 things here sorry I apologize about Tuesday I had a sub lined up but his wife got really sick and so do you know you don't yeah well that's not gonna happen but well what that means is less material on the exam next week so so there you go right there's a compromise there's a we can compromise a little bit okay so yeah let's let's just chat for a second here so I have your homework from before I'll pass back in a class today so this assignment you will get back on Tuesday to have everything before the exam I still plan on doing the exam next Thursday even though we won't have covered a whole lot but that just means you can focus more on some of these sections and I think maybe you noticed this homework was maybe a little more difficult than some of the other ones so you can focus on that without getting overloaded with a bunch more sections so I think this is actually gonna work out okay so that again the exam will be next Thursday same you know very similar kind of format you know again I'm not going to to put you know the most difficult problems on the on the exam but I do I am going to test to see that you have some understanding of what's going on and part of that will include a couple of proofs for sure but they're not going to be the hardest possible proofs but yeah oh so computing things right so the you know the kind of questions you you would you would see would be these these are things that you should be prepared to do from the homework that I've given you okay so let's see aside from that here's another thing I'll tell you what I'm going to do is and this give you some incentive to kind of keep up with things and I'm doing this mainly just because we haven't done a whole lot and we had we didn't class on Tuesday I'm going to put one problem on the exam directly off of the last exam okay so look over the solutions okay I want to keep you guys up to speed because you know this is gonna be helpful for you on the final plus induction GCD all that stuff it just keeps coming up over and over again so it's it's really important to have that down okay when I say that I mean verbatim I'm not changing anything so just all you have to do is make sure that you look over the solutions are all online okay so there's one freebie right away okay so do that definitely do that and let's see other than that I don't think there's anything else I wanted to say we're going to talk about one more section today I'm going to give you one more assignment this assignment I think you'll find is actually easier than the last assignment is kind of was I should say and the just like with the last exam the assignment that I'm going to give you will be due at the exams you have a week to work on it Tuesday when we review I'll talk about a couple of the problems and such to you know to help you out with that yes yes it will be it's a week though it's a week you got yeah I mean everything keeps getting pushed back in the 60s you could have something new covered and tested on the next day in 2050 oh you can't put anything on the exam that was covered you know any any more than three months before the test you know I have now a week is the reasonable amount of time okay it is okay so what did I want to say there was one other my dad no I was not in school in the 60s I was not born in the 60s my my dad was in school in the 60s and he told me all these things were things were um no no I hang on a second here when you get this back again sorry okay so I just click this once now right is another okay one time I don't need to go to this second option here and I okay let's see or do I well because now it's not there we go now that's not right though is it there we go okay did I want did I want to yeah okay let's see I'll yeah this is the one I think I had before was that okay alright here we go okay thank you okay so 3.2 okay and you'll actually be glad to know that a lot of this you've already done if you those of you that took the screen which is most of you maybe all of you at this point those didn't take it dropped out a while ago I think but um so you've you've probably you've probably seen this before this is going to sound really complicated you made me self-conscious you made me self-conscious today and so I left it in my office yeah it's black no no no my yeah my my well my girlfriend has changed my wardrobe completely now so I didn't I didn't used to know not exactly I mean she's bought me a lot of stuff you know she doesn't literally put the clothes on me no I would no I don't have I have no complex about this I gotta be careful what I say cuz she could she could watch these these lectures here not that she would she's not that bored but she could see this I gotta be careful what I say okay we'll see about that yeah okay so we're gonna look at a few questions here sorry this is a little screwed up here but this this is called the sieve of Eratosthenes and it's actually the section is this is only about one fifth of the section this is in fact you probably might have learned you literally might have learned this in third grade it's actually possible you learn this in third grade so here are the questions that we're gonna be looking at so the first one and most you probably already know the answer to this no no okay how many primes are there okay you've you've probably seen this before most of you probably know this right and feeling many we're gonna talk about this today probably you did this in discreet at some point I'm guessing okay something that's a little harder is the following what does their distribution look like it has something to do with it yeah we're not gonna talk about that in here well so there's yeah there's there's something called the the Riemann hypothesis and it has to do with solutions to the certain called zeta functions but we're not we're not going to go into any of that stuff in here no trust me that is not that's not fun you do that's that's true yeah if we all worked on it for the rest of our lives there's no chance that we would solve it that is that is the truth no well people have people have worked on this for a long time and it's very very very hard okay yeah it's now it's crap it's a little it's a little tricky okay and then the last question is how can one determine if a given natural number is prime right these are very general questions and so we're gonna look at a few of these today and I'm just gonna kind of it's gonna be maybe a little more informal than normal but I'm just gonna start this way I'm just say here's some answers these answers are not the best answers these are kind of weak answers but at least gives you some ideas to you know just kind of an introduction to how you know some of these questions might be solved okay so the first I'm not gonna spend a lot of time on this even though that's the title of this section but I don't like writing this this is very long word okay that's what I'm gonna do that's exactly what I was going to do actually the s of e yes that is very very nice okay so here's the the idea is I'll just put this kind of parenthetically here the idea is to roughly just you know find the primes which you know inside of the natural numbers okay which erotophanies yeah if this is I guess that the pronunciation is sieve although maybe some people would say sieve but everybody in the States this is I don't understand this but everybody in the states that calls a picture says picture how many of you say picture for picture probably a lot of you do no not everybody does that okay it's the Colorado thing I don't know is that right okay yeah it's I never heard that before I never heard that before I moved here I yeah I don't yes thank you okay thank you right picture exactly a beer yeah all right okay so this is really simplistic it's all it's so simplistic that it's almost kind of silly that there's a name for this like it's some big famous brilliant idea but it's really not at all okay so of course we can't you know we can't actually write all of these down but we'll just write down some of them so you can see how this goes yeah although that's not that's not really gonna help us here but okay so I'm just gonna go up to 20 there we go okay and here's what you do you so you basically just want to see which ones of so the point is what which numbers are prime and of course you can you know you probably just know off the top of your head what what they are but suppose we were to go up to you know 500 or something like that now there's a nice simple sort of algorithm that you can teach a very young child to run through and be able to actually figure out what the primes are and so here's the algorithm clearly right to is definitely prime so we're gonna I'm adding something to this a little bit so we're gonna circle number two okay now what we're gonna do is we're gonna cross out all I'm running out of room here so all x bigger than two which are divisible by two in other words the even integer is bigger than two okay so that's easy right we're gonna cross out four six eight ten twelve on down okay so the idea is that these are not prime of course because they have a factor of two two is prime you know it also has a factor of two but there's no other factor right between one and itself so since these other guys that we crossed out all have a factor of two and they're all bigger than two there's a factor that's not equal to one in itself so they're not prime right okay so now what we're gonna do is we're just gonna go let me say it a better way sorry I don't want to say it this way so circle the next if you want to say first integer which isn't crossed out and of course in this case it's the number three and again three's prime of course not divisible by two it was we would have crossed it out so it has no factors other than one in itself of course this is obvious and then what we're gonna do I was gonna write more down than this but you know you you folks are all smart and I don't think you need me to we're just gonna repeat this this process and this will generate the list of primes okay and so let's just go ahead and do this I'll wait till you have this written down but if you pay attention this is gonna be very easy you could all reproduce this easily I don't think there's gonna be any issue with this so okay so when I say repeat the process what do we do before we we circle to then we crossed out all the ones bigger than two that were divisible by two so now we circled three so we're gonna cross out all of the integers bigger than three that are divisible by three okay so the first one we come to a six well that's already crossed out so we don't need to do it again it's already gone okay so imagine the number being crossed out it's just kind of erased I mean you sort of think of it that way because we're interested in finding the primes okay so what's the next one nine right so we're gonna cross this out 12's already crossed out 15 we're gonna cross that out 18's already crossed out and we're stopping at 20 now so this takes care of all of them okay so what's the next number that's not crossed out it's five right so we're gonna circle that that one's prime and we're gonna cross out all the ones bigger than five that are divisible by five so 10 which was already crossed out in the beginning because it's even 15 is crossed out because it's divisible by three 20's crossed out of course because it's even of course if we continue to go you know down the list we would we would get to 25 and 25 would be not crossed out we would actually have to cross that one out because 25 is not divisible by two or three okay so now we just keep doing it so okay so the next one seven circle that okay so again because this list is really small the only other number that's divisible by seven that's bigger than seven on this list is 14 which is crossed out in the beginning because it's even do you have a question yeah that's yeah well yeah so that that is we're gonna get to that yeah I just haven't talked about it yet but yes that is correct yeah absolutely okay so then now notice this this is kind of interesting I suppose now we come to 11 well what else do we have to cross out well there's nothing left because 20 22 is the first one bigger than 11 that's divisible by 11 it's not on the list so everything else that we've got left over we automatically know are these are primes right okay so 11 13 17 and 19 okay does this make sense this idea okay so I'm not gonna say a whole lot about this other than to say that the ones that we have circled are in fact primes I'm not gonna try to write a proof of this I think this is probably pretty pretty self-explanatory but you might say well how do you know for example that you know one of these numbers that were circled isn't divisible by one of the numbers that were crossed out before okay so why is that well of course this is a really small list here but you know for example I mean you know 17 is prime but but how do you know that maybe it was divisible by one of these guys that was already crossed out and it's not actually prime 17 is a small number but what if you get to 509 or something like that right well the answer is 17 can't be divisible by any of the numbers that were crossed out because the numbers that were crossed out were crossed out because they were divisible by something so whatever that thing is that divides number that was crossed out that number that was crossed out can't divide the number that was circled because then the original thing would divide it to okay there's a theorem that says if a divides b and b divides c this transitive property than a divide c okay so a number that was crossed out can't divide one of these guys because it was crossed out because it was divisible by something and that thing would have to divide this new guy too okay and so that's roughly the idea behind it okay and and just by the algorithm every time you circle a number you automatically cross out everything else bigger than it that that it divides so those are all gone okay so if it can't be divisible by one of these circled numbers because it would be already crossed out and it can't be divisible by one of these crossed out numbers because it would have been divisible by one of the circled numbers so it does generate the list of problems it does work yeah oh well you can't you couldn't I mean this isn't something that's feasible to do in practice anyways though I mean you would never finish I mean you could you know you could oh okay yeah well yeah I mean no you can I mean certainly you can generate you know algorithms can can can generate the primes up to any you know given point if you have enough time you can certainly do this algorithmically really just by going through it like like yeah I mean so but of course the thing is you couldn't actually program the computer to do it for everything because it would never terminate right but you could program a computer to find all the primes up to you know whatever 10,000 or something like that and it could should be able to do it very very quickly you know I mean because you know you could you could just again it could just strike out all the evens first then it goes to the next to me you could make it very precise just like this and it would it would work for sure but there are actually more this is a very naive approach there are much more efficient ways to do it than this but this is just kind of give you a background as to kind of you know sort of the setup as to where we're going to go from here this is sort of the primitive the very primitive love this guy you know this guy is an old old dead guy now I mean you know this is ancient stuff okay we're going to get to slightly more modern stuff but the really modern stuff is is hard is really hard and it's way beyond the scope of this this course but we're going to get to some things that are a little bit more efficient and she had mentioned something which we're going to talk about in a minute too okay and so the way I've been blabbing now for a while so I assume you all have this down right okay okay so what I'm going to do to kind of introduce the next idea is we're just going to go and look at a very specific problem and again this is the kind of problem that you may have seen a while ago in elementary school maybe is the number 509 prime okay so I'm going to sort of lead you into the next result here solution so the question I want to want to ask you is question must we check every integer x with one less than x less than 509 to answer this question okay that's right that's right so this was a little bit vague what I'm writing down here but what I mean is do we have to check every integer x strictly between one and 509 to see if x divides 509 right because if there is one then it's not prime do we have to actually check all of those is the question and the answer is no and so the first proposition this is kind of what you both were alluding to and this is this is in the book I don't know if they actually listed as a proposition although I want to offset it this way because it's you know it's kind of useful and it's really not hard to prove first proposition just says that suppose C is bigger than one is a composite integer then there is a prime divisor P of C such that P is less than or equal to the square root of C I think this is what you were you were asking about before okay so I'll show you why this is true okay so what is for I write proof what is this really saying it's saying that okay well suppose you have a number you want to check to see if it's prime well if you know it doesn't have any prime factors less than equal to the square root of it then and then it is prime then in fact it is prime so we're composite then it would have one that's what this proposition is saying okay that's right yeah I'm just trying to be completely clear here but yes it's definitely implied by composite yeah it's a little bit redundant okay so let's assume that C is composite then okay there was a lemma I proved about about this this was a it was the last section we were talking about the fundamental theorem of arithmetic and I defined what prime and composite what these things are and then I proved something about composite numbers so if the numbers composite then it can be written as a product of two integers that are strictly between one and itself right this was actually a lemma in your notes from a while back but this was in the fundamental theorem of arithmetic section if I remember right so then C is equal to A times B some integers A and B I'll just write it this way and B are in Z satisfying one is less than a is less than C and one is less than B which is less than C okay so these integers A and B they compare that under the natural wording of the integers so for sure either A is less than or equal to B or B is less than or equal to A one of those is definitely they may be equal to each other for all we know but the definitely one is less than or equal to the other one there's no question about that right so we may assume that A is less than or equal to B okay now you might say well what how do you know that what if what if B is bigger than or equal to that well that's that's okay if it is and we'll just we'll just rename everything and just flip it everything around so the point is just one of them's less than or equal to the other one and so since multiplication is commutative it really doesn't matter which one we take okay so what we're going to do now is we're going to multiply through by A and so what do we get when we multiply this inequality through by of course we can do this because A is positive so we don't have to worry about you know flipping inequality signs and that kind of business so A squared then is less than or equal to AB i.e. A squared is less than or equal to C right because by what we have above this this is C right I have that up here C is AB so we just replace AB with C this is what we get everything's positive right A and C everything's positive here so if A squared is less than or equal to C then if we take the square root of both sides that's legal to we can we can do that we have two positive numbers one's less than the other than their square roots are also the square of the first one's less than or equal to the square of the other one right so if you want to write this down I'm just taking the square of both sides that's all I'm doing here okay now this isn't we haven't actually proved what we want to prove yet though right we just prove that A is less than or equal to the square of C we need to get a prime divisor a prime number that divides C that's less than or equal to the square to C okay what we do know is that by the okay I'm running out of room here fundamental theorem of arithmetic that okay fundamental theorem of arithmetic okay yeah I should have done that but let me finish this here there is a prime number P such that P divides a right that's part of the fundamental theorem of arithmetic arithmetic that every integer bigger than one is can be written as a product of primes and then there was this uniqueness component and that took a lot of work to verify but the easy part is just that there's a prime divisor okay and let me try to squeeze as much as I can down here okay so if P and A are positive and P divides A that means that P is definitely less than or equal to a right which is less than or equal to the square root of C so P is definitely less than or equal to the square to C you all see this you see and you see why P is less than or equal to A right because P is positive and it's a factor of A there's a there's a theorem back in 21 or something about this right it something divide something else than their absolute values there's something about it you know comparing the absolute values well these are all positive so if P divides A then P is certainly less than or equal to it I mean it's just every factor right if you have all positive you know these are all positive a factor is definitely less than or equal to the number itself for sure right so P is less than or equal to the square root of C so what is it that we need to there's still one other thing we have to check in order to finish the proof here what is it that we actually haven't shown yet anybody see this yes that P is actually divisor of C okay we've got everything else but we need to check that P is actually a divisor of C that's not hard I'll just say it right now and then I'll write it down P is a divisor of A right P divides A A divides C therefore P divides C there's this transitive property this is somewhere in two one probably if A divides B B divides C then A divides C okay if you have a number that's a factor of another number and that number itself is a factor of another than the first one was a factor that's not hard to see really I don't think and so that's it I mean that's really the end of the proof then let's see I'm gonna be no I shouldn't do this but so we have let's see P divides a and a divides C so P divides okay I finished it no no no no you would be you really honest to God he would be shocked at how good my writing is on the board if I was doing that somebody like wow I can't believe it's the same guy but I think this is actually improved a little bit from the very first day I think it's gotten a little better yeah yeah right that's true okay so that's it that's it so point is if you have a composite number then it has a prime divisor that is less than equal to the square root of it so if there is no prime divisor less than equal to the square root then it has to be prime and so now we can go back to this first problem and we can answer the question much more easily than just plugging everything in from from you know two to five hundred eight to see if their factors we can actually just take the square root and see if any of the primes below that are factors and if the answer is no then we know it's prime okay everybody have this down now okay okay so let's let's go back to the first problem okay so what we're gonna do is this let's take the square root of 509 this is roughly equal to 22.5 this is an approximation it's it's definitely irrational so it goes on forever but this is all that we really care about right now the prime numbers less than or equal to and this isn't hard of course you can easily find these two three five seven eleven thirteen seventeen and nineteen yeah there's no point there's no point no the author likes to make things unnecessarily complicated for no reason okay none of these you can check this divide 509 this is easily checked well yeah well okay so I mean this is not really this is more the point this course is more theoretical but you know course two doesn't about it because 509 odd right three doesn't divide it we may prove this at some point but yes so you may have learned this a while ago that you take a positive integer is divisible by three if and only if the sum of the digits is divisible by three maybe some of you have heard of this before 509 the sum is 14 it's not divisible by three so 509 is not divisible by three certainly not divisible by five because it doesn't end in zero or five right yeah there's a rule for seven's although in this case I mean you can just kind of do it just you know seven and a five and I just do the long division goes in seven times with nineteen goes twice and there's a remainder of five so it doesn't work and you can do something similar for the rest of them but yeah I mean really you can you can do this in a matter of less than a minute probably just to you know go through it so by the by this proposition 509 is prime has to be right if it were composite proposition says that some prime number less than equal to the square root of it would divide it but it doesn't so it has to be prime okay now we're going to look at another problem most of you have probably seen this before maybe a couple of you haven't I was going to ask the question first but I'm just going to go straight into the proposition I already asked it before anyway so this the one of the questions I had that we were going to consider today is how many primes are there right and most of you know that the answer is that there are an infinite number of primes since we're doing number theory shouldn't be that shocking that we're going to write a proof of this for you even though you may have seen it in discreet how many of you actually did see this in discreet okay most of well maybe not most but over half of you what's that you don't remember okay well okay well there's at least one person that will benefit from seeing this again okay okay so this is anybody remember who this so the proof that there are an infinite number of primes is sort of the standard proof who this is attributed to anyone remember this old guy just that yeah it's you you clad is that he usually gets credit for this okay there are in fact infinitely many primes okay so I'm gonna try to just I'm just gonna squeeze this here on this page I think here's the proof before I do write the proof let me let me say something about why this is kind of a nice result in math it's not stop drinking before class it's not that would be horrible I hope that's not true it's not obvious right away that that there are an infinite number of primes and the reason is if you just kind of think of this naively just think of all the numbers well okay this is kind of night but just you know imagine you you just you know have all the numbers in a big bag and you pull out a big huge number with you know a thousand digits in it there's a sense in which it's you know maybe likely that it's not prime because you because you have so many numbers before that could possibly divide into it right so the bigger the numbers get in some sense you might think that okay well it's less likely that they're going to be prime because there's just too many possible factors below right so maybe there's a point that comes where there aren't any more primes certainly it's not unreasonable to think that if you if you're not really well-versed in this stuff that maybe because there's just so many possibilities that every number after say some number with 50 billion digits after that there aren't any primes left anymore because there's so many possible divisors before that in fact one of them actually divides everything after that but it doesn't actually that's not the case so the fact that they're an infinite number of primes is not totally obvious that this would be the case and the proof and it's really nice it's very short and again I know a lot of you've seen this before but the idea is is this it's approved by contradiction in fact there are several different proofs of this this is just one of them did you have a question yeah well yeah I think I know what you're saying and we're going to get to that I probably won't spend time on this right now but there there's definitely some some some theorems about the distribution of primes and I think this will at least give you kind of an indirect kind of answer to what to what you're saying but I'm gonna I'm just gonna hold off on that for just a minute but depending on exactly what you mean yes your yes the answer is sort of yes let's suppose we have contradiction only finitely many and let's just list them out okay there's finitely many say p1 comma p2 comma p3 comma on down to say p sub n right we don't know exactly how many there are even we assume they're only finally many I mean we know that there's some I mean we just found I mean we have a list of primes up here so we know that it's definitely there's at least this many for sure but it stops at some point that's the main idea okay so and maybe ends a million a billion whatever but it's just a finite list of primes and then say after that there aren't anymore okay and so here's this this kind of a standard trick what we're gonna do is we're gonna let capital and maybe I shouldn't do this let me just call this m okay let m be p1 p2 multiply together all the way down to pn just to be completely clear here because it's gonna be ambiguous plus one okay so it's all the primes multiply together and we're gonna add one at the very end okay okay okay so what can we say about m well of course it's an integer that's for sure it's definitely a natural number m is definitely bigger than 1 right because we know there are at least some primes for sure I mean even just tinkering a little bit even if you know nothing these are primes so there's you know there's at least 8 of them that's for sure so this number is definitely m is definitely bigger than 1 for sure maybe maybe it's 9 but yeah we're gonna show it's not possible but m is bigger than 1 so what do we know go back to the fundamental theorem of arithmetic what do we know about a natural number that's bigger than 1 it can be it's either prime or it can be written as a product of primes that that was part of the fundamental theorem of arithmetic right and then again there was the uniqueness part we're not interested in that right now point is we have an integer bigger than 1 it can be written as a product of primes might say product I mean maybe it's just one prime maybe it's prime itself but there's definitely a prime divisor that's the that's the main idea this is from the fundamental theorem of arithmetic okay so there is some prime divisor of m okay right because it's bigger than 1 well remember what our assumption is we're assuming that there are only finally many of them so these are all of them I mean of course this isn't true that this is our assumption right that these are all the primes so if there's some prime divisor of m it has to be among this set because those are all the primes we're assuming those are all of them so m has to be p5 or p100 or whatever it's got to be one of them for sure right so hang on just one second so this prime whatever this prime divisor is it has to be in the set p1 p2 down to pn okay let's let's say it is piece of we don't know what it is you know piece of I say right okay so this what was your question that is that's definitely composite yeah now we're adding one and you'll see why we're adding one here that's gonna come up in a second but the point is just all that all that we care about right now is that this number is bigger than one and so it has a prime divisor and since these are all the primes that prime divisor has to be one of one of these okay so so what do we know we know one pi divides m right but we also know that p sub i divides p1 p2 the product all the way down to pn okay I'll pause for just a second about this well this just comes from the fact that it's a prime divisor it has to be one of these guys this is what I was just saying but since it's an element in this set right it certainly is a factor of this product you have a product of a bunch of numbers just take any number in that product that's definitely a factor of the whole thing right because you just take it and you just multiply by the by the remaining numbers in the product okay right two times three times seven seven is definitely a factor of that because it's seven times two times three times nine okay that whole product so is this okay make sense okay can I go ahead and go to the next page so since pi divides m and pi divides this product p1 p2 on down to pn we get this is part of a part of a theorem that I prove for you but pi has to divide m minus this product okay if a number divides two different numbers then it's going to divide their sum their difference I mean it's going to divide all these and you can you can get that just by writing it out directly right if I'll just say this right if pi times x equals m and pi times y is equal to that then pi times x minus y is going to be equal to m minus that it's very easy to just see that directly just if you just write it out I'm not going to do it because it's part of the theorem that I prove for you but this is definitely the case but okay forget about this period at the end good if you go back to the definition of m okay I'm just going to go back one so remember what m is m is the product plus one m is the product plus one so just remember that's what it is of course it's in your notes too so what is the product plus one minus the product it's just equal to one right here's the period okay can we have a prime number being a factor of one it's three a factor of one no of course it's not the only ones are one and minus one and those are definitely not prime so there's your contradiction you can't have a prime being a factor one one and minus one of the only ones so because that's a contradiction the assumption that there only finally many has to be wrong and so there has to be an infinite number of primes so this is use this word so this is absurd that's not the case no prime divides one thus there are infinitely many okay so that is equal to prove there are other proofs of this in fact there are there's probably 10 or 12 of these that have you know shown up over the years this isn't the only way to do it but this is this is kind of the classical way to do it and it's certainly a fairly straightforward proof okay so what else we're going to do yeah so they're going to prove one more proposition and there's going to be a really quick corollary of that and that's what we're going to hopefully I can get this done so now we're going to talk about the other question about the distribution this is a kind of a weak result but the question is okay well you know say you give me a big number and you know say we want to know okay well how many primes are there that are less than or equal to that number there's something called the prime number theorem which I don't know if we'll get into in this class or not but it gives kind of a estimate of this but this is going to be a very crude estimate but we can actually say a few things about this so we're going to kind of change gears now and we're going to we're going to order from now on we're going to order the primes we're going to kind of fix this okay and so what I mean is p sub 1 is going to be the first prime p sub 2 is going to be the second prime p sub 3 will be the third prime of course which is 5 and on down right so if I write down you know p sub 200 that's just the 200th prime whatever that happens to be we don't know what it is necessarily of course we don't have to top of our heads but we know there is a 200th prime because we just prove that there is an infinite number so we never run out of them right okay so what we're going to do now is we're going to prove this second proposition and I'm going to give you a very quick corollary of this and that's actually all I'm going to do in this in this section I guess this is not this is proposition 2 isn't it yeah is it 3 oh oh wow okay I am all right and I'm just going to be proposition 3 well okay yeah I think I have to do disagree with you there I think it's proposition 2 if we I only had one proposition for that we had a theorem before that but there was only one proposition right there was two problems oh I didn't call that a theorem okay I see okay okay all right got it yeah I'm going I'm going nuts today okay okay yeah you're I know you're just concerned about me I appreciate that every positive integer n okay this is going to look kind of weird I guess but p sub n is less than or equal to I'm going to try to make this as clear as I can 2 to the power 2 to the n minus 1 okay you can read that so remember I just talked about what this was right p sub n is just whatever n is it's the nth prime if n was 7 it would be the seventh prime right and this is true for every n so what this is saying roughly is that you know in some sense the key there can't be gaps that are huge huge huge gaps that they're somehow bounded by this kind of exponential function in some way okay no not that I know of anyways no okay so here's what we're going to do and this is kind of nice because it puts together a few things that we learned a long time ago and this is it's also nice because I'm going to do another induction proof which which I think you know a few of you kind of need to get some extra practicing some of this we're going to do is we're going to proceed in this case by strong induction okay we didn't do as many of these back in chapter one so now you get a chance to see this again and this was again the second principle finite induction I asked you this on the first exam right okay so that's what we're going to do so let's look at the base case of this shouldn't be very hard the base case is just the assertion that that p sub 1 is less than or equal to 2 to the 2 to the 1 minus 1 right i.e. what does this reduce to okay well what is what's p sub 1 okay it's equal to 2 this is just the assertion that 2 is less than or equal to and what is this equal to what number is that 2 right 2 to the 1st 1 minus 1 is 0 2 to the 0 is 1 so the right side just becomes 2 to the 1st which of course is true and now we need to do the inductive step so I think yeah I need to start getting a spacing better on the screen here but oops okay I didn't mean to do that let's see where my oh now I'm screwing it all up the blue circle which down here that's not blue okay that's blue this color is blue this is like lavender purple something like that yeah sorry you lose okay alright so you guys have this you guys have this down okay okay I can go to the next page all right okay so here's the inductive step inductive step what we're gonna do is we are going to we're gonna let n be a natural number and we're going to assume I'm just gonna write this out explicitly here so there's no confusion we're gonna assume that p sub 1 is less than or equal to 2 to the 2 to the 1 minus 1 of course we already verified that p sub 2 is less than or equal to 2 to the power 2 squared minus 1 and on down right so p sub n is less than or equal to 2 to the power 2 to the n minus 1 yes okay so yeah you're talking about this power up here right yeah okay so the power is oh oh sorry sorry sorry yeah I I thank you I I screwed that up okay yeah it's it should be up here yeah it should be up here sorry about that minus 1 okay yeah thank you for that that would have really screwed the proof up okay everybody clear on this all right it's just what I had written before on the on the previous page I just I just screwed up the the spacing okay so what we need to prove we will show we've got to show this is true for n plus 1 right so we're gonna show that p sub n plus 1 is less than or equal to 2 the power 2 to the n plus 1 minus 1 right which of course simplifies to just I mean the part in parentheses simplifies to just 2 to the n right at the top okay and so here's here's what we're gonna do this does require a little bit of work so this is gonna be very similar to the proof that we we did that they're an infinite number of times we're gonna consider the number p1 p2 p3 you know on down to pn and then I'm not gonna put the parentheses in this time plus 1 okay so this is kind of like what we did right in the the proof of the last theorem we're gonna let p sub j be any prime divisor okay I want to be very clear about this j now we've used subscripts before just as dummy variables this j literally means we're staying with this notation that I introduced on the previous slide with the ordering so what I mean here when I say p sub j I mean the j th prime right it's not just some some prime the j actually means that it's the j th 1 as we go in order okay it certainly has some prime divisors bigger than one so there's something but the j does does represent the order here too okay here's and I'm gonna have to kind of gloss over this a little bit I'm just gonna have to say because we're gonna run out of time here but what can we say about this index j well here's my question can j possibly be one in other words can this prime be p sub 1 well if it was if it was then p sub 1 would divide this but it also divides this so would divide one by just what we did in the last proof so it can't be p sub 1 because it would divide it would then be a factor of 1 which it can't be similarly it can't be 3 because if it were it would divide this and it would divide this and so again it would divide one so the point is that j has to be bigger than it has to be can't be any of these guys because you get that contradiction we got in the last proof that would have to be a factor of 1 okay so j has to be bigger than or equal to n plus 1 so I'll just call this equation 1 I guess or inequality 1 I'm gonna say I'm there's another part of this inequality which I'm gonna get to if I don't finish this now I'll finish this up on Tuesday by the way this is not you're not gonna really need this the rest of this in order to be able to do the homework so that's it's not really gonna affect your your homework anyways plus it's not due until next Thursday but why is this true well okay this is just it just comes from from this really it's just that if j is bigger than or equal to n plus 1 well what is what is p sub j and what's p sub n plus 1 p sub n plus 1 is the n plus first prime p sub j is the j th prime if j is bigger than or equal to n plus 1 then that means that p sub j the j th prime of course is further down the list than the n plus first prime right so the n plus first time is smaller than p sub j right just think about it right the second prime is certainly smaller than the third prime because that's how we got the mortar okay so if we know that j is bigger than or equal to n plus 1 then p sub j is bigger than or equal to p sub n plus 1 okay and what can we say about p sub j well p sub j is since it's a factor is definitely less than or equal to p1 p2 p3 on down to pn and then plus 1 on the outside so just to make this totally clear we know that p sub n plus 1 is definitely less than or equal to this right guys by that transitivity right one thing is less than equals something which is less than equals something that first thing has to be less than equal to the third thing right for sure so okay I think what I'm going to do at this point I'm gonna make a note here I think what we're gonna do is we're gonna stop at this point for today otherwise I'm just gonna get stuck in the middle of something and I don't want to do that so this is where we will end I'm gonna still give you the homework and here's what I'm gonna tell you the assignment you just turned in because of they're not being classed on Tuesday and because I wasn't really available as much I'm going to award you more completion points for this than I would have otherwise done okay I'm still gonna grade some but you'll get I'll give you a little bit of a break on this assignment as a result of that okay and then some a few of these problems a couple of them are gonna definitely require a little bit of thought but I still think you're gonna find that this assignment is not quite as bad as the last one was and again on Tuesday we'll spend some time we'll talk about some of these so you'll still have a couple of days to finish the assignment I'm not giving you that many problems one three four just part a seven nine B and 12 C so you only have six problems and you've got a week to work on these okay okay so just to remind you Tuesday I do recommend that you come to class we'll we'll review what little we've done since the last exam I'll give you again an idea of kind of the things to be looking at remind you that one of the problems will come up directly off the second test right I'll have this homework assignment back to you so you have all your homework before the exam and so just just to be clear this is probably something else I should say don't leave just yet because I'm gonna pass your homework back okay let me let you know what sections are gonna be on the test right you might want to know that so we finished off with if I were yeah so the first test covered up through section let's see where were we at here so section 2 3 2 3 yeah and so we did after that we did 2.4 it's called the Euclidean algorithm in the lecture I just called it I think GCD and LCM is all I said because I didn't actually cover the Euclidean algorithm so 2.4 and then we skip 2.5 then we just went to 3 chapter 3 so it's gonna just be 2.4 3.1 3.2 that's it that's it okay you don't have a ton of stuff really so you definitely want to be looking over that I will have the solutions to this homework that I'm passing back will be posted this afternoon or early evening so you'll have they'll well no they're just gonna be the ones that were grading so I'm not gonna I'm not gonna write out the ones to the other ones but of course you always welcome to come and talk to me or ask me I can tell you but yeah okay so let me go ahead and turn this off and let me pass these back to