 Grading for me though, so that's all right So let's just talk about a couple of Things here before we get started Your test is starting to Come up. It's coming up here pretty soon now. So first test is next Thursday. Okay? Yeah That sounded almost like you're gonna vomit. I appreciate you not doing that in class Yeah, please please don't do that I've had that happen before and it's not it's not pleasant for anybody Okay, so next Thursday's first test. So here's what here's what we're gonna do we're gonna cover one section this week and You will have another assignment the assignment why should we do at the test? So I'm gonna give you nine days on the next assignment Otherwise if you turn it in Tuesday, then I won't have it done And it's just it's not really advantageous really for you to turn in on Tuesday rather than next Thursday So I'm just gonna give you two more days for it. Okay, so you can turn in next Thursday I'm gonna give you a few more problems, but it's like says only one section you have nine days to work on it So I think you should have plenty of time to finish it So we're gonna spend two days on that next Tuesday. We will review and I'll also right now I'll go ahead and just kind of give you a very rough idea of kind of things I would be expecting on the exam Remember to listen up to this. Okay? I'm not going to for example ask you to prove that the division algorithm on the exam. Okay I'm not gonna ask you to write Two-page proofs on the test what I'm gonna be doing is roughly here's what I'm gonna do on the exam I'm gonna ask you I may ask you the statement of the division algorithm state exactly what it says, right? This is what we did last time this big long proof. I'm not gonna have you do that I certainly ask you statements of theorems You know state the binomial theorem for example, right? There's another thing I could I could ask you I could ask you to state the first principle of mathematical induction, right or the second things like this. These are fair game. I Could ask you definitions and as far as proofs go You will see that will show up, but they will not be overly complicated. Okay? They will certainly be of the form of the flavor that you've seen before and Probably they will not be among the hardest problems you've done. Okay, there will be things that you really should be able to do So that's that's just very roughly what you should be looking at. All right Since you've been that's what we've been doing is proofs. That's what the point of the course is So of course you should expect that you're gonna have a proof or two to do on the exam But it will be reasonable. It'll be reasonable I don't expect you to be clever and come up with all these good creative ideas on the spot when you're nervous Okay, I don't expect you to do that. So You don't have to worry about that Okay So And my plan right now we'll see how this goes my plan right now is what you're turning in you've turned in two sections So I'm my plan as of now is to count them both as separate assignments. Okay, so you'll get two grades for this assignment That may not do that depending on how The grading goes if I Start to get Really really upset then maybe I won't do that but now I'm just I'm I won't that I'm sure that's what I'll do So All right. Well, let's go ahead and move on then to 2.3. See this is the better All the homework you're turning in today. Yes for sure you definitely will you'll have it back before the test Yeah, you will not have it back on Thursday. I can tell you that there's no way I'll get through those two by Thursday, but you'll have it back next Tuesday. Yeah No, no, no two three will be on the test. Yes, we're because we're not doing anything new next week next week It's just review, but but this it's yeah, I mean, it's it's a week and a half before the exam So this this will definitely be on the exam. So everything that we covered through this section will be fair game on the test Okay, so This section this is something you you actually learned about these things a long time ago You might have learned about this in fifth or sixth grade. We're gonna take a much Different approach to this is not gonna be just what's the GCD of seven and nine? It's it's not gonna be like that, but I'm sorry. Yeah, it's gonna get a little just a little more complicated than that Yeah, right Okay, I'm gonna I'm gonna have to bring in rocks then I guess okay So here's what I'm gonna do I'm just gonna go straight into the definition because you've you've all seen this for I'll do it a Couple examples for you here in a second. We're gonna start off with something even more basic than than this Okay, so a and b are integers Okay, divisor a is a divisor of b. There's about a thousand and seventeen names for this I'm just gonna give you a few of them or a factor of B or B is a Multiple of a Sorry, this is really tedious, but I'm just gonna cover all my bases here Yes, there's more Okay This is the last one. I'm gonna tell you a divides B if So what does it mean? Well, there are a lot of ways you can say this but the easiest way to say it is just that Let me just put an asterisk around this Just means that when you divide there's no remainder of course the division algorithm what we're dividing by was assumed to be positive, but That's the rough idea. So just to say it formally. It just means that a times x equals b for some integer x so Just do a quick example of this I don't think I need to belabor this too much but Just to make sure that you all are aware that you all know what this is. This is very very basic stuff for I'll probably usually default to saying divides for divide 16 because it goes into 16 with no remainder. That's all it means Okay. Yes. Yes. Yeah, so another way of saying it is shorthand would be a x equals b for some x epsilon z Yeah Is that what you're asking? Yes. Yeah So for divide 16 since you just have to get used to this notation You may not have seen this before since four times an integer namely four Equals 16. All right. Yes. Yep. Yep, just a vertical bar but And also just to introduce this notation in which you'll see a lot two does not divide three So that's just a bar with a little slash mark through it I don't think I'm even going to to this You could certainly prove this but I'm not gonna do that I Think you all believe that two is not a factor of three, right? I mean if it was then you'd have two x equals three for some integer x, but then x would be three halves That's one point five. That's not that's actually not an integer. So you can't do it, right? Okay So here is what I'm going to do now the book presents this theorem which I'm going to prove here Basically, all I'm going to do today unfortunately, these these things are going to take a little bit of time is prove two theorems and If we have some more time, I'll go into some examples But Thursday what I'd like to do is do mostly examples and then Tuesday same thing Okay, so now we've talked about this idea of multipolar factor or divisor and So now we can prove some basic properties of this Relation and we'll actually use some of these in a proof that's coming up Which is now we're going to start to get into things that maybe you haven't seen that are not as obvious Some of these I think you'll find to be pretty intuitive. So we're going to let a B and C be integers So instead of stating everything at once and having you fall asleep even more quickly than you already will I'm just going to break these down into pieces and I'm going to prove each of them separately I think it'll just be easier to follow and stay awake this way Okay, so the first thing Is pretty straightforward. I think Yes, we're in two three the GCD and now we're on theorem one and the setup is just that a B and C are integers Okay, so the first part just says that we don't know what a B and C are they're just they're just integers a is a factor of zero one is a factor of or divides a and A divides a Doesn't matter what a is this is always true this is something that I Think that you can prove yourself probably Okay, well, I'm just trying to again get you familiar with this notation Remember this this vertical bar just means that it goes it divides into it evenly without a remainder Okay, in other words, we're so the first Proposition or the first thing we have to prove here is that a is a factor zero What do we have to prove we have to prove that a times an integer equals zero That's what it means to be a factor right what integer would we multiply a by to get zero? Zero that's easy right. It's very simple. Okay So all we have to say is that's a times zero Equals zero to prove that one divide day. What is it? What do we need here one times? what a equals a and The last one what do we want to write here? a times one Equals a That's it. All right, that's easy Okay Some of you are chuckling like well, I know I know I'm aware that this is pretty easy, but It's just good to get used to this Notation, okay Okay, the second one. Let's see if I'm gonna have room to do this All right, I guess I can maybe I can Know what's gonna happen? Okay, do you guys have this down? Anyone need some more time? Okay Yeah, I'm gonna do that on this. I'm gonna do that on this one Okay, because we don't really need this for the next part, but Okay, so the second is to prove that a is a factor of one if and only if Okay, some of you have the book open and you're gonna cheat here, but Those of you that don't have the book open what can you say about a if a is a factor of one Then what does a have to be? Can it be five? Can it be zero? What does that have to be? One or One or minus one, right? Okay, so the proof here. I'm gonna give is Not extremely formal, but I'm just really trying to convince you that this is the case And I'm also trying to show you that okay the technique for proving an if and only if statement We talked about this before if and only if there are two sub parts The first part is you prove whatever the left Statement is you prove that that implies the right statement Then you assume that the right statement is true and you and then you deduce the the other left statement Okay, you have to do two things for lift and only if statements If you do it directly that is okay, so and here you're gonna get used to this I'm gonna just start introducing it now. What does this mean? The arrow going to the right means that we're going to assume the left-hand Statement and we're going to deduce the right-hand one and then the arrow going the other way means we're gonna assume the right-hand statement We're gonna deduce the left one, okay so let's Assume that I'm not even gonna do the other implication because it's just so obvious but Let's assume that a divides one and I'm gonna Keep doing this because this this will help you follow in I think it'll help you in your home or two We want to prove That a is plus or minus one No, I'm just I'm I mean, okay, so yes, it is it is redundant So for now, I'm just gonna write it all out but in the future. I probably won't but yeah This is in general you would just write the right arrow and that just sort of means you assume the left-hand side. Yeah Okay, so What do we know because a divides one? What do we have we know that? a times x Equals one right for some integer x and I'm gonna abbreviate this now a little bit for some x in z So here's the thing to note That x is not zero I'm not gonna write out why but I'm just gonna have you think about it. Why can't x be zero because? Well, if x was zero, then you would have one equals zero, which is not true You guys see that x certainly can't be zero because you get it something that's impossible so That means we can divide and we can get a is equal to one over x Right and here's where I'm just I'm because I have more things I want to focus on here I'm not gonna get too precise here if a is one over x and a is an integer remember a is an integer a B and C are integers I was the assumption in the beginning of the statement of the theorem the only possibility for x is one or minus one Right, I certainly can't be zero or a or a is undefined If it's bigger than one or less than minus one then a an absolute value is between strictly between zero and one and no integer has that property right Vex is five or minus ten We get something between strictly between zero and one or strictly between minus one and zero and no integer has either those properties Okay. Oops. Sorry. Yeah. No, that's what I wanted. Okay a is plus or minus one since X has to be plus or minus one. Okay The other implication A is plus or minus I'm not gonna write it out if a is plus or minus one then definitely a divides one one divides one because one Times one is one minus one divides one because minus one times minus one is one. So that's just I Don't think there's any need for me to do that. So I'm just gonna write this Hopefully you don't think I'm being in jerk here, but I Think you should agree hopefully agree with this, right? One and minus one both divide one. That's that's easy. Okay Okay, so that takes care of the second part right any questions about B It's okay. Okay part C if a divides B and C divides D Here okay, so I should I should say this because I didn't I didn't tell you what D is but Yes Okay The theorem we just proved So okay There are a bunch of parts to this theorem and so what I'm doing is I'm just separating each of these improving them individually So this is sort of a conglomeration of a bunch of Yes, yes, it's that's all of it everything I'm doing here is in there. Yeah, absolutely No, I understand. Yeah, that's that's fine Okay, well if you have any questions, just just let me know Okay, so I didn't tell you what D is we just assumed a B and C were integers, but D is also an integer Yeah, okay, so if a divides B C divides D then AC oops, that's not what I meant. Sorry. Yeah. Yeah. Yeah, sorry about that Okay, I didn't mean to write equals. Sorry Okay Sorry. Yeah, I know I'm a jerk. I suck. I should be fired Okay, okay, so let's see. What do I what do I want to do here this? What's that? Oh Okay, well, I guess I can do that Yeah, let me see You know this thing is there for some reason that the eraser is here. There we go Let me do the small eraser There we go Yeah, that's the problem is if you do the big one it'll race everything. Yeah, so okay Sorry about that Okay, AC divides BD. That's what I wanted. Okay So why is this true? Well, this this is actually not that hard if you just write down what the definition is Okay, so we assume that A divides B and C divides D and we're gonna show that AC Divides BD. All right. Okay. So what do we know? So we have since a divides B. We have AX equals B Yeah, we're just gonna end up substituting and since C divides D C Y equals D, right? By the way, don't use don't be careful about reusing variables You would do not sit You do not say AX equals B and CX equals D because that that implies that you mean that this that the X is the same If you use the same letter Unless the context is very clear the reader has to infer that you mean that X means the same thing Okay, so be really careful about this if a divides B and C divides D Those quotients certainly don't have to be the same right two divides four three divides nine But the quotient for the first part is to the second quotient is three so you don't want to use the same letter Okay, so be careful about this. Yeah A divides B and C. Yeah, C divides D So that means that a times X equals B right and C Y equals equals D For some integers X and Y right, okay, so what do we get then thus? Well, we can just multiply everything together right so we get AX times C Y Equals BD Okay, multiply these together multiply these together. That's where that comes from I'm not going to use parentheses. I mean I will in a second, but in general you don't need to use parentheses and such It's not ambiguous because multiplication is associative. You don't need parentheses. Okay, it's also commutative So we can you know move the order around whoever we want So a C times X Y equals BD and that proves that a C divides BD Okay So you can again you can regroup the terms over you want to you can move everything around because multiplication is Commutative so we can just shuffle things around to get what we want right And now things like that. Okay, this came up before too. I'm going to mention this now again You don't need to even say this it's just sort of understood You don't need to say that the product of two integers is an integer the difference of two integers is an integer The sum of two integers is an integer. You don't need don't worry about that. That's just understood So you don't you don't need to write that down. All right Okay, so sorry Let's see. Okay, so This is similar if a divides B and B divides C Then a divide C. Okay, and this is all I'm going to say here Because I want to get through everything today that I want to get through This is similar to C Okay, you just write it out and it just it just ends up working out. This is this isn't anything. That's really that difficult Okay and E okay, so I'll write this out And this says that a divides B and B divides a if and only if a Equals plus or minus B. Okay So let's go ahead and prove this So we have this we have this now. Okay, so now that you've seen this arrow notation I'm just still going to write this down again just so that's on the screen so that you don't forget what our assumption is Okay, he divides B and B divides a and we have to show that a is plus or minus B so we have again by our assumption a x equals B and B y equals a right for some integers a and x and y okay, so We're gonna look at a couple of cases here case one is that B equals zero You'll see I think why I'm separating these cases here in a second. Okay, so look at what we have up here If we know that B is equal to zero then what can we say about a? Yes It has to be zero also Right. Yes Right exactly. Yep. That's all right. That's right. Very good. Okay, so in this case a has to be zero as well and Here's my question to you So case one is B equals zero then a has to be zero as well We just proved that if B and a are both zero is this true is a equal to plus or minus B Definitely true a equals B and a equals minus B because they're both zero in fact. They're both true in that case Okay, and I'm just gonna write that because it just should be obvious at this point if a and B are both zero That that is clearly true Okay, so Second case is That B is not zero. Whoa, that was huge. Okay. Sorry about that Okay, B is not equal to zero and here's what I'm gonna do just to make this easier to follow Let me let me put a one under the first equation and a two under the second equation You'll see where they're not equal to zero comes into play here in a second Okay, so we have a x equals B and B y equals a so here's what I'm gonna do B y equals a that's the second equation. So I'm gonna replace the a in the first equation by B y Okay, because we know it's equal to a so I can certainly do that. So what we get then is B y times x equals B So what can we conclude think about this in basic algebra now? What can we conclude here if B y x equals B? What can we say? Why y x equals one? Why can we do that? Because B is not zero That's huge, right? That's why I have this two cases if B was zero you couldn't say that necessarily, right? You couldn't necessarily say that y x equals one if B is zero you could have zero times five times ten B times zero. Sorry B times five times ten is certainly equal to B if B is zero They're both zero, but five times ten is not one. So it's important that you distinguish the two cases Okay, so Now what can we say? Well, okay, so we've got y x equals one Do you guys buy this y divides one right by definition y definitely divides one You see that just from from this right here and If you look at part B What can we say about why? Well, not necessarily. I think I have the labeling right we did this in part B, didn't we? Every factor of one has to be one or minus one since y is a factor of one y has to be plus or minus one But if we know look look back up here for a second and try not to break my Microphone look at this Look at what we're trying to prove we're trying to prove that a is plus or minus B, right? We know that y is plus or minus one if y is one then a is B if y is minus one then a is minus B Therefore a is plus or minus B make sense Okay, just just look at whether or not one y is one or minus one and then you get exactly what it is We want to show okay So I'm gonna try to make this as clear as I can by two then a is Plus or minus B, and that's exactly what we wanted to prove. Okay. Let's see We're at F. I guess okay. We're almost done actually Okay, so if it is a factor of B Yes, oh Right right right so there was the other implication Thank you. Sorry about that, but I'm not gonna do it. Anyways, it's fine It's easy the other implication is obvious if a is plus or minus B Then it's easy to check that a divides B and B divides a so if you want to write this in your notes The other implication Easy, okay. Sorry. I mean it really does not be just saying oh, it's easy for me. It really is easy It really is okay, so I'm not gonna I'm not gonna do that, but thank you. Sorry. Thank you for bringing that up I forgot to do them to state that about the other part Okay, so The absolute value of a is less than or equal to absolute value of B. This should be not too surprising Right, I mean intuitively if you have a being a factor of B then a is smaller than B in some sense At least most of the time it is, right? Okay, that weird screaming sound is a stool in case you're wondering what that is like as it goes. It's the room is haunted Okay, so so let's suppose that a divides B and B is not equal to zero Okay, so again At some point I'll probably stop saying explicitly every every time what we're gonna try to prove But I'm just gonna keep doing it now Just for clarity prove that the absolute value of a is less than or equal to the absolute value of B So again We have or we know or what have you? That a x equals B since a divides B ax equals B for some x and z since B is not zero, right? also What can we conclude about a and x? Well, they can't be zero either, right? Because if either of them were zero then B would have to be zero so again I'll Make this very clear okay, so Also since a x equals B. Let's see if you buy this the absolute value of a x We talked about this before briefly is the absolute value of a times the absolute value of x right? Equals the absolute value of B Why is this true? Yeah, that's what we all ultimately that's what we want. Yeah, I mean we're not done with the proof yet We're not done yet though. We're not done Yeah, sorry, sorry about that. Yeah, okay Yeah, this will happen every now and again of course this sorry. Thank you Okay Why is this true? Well? Look at just look at this side for a second the absolute value of a x and this is something We're not gonna prove in here the absolute value the product of two numbers is always the product of the absolute values That's always true So the absolute value of a x equals the absolute value of a times the absolute value of x That's just something that we're assuming why is it equal to the absolute value of B because since a x equals B Just take the absolute value of both sides the absolute value of a x equals the absolute value of B And so that's where the last part comes from Okay Okay, I also I'm gonna ask you this Why is this true? One is less than or equal to the absolute value of x Because x isn't zero the absolute value of x then has to be at least one. It's not zero, right? Okay, so What we're gonna do is we're gonna multiply this inequality through by the absolute value of a Well, one times the absolute value of a is just the absolute value of a it's less than or equal to The absolute value of a times the absolute value of x when I say multiply through of course I'm talking about this this inequality right here Why can we do that? Why can we multiply both sides by the absolute value of a? Because a is not zero. Well, no, it's it's not really that it's that the absolute value of a is bigger than or equal to zero Right doesn't matter what a is the absolute value of a is by definition of absolute value is bigger than or equal to zero So we don't have to worry about flipping the signs, right? This again But should be something you think about every little thing you do you should you should know why you're doing it okay and the last part of this is Just remember that The absolute value of a times the absolute value of x equals b. Sorry the absolute value of b We get That the absolute absolute value of a is less than or equal to the absolute value of b And that's what exactly what it is that we were trying to prove. Okay, right? We just got that from over here So all I'm doing is just replacing this with the absolute value of b, and that's what we wanted to show Okay let's see And then we have one more and that'll be the end of this theorem and we're gonna do one more Theorem, and then we will be done So let's see. I think yeah, I'm gonna go ahead and try to go to a new a new page here So I'm gonna wait till everybody's got this done got this down rather so they have this down anyone still writing Okay, so this is I'll stop boring you after this Well, actually, that's not true. I'm just gonna bore you with something different if a divides b and a divides C then a is a factor of or a divides B x plus C y for Any integers x and y in this case, I'm not gonna Write down what we're gonna try to prove. I'm just gonna write down the assumption Then we'll we'll just go ahead and do it. So we're gonna suppose that a divides B and a divides C and now we're gonna let X and y be integers. Okay, so it's Staring in the face right up here. This is what we're trying to show a divides B x plus C y okay, so since a divides B and a divides C We have yes, I'm not gonna use x and y in this case, but We'll say alpha a times alpha equals B and a times Beta equals C for some integers alpha and beta right okay So again, just to be clear. Let me let me label these two equations. This is one and this is two So just to give you some idea of where I'm going with this and just also to help you maybe figure out some of these proofs What it is it what is it we're trying to prove we're trying to prove that a divides B x plus C y so in other words we want to get a times something equals B x plus C y and What we have is we have these two equations right here So we need to get a times something equals B x plus C y well What if we multiply both sides of this by x then we'll get the B x part and we'll have an a on the left side What if we multiply both sides of this by y then we'll have a times something equals C y then if we add them together Since each of the left-hand sides has an a we can pull out the a to get a times something is equal to B x plus C y Which is what we want to prove That's the idea Basically, yeah, that's right. Mm-hmm. Yes. Yep. That's right Yep, you got the right idea. Okay, so by these two equations We have multiplying the first a sorry a alpha equals B multiplying both sides by x we get a alpha x equals B x and From a beta equals C multiplying through by y we get a beta y equals C y then we just add and then it just falls out right away after that So we get a alpha x plus a beta y equals B x plus C y right Just use the previous two equations add the left sides together. They have to be equal to adding the right sides together because they're equations, right and Pulling out the a we get a times alpha x plus beta y Equals B x plus C y Okay, so again, remember I'm suppressing this but the point is to know that a divides B x plus C y It's not enough just to know that a times something equals B x plus C y you need to know that a times an integer is B x plus C y but because integers are closed under multiplication in addition and these are all integers We know that this is an integer right but just be very careful by the way This will this may happen later on you may have to prove some things Involving divisors and such to prove that a divides B when a and B are integers It's not enough just to show that a times x equals B for some real number x that x has to be an integer and that is going to actually For some of you is going to make you pull your hair out from time to time But you have to remember that because a times B over a is always B you can always cheat and say that Oh a times B over a is B so a divides B. No, that's not the definition. It's a times an integer So I'm saying this for a reason they got it got to be really careful about this This if you're not careful, this will just sneak in because you're just going to say why just want it to work out I want it to be easy. So I'm just going to do that Well, you can but it's not going to be right. So I'll just be warned about that. Okay, so be really careful about that I'll say more about that when we get to problems that involve this this kind of this kind of thing. So That's it So that's the theorem. We're going to use some of this actually in what's About to to come next Okay, any questions so far? Yes Yeah, that's right. Yeah. Yeah, and I'm not going to say, you know on foot on February 12th What was theorem 1c? I'm not going to do that. You know, I'm not going to say that Did you have a question? Oh In this in this proof on this proof. It just doesn't come up. There's just no Something being zero just doesn't create any problems in this case basically because we're not trying to cancel or divide by it In this case the other one we wanted to cancel the B. So we wanted to make sure it wasn't zero Okay, let's see. So Let me go into the next let me see here, what's out there we are. Okay, you ignore that I just couldn't see the little I'm getting old and blind. I can't see things anymore. Okay So we're going to talk about now is the GCD and Before we get into this, I'm just going to give you an example This is something you certainly learned about a long time ago Without defining it yet Let's see. What are we on example two now? I? Think we are right. I'm just going to assume you guys know what this is GCD greatest common divisor right? That's what this stands for greatest common divisor. So what's the greatest common divisor of 12 and 16? Have you all seen this before? You all learn this at some point. Okay, so it's basically it is just the biggest The biggest integer the biggest natural number that divides both the numbers. That's it. That's what the GCD is We're going to go a lot farther with this in class now than what then you did maybe in sixth grade or whatnot but What I'm going to do is Actually, I'm going to go to a new page here because I want all the room that I can get here And I'm going to have to split it up soon anyway, so All right So before I actually define the GCD what I'm going to do first is Give you a theorem and this is this is the last theorem that we're going to do today I don't think the book does it exactly this way, but it's essentially what the book is doing suppose that a and b are integers That are not zero What do I mean by not both zero? I mean One of them could be zero Okay, maybe then they're both non zero or one of them zero. It's just that it's not the case that they are both zero That's all I'm saying Then There is a we've seen this word before a unique positive integer D satisfying these two conditions D is a factor of a and D is a factor of B and the second condition is Is that if C is any integer such that C divides a and C divides B Then C divides D and so C is also less than or equal to D. This this follows right away from this this Assertion okay This D in your mind this D is going to be the greatest common divisor I'm just having to find it yet, but that's we're going to prove a theorem that says that it exists first And then we're going to say that that's exactly what the GCD is This this other condition may look a little bit strange to you at first you might be used to thinking okay The greatest common device okay Well, I know it divides a and divides B but greatest should mean that if C is any integer that divides a and B Then C is less than or equal to D and the book actually states it that way But then at the end of the section they give you a theorem that says exactly this and every other person except for the author He lives in his own world Everyone else defines GCD this way everybody else does because this is a stronger the C divides D is a stronger condition It's just it actually will be more useful to you when you're doing proofs And it's equivalent to just saying C is less than or equal to D if you just assume if you just want to prove that C is less than or equal to D from that you can actually prove that C has to be a fact not only less than equal to D It has to be a factor of D as well So you might as well just say the strongest thing you can when you're when you're doing a theorem You might as well just do it right away. Okay, so that's what we're going to do now All right, so Here's the idea and It gets a little messy, but I would say it's it's probably not as bad as the division algorithm Okay, so we're going to assume That a and B are integers and they're not both zero Okay So we have to show the just like with the division algorithm We're gonna show the existence first and then we'll do the uniqueness I may not even get to the uniqueness I may just do the existence today and we'll do that on Thursday. There exists some in a positive integer D that has these two properties So somehow we have to produce this integer D And the way we're going to do this is kind of similar to how we did the The proof of the division algorithm. So what we're going to do is we're going to set the set S to be Let's see. I'm gonna use the same notation the book does here. Okay S is going to be the set XA plus YB where X and Y are integers and The sum is bigger than zero. Okay, that's the set S. So if you How many of you have taken linear algebra? Okay, so you might recognize what this there's a term for this Maybe that you might recognize from linear algebra if you haven't don't worry about it This set S is just all the integer linear combinations of A and B Right, that should look familiar to you if you remember linear combination from linear algebra. Okay so What are we going to do? Well? Think back to the proof of the division algorithm. What do you if you remember what we did with that proof? I know you maybe haven't memorized it because you don't really have to but What is it that we were trying to prove about S before with the division algorithm? We're going to do the same thing here There's what's the first thing do you remember anyone remember this? It's not empty exactly. It's not empty Okay, so That's our first Claim so we claim that That S is not empty The book does this a certain way, but the way the book does it is not the best way to do it I'm going to do it the best way So what do we know we know that A and B are not both zero at least one of them is not zero So we're just going to assume a is not zero the same argument applies if we assume B is not zero, okay? let's say that A is not zero again. You may say well, what if a is here and what if it's B? That's the one that's not so we just do the same argument apply here except you just do it to be then Okay, let's look at Let's look at this a times a Plus zero times B Look at that expression right there forget about the bigger than zero for now Just forget about that for now Does this fit the form of the elements that are in S is this of the form something times a plus something times B? Yes, and when the somethings of course are integers. Yes, these are integers and What is this equal to this is an attric question? This is a squared and Since a is not zero what can we say about a squared this I mean you can say a lot of things about a squared It's an integer. It's a square It's greater than zero or greater than or equal to one same thing. Yeah All right, you guys by that So what can we say about a times a plus zero times B? It fits this form and it's also positive. So it's an S by definition of S You guys seen this okay has the form it has to have and it's also positive That's what we need for it to be an S and since we know there's at least one thing in there We know that S is definitely not the empty set Okay, so now what can we do again think back to what we did with the division algorithm? What can we say? There's a certain property. We learned way back in the beginning of the course. We have this Subset of positive integers. It's not empty What do we know about it? Yes? It has it's yes, it's well ordered. So the set Yeah, exactly. So the well ordering property. Remember says that any non empty subset of the of the natural numbers has a least element As is a non empty subset of the natural numbers Why is it a subset of the natural numbers because of this condition right here has to be bigger than zero? So has the least element well ordering property Okay, I don't mean this is a derogatory term here. Okay, so don't sell. He's racist. No, I'm not Okay Bless you. Okay, so there's certainly a least element D and S right least element. What do we know about D? We oops Yeah Sorry, okay. Let me go back to this Okay, we know put an asterisk here. Hope I can fit this in D equals X a plus y B right for some integers X and Y. Why is that true? Why is that true? Because why? It's an yeah, exactly. Yeah, it's an S D's an S and in every element of S looks like this. So since D is in there. It's got to look like that Okay, just by definition what S is Okay, so now what we're gonna do is we're gonna show that D has these properties up here in A and B We're gonna show that that's true Okay, so now what we're gonna do is show that D Thank you. I don't know what's wrong with me today. I'm sorry Okay, so let me just write this down. I don't think I'm reusing this character here one We're gonna show that D divides a Okay, so we're gonna prove that Now you can't get that right away from this equation You can't because of this Y B term and showing that D divides a we want to show that D times something is equal to A You cannot get it directly from this I can tell you if you want to just monkey with it You're not gonna be able to get it you got to do a little bit more work than than that So what can we do well? What we can do is we can divide a by D Using the division algorithm. This is good review also just so that you remember what this is saying Okay, so if we're gonna divide a by D using the division algorithm, what does the division algorithm say? It says that a is equal to what? D times q plus r For some integers q and r. What's the special condition on our strictly less than but yeah Whatever you're dividing by the remainder has to be between bigger than or equal to zero, but strictly less than what you're dividing by Okay, what was the assumption that we had with the division algorithm you guys are laughing for some reason I don't know why you're laughing, huh? Okay That's so lame, but okay No, I know I appreciate your honesty. I was just curious What what's that That's we that's what yeah, right, but that is what we're actually gonna try to show actually so yeah That sounds good right now Are there dairy queens in Colorado Springs? I understand. Okay good. Okay. I'm from Ohio, so I haven't seen any here so yeah Someone else from Ohio really seriously, okay, let's talk about that for like 20 minutes Can't do that, but yeah, I'm sorry. I'm sorry for you. You're much Columbus Columbus Dayton. Oh, yeah Dayton Dayton's scary man. I've been to Dayton a few times and I've seen I mean it's a scary place Now I have some not a family around there, but yeah, it's don't go to Dayton Okay, don't do it. You go to Ohio. Don't go to Dayton. Go to Cincinnati. That's there's where you need to go Okay That might actually be a good idea too, but I'm Okay, all right. I would love to chat about this. We're gonna have to get back to this okay well so What we're gonna show is that R is equal to Zero that's what we're gonna show and I'll write this down again for you, but If you what is it? We're trying to prove we're trying to prove that D divides a right if you know that R is equal to zero Then D divides a right D times something equals a because the other pieces is nothing. It's just zero So we can prove what we want to prove if we know that R is zero okay, so What we're gonna do is we're gonna suppose that's not the case suppose not Then what can we say about R if R is not zero then R is then? Strictly squeezed in between zero and D right because you already knew that it was Bigger than or equal to zero, but if it's not zero then it's strictly bigger zero okay, so I Don't want to yeah, I'm running out of space here. So Let me see if I can squeeze this in here We already know that D has a certain form right that D is equal to x a plus y b right so and I'll hopefully I can get this in here a is Dq plus R okay, I want to go too fast, but this part is just coming from just above We already have that right here. Okay, you see that I'm not just pulling this out of nowhere It's just it's we're just two lines above, but if we replace the D with what it's equal to we know that this is equal to x a plus y b Times q plus R right? You see what I did here. I just replaced the D with what it's equal to Okay, everybody have this We okay now okay, so Let's just distribute this so This is x q a plus y q b Plus R right If I distribute this a is x a plus y b times q plus R Let's just distribute the q through keeping the in the b on the right. So we get x q a plus Y q b plus R that's equal to a so solving for what do we get we get? R is equal to Okay a minus x q a Minus y q b right you want to see what I did Just look at the end forget the middle's part be on the left side Just just subtract everything off over to the left so that we can get our by itself And so what is this equal to this is equal to? one minus x q times a Plus minus Y q times b Do you believe that yes believe that So what can we say about our then well Okay, remember the assumption on our R was strictly between zero and D remember that are strictly between zero and D So it's positive ours positive for one thing But what else so the fact that ours positive and look at the form that we've expressed our in what can we conclude about our now? R is in s Right because it's a linear combination an integer linear combination of a and b and it's positive That's the definition of s. Okay, so what's the problem here and now we've now we've got a problem. Yes. Yes Yes, exactly Remember remember D was chosen to be the smallest element in s you might have to go back a page I'm not going to do that because we're running out of time R was strictly between zero and D R is less than D But it's in s. It's a it's an element in s that's smaller than D that contradicts the fact that D was the smallest element Okay, so therefore our had to be zero it couldn't be non-zero because we get this contradiction And then we get what we want. I'll just write this out again. This this is already in your notes We already I already wrote this down, but zero is less than R is less than D and D is the least element of s So this is impossible, right? So what do we conclude then from this contradiction? I just said it I guess so we conclude that Remember what we assumed to that led to this contradiction There's something that led to this contradiction was that R was not zero So therefore has to be zero and if you go back in your notes What do we get? We get that a is equal to what was the original equation a equals DQ plus R, right? We divided a by D, but since R is zero this just becomes DQ and We conclude that D divides a right? Here's where I'm gonna write one more thing down. This is gonna be really really quick and then we're gonna stop I'm not gonna do it because the argument is exactly the same So all I'm gonna say is By a similar argument D divides B It is the same argument. You just do the same thing except with B instead of a that's all Okay, so we still have to prove the other condition which we'll do I'm not gonna try to do it now We'll do it next time then we'll talk about a couple problems and we'll do more problems on Tuesday So you should have time to you know get the homework done I'm gonna give you the assignment now even though I haven't finished the section I might as well just go ahead and give it to you right now Hopefully I don't miss it. I don't skip any this time homework Okay, so this is section 2.3 Most of these have tons of parts to them, which I'm not gonna make you you do most of them I'm not gonna make you do So one two four B six a and B 12 14 a 16 20 B and 21 B Okay, so one two four B six a and B 12 14 a 16 20 B and 21 B Okay, so we will finish this up on Thursday again This is this is all that's due that it'll be due next Thursday. Okay, and that's it I'm not gonna be more homework on Thursday. That's that's it. That's the whole assignment If anyone didn't get