 Hopefully, I know we're getting started a little bit late today, but hopefully we'll have time to go over maybe a homework problem Yes, yes, that's right Okay So let's go ahead and get started here. So again want to remind you Exam next Thursday or review on Tuesday and again the things to be looking at Okay, guys, you'd come down for just one because it's important stuff. You guys want to listen to this exam Questions will be things like statements of theorems definitions and then you have a few easier proofs to do nothing That's going to be unreasonable or anything too messy Okay, it'll be along the level of you know things you've done in the homework, but not the hardest stuff Okay, so they should be things that will more or less work themselves out if you kind of know what you're doing, okay? Pascal's rule you should know what that is. Yeah, you should definitely know that No, no, you don't need to know the proof The theorems that I've done and You don't need to know the proofs of the theorems that I've done in class, but you should know the statements Okay, and we'll talk more about that on Tuesday too Yeah, I'm not going to give you any specific sheet, but I will definitely tell you the kinds of things that you should be looking at Okay, so there's not going to be any big surprises on the exam. Okay Okay, so Where were we? Okay, so we were trying to prove Okay Theorem I think it's going to if I remember right, okay Yeah, so This you already have this in your notes A and B are integers Not both zero there is a Unique I'm going to use a little shorthand here because you already have this from last time a unique natural number D All right Such that and I may be labeling it slightly different than I did differently than I did on Tuesday, but First condition is that D divides A and D divides B right second condition is Whenever C divides A and C divides B also C divides or is a factor of D And then I think I said parenthetically thus C is less than or equal to D. I believe I said that didn't I? Okay, and then there's a third condition Okay, well, maybe I didn't label this three, but it's part of what I said D is equal to x A plus y B for some integers x and y Right, I did say this. I'm sure right Some maybe I didn't label it as three, but this was there. This was there, right? Did somebody say yes I stated this I you didn't have that at all. Are you sure? It's it's okay. So I'm just making it explicit here, but yeah um, so what we did I'm of course. I'm not going to go and do this again, but um, the uh continuation of the proof Okay, so what did we do last time? So we constructed the set S I showed that it was non-empty Then I had a least element and we chose our D to be an element of that set And so we've already If you were following on Tuesday, we've already established three, right? We chose D to be a member of the set, which was the set of all integer linear combinations of A and B So we already know that three holds from from what we did on Tuesday And I also showed you that D was a factor of A and then I said similarly D is a factor of B So so we've got one and three down already now Okay, so we just have to establish the second condition and then we're going to establish uniqueness after that Okay, okay, so That's this is as I said the last thing we have to do is to check the second condition So let's and this isn't actually not that hard. So let's suppose that um C is an integer and C divides A And C divides B Okay, so we will show That C divides D So what do we know? Okay, I'm going to try to go a little bit more quickly here so I can get through everything today So C is a factor of A so that means that C times Say alpha equals A And since C is a factor of B, we know that C times beta equals B For some integers alpha and beta, right? Okay, and so we're almost done Okay, so everybody with me so far We're assuming C is the fact of A and C is the fact of B So by definition that just means C times some integer is A and C times some other integer is equal to B And so all we have to do is just see that I'm just writing here D is equal to By this third condition which we've already checked All right, XA plus YB All right It's just from up here which we already got And so let's just substitute this in now So now this is x times C alpha Plus y times C beta Okay, and just for clarity here Here's one, here's two By One and two Okay, does that make sense? Just substituting in So what is it we're trying to prove? We want to show that C is a factor of D You can see how we do that how we can do that now, right? We want to show that C times an integer is D That's what it means to be a factor We can just pull out a C here and what we have left is an integer So C is a factor of D Okay, so So thus C is a factor of D Okay So there's one other thing This parenthetical comment here, I would like to prove that as well So why does that imply that C is less than or equal to D? So all I'm going to say is Let's see if you guys buy this Do you believe That? C is less than or equal to the absolute value of C, right? Okay, now This is less than or equal to the absolute value of D And I'll tell you why that's true Because of Theorem one part F This is in your notes If a number divides some non-zero number Then that first number in absolute value is less than or equal to that second number in absolute value This is theorem one part F So let me just write that down This is by Theorem One part F All right So what's the assumption though for part F if a divides b and b is not zero? Well We know that Okay, we've got C's We just proved that C divided D We know that D is not zero because it was chosen to be an element of s which was a set of natural numbers D is not zero, right? Of course D and n right that means it's positive So we know D is not zero And so what's the last thing we need to do to show that C is less than or equal to D? Well Very simple absolute value of D is equal to D because D is positive D's positive. Yeah Oh, it excludes it excludes it. Yes, definitely So if you kind of just put this chain together going from left to right Then you've got C is less than or equal to D, which is what we wanted. Okay All right, so As far as the uniqueness part is concerned I'll just do this kind of quickly here In fact, I'm gonna I'm not going to write a lot here because I'm not I'm not really that concerned about the uniqueness part But let me just say it really briefly Let's suppose that D1 and D2 satisfy 1 through 3 Okay, so we'll show I will show you that They're the same Okay, so I'm going to do this I'm not going to spend a lot of time writing all this stuff out, but I'm just going to say a couple of these things in words Okay, D1 and D2 satisfy these conditions. So notice then that D1 divides A and D1 divides B And whenever some integer divides both A and B, then it has to divide D1 Okay, well, I'm going to say it first and then I'll write it down D2 Satisfies one as well. So D2 divides A and B so D2 has to divide D1 And then you can reverse you can sort of transpose that argument to get that D1 divides D2 So D1 is less than or equal to D2 by this parenthetical comment and D2 is less than or equal to D1 after the same So This is all I'm going to say then since you can get this over here since D1 divides A And D1 divides B Also D1 divides D2 And this is just by the second condition. Well, really this is by 1 and 2, but just applied to One's applied to D1. One's applied to D2 so D1 is less than or equal to D2 Okay Remember D1 and D2 both were assuming they both satisfied all three conditions Okay, D1 divides A and D1 divides B and the second condition says Since it also holds for D2 that since D1 divides A and D1 divides B D1 divides D2 That's where I'm getting this in conclusion And then you can just flip the roll of D1 and D2 to get That D2 is also less than or equal to D1 And if you know that one number is less than another and that number is less than the first they have to be equal Okay, so therefore D2 is less than or equal to D1 and Okay Okay, I'm sorry this I just barely squeezed that in here. Hopefully you can all see this So D1 equals D2 and that shows that there those two numbers are unique All right And you guys all can you see this anybody need more time writing this down? This brings us then to our next definition Now that we went through all that work You're going to be happy to know now that Everything else we're going to prove here the proofs are going to be very short comparatively speaking And this is the last part of the section A and B are integers Which are not both zero then this unique integer D that we just established from theorem two is called the greatest common divisor Okay greatest common divisor of A and B And it's denoted Just gcd and then left parentheses a comma b right parentheses okay, so for example, and this this Of course, you probably don't need me to do this, but I'm going to do it anyways Just to make sure that everybody's aware that this is exactly the same thing that you've already learned a long time ago the gcd of Two and four Is equal to two Okay, let me ask you this What's the gcd of minus 10 and minus 15? I'm hearing a lot of muffled Okay, that's yeah, it's positive five Definition of gcd is that it's a positive. It's a natural number. It's positive. Okay. Don't forget that. That's that's why I gave you this example It's always positive If we don't define that it there would be positive and the gcd is not uniquely determined Okay, so we don't we don't want to say it's five or minus five and there's a sense in which you could have chosen that too, but Um, we want to make it unique so that we know exactly what we're talking about. So this this is equal to five So it's always positive. Remember that Okay, so What i'm going to do now Um, i'm gonna just because I want to get through all of this there's there's a corollary in the book that i'm going to skip Um No, i'll tell you what i'm not going to skip it, but i'm just not going to write out the proof of it Okay, you excited about that? Okay um And it just says this basically it just says that Every linear combination of a and b is a multiple of the gcd of a and b So suppose that a and b are integers Which are not both zero? And what we're gonna let me just go over here. It's capital t Okay, so t is just this set this should look familiar to you because we used this set when we were establishing the existence of d right then This set t is precisely the set of multiples of the gcd. Oops, that's Too many hams there, so And this more or less follows from the previous theorem So i'm not i'm not going to say anything about this really the ideas are already there and what i did before and the book proves This too, but it's it's actually not that hard so um What we're going to do now i think Hopefully, um You'll find this to be a little bit more interesting. Um, we're going to talk about something this condition Called being relatively prime. This is something that will definitely be studying a decent amount in here You've all heard of prime numbers before i'm sure. Um, it's not exactly what we're going to talk about yet Uh, yeah, we'll probably get to that later at some point You love to make fun of my writing well I don't know that that's going to happen honestly Oh, I okay, i think the fee is going to look okay. Um, well that remains to be seen Well before we introduce that i'll have had time to practice. So hopefully it'll be better No, i'm not going to do that. Okay, so Uh, there's before i before i finish this definition. I want to ask you a question Okay, if you notice, uh, when we defined the gcd. I was always writing a and b or not bow zero They're not bow zero. Why am I writing that? Well, what if they were bow zero? Well every number divides zero So there is no g there is no greatest common divisor because everything divides it. That's why we need to impose that condition Okay, I should have mentioned that before maybe but um Well, whenever you're talking, okay So whenever you see gcd if you ever see that in the homework and the book doesn't isn't explicit about it You always assume that a and b aren't bow zero one of them might be What's that? Yeah, I mean it's just sort of a yes. I mean, yeah, it's sort of build build into the definition and I may not always write that either But it's always assumed that a and b aren't bow zero. Yeah a and b are Said to be relatively prime Certainly that doesn't imply that a or or b has to be prime itself This so when you when you hear when you see the word prime Don't immediately assume that that a prime number is going to come to this definition because that's not the case If And it's actually a very simple definition if the gcd is equal to one So let's just do a quick example of this so Um, how about we do this and I'm just I'm I'm doing this for a reason Here's something I want to be very clear about here and you're gonna say well, of course, I know that it's in the definition but Uh, you're used most of your are used to the concept of you've seen this concept of a prime number You've seen that defined you can all point out what the primes are Relatively prime modifies two integers not one. Okay, so you say that seven is prime But it doesn't make sense to say that 11 is relatively prime doesn't make any sense two integers you say are relatively prime always modifies two integers not one So be really careful that you kind of get that down So why did I choose 10 and 21? Well 10 and 21 are certainly not prime right neither of them is a prime number They're relatively prime because they don't share any factors in common except for one and minus one Those are the only integers that divide both of them at the same time Okay, so think of Kind of intuitively think of two integers being relatively prime if they if they don't share any common factors Okay, except for the trivial ones right Okay, so is this example makes sense Okay, so I don't think I'm going to do another one Well, like yeah now I I'm not going to but So there's a really not and this is actually very useful Um as well. So this theorem and it's also very easy to prove fortunately given what we already know Theorem three Okay, and I'm just for the sake of time. I'm going to suppress writing out A and B are integers that aren't bow zero when you see A and B It's it's assumed that they're integers and they're not bow zero. Okay Yeah, I will yeah So integers A and B which are now bow zero are relatively prime If and only if there are Integers again, I'm sorry for being a little sloppy here here. There are x and y in z in other words integers x and y such that Sorry, and I'll put an asterisk around this such that x a plus y b equals one Okay, so this is a necessary and sufficient condition for two integers to be relatively prime There's some linear combination of them that gives you one. Okay, so if you notice By the way, just if you look over here, you can see if you this isn't too hard to establish, right? That you can find a linear combination of 10 and 21 that gives you One right minus two times 10 plus one times 21 gives you one Okay, and you can always do that for any integers that are relatively prime Okay, I'm just going to use the bar here. I'm sorry. I didn't do that Okay, does everybody have this Are we okay? Okay, and now like I said The the proofs that I'm going to do now are going to be very short and very direct and they're not they're not going To be long and actually in the homework you're going to have a couple of proofs to do I really would encourage you I always say this but to try to follow along and understand the proofs Because there are a couple of homework problems where you're going to have to sort of think this way Okay Okay, so again, they're sort of if and only if statements direct proofs or proceed in sort of two steps We're going to I'm going to suppress the arrow that I gave last time but the first part is we're going to suppose that a and b are Relatively prime, right? There's not much to say here really but what does that mean? So by definition That just means that the gcd Of a and b Equals one right, so it means to be relatively prime And I label these statements one two and three. This is in your notes now. I'm not going to go back to it But um So if you so theorem two this is from today theorem two part three Says that the gcd is a linear combination of the of the two numbers Right, this is this is this was there a couple slides ago So that means that we know that x a plus y b equals one for some x and y that's just the third part of the theorem You guys understand what i'm saying here And That's all there is to the first part. Okay, it's just we sort of just get it for free by the theorem. Okay Uh, and then the second part of this Let's suppose that x a plus y b equals one For some integers x and y What is it that we want to prove we want to prove that that uh a and b are in fact relatively prime that If some linear combination gives you one that that they don't have any common factors other than plus or minus one Yes You're basically when you write that that x a plus y b you're powering out it You're basically saying that the gcd write that No, because just knowing that a linear combination is equal to of two integers is equal to a certain number does not mean that That number is a gcd right away for example Two times five plus one times ten is 20. That doesn't mean that 20 is the gcd of five and ten No, but i'm saying if it happens to be one Then that does happen to be the gcd Okay, so there's something special about one that forces it to be the gcd But in general just because you know a number is a linear combination of two numbers That doesn't mean that that number has to be the gcd, but because it's one it forces it on you and that's kind of the point of this theorem okay So here's all we're going to do so we're going to let d be The gcd of a and b right So what do we know there are a couple of things that we know about d We know by definition of it being the greatest common divisor that is in fact a divisor of a and a divisor of b All right, that's just by definition So we know that say d alpha equals a and d times beta equals b for some Alpha and beta in z right okay, so here i'm going to go back up here Let me uh, let me put an asterisk here nick to this equation that we've got x a plus y b equals one So by This starred equation x times d alpha plus y times d beta equals one You guys see what I did here? We already have that equation x a plus y b equals one And we know by definition of d being the gcd of a and b that d is a factor of a and d is a factor of b So i'm just replacing a and b in that equation with d alpha and d beta okay So What if we pull out a d here? We get d times x alpha plus y beta Equals one okay you guys by that It's not too bad And so let me just make this a comma here And what's the conclusion? Well D then is a factor of or d divides one right? And if we go back And this is partly why I did this theorem one all these sort of simple things This is b i'll write this down in a second, but this is b of theorem one if an integer divides one that integer has to be plus or minus one Since d divides one d has to be plus or minus one. Why is it not minus one? By definition of gcd it's positive So the only possibility left is that d is equal to one Okay By b of theorem one We know that d equals plus or minus one Since by definition d is bigger than zero We get d equals one and that's exactly what we wanted to show Is that the gcd was one? Okay, so there and this is very useful. You'll see how this is applied To get some some results that are not obvious or you might guess that they're true But they're not it's not obvious how you would possibly prove that they're true This fact is very useful in elementary number three and you'll see an example here in a second Okay So Where else do you want to go here? Okay, well luckily we don't have a whole lot left and then I can spend hopefully I'll get you started on a couple homework problems before the end of class today All right, so Here's a corollary Okay, I think I already gave you a corollary one if I'm not mistaken Yeah, okay corollary two This shouldn't be that that's surprising If the gcd of a and b Equals d so d is the biggest Common divisor of a and b then The gcd of a over d and b over d. What do you think this should be? one If we divide through by the greatest common divisor There shouldn't be anything left that will divide them both right except for one if there was then We it wouldn't have been the greatest common divisor it would be something bigger Okay, and this is very short Okay So I want you to think through this this is not going to take long at all Okay, so there there are several things that we learned In theorem two several conditions that this Greatest common divisor has The third condition is that and this is very important. Don't forget this You know you learned a long time ago that the greatest common divisor is essentially the greatest Integer that that's a factor of a and b what you probably didn't learn is that it's a linear combination of a and b That's something that most people that that's not something you hear in in fifth grade usually Um, but that's huge for us now This this this makes a big difference It helps us to prove a lot of things that we otherwise wouldn't know how to prove So if the gcd of a and b is equal to d then by three of theorem two Three from today of theorem two d is equal to x a plus y b for some integers x and y Right, and I'll just put this parenthetically right. This is three of Theorem two Okay We want to see if you can see how to how to proceed from here We want to show that the gcd of a over d and b over d is one and so In order to prove that by the previous result all we have to do is show that Some linear combination of it is equal to one right The previous theorem says that then the gcd is one we just proved that So how can we get a linear combination of a over d and b over d to be one? You see this All we have to do is divide everything through by d and we're done. That's it. Okay, so again I'll put a little asterisk underneath this And then we get one Is equal to x times this a over d. This is one way to write it Plus y times b over d, right? That's what we get if we divide everything through by d So we're now this is it now. We're done by What was it theorem three? Is that the last result the numbering theorem three? Yes maybe That was the last I could go up if I wanted to guess I'll stop being lazy Yeah, okay theorem three Okay, you see that The theorem three says as long as some linear combination is one then they're relatively prime And conversely okay So I'm going to ask you a question And there's just two more results that we're going to do and then and these proofs are very short and then we'll be done question one suppose we know that a is a factor of c and b is a factor of c um The question is must a times b be a factor of c This is just something for you to think about for a second before I go into the next Result, okay, so in other words, we know Some number divides c and some other number divides c does the product of those two numbers necessarily have to be a factor of c What do you think anybody want to hazard a guess here? Right? Yes Yes You got your book open, but uh, yeah, you're you're right Can you can you give me a counter example? Can anybody give me an example where? Uh a divide c and b divide c, but the product doesn't divide c anybody think of something where this uh doesn't hold Yeah, okay, so Yeah, yeah Right, so what he what he's saying is um If I heard you right Six divides 24, right? And 12 divides 24 But the product is 72. That's too big and it doesn't divide 24, right? So yeah, so that's a That is a perfectly good example Yeah, Joe Right Okay, so what we're going to Um Okay, so what is true is that Yeah, so If your product is smaller than yes, it should work although we're going to go in a different direction here um So, um, let's see So let's see. What do I want to say here? Um Yeah, so there's a condition though that guarantees that The um the product will will divide it and it actually has to do with being relatively prime So if you know that there's so if you notice this example six and 12, they're not relatively prime, right? Six is a gcd It's not one so if The gcd happens to won't be one and that will guarantee that the product actually divides Okay So let's see So and so sorry what I said before is not right So for example four divides 12 and two divides 12 the product is eight, but that does not divide 12 So the fact that the product is less than or equal to does not doesn't imply it Yeah, I thought I had that wrong, but yeah, so that's not good enough actually the the relatively prime condition You really need something a little stronger than that Okay, um So It just says this Yeah, sorry for miss speaking before so suppose that um a divide c And b divide c and This is just shorter than actually writing everything out the gcd of a and b Is equal to one in other words a and b are relatively prime then ab divide c I'm only going to torch you with after this with one more proof um So let's see how this goes This actually once we have some of these these theorems we're starting to get some tools some machinery Proving these things is actually not that hard really um So let's assume that um Let's see. Um, sorry. Let's assume that alpha a equals c and beta b equals c And that the gcd of I'm just trying to minimize my writing here to save time for homework Gcd of a and b is equal to one Here um alpha and beta are integers or some integers so What else do we know? The gcd of a and b is one That means that some linear combination of a and b is equal to one All right, and again just goes back to theorem two the last part of theorem two So we also know that x a plus y b equals one or some integers x and y Okay, so let me give you an idea about how you might proceed with this We need to show that ab is a factor of c Okay That's what we need to prove Well, we have this equation over here What is it? We're trying to show we're trying to prove that ab times something is equal to c So We've got three pieces of information, right We've got this We've got the second equation here, and then we've got the asterisk equation Any idea? So what we're going to do is we're going to look at this equation right here There's something that we're going to do Notice that what we have to assume both of these equations involve c Okay, so we're going to want to use those somehow Any idea what we might be able to do with this in order to Be able to get a a b times something equal c There's just one thing we can do to both sides and then if we use one and two it's going to pull it's going to come out right away Multiply by c exactly and once you do that you substitute done It's really not that bad So we're going to multiply this start equation by c And once we do that Let me just put two stars here two asterisks here We're going to get x a c plus y b c equals c All right, everyone okay with this I doubt that wasn't too hard I just multiplied through by c and I want I want you to see how we can get things to work out here We want to get a times b times something to bc So what we'd really like to do is this we'd really like to have an a and a b here and an a and a b here Because if we do that we can pull it out Then we'll have a b times something and c and that's what we want to prove right So the question is which of these two We we actually there's there's a correct choice and an incorrect choice really to use here We want an a and a b to pop out here So we want to you replace the c here with one of these two things Do we want to use the first one or the second one? The second one because that's the one with the b in it You see and then here we want to use the first one to replace the c without the a because this one needs the a Then we'll have an a and a b in both and we can pull it out Okay good all right, so By one and two this equation becomes x a times beta b plus yb times alpha a Equals c okay, so everybody see what I did here. Is this okay? Pull out the a b and then we get a b divide c we're done so Pulling out the a and the b a b times the quantity x beta plus y alpha equals c Since a b times some integers c by definition a b divide c Okay, and the last thing that we'll do And again for the sake of time here what I may do Because the the idea is very very similar. Um, I'm going to state this because this is very important You actually use this in your homework. I'm going to state what's called ukulelema. I'm not going to prove it. Um It's in the book and it's very similar to this in fact if you understood this I think most of you could prove it yourselves without me even doing it Um, so let me just do that. I'll say ukulele excuse me ukulelema and then we'll talk about a couple homework problems Okay, all right This will probably use quite a bit throughout the semester if a divides bc And um the gcd Of a and b is one in other words a and b are relatively prime Then a divides c And you do the same kind of thing you assume a divides bc So a x equals bc and you've got some linear combination of a and b that's equal to one Then you just multiply through it the proof is very very similar to what we just did So the same, um I should have maybe asked the question before I don't want to be very clear about this too I'm just going to say it now because I think you can follow it just with me saying it Um, well, what if we don't have the that uh a and b are relatively prime is this still true If a divides bc Um, then does a divide c Well, no, right? No Well, yeah, I mean so for example six Certainly is a factor of two times three But six is not a factor of three Right, so it definitely doesn't work in general and so you can't just Do away with this hypothesis you actually need it Okay Okay, so how about we talk about a little homework, okay? How's that? In fact, I'm feeling kind of generous today. It's a nice happy day. It's valentine's day. So my valentine's day gift To you will be No, no, no, no, um, not that Um No, what i'm what i'm going to do is um Um, yeah, I'm not I'm not yeah, I've gotten through 14 of them already. I only have three more to go I know I watched cartoons this morning with one of them, but I don't know if that counts as a date or not, but um Just kidding. I didn't Okay, everyone got uh got this down now. I'm okay with this. Okay So why don't I do uh, I will get you started on the homework by doing one of your homework problems for you, okay? and um And of course since this isn't due until next thursday Um, we'll also have time on tuesday to talk about some more of these And so I want to there there are a few and maybe you realize this in the last homework too um That you know some of these are a little bit tricky or at least you have to kind of see a of A trick to do some of these but for B Okay, so yeah, you I know you you've already asked about this that I'm just going to go ahead and and talk about I know I changed my mind I changed my mind. Um Hey, but that's good. That's good for you though. It's good for you. Um prove that 15 divides Two to the four in did you uh Did you look at it after I gave you that hint? Yeah, did you get it? Yeah, okay? Okay, let's see now you have the satisfaction everybody else here is just going to go Oh And see now you can actually appreciate this So You're better off. That's my point. Okay. Yeah um So what does it say? for for All And bigger than or equal to one By induction. Okay. All right, so probably will not end up grading this problem. Um But But but hey, well at least hey at least, um You know, uh, if you had no idea what you were doing and you left a blank now You'll have something to say and so at least you won't get dot completion points for this Uh, you could you could you could you could do that and And Well, but also this is gonna Let's let's see. So what else what else did I give you? Yeah, exactly. Yes. Thank you. Okay. At least one person appreciates me. Thank you. Okay. I appreciate that. Okay Okay I'm just gonna do the base case you guys upset me now. I'm not Just kidding. No, I'm not. I'm just kidding. I'm gonna do the whole thing. Okay So let's let's just go back through the the steps. We're in this is this is the first principle. We don't need strong induction for this. Okay Okay, so um Again, we're going to sort of split this up into two pieces. So the base case of the induction we just have to Check and there's not usually a lot to write here a lot of you're writing more than you need to that Well, we just need to see that this holds for n equals one, right? So we need to check that 15 divides or is a factor of two to the four times one minus one right And so all I'm going to say here is that this is clear since Well, what is two to the four times one? minus one Is 15, right? And that's really all you have to say. I mean if you want to go ahead and say 15 times one equals 15, then that's fine Um, but I know I trust that you know that a number divides itself. Of course, it's a theorem, but you you all know that of course So this is fine. This is okay okay, so inductive step Okay, and I'm writing it out this way just because I want to there was a little confusion about terminology and and in the first assignment So I want to make sure that everybody's aware of what this this means some some people were calling the the base case The inductive step and vice versa so So what do we do now? So we assume for some natural number n That and I'll put um Put an asterisk here 15 divides two to the four n Minus one That's the inductive hypothesis okay, and so We will show or we must show that 15 divides Two to the four times instead of four n is four times n plus one minus one Right, that's what we have to show okay Everyone clear with on this part So what is this really saying? This just means we need to show that this expression here is 15 times some integer That's what we need to show And we have assumed that 15 times some integer is four is sorry two to the four minus one okay, so By our inductive hypothesis What do we know? We know that 15 times say k equals two to the four n minus one for some integer k right Whoops, sorry. I meant to write z there. Sorry about that Okay, um, oops Just to be clear also on your on your letters the visibility is a relation on the integers So saying that 15 divides something of course in this case. These are all positive. Okay, so I mean You do already know that k is going to have to be positive So it's really in some sense. It's not wrong to say that k is in it But in general when we're talking about the visibility, this is something we define on the integers not just the natural numbers, right? okay, so All right, so I want to try to give you a way of thinking about this Okay, and this this for some of you some of you are going to get stuck here at this point say I'm not really sure what to do next Here's the way you want to think about it. All right somehow You need to use this inductive hypothesis somewhere. It's going to have to show up somehow So you need to show that 15 is a factor of this So what is it that we're going to do? Well, what here's what we're going to do We're going to we're going to take this we're going to mess with it a little bit And we're going to hope that the inductive hypothesis can come into play somehow And we want to mess with it enough invoking the inductive hypothesis to get a 15 to pop out That's the idea Okay, and that's that's generally how you're going to do this if you want to show that 20 divide something You have your inductive hypothesis and what you need to prove Take whatever you want 20 to divide and mess with it until you can get a 20 to pop out of the whole thing That's the strategy Okay, so this is all that that um you have to do so just simply note this two to the four Times n plus one Minus one Equals and again remember what our strategy is we need to show that 15 divides it So what we want is to be able to to mess with it to the point where we can pull out a 15 from the entire thing Okay, well, what's this equal to first of all? So this is And by the way, none of this none of this is actually that hard You just sort of have to know where you're going here. So this is two to the four n plus four Minus one you'll you'll buy that Okay, and I again, I really encourage you to follow along here Okay, you believe that So sorry, yeah Two to the four and all I did was so I've got two to the four times n plus one Minus one is two to the quantity four n plus four just to do distributing the four through minus one Which is two to the four n times two to the fourth minus one Okay And you guys believe this You guys buy that two to the fourth is 16. So I'm just rewriting this as 16 times two to the four n minus one Okay, now here's where If I don't say anything you'll just sort of you might some of you may be left wondering how you would know to do this Remember somehow we're gonna we really need to use this somehow That two to the four and minus one is 15 times something well We don't have two to the four and minus one here. We have 16 times it Okay, we want a 15 to be able to be pulled out of this whole thing and notice that 16 is Really close to 15 right So this is 15 plus one times two to the four n Minus one Right, that's not that's not too hard. And so if you distribute this this is 15 times two to the four n plus two to the four n Minus one you buy that you see that Now you see why I did that we need 15s That's why I did it We changed the 16 to 15 plus one because then we get a 15 and then once we do that We can we can pull out a 15 here The only other thing we need is that there's a 15 here, but that's just the inductive hypothesis And then we're done. So Uh normally I wouldn't litter this up this way, but um just so that everybody sees where this is coming from Is this clear now? You see where I'm getting this? Okay so Let me just recap this in case anyone's lost the their place Remember what we want to do is show that 15 is a factor of this Okay, so what we're going to do is we're going to take this which we have right here And we're going to hope that a 15 pops out So this by distributing the four, this is just two to the four n plus four minus one If you if you're lost it, I'll just just try to follow each of these steps Which is equal to two to the four n times two to the fourth because when you have the same base Of course, you had the exponents you learned that a long time ago. Okay Two to the fourth is 16 and we're just going to put that out front like this So we get that 16 is 15 plus one So we've just changed it this way and now we're just going to distribute this 15 times two to the four n which is this Plus one times two to the four n which is just two to the four n then we saw the minus one on the outside And then this becomes 15 k by the inductive hypothesis Right there And now you can see how this is going to end right so we're not quite technically we're not quite done yet oops So we still need to show that this is Divisible by 15, but that's easy This is just 15 times Two to the four n plus one Oops, sorry four k. Yeah Plus sorry plus k. Yeah, thank you. Sorry about that. Okay so What did we just show? thus 15 is in fact a factor of Two to the four times n plus one minus one yes because Because 15 okay two to the four n minus one is 15 k The 16 is only multiplied through by two to the four and if this was in parentheses We could have done that but because the 16 is only only multiplied by two to the four and we can't we can't do it Any other does this make sense? You guys see how this goes? Okay, so that's maybe a little tricky But you're going to get used to these kinds of things as the class goes on So now you have one less problem to do and then we can talk about more of these on Tuesday 15 k plus one Okay Okay, yeah. Yeah. Yeah. Yeah. Yeah, sure. Yeah, you can do that too So yeah, so the other point is that and it's kind of how you know six one half dozen of the other What she's saying is you just solve this for two to the four n Okay, so it's just kind of shuffling it around a different way So yeah, you could do that too Yeah But um, so whatever. Yeah. Well, okay. I mean whatever seems more natural. It's going to be different for different people Um, so I think we're going to run out of time here. So I think I'm going to stop there Um, so like I said, Tuesday we'll review. We'll do a couple more of these But try to get some work in on these problems before Tuesday if you can We'll talk more about these and then this is due on Thursday next Thursday with your exam. Okay