 Hemsor, welcome to Unizor education. Today's topic will be induction, mathematical induction. Induction is a method you can use to prove certain formulas. And let me start from a very simple problem and how this particular method is applied to this. There is a legend about German mathematician Karl Friedrich Haus. When he was a child, apparently teacher asked the class where he was to add numbers from 1 to 100. Maybe the teacher wanted students to keep quiet for a while, so it would take them a while to add all these numbers. And quite suddenly the teacher saw that young Karl raises his hand and basically gives the right answer. Apparently what he did was he noticed that if you will take this sequence of numbers and write it backwards, 100, 99, 98 plus 1. Well, basically you have exactly the same sum of numbers. And if you add them all together, it will be double the sum which was originally supposed to be calculated. But if you instead add it vertically using these pairs first, you notice that the sum of these is 101, the sum of these will be 101, 398 is 101, etc., and the last one will be 101. And how many times 101 participates in this sum? Well, obviously 100 times. Well, using the multiplication, the total sum will be 10100. But then again, this is the double of the original sum. So what we are looking for is half of this number, which is 5052. And that's the answer. How is it? Well, that's the right answer. Now let's try to generalize it a little bit. Instead of adding numbers from 1 to 100, we will use the same methodology that Carl Hauss was using. And let's try to add numbers from 1 to any arbitrary number n. Well, we'll do exactly the same thing Carl did. He was a genius by the way. Let's write them backwards. And again, sum of all these numbers is double the sum which we are looking for. But instead of summing it horizontally, we will sum in this direction vertically and then multiply it. And as you understand, the sum of these is n plus 1. Sum of 2 and n minus 1 is again n plus 1, etc. And sum of these two is also n plus 1. Exactly the same logic. n plus 1 participates n times. So the total will be n times n plus 1. And since this is double the whole sum which we are looking for, the real answer is this. This is the formula. Okay, fine. Let's write it down again. 1 plus 2 plus, etc. plus n equals. This is a nice formula. Whenever we want to calculate sum of numbers from 1 to 1000, let's say, we can use the formula, it's 1000 times 1001 divided by 2, whatever it is, very easy to calculate. Great. Now, can we really use this formula for any n? Well, we didn't prove it for any n. We kind of guessed that this would be exactly the sum. We basically noticed that there is a vertical sum of n plus 1 in each pair of vertical numbers. But then it's really a guess that it would be exactly the same for all these intermediate members as well. And we cannot obviously check that this particular vertical sum will be equal to n plus 1 for each number. We just did it for a couple of numbers in the beginning and a couple of numbers at the end, and that's it. So mathematically speaking, this is not a rigorous proof. And the method of mathematical induction is exactly this rigorous proof. The method we can prove the formula is really valid for any n. Now, here is the method how it's supposed to be used in the rigorous sense, so to speak. This method contains actually three steps. Step number one, check for n equals to 1. So we have to check this formula for some initial number. Usually it's number one, for instance. Well, for number one, the sum on the left will be from 1 to 1, which is 1. And the number on the right will be 1 times 1 plus 1, 2 divided by 2, also 1. So 1 equals 1, formula checked. Okay, we definitely know that this formula is valid for certain initial number. In this case, it's number one. Number two, assume, and this is a very important word, assume that the formula is valid for number n equals to k. Where k is any number, whatever you want. So let's assume, let's assume for a second, that 1 plus 2 plus, etc. plus k equals to k, and the k plus 1 divided by 2. Now, this is an assumption. We are not saying this is the truth. We are assuming that this is true. And based on this assumption, number three is, let's prove it for n equals to k plus 1. So let's assume that the formula is true for n equals to k. And using this, let's prove that the formula is true for n equals to k plus 1. Let's assume that if we will add k numbers from 1 to k together, we will get this number. Now, let's add k plus 1 numbers. What happens in this case? Well, let's calculate now. If you take numbers from 1 to k plus 1, it will be 1 plus 2 plus, etc. plus k plus k plus 1. This is the sum which we are trying to prove is equal to the same formula, but with n equals to k plus 1. Okay, let's do it. Now, using our assumption that the formula is true for n equals k, instead of this sum from 1 to k, we can write k times k plus 1 divided by 2. So this is our expression now. This is exactly where we are using this assumption. Now, let's just change it a little bit. It's k k plus 1 2 plus 2 k plus 1 divided by 2. So instead of k plus 1, I multiply it by 2. So we will have a common denominator. So k plus 1 can be factored out. k plus 2 remains divided by 2. So again, using our assumption that's extremely important, we have come up that this particular sum from 1 to k plus 1 is equal to this. Now, let's think again what we are trying to prove. We are trying to prove that n plus 1 plus 2 plus etc plus n equals to n times n plus 1 divided by 2. We are trying to prove that this formula is valid for any n. We can assume that for n equals k, it's true. Now, how does this formula look for n equals to k plus 1? Well, on the left it will be 1 plus 2 etc plus k plus 1, which is this. On the right, if n is equal to k plus 1, n is equal to k plus 1. n plus 1 is obviously k plus 2 now, so n divided by 2. So this formula with n equals to k plus 1 is exactly what we have just received. Sum from 1 to k plus 1 on the left and k plus 1 times k plus 2 divided by 2 on the right. So again, what we have done, we have proved that the formula is true for n is equal to k plus 1, assuming that the formula is true for n equals k. Where the assumption was used, again, whenever we broke this sum into two components, we got one where we can use an assumption plus the right one, the k plus 1, which is kind of a small one. Then after small transformations we came to this thing. So basically we made all these three steps. Okay, is that the end of it? Well, obviously not. We have to say something to basically prove that the formula is true for any number. How can from these three steps we can derive that the formula is true for any number? Very simply, formula is checked for n is equal to 1, right? We definitely know it's true. Then here is the assumption, if formula is true for n equals k, then the formula is true for m e is equal to k plus 1. Great, but we know that for k is equal to 1, formula is true. We checked it, all right? So first step is k equals to 1, and we use this transformation. If it's true for k, it's true for k plus 1, which means if it's true for k equals 1, it's true for k equals to 2, k plus 1, right? Great, so we basically prove that the formula is true for some of two numbers. Okay, since it's true for 2, again using this transformation from k to k plus 1, where k is equal to 2, what follows is that the formula is true for k is equal to 3. And obviously this process can be continued indefinitely, and for any number, any natural number, 100 million or whatever, using this logic step by step from 1 to 2, from 2 to 3, from a million to a million and a half, we are actually making these logical assumptions, and therefore it proves that it will be true for any number. Because for any number we can repeat these steps whatever number of times we want, starting from 1 up to that particular number, and the whole thing will be proven. Again, what's very important is, number 1, we checked it for some initial number, usually 1, then the formula can be assumed for a certain number. Let's assume it's true for k. Then using this, we prove that the formula is true for k plus 1. Right, okay, so this is the methodology, and what probably seems to be a very important step is this step, because we spent basically most of the time proving that if the formula is true for m is equal to k, then the formula is true for k plus 1. Now the fact that we have spent a little bit more time on this than checking that the formula is true for the beginning, for the initial number 1 doesn't mean that this is unimportant. Actually, if the first step is not made, then we cannot make the first and then all subsequent up to some steps in this logical step. So it's very important to check that the formula is true for the initial value of m, usually, usually it's 1. Now if we don't do this, if we just make the proof that if some formula is true for m is equal to k, then it will be true for m equals k plus 1. If we just make this logical step, it doesn't prove anything without this checking. As a matter of fact, you can prove a lot of absolutely wrong formulas using this methodology. And again, since it's not checked for the beginning for the m is equal to 1, then this logical transformation from assumption to the proof doesn't really mean anything at all, because you can't use it. Here is an example. Let me just make a little bit more space. What I would like to explain right now is that this transformation from step 2 to step 3 from assumption that the formula is true for m is equal to 1 to prove that it's equal to m plus 1 is not sufficient. Here is my formula. m equals m plus 1. For any m, obviously the wrong formula. Alright, but let's try to prove it using the method of mathematical induction. And what's very important, again, the number one step, check, that the formula is true for some initial number, let's say for m is equal to 1. Let's say we do not do this. Obviously if you put 1 equal to 1 is equal to 2, it's definitely wrong, so the whole method of mathematical induction is not fully implemented. However, step 2 assumes that k is equal to k plus 1. For some k, now prove for k plus 1. Let's n to be k plus 1. On the left we will have k. Using assumption, we will substitute it to this. We will see that k plus 1 using this assumption plus 1, which is equal to k plus 2. Which is exactly our formula for k is equal to k plus 1. k plus 1, k plus 2. That's exactly what we have proved, k plus 1, k plus 2. Using this assumption, we have proved basically the transformation from k to k plus 1. Obviously, using just these two steps, the assumption for n is equal to k and proving that the formula is also valid for n is equal to k plus 1, this works. But again, without the checking, it's absolutely worthless. You can't really say anything because logical transformation is based on steps from 1 to 2, from 2 to 3, etc. So the initial step, like 1 for instance in this case, is extremely important. Well, basically that concludes explanation of what is method of mathematical induction is. And there are a few problems which I have also put on this page. And you definitely are invited to click on this and try to solve them yourself. And if not, just look at my explanation of what these particular problems are about and how to solve them. Thank you.