Where the correct answers are this one, this one, and this one. If you check back at the description I gave of the method of induction, you shouldn't have any difficulty realizing why this is the correct answer, why this is the correct answer, and why this is the correct answer. So if you got any of these wrong, you definitely need to go back and look at the lecture again and make sure you really understand induction. It's a very important technique in mathematics. Well, let me give you another example of an induction proof now. And this time, I'll lay out the proof in a way that makes explicit the connection between the argument I give and the description of induction I gave a moment ago. And the example I want to give is the following theorem. Well, calling it a theorem is actually a bit of a stretch. But here's what I'm going to show, that if x is a positive real number, then for any natural number n, 1 + x to the n + 1 is bigger than 1 + ( n + 1) x. Okay, this result is not going to set the mathematical world on fire, but it is a good example to illustrate the way mathematical induction works, and in particular, to connect the arguments that I'm going to use to the earlier description I gave for mathematical induction. So here's the proof. Well, I'm going to begin by stating the method I'm going to use. It's mathematical induction. Since one of the purposes of proofs is to explain why something is true, it's always a good idea to begin by stating the method you use. That orientates a reader to the argument that's about to come. And to connect it to the description I gave of induction before, I'm going to let a of n be the statement (1+x) to the n+1 > 1 + (n + 1x). So by induction I'm going to prove. For all n A(n). Well, the first step is to prove A(1). And A(1) is the statement. 1 + x squared is bigger than 1 + 2x. We put n equals 1 here. We'll get 1 + x squared greater than 1 + 2x. Is this true? Well, the answer is yes. It's true by the binomial theorem. Which tells us that (1 + x) squared is equal to 1 + 2x + x2, which is bigger than 1 + 2x. Because x squared is strictly positive. Notice that x is positive, so in particular, x is nonzero. Hence x squared is strictly positive. Hence we have a strictly positive term here. Hence that expression is strictly bigger than 1 + 2x. So A(1) is true. The induction step Involves proving for all n, A(n) implies A(n+1). Well, how do we typically prove a statement like this? We pick an arbitrary n and prove A(n) implies A(n+1). And how do we prove an implication like this? We assume the antecedent and deduce the consequent. Well, A(n) is the statement 1+x to the n+1 greater than 1 + (n+1)x. And A(n+1) is the statement 1 + x to the n + 2 is greater than 1 + n plus 2x. So the induction step boils down to assuming this and deducing that from it. So I'm going to focus on this now. And I'm going to begin by looking at the first expression. (1+x) to the n+2. In order to use the induction hypothesis, I'm going to pull out one of these terms and write that as (1+x) times (1+x) to the n+1. And by the induction hypothesis, (1+x) to the n+1 is bigger than 1+(n+1)x. And that equals 1 + (n+1)x+x+(n+1)x squared. Which equals 1 plus I've got (n+1)x and another x, so altogether I've got (n+2)x + (n+1)x squared, and this is positive. So that expression is strictly bigger than 1 + (n+2)x. And that part of it and the part here connected here with this is this. In other words, this proves n + 1. And the theorem follows by induction. Okay I've laid this out in gory detail here, but my purpose was to connect the kind of argument that you give when you do an induction proof with the description of induction that I gave a moment ago in terms of predicates A(n). So this is, so what I've been doing here, is really emphasizing how my area description of induction connects to an actual proof. If you check out all of the pedagogics, stuff that I've put in here, you've really got a two or three line argument. I mean, it really boils down to that step, and then this step here. Those are the two key steps of the induction proof. Okay, well there we are. Let me summarize how induction works. You want to prove that some statement A(n) is valid for all natural numbers n. You first prove A(1). That's usually a matter of simple observation. The next step is give an algebraic argument to establish the conditional A(n) yields A(n+1). Usually you assume A(n and then you deduce A(n+1) from it. And typically, you do that by looking at the form of A(n+1), manipulating it into such a form that you can apply A(n) in order to carry out an argument. In other words, reduce A(n+1) to a form where you can use A(n). In conclusion, by the principle of mathematical induction, this proves for all A(n). Occasionally, you might have to work pretty hard to establish a first case. Generally, this step is easy, and this one is where you have to put in the effort. But neither of these two steps on their own constitutes an induction proof. It's the whole package that constitutes induction. My illustration of induction as knocking over a row of dominoes, while it's intuitively helpful, obscures the fact that this principle of mathematical induction is actually a pretty deep principle of mathematics. The reason it's deep, by the way, is because this is talking about an infinite collection. It's talking about all of the natural numbers and there are infinitely many of them. And the moment you start talking about infinity in mathematics, there are subtleties and complexities lurking on the wings. Into that, and that's why we usually complete a proof by writing a sentence something like by the principle of mathematical induction, the result follows. Because this is heavy weight, and it's usually good practice and a shrewd move to mention heavy weight, it there's a heavy weight in the proof. Okay, well that's induction I'd better tell you about a fairly common variant of induction. We sometimes need to prove a statement of the form, for all n greater than or equal to n0, A(n). Where n0 is some fixed number, 5 or 10 or 20 or a million of something like that. In this case, the first step is to verify A(n0). A(1) may not be true. In fact it's usually the case when we are trying to prove something of this form, the reason we have a n0 as the starting point is precisely because the result doesn't hold at the beginning. We have to go out somewhere through the natural numbers until the results starts to hold. General induction starts at one, this variant starts at some point beyond one. But other than that, the argument's essentially the same. In particular, the induction step is to prove for all n greater than or equal to n0, A(n) implies A(n+1). In fact, the example of induction I want to give you next uses this variance of induction. And that example is part of a famous result of mathematics called the Fundamental Theorem of Arithmetic. Before I do, let me give you a quiz. Is 1 a prime number? Okay, I know this isn't the focus of the course but there is a lot of confusion about this. And if we don't clear it out of the way now, it's going to keep haunting us as we go through the material that's coming up. Now I could, of course, simply give you the answer. And in previous terms, when I've given this kind of course, both in a book and elsewhere, I've given students the answer. But it is one of these things that people just don't sort of really pay attention to, and then it comes up again when they have homework exercise and discussions. So, what I'm going to do, is give you this as a quiz. So, I want you to answer whether this is a prime number or not. Whether 1 is a prime number or not. Okay? Well, the answer is, 1 is not a prime number. And the reason is the definition. Here's the official definition of a prime number. It's a positive integer n, greater than 1, whose only exact divisors are 1 and n. That's all there is to it. It's a definition. So from now on please be very careful that you really know the definition of a prime number. 1 is not a prime number. Incidentally, whilst we are still on the topic of getting definitions cleared out, let me make another remark. 0 is not a natural number. Again, this is by definition. 0 is an integer, but it's not an actual number. Historically, the natural numbers are the counting numbers, and people historically started counting at 1, 2, 3, et cetera. 0 was very much the Johnny come lately, okay? So these come out all the time. They cause confusion, unnecessary confusion, so I thought I would get it out of the way now. And I would draw your attention to this definition by putting it in as an equation, okay? Now let's go on to what I was planning to do. Here's the part of the Fundamental Theorem of Arithmetic that I want to prove. It's the following theorem. Every natural number greater than 1 is either prime or a product of primes. Now we have to use that variant of induction for this one because it's not 2 at n equals 1. The theorem only holds for n equal to 2 or more. In fact one is the only number for which it doesn't hold. So we're only adjusting the variance for just this one point. But we have to start at n equals to two. In other words, the first step of the induction is going to involve looking at n equals to two, not n equals to one. But with that one change, it's going to mean induction proof, so proof by induction. Well, the induction statement then is what you might think it's going to be, n is either prime or a product of primes. Okay, based on the examples we've seen so far, you might imagine that A(n) in the induction statement is this. A(n) is either prime or a product of primes. Well, that turns out not to be the case. Now that just doesn't work. We have to pick something else. And this is what we'll pick. For all n, if 2 is less than or equal to m, is less than or equal to n, then m is either a prime or a product of primes. So the induction statements A(n) talks about All the numbers between 2 and n being prime of products or primes. So it's a sort of cumulative statement if you like. As we go up through the natural numbers, we iterate on the cumulative fact of things being primes or products of primes. And the induction proof establishes that A(n) is true for all n bigger than 1. So, we have to start with the first step, namely, n=2. A(2) says what? Well, if n = 2, there only is one m between 2 and n. So, the universal quantifier here is essentially vacuous. I mean, it's just a special case of a general case. There is just n=2, so this statement when n=2 simply says 2 is either prime or a product of primes, which is true. 2 is a prime. Now, we're going to assume A(n). And deduce A(n+1). Well, A(n+1) is going to talk about all numbers m between 2 and n+1. So let's look at one of those ms. Let m be an actual number, 2 less than or equal to m, less than or equal to n + 1. Well if m is less than or equal to n + 1, m is either less than equal to n, or m is equal to n+1. If m is less than or equal to n, then by A(n), m is either a prime or a product of primes. The only other possibility is that m equals n+1. Well, If n + 1 is prime, Then m is prime. So what happens if m = n+1, and n+1 is not prime? Then there are natural numbers p, q such that 1 less than p, q less than n+1 and n+1 = pq. If n+1 is not prime, That it can be obtained as a product of two smaller numbers. Not being prime means you can find two smaller number whose product equals the number, two smaller numbers other than 1, that is. So this is just a definition of not being prime. Well, since 2 less than or equal to p and q less than or equal to n, by A(n) p and q are the either primes or products of primes. These two numbers p and q are less than n + 1, so they are less than or equal to n. That means these numbers are ms that fit into the induction hypothesis. They are less than or equal to n, so by A(n), each of these is either a prime or a product of primes. But n+1 is a product of them. So n+1 is a product of two numbers, each of which is either a prime or product of primes. n+1 is a product of primes. The theorem follows by induction. If you look back through this proof now, you'll see why we took our induction statement to be this somewhat more complicated looking statement. We wanted the induction statement to talk about all the numbers up to and including n being either prime or products of primes. So that when we went through the proof and we came down to a pair of numbers like this, we could immediately apply the induction hypothesis, A(n), in order to conclude that each of those two were either prime or products of primes. And therefore, that the n + 1 was either a prime or a products of primes Incidentally formulating induction statement that are cumulative in this function is quite common. It's often the case that an induction proofs you end up dropping down to some number or some pair or a group of numbers smaller than the one you've got and needing to apply the induction hypothesis to those smaller numbers, as we did here with p and q. Okay, how did you like that one? Okay, so, now, you know about induction. And with induction, we're starting to get into some serious university level mathematics. Though the general idea is simple enough, if you can always go one more step, then you will eventually reach every natural number. In practice it can require a lot of effort and ingenuity to construction an induction proof. See how you get on with assignment eight.
