 Welcome back to our lecture series Math 4220 Abstract Algebra 1 for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Missildine. Lecture 6 in our series is going to be based upon section 2.2, the division algorithm from Tom Judson's abstract algebra textbook. This will be a continuation, actually, of our discussion of the induction axiom and the well-ordering principle associated to whole numbers, right? We're also going to be focusing on the idea of divisibility in this lecture. And so to start off with, we introduced the so-called division algorithm, which is kind of a funny little name because it turns out as we prove the division algorithm, we won't actually have an algorithm in hand. The algorithm one would typically use when it comes to division of integers is typically like the long division algorithm we learn in school, like primary school and such. We'll also a little bit more about this at the end of this proof. But the division algorithm, as it's commonly referred to in the literature here, the division algorithm, well, this is a statement about integers, right? We have two integers a and b such that b is a positive number. It's strictly greater than zero. So given these two numbers here, there exist unique integers q and r such that a equals qb plus r, where r is some number between zero and b, where it could actually equal zero, but it won't ever equal b. It'll be strictly smaller than b. So give me some motivation behind this equation right here. We have these two numbers a and b. We want to consider what happens when we divide a by b. But the thing is, if we're living in the realm of the integers, division isn't exactly always doable, right? I mean, sure to four divided by two is equal to two, but one divided by two is it's not an integer. It's a rational number. But without, you know, within the realm here, how do we describe what division actually means, right? You know, as we as citizens who belong to a three dimensional manifold, we look out in the sky, we see the three manifold, but is it Euclidean or is it some type of non Euclidean geometry? Well, if we could step out of our geometry and look at it, you're like, oh, we can see the curvature. But as we live inside of it, you know, things look like straight lines. So how do we see curvature when lines by definition is what straight means? It can be difficult to see the experience while you're inside that specific set. So how does one describe division to the citizen of the integral world, right? Well, we can't talk about fractions. And so instead, we talk about this type of statement right here, where these numbers q and r, as the mnemonic device might suggest, these will take the roles of quotient and remainder when we do division. So division in terms of integers means that we're going to basically look for the largest multiple of b that's still less than or equal to a and then a r here, the remainder will compensate the difference. How many things are left over? If we want to pass out all of our little candies amongst our friends here, right? We have four friends, but 15 candies. It's like, oh, everyone gets three candies, but then we're left over with how many did we have there? I said 15 candies and four friends, we're gonna three times two. Sorry, not three times two, three times four, that would give us 12, but then there's still three left over. That's the type of thing we're trying to describe in this in this lecture right here. So let's let's first make it clear that the division algorithm is in fact a valid a valid statement here. And I want to go through the proof of this because actually is an application of the well ordering principle we've learned about in the previous lecture here. So to apply the well ordering principle, we have to come up with a set of natural numbers. And then we can invoke the well ordering principle to get a minimal natural number inside that set. So the set that we're going to play around with is we're going to take the set of all differences of the form a minus bk. So if you think of our equation right here, the idea is we're going to move the qb to the side a minus qb is equal to r, and we're going to allow this q here to vary. So a and b are going to fix numbers. We're going to allow q to vary. That's where the symbol k comes into play. And so then we're considering elements of the form a minus bk, like so. Now we have to be careful and make sure that this is a set of natural numbers. So to accomplish that, notice that while k is going to be any possible integer, we do restrict our choice of k so that a minus kb is greater than equal to zero. A non-negative integer is a natural number after all. So s is going to be a subset of the natural numbers like so. Let's see. So what else should we say? So it's a subset of the natural numbers, but is it empty? Is there anything that actually belongs to the set s? Well, we could double check here, right? So if a is itself a non-negative number, notice if you take a minus nothing, you know, a minus zero times b, since a was non-negative, that would be a that would be a loud combination inside our set s. So therefore s would contain something. On the other hand, if a is negative, right, if a is negative, what we're going to do is we're going to take a minus b times 2a. So think about what's happening there for a moment. You could factor out the a, a it's itself a negative number, but then look at one minus 2b right here by by assumption b was a positive number. So if you take one minus 2b, that is certainly going to be a negative number as well, right? Because b itself has to be a positive integer. So if I subtract two times it, you're going to get something bigger. The main reason we need a two here is because what a b is one itself, right? We want to make sure this is zero per se. We're trying to, I guess that might be okay. But anyways, we want to make sure that this product is non-negative for which those conditions we have right here. So irrelevant of the assumptions of a, we can guarantee that there is something inside of s, right? s right here is not the empty set. So we have a non empty set of natural numbers under these conditions, we can now invoke the well ordering principle. Every non empty set of natural numbers has a minimal element for which we're going to call that r. Now I want to mention for the sake of students watching this video, right? When one uses the well ordering principle, we're usually pretty good about recognize, oh, yeah, it has to be natural set of natural numbers, right? I mean, because we have a set full of the color blue and porcupines, then clearly it's like, oh, the well ordering principle doesn't apply to that set. We make it very clear that we should be having only natural numbers in consideration, but it's very quick to forget many of us are that the set needs to be non empty. Because most of the time when we talk about sets, they're naturally non empty. I mean, there's only so much you could say about the empty set. So we often take for granted that there might be anything in the set, because one has to remember that the definition of a set doesn't necessarily exclude the possibility that it's empty. We have to verify that something actually belongs in belongs in the set, which we've done. And therefore we can call that element r. So now we have a candidate for this equation right here. We there's got to be some q that produce that r, right? Because of the set, there's going to be some k value, which we'll call it specifically q. So that r equals a minus bq notice if we solve for a, this will give us the equation that the division algorithm requires, right? So that's that's a good place to be in right now. So that gives us existence, right? But the division algorithm has more than just existence. We say that there exists unique elements, unique elements p and q such that these conditions happen. So now that we have existence, we need to show uniqueness of r and q in this consideration. That's what we're going to do next. So I guess I guess the other thing we haven't done yet is we haven't established the fact that r satisfies these inequalities that r will be between zero and b. The minimality of r is going to be useful in that regard. So by definition, r itself does have to be a non negative number, because s only contains non negative natural, it only contains natural numbers, right? Only non negative elements there. So the fact that r exists means, and since it's an s, it will be non negative. But why can't it be bigger than q? I mean, why can't it be bigger than b? Well, let's suppose it were, right? So let's consider the number a minus b times q plus one, which you can argue that this number is going to be smaller than r, right? Because if you distribute the b, you get a minus bq minus b, a minus bq is equal to r, and then minus b here. Since b is positive, subtracting will make you get smaller. So this number is going to be strictly less than r. That's an important thing. The minimality of r then suggests that this element right here cannot be in a side of s, which would then support, that would support the fact that r minus b actually is a negative quantity, which we're mentioning right here. And so since r minus b is negative, since r minus b is negative, right, r minus b is less than zero, then add b to both sides and you see exactly that r is less than b, which is what we're trying to get right there. So that verifies that r then, the r we've constructed satisfies its inequalities, zero less than equal r less than b. So now we have existence, there is an r and a q that satisfy the conditions given by the division algorithm. Why is it then unique? Well, to show that something is unique, typically, the proof's the following, we're going to do a proof by contradiction, where we're going to assume two elements satisfy the condition and then derive a contradiction. A slight modification of that is we'll just assume two elements satisfy the assumptions in play here without any extra assumption that they have to be distinct elements, you just have two two elements that satisfy the conditions, right? So we'll assume there exist elements q and r so that a equals bq plus r and there are just other elements q prime r prime so that a equals bq prime plus r prime and then they satisfy the inequality conditions required. We're not necessarily going to assume that r and r prime are distinct, then we can argue they're equal to each other. It's basically the same thing as the proof by contradiction. We just have to show that if there's two things that satisfy the same condition, they're actually equal, thus proving there was only one of them. So we will make that assumption. There's a q and r and there's a q prime r prime. These are both quotient remainder pairs. So if they were different, right, one would have to be bigger than the other in terms of the remainder. So without the loss of generality, we can assume that r prime is greater than equal to r. Again, I didn't construct this as a proof by contradiction. I'm not I'm not assuming that they're different. So it actually could be allowing the fact that r prime equals r. But if one would if they are different ones bigger than the other, so we can assume that r prime if there is a bigger one is the bigger one. All right. And so with this right here, these two equations, I mean, we could put these together and have like a system of equations, right? You know, we have a equals bq plus r, we have a equals bq prime plus r prime. This is not a very difficult system of equation to solve here, right? Because both left hand sides are equal to zero, we could substitute out the A and just make them equal to each other. We end up with bq plus r equals bq prime plus r prime. So we can remove the A from the equation and get the following said statement. Now noticing that both equations have a multiple of b, let's move them together. And so I'm going to I'm actually going to move the bq prime to the left hand side. That just feels good, I guess. So you get bq minus bq prime, we're going to move the r to the other side as well, r prime minus r. And then the left hand side would then factor, you know, there is a multiple of b right there. The mathematician, I can't help but factor out the b. You get b times q minus q prime is equal to r minus r prime. Now you can actually see why we move things the way we did. Since we're assuming that r prime is greater than equal to r, what that tells us about the left hand side here is that r prime minus r will be greater than equal to zero. So that's the thing we want to say about this right here. Now in terms of divisibility, we see that b divides r prime minus r. We see that b divides r prime minus r because we have a factorization of r prime minus r, not a big deal right there. So we get that pretty quickly. We also from there, I mean since b divides r prime minus r, and clearly r prime minus r is less than r prime. r is non-negative number as well right? It's greater than or equal to zero. So b, it divides r prime minus r which is less than r prime. But r prime itself by assumption is less than b. So think about that for a second. We have a positive number because b itself is positive. We have a positive number that divides a number less than itself. The only multiple less than b that's greater than or equal to zero is zero itself. So these inequalities combined with the divisibility argument right here, I mean basically what we're saying is since b divides r prime minus r and these are both positive numbers, we basically can force an inequality right there. The only way we're going to get something like this happening otherwise in terms of the divisibility here is that we have to get zero right? B does divide zero and that wouldn't violate this inequality here right? We don't have that b as less than b. The only way to escape a contradiction here is to assume that r prime minus r equals zero. But that then gives us that r prime equals r and then plan with the with the previous equations right? That would then force that q equals q prime. And thus we see that the q and r from the division algorithm are in fact unique. That there exists only one pair of numbers that'll satisfy both this equation and the inequalities, this equation and inequalities. And so like I said, while we are able to prove the division algorithm from the well-ordered principle, unfortunately this argument is a non-constructive argument. That is although we know q and r exist, we don't have any idea what these values are, you know, for any specific numbers. Now fortunately, like I mentioned earlier, the long division algorithm that we learned from grade school, which is actually why we call this theorem the division algorithm because you didn't actually see an algorithm in play here, that's actually what helps us compute these things. So in fact, if we were doing something like, oh let's take 123 divided by 5, you know, what does one actually do? It's like, well I searched for a multiple of 5 that goes into 12. That would be 10, which is 2 times 5. You subtract it, you get a 2, bring down the 3. 5 goes into 23, well the biggest multiple I can think of would be 20, which is 4 times 5 like that. You subtract, you get a remainder of 3. So you get something like this. So we're saying that 123 is equal to 24 times 5 plus 3. We can do this algorithm. The algorithm itself isn't too difficult to do, but this proof, the fact that we know this algorithm will work for any two numbers is a consequence of the well-ordered principle.