 In this video, we're gonna introduce one of the most powerful proof techniques in all of mathematics. I mean, this technique is used all over the place in abstract algebra, number theory, topology, geometry. I mean, you can use this all the time. Whenever a set is indexed by natural numbers, you basically can use the principle of mathematical induction to help you prove things about the growth of that sequence of the index set with regard to that natural numbers. Basically, whenever the natural numbers come into play, it's very likely you might prove something using induction. Now, it does turn out there's commonly referred to as two different principles of mathematical induction. In this video, we're gonna refer to the first principle of mathematical induction. Sometimes it's referred to as the weak principle of induction, since the second principle is commonly referred to as strong induction. But honestly, this adjective first is generally omitted, and so we just call this the principle of mathematical induction for which people might just say the principle of induction or heck, people will just sometimes say by induction, right? One of my proudest moments as a graduate students when I was writing the proof of a theorem, right? So there was a theorem, and then my proof was the following. It follows by induction. Okay, that's like verbatim what I wrote down for this proof. I don't remember the exact theorem itself, but it didn't matter at the time. This was a homework exercise for a graduate class in abstract algebra. And my professor, when he wrote, when he read the proof, he gives me a nice looking red check mark to suggest that I got full marks on this exercise here. He accepted my proof. And the idea was that at this point, he trusted me as a mathematical student that I said it, the proof follows by induction. I didn't provide all of the details. I didn't provide any of the details. I just said that it could be done. And he trusted me as a student that yep, it does follow by induction. And I bet you could provide all the details that would be necessary, but I don't need to see these details. And since I believe that you can provide them, you don't need to see these details. It's a super, super proud moment. Now, for most of us watching this video, we're not at that moment in our mathematical careers. This might be our first exposure to mathematical induction. So let's actually talk about all the important steps that go into mathematical induction. I do want to mention before we start that though, in the previous video for lecture 19, we introduced the so-called well-ordering principle. And what we will actually see later on in this unit about integers is that the well-ordering principle actually provides to us a proof of mathematical induction. So because every set of every non-empty set of natural numbers has a minimal element, we actually get the principle of mathematical induction. So we can prove this as a theorem of that. We will postpone that proof until we're a little bit better at induction first, if that's okay. So what does the principle of induction tell us then? So let S of N be a statement about integers for some natural number N. So this is some statement, which we'll see some examples of this later on. Oftentimes these are equations, for example, about integers, but they're indexed by natural numbers, which remember, zero is our smallest natural number in consideration here. Now suppose that the statement S of N sub zero is true for some integer N sub zero. Now these statements do not necessarily have to be indexed for all natural numbers, but there's some first natural number that is considered, that the statement is true for that one, N sub zero. Now oftentimes, that'll be the number zero itself, because the statement will be true for all natural numbers. Sometimes N sub zero, so N sub zero could be zero when you're considering all natural numbers. But oftentimes N sub zero might be one because we're only considering the set of positive integers for which the statement is true. It might not be the case that the statement makes sense for zero, like if there's division involved, you can't divide by zero. So maybe zero is exceptional, but the statement's true for all others. But you are gonna see situations where this number N sub zero could be two, right? The statement could be true when N equals two, and we're just considering integers for which N is greater than equal to two. I mean, why not start at 17? There will be situations where a pattern might not begin until the 17th number, and then after that it follows, okay? So this first number is an integer, it doesn't matter which integer it is, but there's some initial value for which the statement is known to be true at the number N sub zero. It could be zero, it could be one, that's likely what it is, but it could be honestly anything, all right? And then, so we have that, so we have a statement that's indexed by natural numbers. We know the statement is true for a specific integer, which again, N sub zero is indexed, it's indexed by the natural number zero, it doesn't have to be zero itself. If for all integers K that are larger than or equal to N sub zero, if the statement S of K implies that S of K plus one is true, then, so if that's true, then S of N is true for all integers N greater than or equal to N sub zero. So let me try to unravel what's going on right here. So basically what we're saying here is we have these statements about integers, again they're indexed by natural numbers. And so you have some first number which you're gonna call N sub zero, and I'm gonna draw it like a little domino. Then you have your next number which is N sub one, and which of course is just N zero plus one, just to be clear there. You have this next number N sub one, and this is gonna be our domino as well. And then you have another number N sub two, you have another number N sub three, and be well N sub three is just N sub zero plus three. That's all the number is here. And then this continues on for a while, okay? So this condition right here, if for all integers K, S of K implies S of K plus one is true, I want you to think about with our dominoes here, we have N sub K, and then the next domino is gonna be N K plus one. So what we're saying here is that, hey, if this domino falls over, that domino will fall over too. So when you push the domino over, it hits the next domino, which then hits the next domino, which then hits the next domino, which hits the next domino, and they're all knocking each other over. You know, you're watching this video, it's posted probably on YouTube. There's lots of other cool videos you can probably find on YouTube about domino sequences. Feel free to look up one, right? You know, these people who set up these domino arrangements, they push over the dominoes and this whole cascading domino sequence happens, they're cool to hold, to behold, right? There's two important things that you need when you have these domino sequences that make it cool. One, you have to have the case that when one domino falls, the next domino falls. If your dominoes are so far apart that when they fall over, kerplop, it doesn't hit the next one, that's a bummer when it comes to this domino sequence. So they have to be close together so that one will hit the next, okay? But then also what's important is that someone has to push over a domino. If you just see all of the dominoes completely standing up and nothing else is happening, you're like, well, that's kind of cool but it's really no different than watching a painting or a mosaic or something. It's when they fall over that you really see the art form happening. And so that's what an induction proof is. It's dominoes falling over. This first statement is that there is a domino falling over, the first domino. So one pushes the first domino, then they fall over. Then the second statement right here, which is commonly referred to as the inductive hypothesis, you have to prove that if S of K is true, then S of K plus one is also true. That is, if this domino falls, the next one will fall as well. They're close enough together to knock each other over. And as this is an arbitrary case right here, this will actually show all of them falling over at once. Induction is a very powerful proof technique because it actually proves an infinite number of statements about integer simultaneously. You prove that it's true at some point and then if it's true at some place in the middle and the next one falls over, then all of them have to be true. So an induction proof, again, we'll prove why induction is a valid proof technique later on in our lecture series, not in this video here. But what I can tell you is the strategy of induction proof. It always comes in three phases. There's the so-called base case. You have to prove that the statement is true for your initial value. Whether that's n equals zero and equals one and equals 17, it's true for some initial value because someone's got to start pushing over the dominoes. Second, you have this inductive hypothesis where you assume it's true for S of K because when you look at the statement right here, S of K implies S of K plus one. That is a conditional statement. S of K implies S of K plus one. How do you prove a conditional statement? Well, by the method of direct proof, you assume the hypothesis is true and then you conclude using laws of inference and things. You conclude that the conclusion is true, okay? So the inductive hypothesis is the assumption of S of K. So you assume S of K to be true and then you proceed to prove that S of K plus one is true. I call that the inductive step. So here you prove it for S of N zero. You assume S of K is true and then you prove S of K plus one is true. Those are the three steps to any good induction proof and I want to now show you a few examples in this video of how one can prove things by induction. Perhaps the classic example is the following. Prove the identity of one plus two plus three plus four plus five all the way up to plus N is equal to N times N plus one divided by two, okay? Now this is actually an identity we've already seen before in our lecture series, we've proven this true using combinatorial proof. So it might be interesting to compare this proof technique versus the combinatorial proof we'd seen previously. It's a very different technique and it still proves the statement true but in a very different way. And so we're gonna proceed to prove this by induction. When people write induction proofs, they often declare this. They don't always do it at the beginning but we're very much practicing. So I'm actually gonna tell you that we're gonna prove it by induction. But of course, as you get more experience with writing proofs and reading proofs, it turns out that these declarative statements about what proof technique you're using become happening less and less and less because it's understood that the audience that if I start describing a base case and I start describing an inductive hypothesis and inductive steps, then clearly I'm doing improved by induction. But for beginners who are not used to the template, it's sometimes useful to say it. But like I said with an example earlier, if I'm in a homework problem as a graduate student, there was one point I basically said, we prove the statement using induction and that was the whole proof. Sometimes that's all you say, it follows by induction. Again, we're not at that level yet but maybe one day we can be there. You can have that proud moment too. All right, so to prove this by induction, what we have to do is we have to provide a base case. So when you look at this thing, like what is the smallest integer for which this makes sense? You actually can get away with this identity being true for n equals zero but you have to make some sense out of the left-hand side. What does it mean from one plus two to plus three all the way up to zero? It feels a little exceptional. So if you wanna start with n equals one, I'm perfectly fine with that. A lot of people prove this identity with n equals one as the base case. But like I said, you could get away with n equals zero because when n equals zero, you can think of the left-hand side as this empty sum. There was nothing over there, empty sum. With the empty sum, it's understood to be zero. So the left-hand side is zero and then when you put zero in here, the right-hand side's equal to zero. Not a big deal but like I said, the most standard way people approach this is they prove this identity for positive integers and therefore our base case is gonna equal one. So we can declare that. We will show the identity holds for n equals one. Notice that the left-hand side of the equation will be one in that situation because if you take the numbers one plus all the way up to one, that's just the one itself. Which again, I would argue that that case is just as, when you write it like this, that's just the same thing as the empty sum. It's no more exceptional but like I said, we'll stick with n equals one here. So the left-hand side will just be the number one. On the right-hand side, you end up with one times one plus one over two, which gives you two over two, which equals one. And so this tells you that when n equals one, this identity holds. The left-hand side's one, the right-hand side's one, that then proves the base case. The base case might be trivial but it is necessary. If you don't have a base case, then the whole process might not get started. It's an easy step but it's a critical step for induction. Sometimes there's so easy people limit them that is bad practice. If you're gonna provide the details of an induction proof, make some statement about the base case, even if you're like, oh, the formula is trivially true for, is obviously true. It's clearly true when n equals one. You could say something quite pompous like that. It's like, duh, it's true for n equals one. I'm okay with that but you should say something, right? But of course, provide enough details for your audience so that it's appropriate. Then the second step is the inductive hypothesis. All this is, this is an assumption. We assume that the statement holds for some value k. Generally I will use a different symbol than n which would appear in the formula. So that's like, oh, this is some fixed arbitrary number. It's true at that step, okay? So then you just assume it. You assume the inductive hypothesis. Then in the inductive step, you're gonna prove that so because we've assumed, we assume the statement holds for S of k. We proved it holds for S of one. We're assuming it's true. Now we have to prove that the statement is true for S of k plus one. So how do you approach that? Well, I'm just gonna take the left-hand side of the equation for S of k plus one. So the left-hand side would just be one plus two plus three plus four plus five all the way up to plus k plus k plus one. And this is then how you typically can then bring in your inductive hypothesis. When you look at this sum, look at the first k terms in the sum. By the inductive hypothesis, the first k terms is equal to k times k plus one over two. So I'm gonna substitute that in for the first k terms of my sum. And you should make mention that you're using your inductive hypothesis here. By the inductive hypothesis, you get that one plus two plus three plus k is equal to k times k plus one over two. But you also have this extra baggage of k plus one. But hey, let's just work with this for a moment. k times k plus one over two. I wanna add to it k plus one. I can find a common denominator times two over two here. We get the following statement. Now notice that both of the terms have a k plus one in it. If you factor it out, you'll be left with a k plus one times k plus two over two. And I'm gonna rewrite k plus two as k plus one plus one. Cause then this right here, this is the right hand side of the equation associated to S of k plus one. We started with the left hand side of the equation associated to S of k plus one. So those two things are equal to each other. So we've now proven the identity for n equals k plus one. That then means that the identity is true for all integers greater than or equal to one. And that's by induction. So we scream induction at the end of this proof. And that's then it follows by induction. This thing is in fact the case there. Now I do wanna make a comment about our commentatorial proof here that we had mentioned earlier. That when you look at this induction proof, this shows that the formula is true. But it doesn't at all tell you where the formula came from. That's the interesting thing about induction is that if you have a formula that you think it's true, you can prove it by induction and boom, there it is. As opposed to the commentatorial proofs we saw previously for which in the commentatorial proof it actually gives us not just that the identity is true. It gives us good reasons to why we think it's true because of its commentatorial relations. And so there's some benefits and weaknesses to induction because of that. It gives us no intuition to why it's true other than it's true. But the advantage is if you're like a researcher, a mathematical researcher and you're exploring some new formula, if you think the formula is true because you've collected data and it seems to be working for a while, if you can prove it by induction, you don't need a reason why it's true. It's like, oh, I now know it's true by induction. And then like all of the reasoning which you did secretly in your ivory tower doesn't have to be published because induction gives the truth of full of it, the truthfulness of it. And even though you sort of discovered it because of data collection or something, that actually is kind of an asset when you think of it that way. All right, so let's look at some two other induction proofs before we end this video. We're gonna go through these ones a little bit faster this time. Remember three phases, base case, inductive hypothesis and the inductive step. This time we're gonna prove that every integer of the form 10 to the n plus one power plus three times 10 to the nth power plus n is divisible to nine. And we're gonna do this for all natural numbers this time. So this will include zero as well. So our base case, we're gonna say that we prove this thing by induction. Okay, so we declare into the audience how we're gonna do it. Then for our base case, we take n equals zero and just look what happens when you plug in zero. You're gonna get 10 to the zero plus one power plus three times 10 to the zero plus five. Now, of course, zero plus one is one, 10 to the first power is 10. Here, 10 to the zero is one times three is just three. So now we have 10 plus three plus five, which is equal to 18, 18 is nine times two. So it is in fact the case that when n equals zero, that number is divisible to nine. So the base case is then satisfied. Base case is done. All right, now for our inductive hypothesis, assume that the number 10 to the k plus one plus three times 10 to the k plus five is equal to nine m. So we're saying that this number is divisible by nine and that's exactly this number when you plug in a k. So this gives us our inductive hypothesis. We're assuming it to be true. All right, so then of declaring to the audience right here, next we will show that the number 10 to the k plus two plus three times k to the 10 to the k plus one plus five is divisible nine. So I'm telling you what I'm about to do. And where do these numbers come from? Now the exponent of the 10 is k plus one and this exponent right here is k plus one plus one. I've already took the liberty of simplifying, no big deal. So let's start with the number. I wanna argue that divisible by nine. So this is where the magic thing comes into play. How do I show that this is equal to nine? Well, it takes a little bit of, I'm gonna show you what's done here, but you'll see that it's valid. Coming up with the exact calculations here can be a little bit more tricky. But the idea is like, what could you do with this? I have to somehow produce this number in my calculation cause I know that's divisible by nine. Well, you have these 10 to the k plus twos and 10 to the k plus ones, but here you have 10 to the k plus one and 10 to the k. If I could factor out 10, that might be helpful. Not sure what to do with the five yet, but these first two numbers are divisible by 10. You could factor out the 10 and therefore you're gonna get 10 to the k plus one plus three times 10 to the k. That's all times by 10 plus five. This is almost the number I'm looking for, but it's missing the five. So a nice little maneuver, what we can do here is we can add five, but we have to also subtract five so things are balanced. Distribute the 10 onto that negative five and then separate it. This is the same thing as taking 10 times 10 to the k plus one plus three times 10 to the k plus five, which notice that by the inductive hypothesis, I know that number in the parentheses is divisible by nine. You have a plus five, which is already there and then you have this minus 10 times five, okay? Now, like I said, by the inductive hypothesis, this number I know is divisible by nine. It's equal to nine M, which if you times it by 10, you're gonna get 90 M. But then what happens here? 10 times five is 50, five minus 50 is equal to 45, which hey, 45 factors is nine times five. You could factor out the nine so you get nine times 10 M plus five. So notice that this last equation, because this equation right here used, excuse me, this one right here used my inductive hypothesis. This last equation shows that the number was divisible by nine. So that then completes the inductive step. And so by induction, these numbers are always divisible by nine. I had all of the steps there. We had the base case, the inductive hypothesis, the inductive step. So we proved because we assumed S of k holds, we then proved that S of k plus one holds. And that's all that's necessary for the inductive proof. It then follows my induction that for every natural number, this bozo right here is gonna be divisible by nine. To complete this video, let's do one more example here. Again, we're gonna go through this one even faster. And also what's different about this one is let's do an inequality as opposed to an equation. The last one we did have to do with divisibility, but divisibility does turn into something about equations because the number equals nine times something. Let's try an inequality to show you how you can prove an inequality using induction. We're gonna prove that two to the n is greater than n for all natural numbers, which again, that includes zero here. So we'll prove the statement using induction. Let's consider the base case. Two to the zero is equal to one, which is greater than zero. So it's satisfied in that situation. Now some of you, when you think of the actual numbers, we erroneously think that doesn't include zero. So you might've started with one, which note here, two to the one equals two, which is greater than one. So it's satisfied there as well. So if you started at zero, started at one, you would both be true. I actually did two, because sometimes in induction proof, you actually have multiple base cases. I've now proven the base cases because what I wanna do going forward is I actually wanna assume that k is greater than one. I included zero because that is a natural number. I proved explicitly that it's also true for one. So then I can assume k equals one, that it's going forward that k is greater than one, okay? And that's what I'm gonna do with my induction hypothesis. I'm gonna assume that two k is greater than k for all k greater than zero. So I had to specifically prove it for zero and one if I wanted to be true for all natural numbers, okay? Because when you do your inductive hypothesis, the k is assumed to be larger than all of the base cases you've considered already. So I'm gonna be larger than one in this situation. All right, you could do larger than or equal to your base case of course, but I specifically do need that k is larger than one here. So that's at least two. And so how are you gonna prove this inequality? Let's play with around with this. Notice what we have to prove. We have to prove that two to the k plus one is greater than k plus one. That's our plan of attack. So what can we do with two k plus one? Well, I noticed that two k plus one factors as two times two to the k. I like that factorization because that's been the left-hand side of my inductive hypothesis. So then I can use my inductive hypothesis. Two times two to the k is larger than two to two times k, right? Since my inductive hypothesis two to the k is larger than k, that then gives you the inequality we're looking for, okay? But then also note here that k is greater than one, which means k is greater than or equal to two, like so. So what we get here is that two to the two times k, of course, this equals k plus k, but since k is larger than one, remember there, we can get another inequality here. It's greater than or equal to one in that situation. And then notice when we put this all together, we now have two to the k is greater than k plus one. All right? Now notice, yeah, you do have equality, but you also have it greater than or equal to. So that's more strict. The strictest symbol we used here was greater than, there was a strict greater than there. So that then when we put those together, we get that two to the k plus one is strictly greater than k plus one, finishing the inductive step. So we have all three cases, the base case, the inductive hypothesis, and then the inductive step. Since we finished all three phases, the result then follows by induction that this inequality is valid for all natural numbers. We'll do some more examples of inductive proofs inside of lecture 20. So take a look there if you wanna see some more examples, but we'll sign off for right now. See you next time.