say it's 3 to the power of 10 minus 2. Now, next 0.03, obviously, is what? It's 3 times 10 to the minus 4. Next 0.03 is 3 times 10 to the minus 6, etc. And altogether, it's basically, as some of these numbers, this, this, this, and all others. So, it's basically, it's 3 times 10 to the minus 2 plus 3 10 times minus 4 plus 3 10 minus 6, etc. etc. equals the first member, which is 3 10 minus 2 divided by 1 minus, and what's my multiplier? Obviously, it's 10 minus 2, right? So, that's what it is, which is equal to 3 hundredths, 10 minus 2 is 100, 1 minus 100 equals to 3 over 99, as I was saying, 1 third to 3rd. So, there is a very close relationship between infinite periodical decimal fraction and sum of the geometrical progression. So, this was used when I was proving to the first part that any periodic decimal fraction is actual irrational number, because this is actual rational number under any circumstances. And now, I will use the same concept of infinite geometrical progression to prove the converse theorem that every rational number can be represented by either periodic or final or periodic decimal fraction. Okay. Now, before proving this, I will simplify my task. If I have a rational number m over m, if I will prove that 1 over m is a periodic decimal number, then how can I say that m over m is also periodic? Well, it's based on something which I presented as a problem one for this particular theme of rational numbers. Problem one is prove that sum of two periodic decimal fractions is periodic. I would like you to think about this as a very simple problem. And I will also make a small presentation about how to solve this problem, but I really don't want to talk about this right now because it kind of diverts us from the main goal. But I hope that intuitively you feel that if you have one periodic decimal number and another periodic decimal number, then their sum will also be periodic. It's important for this particular theorem. And so instead of proving for any m over m, I will prove only for 1 over m, because if I will do this, then I will add my periodic number which is the result of this to itself a couple of times. For instance, how about 2 7s? Well, 2 7s is 1 7 plus 1 7, so I add these two together. Now, this is periodic and that is periodic, and the sum will be periodic as well. That's the problem one, try to do it yourself, and I have, I will present my own decision as well. Okay, so I will concentrate only on 1 over m type of rational numbers, but even that I will try to simplify a little bit more. And here is what I'm going to do. I would like to say that instead of proving for any m over 1 over m, I will prove this theorem only for those m which are not multiples of 2 or 5. And let me explain why. For instance, n is a multiply of, let's say, 2. It's 2 times k. Then I can always write that 1 over n is equal to 1 over 2 times k, or it's 5 over 10 times k. Okay, to prove the periodicity of this decimal representation, it's enough actually to prove the periodicity of this, because this is, it's basically the same thing as I started. Instead of proving for m over n, I can prove for 1 over n and then add it up. Same thing here. I can prove the periodicity of the decimal representation of this and add it to itself 5 times. Now, how about periodicity of 1 over 10k? Well, here it is. If 1 over k is periodic, let's say 0. Well, let's just use this example. 1 4 2 8 5 7. In this case, k is equal to 7. So 1 over k is this. Now, what is 1 over 10 times, which is 10 times smaller, very easily. It's 0.0, 1 4 2 8 5 7. Every decimal digit, I shift it to the right. Since the weight of this was, let's say, 1 tenths, shift it to the right with 1 over 100, which is 10 times smaller. Same thing with this one. 4 used to have a weight of 1 over 100. Now it's 1 over 1000, 10 times smaller. So every number, when I shift the whole set of digits to the right and put 0 in front of it, it's actually divided by 10. So division by 10 in the denominator of this rational number does not really change the periodicity. So instead of proving the periodicity of this, I can always prove the periodicity of this. So it's very important that from the very beginning, I consider n those which are not multiples of 2 or 5, because these can be converted into 10 times something. By the way, similarly, if I have n equals to 5 times k, I can always say that 1 over n is equal to 1 over 5k, which is 2 10 over k. So instead of proving the periodicity of 1 over n, I can prove the periodicity of this. Instead of this, I can put 1 over 10 over k, and instead of 1 over 10 times k, I can prove 1 over k again. So any number n, which contains 2s or 5, can be actually reduced to whatever is the smallest number, which is not divisible by 2 or 5. For instance, instead of proving the theorem for let's say 1 over 35, I can say that 35 is divisible by 5. So it's actually 1 over 5 times 7, which is actually the same as 2 times 10 over 7. And if I prove the periodicity of 1 over 7, which is this one, it's sufficient to prove the periodicity of 1 over 35. Okay. So enough that we are considering the rational numbers 1 over n, or the more formal representation, as I used to use when I was defining rational numbers, it's this representation. And n is not multiple of 2 or 5. Okay. I just invented this sign, not multiple of. There's probably something else that doesn't really matter. So n is not multiple of 2 or 5, and therefore not multiple of 10, of course as well. And now the proof. Okay. I will use this 1 over 7 as an example, I think it would be easier. So in general, if I have 1 over n, I will consider n numbers 9, 999, 999, et cetera, up to 9999 n times. In my particular case of 1 over 7, I will consider 9, 999, 999. Okay. This is boring. 999, 999, 999, 999, 999. And one more, 7 numbers, from 9 to 7 times 9. These are just different numbers. Now, why could I choose them for very simple reason? You see this formula for geometrical progression? No matter what my period is, let's say in this particular case my period is, has the length of 6 decimal digits, then the next number will be this one multiplied by 10 to the minus 6 degree, because we're shifting everything to the right by 6 positions, which is 1 millionth, right? So the next 1, 4, 2, 8, 5, 7 will be 1 millionth of this one. The next one will be 1 millionth of that one, et cetera. So always my next member will be 1 over 10 to the sum degree to the power of 6 or 7 or whatever it is, smaller than the previous one. So Q for these geometrical progressions is always 10 to the minus something, like minus 8, I will put it. So for this particular case, it's 10 to the minus 6, because every new period is 1 millionth of the previous, because it's 6 positions to the right. So if Q is 10 to the minus something, then obviously 1 minus Q would be, for instance, 1 minus 1 millionth. What is this? It's 9, 9, 9, 9, 9, 9, 9 over millionth, right? So these are, that's why my 9, 9, 9s are come into play. Obviously I took it from proof of the direct theorem that every periodic rational number is rational. And now I'm just trying to reverse the logic, basically, trying to prove that every rational number is periodic. Okay, so I took these n different numbers, 9, 9, 9, et cetera. Now there are two cases. Either I will find among them a number which is divisible by n, or I will not. So let's just consider these two cases. Let's take the case number one. So there is some 9, 9, 9, 9, whatever, 9, 9, whatever, which is divisible by n, which is equal to n times k, let's say. Now what is this 9, 9, 9, 9, 9? Well it's 10 to some degree a minus 1, correct? Obviously. Every number which is represented as certain number of nines, well a number of nines in this case can be represented as 10 to the power of a, the a's power minus 1. Because 10 to the a's power is 1 and a 0's. So now we subtract 1, we have 9 nines, the a nines. All right, so having said that, I can say that n is equal to 10 to the a minus 1 divided by k. And we are considering this in a general case, so I will put it here, n to the a minus 1 divided by k. Now what will be in our particular case of 1 over 7? Well I made an observation that 6 nines is divisible by 7. 6 nines is divisible by 7 and it's equal to 1 over 2, 8, 5, 7. You see? This is the same number. So I've noticed that we are considering the case when one of these numbers is divisible by 7, right? So I indeed found one. So this is exactly the case when you have this particular number. When there is none, it will be another case. But right now we have found something which is divisible. And therefore I can say that 10 to the minus 10 to the 6 degree minus 1 divided by k which is 1, 4, 2, 8, 5, 7 is equal to 7. This is exactly the same in our particular case. Okay. Now what can we do about this? In a general case we can say that 1 n, 1 over n is equal to k divided by 10 to the a minus 1 which is equal to k times 10 to the minus a and 1 minus 10 to the minus a. I multiply by 10 to the power of minus a, numerator and denominator. So this will be in the numerator obviously. And the denominator, if I multiply 10 to the a times 10 to the minus a, I will have 1. And if I will have 1 multiplied by 10 minus a, it will be 10 minus a. Now this obviously is exactly like this formula. Now in our particular case, well that will be 1 7 equals to 1, 4, 2, 8, 5, 7 times 10 to the minus 6 divided by 1 minus 10 to the minus 6. Well that's it. The theorem is basically proven. Because as you see, since this is exactly this, I can say that k times 10 to the minus a, k times 10 to the minus a is the beginning. That's the first member of my geometrical progression. And 10 to the minus a is its multiplier. So next one will be k times 10 to the minus 2a plus k times 10 to the minus 3a, etc. Which express is a decimal number, k is some integer, right? So k times 10 to the minus a will be something like 0 point and some digits here. How many digits? Eight digits, right? Then I have the same k shifted to the right by again taking eight digits. So it will be the same digits which are here, will be the same on the second group of eight digits. And then another eight digits, etc., etc., after infinity. In our particular case, what will be is k times 10 to the minus a, which is this, is 0 point 1, 4, 2, 8, 5, 7. Next one will be shifted by one millionth to the right, which is 0 point 0, 0, 0, 0, 0, 1, 4, 2, 8, 5, 7. Next will be 0 point 0, 0, 0, 0, 0, 3, 6. And then again, 1, 4, 2, 8, 5, 7. And when we summarize them all together, it will be 1, 4, 2, 8, 5, 7, 1, 4, 2, 8, 5, 7 because the rest are zeros, etc., etc., etc., which basically forms the periodic decimal number. So we have proven the theorem for a case, except one small detail, but I don't want right now to talk about this small detail. You can probably read about this in the written part of this lecture. It's really about the size of k. So the k to the 10 minus a is within the eight digits, no more than that. So it doesn't overflow the period. But that's minor point and you can read about this yourself. So anyway, we have proven the first case when 1 over n is a rational number and then we consider n numbers 9, 99, 999, etc., up to 999 n times. We have assumed that there is one of these which is divisible by n. What if there is none? Well, that's actually quite easy to deal with. You see, these are n different numbers. If you divide by n, n different numbers, how many different remainders you can get? Well, if you divide by a number and it's not divisible, then you can have a remainder 1 or 2 or 3, etc., or n minus 1. So different remainders are from 1 to n minus 1 when you divide something by n. These are your remainders, 1, 2, 3, etc., n minus 1. There is no remainder n because it means it's divisible. And then n plus 1 remainder is actually the same as 1. So obviously, there are only n minus 1 remainders which can be obtained if you divide all these n different numbers, which means what? That some of these numbers, at least two, have exactly the same remainder. Remember this famous principle of Dirichlet, if I'm not mistaken. If you have 101 rabbits and only one cages, you cannot place them in each individual cage. At least couple of rabbits should be in the same cage. So here is the same. We have n numbers but only n minus 1 different remainders. So at least two of them must be the same. So let's consider that you have two different remainders which are the same. And let me exemplify it again with some numerical example. I will use example 1 over 70. I know this is not exactly what I was considering in the beginning that the number n does not have divisors of 2 nor 5. And this does have it. But I needed just to exemplify about remainders this thing. So considering this principle I have to take 70 different numbers, 9, 99, 999, etc., up to 79 in a row as a big, big, big number. Now, if I will start dividing it by 70, I will start getting certain remainders. But 9 gives a remainder 9 if you divide it by 70. And 99, 99, 99, 99, 9 will give a remainder also 9. Don't ask me how I know about it. Actually, it's very simple. If 6 nines is divisible by 7, then 6 nines and 0 is divisible by 70. So 9 extra and that's that's why 9 is a remainder. So these two numbers represent numbers which have the same remainder when divided by 70. So if two numbers have the same remainder, then their difference is obviously divisible by the number. So this minus this is divisible by 70, which is what, 999, 999, 990. In the general case, I can also say that something like 999, 999, 990, and then 0, 0, something 0, you have, let's say, K9s and L0s. Would be if I subtract these two numbers, which have the same remainder. So one number would be 999, 999, all of them, and another would be 999, the smaller one. So if I subtract, I will have number of 9s and number of 0s. Same thing here. Now, so my statement right now is that if there is none which is divisible by n, then there is definitely a pair which has the same remainder, and their difference looks like this. OK, so this number is divisible by n. So we can say that this number is equal to n times K. OK, what's the result of this? Well, obviously, we can say the following. This number represents something like 10 to the power of m minus 1. That's the 999, 999 minus a smaller 999. 10 to the power of, let's say, I probably have to use different letters here. I'll use lowercase m and n minus 1. That's what this number is. All 999s will do this, and smaller 999 will do this, which is obviously equal to 10 to the m minus 10 to the n or 10 to some number, well, let's say n is smaller than n. And you can have that m times 10 to the m minus n minus 1. And this is equal to still the same thing. Now, I shouldn't do it good. Yeah, we can do the n times K. Now, here is what's important in this particular case. Let me just leave the very last formula. We have, from the very beginning, assumed that n is not divisible by neither 2 nor 5. What does it mean? It means that this 10 to the n degree has no common multipliers, no common divisors, with n, because this has only 2s and 5s, which means what? That this number should contain all the prime numbers which comprise the number n. So no matter what number n is, if you will represent it as the product of certain prime numbers, let's say, then all these prime numbers should be here. None of them is here, because there are no 2s and no 5s, which means that 10 to the n minus n minus 1 is actually divisible by n times some number l. So my initial suggestion that n has no divisors of 2 or 5 actually means that this, my initial suggestion of the case 2, that there are no numbers among these n, which are divisible by n, really cannot be true. If I assume that this is a true statement, that there are two numbers in this group, which give the same remainder, then I will come up with this formula. And from this formula, what follows is that since 10 to the n degree has nothing to do with n, then this number should be divisible by n. And this number is definitely among these. So there is no such case when the number n, which is not a multiple of 2 or 5, has no divisors, has no numbers among these which are divisible by n. So this situation is just impossible. Case 2, when there is no number among these, which is a multiple of n, it's not possible at all. So that basically ends up the case 2. And everything goes back to the case 1. And the case 1, we have already proved. That proves that every rational number can be represented as a finite or infinite decimal periodic fraction. Well, that's it. Now, we still have the problem number 1, which is about summing of two different periodic numbers. You have to prove that this is also a periodic. And I would definitely suggest you to do it yourself. It's not a difficult thing. But anyway, I will provide a small lecture about this particular problem. OK, that's it for today. I hope you enjoyed it. Thank you very much.