 Hi, I'm Zor. Welcome to Unisor Education. I would like to present one more problem on induction, mathematical induction, just to prove one more formula. It's a good exercise and probably it will just demonstrate one more time how the induction approach helps you to prove certain formula. The formula by itself is very helpful. I'm sure we will use it throughout our algebra course in many different ways whenever it's appropriate. So, well, here it is. What I would like to prove is the formula we can represent the difference between two powers. N is an integer number, obviously a natural number greater than or equal to 2. The representation of this is a minus b multiplied by a sum of products of a to certain power and b. What's interesting here is I'm decreasing the powers of a and I increasing the powers of b. So, a is decreasing from a to the n minus 1 to a to the power of 0. That's why there is no a here. It's 1, it's constant. B is increasing from b to the power of 0 to b to the power of n minus 1. In this sum, every product of a to some power and b to some power has a sum of powers always equal to n minus 1, which makes the whole multivariate polynomial as this is called. In this case, it's two variables a and b, and it's polynomial because it's a sum of a and b multiplied in certain powers. So, it makes the power of every element to be equal to n minus 1, which is always a sum of powers a to the m times a to the n or rather b to the n. This element always has a power of m plus n. So, the power of every element in this sum is equal to n minus 1. So, it's n minus 1 and b to the power of 0, n minus 2 and b to the power of 1. So, again, the sum is n minus 1, etc. So, the whole polynomial is of the power n minus 1. This is polynomial of also multivariate polynomial of the power of 1. And when we multiply two polynomials, again, the powers are added together. So, 1 and n minus 1 gives you n, and that's exactly what's the power of this particular polynomial. So, everything fits. But now we have to prove this formula, and I'm going to prove it by induction. So, first what we do, we check it for beginning value, which is n is equal to 2. Well, if n is equal to 1, it doesn't really make much sense. We don't really have it. It's already a minus b. So, starting from n minus 2, what will it be? It will be a square minus b square on the left. On the right, this is a minus b times. So, this is a polynomial of the power of n minus 1. n is equal to 2, so it's just 1. Which means a to the power of 1 times b to the power of 0. And then the last one. Now, there is no n minus 2. These are all non-existencing. Only the last one will be present, which is b. And this is a known formula. I mean, if you don't believe that this a square minus b square equals to a minus b times a plus b, let's just multiply. It's equal to a minus b times a plus a minus b times b, which is a square minus a b plus a b minus b square a b minus n plus a square minus b square what's necessary to do. So, n is equal to 2 is fine. Now, the next step is we assume that this formula is correct for sum n is equal to k. Which means we assume that a to the power of k minus b to the power of k is equal to a minus b times a to the power of k minus 1 plus a k minus 2b plus et cetera plus a b k minus 2 plus b k minus 1. So, we assume this is correct. And now, let's substitute n k plus 1. So, we have to do a k plus 1 minus b to the k plus 1. Okay, what will that be? Well, how can it be proven using this? Well, let's think about it. So, we know that this is true. So, why don't we do this? Yeah, I think it would work. Let's do it this way. I subtract a to the k's b and add a to the k's b minus b to the k plus 1, right? I don't change this. I add it a minus a to the power of k times b and plus a to the power of b plus 1. All right, it's equal to, well, we can factor out a to the power of k. From a to the power of k plus 1, we still have a left. a to the k's and a would be exactly a to the k plus 1. And from there, it would be minus b since I factor out a to the power of k. Here, I will factor out b. So, it will be plus b a to the power of k minus b to the power of k. Okay, why did I do it? Well, because this is something which I already assumed can be represented as this sum. So, let's do it. So, I leave this one without any change. Now, this I will represent as a minus b times this sum. a k minus 1 plus a k minus 2 b plus, etc. plus a b k minus 2 and b k minus 1. Okay, now, as you understand, I can factor out a minus b from both sides, from both terms, right? So, it will be a minus b and what will be left? a to the power of k plus b multiplied by all this. So, a to the power of k minus 1 b a k minus 1 b. Then, we are decreasing a and we increasing b. b and b b square a to the k minus 2, etc. And two last members will be a b k minus 1 plus b k. Lo and behold, this is exactly what this... What did I write here? I'm sorry. This is not... Yeah, I didn't really write my expression. Okay, this is exactly expression of this kind. This is exactly expression of this kind with n is equal to k plus 1. Look at this. This formula with n equals to k plus 1 should be a to the k plus 1 minus b to the k plus 1, which is what I have started with, equals to a minus b, which is this. Now, instead of n, I should substitute k plus 1, so it will be a to the power of k. That's a to the power of k. Then, a to the power of k plus 1 minus 2, which is minus 1 times b, which is exactly this, etc. My power of the a is decreasing and the power of the b is increasing. Until the last one, b to the power of n minus 1, where n is equal to k plus 1, is k, right? b to the power of k. And that's exactly what we have. So, by doing this manipulation, I have proven that assuming that this is correct for n is equal to k. And I'm explicitly using it here, from here to here. I substitute it from there, from this expression, the whole expression, which I can assume is correct. I have come to the same formula for n is equal to k plus 1, which basically concludes the proof. So, this proof, this formula is quite useful actually. You can have some examples of this. Let's do a couple of examples of this formula. You don't have to really remember every example as long as you remember the principle that if you have a difference between two powers, then you have the difference between the bases, and then times the very symmetrical polynomial of two variables, which is multivariate polynomial, with the power, which is basically this power minus 1, with gradually decreasing the first component and increasing the second component as we move along the sum towards the end. So, in case of a square minus b square, we already know this is a minus b times a plus b. This is probably everybody remembers by heart. Well, with a cube, it's a little bit more complex. Now, the polynomial should be of the second degree, right, because this is the third degree. This is the polynomial of the first degree. So, this is supposed to be a polynomial of second degree, which is a square plus we are decreasing a and increasing b. So, after 2, I go with the power of 1. After b to this zero's power, I have b to the first power and b square. And that's basically the formula. Similarly, a to the fourth minus b to the fourth is equal to a minus b, a minus b times. Now, we have to have a polynomial of the third degree, right, because this is fourth, this is one, this is the first degree. So, this is supposed to be third degree. So, we start from a to the third and we will finish with a b to the third, but we are decreasing a and increasing b. Well, that's it for today. Thank you very much. That was problem number seven of induction. Thank you.