Whenever that we have the division theorem available, we can look at the important mathematical property of divisibility. If division of a by b produces a remainder r equal o 0, we say a is divisible by b. Hence, a is divisible by b, if and only if there's an integer q, so is it a equals bq? For example, 45 is divisible by 9, but 44 is not divisible by 9. The notation that we use for divisibility is this, b vertical line a denotes a is divisible by b. And let me give you a warning at this point. B divides into a, or a is divisible by b, Is not the same as b divided by a. With a slanted line. This guy. Is a relationship. Between a and b. It's true or false. This guy, Denotes a rational number. The result of dividing b by a in the rational numbers. So, we have a property relationship which is either true or false. And we have a notation for a rational number. Now, these are quite distinct concepts. This is something to do with two integers, tells you whether two integers are in a certain relationship, this is a notation for a specific number. The reason I'm issuing this as a warning is that beginners in particular often confuse these two things. They both, after all, have to do with division. But in this case, it's a property, has nothing to do with fractions with rational numbers. This is a specific number. It's a specific, rational number that we're referring to. Now, there wouldn't be as much of a problem if the notation wasn't very similar. And it gets worse when people with sloppy handwriting, and you've probably noticed that I'm one of them, often don't distinguish clearly between a vertical line and the slanted line. In fact, this one is already almost vertical. Maybe it would have been better if I'd written it like that. Okay, that's better. But because the notation is similar, it's often very easy to get confused between these two things, the context does disambiguate. If you understand what's been discussed, you shouldn't have any problems. But bear in mind, especially when you're meeting this material for the first time, that there is an important distinction here. And in fact, I've got a quiz coming up in a moment, which I hope is going to help cement your understanding of this notion, so that you'll be able to distinguish it from this one. Okay? And once we've got divisibility lined up, we can define the all important notion of a prime number. A prime number is an integer p, greater than 1, that is divisible only by 1 and p. Notice that we explicitly exclude 1 from the prime numbers. So, the first prime is 2. The next one is 3, then 5, then 7, then 11, and so forth. Now for that quiz about divisibility. Okay, how do you do it? Well, remember the condition we have to check is this one. B divides a if and only if there is a q, such that a equals bq. Well, let's see what we have. The correct one's, B, d. F, g and h. Why don't I have a? Well, remember the whole definition if divisibility was under the assumption that b is non 0. We can't consider divisibility by a 0 number. So that one doesn't work. And likewise, that one doesn't work. And of course, the reason this one doesn't work is because 7 simply doesn't divide into 44. Okay, do be careful about this, though. Division by zero is problematic under any circumstances, and so we rule it out. In this case, it was ruled out in the very statements of the divisibility theorem. Okay, how about this one? Well, again, the cognition we have to check is that b divides a, if and only if there is a q, such as a equals bq. And again, this is under the assumption b non zero. Everything is an integer here. Well, the correct one's, that, that. And that. So, why are the others false? Well, this one is false because that simply doesn't divide that. However, if you sort of got out a calculator and actually tried to divide it, then you were missing the obvious point. This is an even number, and that's an odd number. And we cannot have an even number dividing an odd number. So if you have to do any arithmetic with this, then you're making the mistake that I've been warning you against of acting as if you were in high school. This is a course about mathematical thinking. And what I hoped you would've done is to think about this mathematically, and realize that you can't have an even number divided into an odd number. So, this is a cautionary reminder that this is not about jumping into calculations and applying procedures. This is all about thinking about a problem. Okay, so it's false. Well I would hope that you would recognize it's false without doing any work whatsoever other than a little bit of thinking, okay? [LAUGH] I occasionally like to stick things like that into quizzes, just to keep people on their toes. Well, this one is false of course, because that's simply not the case that 2n divides n squared. The reason the others are false, well, this is false for various reasons. I mean, it's the same, 2n simply doesn't divide n squared for various reasons. The reason this is false is because if n can vary over Z, then that would include n equals 0. And so, we can't have this because we can get a 0 coming in. That includes n=0. Likewise for this one, okay, same thing. So this one fails because 0 was included. If I excluded 0, it would be fine. The only difference between these two is that this is the positive integers, 1, 2, and 3 and so forth. This one because it's all of the integers includes 0. Okay, well let's move on now. I want to prove a theorem now that gives the basic properties of divisibility. So, theorem, let a, b, c, and d be integers with a non-zero. Then, the following all hold. (i), a divides 0, and a divides a. (ii), a divides 1 if and only if a= + or -1. (iii), If a divides b, and c divides d, then ac divides bd. This is under the assumption c is non-zero. Because we can't have divisibility by 0. The division theorem itself excludes the case of division by 0. (iv), if a divides b and b divides c, then a divides c. And this is under the assumption b non-zero. Again, we have to exclude the possibility of division by 0. (v), a divides b, and b divides a. Those two hold if and only if a = + and -b. (vi), if a divides b, and if b is non-zero, then absolute value of a less than or equal to absolute value of b. You can only divide smaller numbers into bigger numbers, or at least numbers that are no bigger than. You can't divide a number b by something that's bigger than it in absolute value. One more? (vii), if a divides b, and a divides c, then a divides (bx + cy) for any integers x, y. Okay, you prove all of these by going back to the definition of divisibility. What I'm going to do is just, I'll give you, I'll prove two of them as examples. Let me just pick number (iv). Let's prove (iv), that's this one here. All the proofs are essentially the same idea. If a divides b, and b divides c, then that means there are integers d and e, Such that b = da, and c = eb. That's the definition of divisibility, in which case c = d times e times a which again, by the definition of divisibility, means a divides c. Let me do one more, let me do (vi). Okay, so a divides b so that means, since a divides b, it means there is a d, Such that b = da, in which case taking absolute values, absolute value b equals absolute value d times absolute value a. And since b is non-zero we know that the absolute value of d will have to be greater or equal to 1. So the absolute value of a is less than or equal to the absolute value of b, okay. B is non-zero, so d can't be 0. So its absolute value is greater than or equal to 1. And if the absolute value of d is greater than or equal to 1, then a has to be less than or equal to b. The other statements in the theorem, Are proved similarly. You just take it back to the definition of divisibility. And when you know that remember the definition is b divides a if and only if there is a q, Such that a = bq. And in all cases you go back to the definition and then the thing drops in a couple of lines. Okay, well that lists all of the basic properties of divisibility. Okay, it's time to prove the fundamental theorem of arithmetic. Theorem, Every natural number greater than 1 is either prime or can be expressed as a product of primes in a way that's unique except for the order in which they are written. For example, 4 is 2 x 2, or 2 squared. 6 is 2 x 3. 8 is 2 cubed. 9 is 3 squared. 10 is 2 x 5. 12 is 2 squared x 3, and so on and so on. 3,366 = 2 x 3 squared x 11 x 17 and so forth. I worked that one out in advance by the way. Okay, so 2 itself is prime, 3 is prime. 4 is a product of primes, 5 is prime. 6 is a product of primes, 7 is prime. 8 is a product of primes, 9 is a product of primes. 10 is a product of primes, 11 is a prime. 12 is a product of primes, and so on and so on and so forth The expression of a number as a product of primes is called its prime decomposition And when you know the prime decomposition of a number, you know an awful lot of information about that number and how it behaves with relation to other numbers. Well we proved part of this a little while ago, proved the existence part. Though I'm going to give you another proof of existence in a moment. Well the new part is to prove uniqueness. The uniqueness proof will require Euclid's Lemma. That says if a prime p divides a product ab, then p divides at least one of a, b. The proof of Euclid's lemma is not particularly difficult, but it would take me outside the scope of this course. Remember the focus of this course isn't to teach you number theory and number theoretic techniques, it's to develop mathematical thinking. And it would be too much of a detour from that goal in order to prove Euclid's Lemma. But if you just Google Euclid's Lemma, you should be able to find some references that would lead you to a proof. Let me give you the existence proof. More precisely let me give you a new proof of existence Remember this is the statement of the theorem. Any natural number greater than 1 is either prime or can be expressed as a product of primes in a way that's unique except for their order. And we're going to prove existence. The previous proof of existence that I gave you used a method of mathematical induction. And I gave it as an illustration of the method of induction, but I'll give you a different proof now. I'll prove it by contradiction. Suppose there were a composite number, that's a non-prime, that could not be written as a product of primes. Then there must be a smallest such number. Call it n. Since n is not prime, there are numbers a, b strictly between 1 and n, such that n=ab. If a and b are primes, then n=ab is a prime decomposition of n, and we have a contradiction because n was chosen so as not to have a prime decomposition. It was actually the smallest number that didn't have a prime decomposition. Well if it's not the case that a and b are primes, then at least one of them must be composite. But if either of a, b is composite, then because it's less than n, it must be a product of primes. So by replacing one or both of a,b by its prime decomposition in n = ab, we get a prime decomposition of n. And again, we have a contradiction. That proves existence. That's first part of the proof of the fundamental theorem of arithmetic. Now let me prove uniqueness. So what I have to prove is that the prime decomposition of any natural number n greater than 1 is unique to the ordering of the primes. And I'll prove it by contradiction. So assume there is a number n greater than 1 that has two or more in fact different prime decompositions. Let n be the smallest such number. Let n = p1 times p2 times da, da, da times pr = q1 times q2 times da, da, da, qs, be two different prime decompositions of n. So we have n as a product of r primes. And we also have n as a product of s primes, some of the primes being different Well, p1 is a prime factor of n. So p1 divides n. So p1 divides this product. Since p1 divides q1 times q2 through qs, and let me stop at this moment and see what I've done. I've said p1 divides n. So p1 divides this, and I'll split that into two that way. And notice that p1 divides q1 times that. And I've done that in order to apply Euclid's Lemma. So now let me continue that sentence. By Euclid's Lemma, either p1 divides q1 or p1 divides the product q2 through qs. Remember, Euclid's Lemma says that if I have a prime number p, and if p divides a product, a times b, then p divides at least one of ab. So Euclid's Lemma was that if p divides a product, ab, it divides at least one of those two numbers. So from here, p1 divides this product. I've now written it as a product of two numbers. And I can now apply Euclid's lemma and say that if p1 divides that times that number, it must divide at least one of them. Either p1 divides q1, or it divides the other part, or maybe both of them Okay? Hence, either p1 = q1 or else p1 = qi for some i between 2 and s, okay? If p1 divides q1, then they're both prime, so the only possibility is that p1 equals q1. Or else p1 divides this product of primes, which means that p1 actually is one of those primes. So first case, I know that p1 = q1. Second case, I know that p1 = qi for one of the is between 2 and s. But then we can delete p1 and qi from the two decompositions in star, which gives us a number smaller than n that has two different prime decompositions, contrary to the choice of n as the smallest such. Okay, having established that pi is equal to one of these qs, we don't know which one, but it's going to be one of them. I can delete the p1 form here. I can delete the appropriate q from there. And when I delete p1 from there and some q from there, I still have equality. I still have a prime decomposition of a number. But having deleted that common prime, the number that I get is smaller than n. So I get a number smaller than n that has two different prime decompositions. The number smaller than n will be p2 times da da da times pr. Also, that will be a number smaller than n. And I've deleted a number from here, and what's left is a prime decomposition here. So the proof involves simply identifying the fact that, or while recognizing the fact this uses Euclid's lemma, that you're going to have the same prime on both sides. You can delete it and get two different decompositions of a smaller number. And that proves uniqueness. Okay, well now we've proved the fundamental theory of arithmetic, cool? So did that make sense? Chances are you're going to have to spend some time trying to come to grips with this. Though I know you're familiar with arithmetic, especially whole number arithmetic, this is probably the first time you've tried to really analyze numbers. See how you get on with assignment nine.