 This video represents the very end of our unit on commentatorics in our lecture series math 3120 now Just to be clear that does not mean this is the last video in our lecture series about commentatorics In fact this very last topic we're going to talk about we're going to spend Well time in this video of course, but we'll also spend some time in lecture 18 talking about it as well And that's because the topic itself can be very difficult for people as they first learn about it And this is the topic of a Commentatorial proof and after all this is the second video in lecture 17 So by our tradition this should be about something logical and as such We're going to connect it to the topics of commentatorics We've been studying and so what is a commentatorial proof well It turns out that the proof technique is fairly simple to explain Imagine that you have a set a set a and typically in commentatorics what we want to do is we like to Compute the size of a set so we're trying to compute the cardinality of some set a and let's say that after some calculations we argue that the That the set the cardinality of set a is equal to some function f of n Where n is some parameter and a itself is a set that will depend on and its size is dependent upon some variable in play here So let's say that we show that The cardinality of a is equal to f of n But let's say we also can show that the cardinality of a is equal to g of n Well as we have two different formulas that then count the same set We would have to then conclude that the two formulas are equal to each other and This is at the heart of what a Commentatorial proof is we count the same set in two different ways And then we can conclude that the two different formulas we came up that counted the same set have to be equal to each other Simple, right? Well, why am I spending so much time on such a simple proof technique? Well, it turns out commentatorial proofs are easier said than done It turns out that in order to get good at commentatorial proofs because the concept is simple It's coming up with the right with the right set to count that is difficult And honestly experience and gut feelings are what we want to develop right now So let me begin with an example here a very classic example Let n be a positive number Although honestly this statement is true if you take n to be zero as well But we'll just focus on on just positive numbers here really doesn't change the proof that much And let's prove the identity 1 plus 2 plus 3 plus 4 plus 5 plus 6 up to Plus n is equal to n times n plus 1 over 2 Now this right here is a classic. I'm talking about classic formula and typically this shows up students to see this for the first time inside of calculus 1 Which for SUU students this one sides with math 1210 and this is typically introduced as students are preparing to learn about our or already learning about Riemann sums an object we would study before we introduced the definite integral as a Limit of a Riemann sum is where we get the integrals at all now it depended upon your Calculus 1 class you may or may not have seen a proof to this formula Again, it depends it depends on the curriculum and the university or the school whatever you but oftentimes when this formula is presented Whether it's in calculus or whether it's presented to another setting It's proven using a take a technique known as mathematical induction And we will actually start learning about induction in our very next unit of this lecture series about integers Induction is a very very important Mathematical tool and so mathematical induction deserves our attention Mathematical induction is very very useful at proving formulas like this. It's really good at it It's super nice super clean But one problem about induction is it doesn't really explain why the formula holds It just proves that it's true which is something we'll talk some more about in the future, right? We'll talk about what's the difference between a constructive proof and a non-constructive proof a proof can prove that something exists But we have no idea what it is or Conversely, you can have a proof that proofs that something exists because it provides it to you in front of you Induction is kind of like that non constructive proof. It can prove that a statement is true, but it's true Who knows why I mean we know it's true But it doesn't give us any idea where the formula came from like how did you discover it? conversely a combinatorial proof actually provides a Why it gives us some intuition of where the formula comes from and so the reason I want to start with this proof right here Is there's a classic myth that 10-year-old Gauss Gauss is you know known commonly as the prince of mathematics? Frederick the Carl Frederick Gauss here it's supposedly as a 10-year-old proved this formula and Basically the proof that Gauss is attributed to have given is essentially the proof that we are going to provide right now Although we will provide a more modern flair to it using the notion of a Ferris diagram But it's essentially equivalent to it and again in addition to proving the formula We also know where the formula came with because as the legend goes Gauss is supposedly in his grade school math class given the task of adding 1 plus 2 plus 3 plus 4 plus 5 all the way up to 100 I believe was the number you're supposed to do supposed to add together the first 100 integers and so while most of the grade school students Then we're like 1 plus 2 is 3 plus 3 is 6 plus 4 is 10 plus 5 is 15 going through that's easy when you have single digits But eventually start getting double digits triple digits additions. You're likely gonna make some mistakes along the way There's a lot of places where human error can happen But Gauss actually took a combinatorial proof approach Which he that that is he he decided to count this number in a different way Which was equivalent to what he had to do it ended up being this formula right here And times n plus 1 over 2 and therefore he could very quickly count the number and also is the only student who didn't do it Erroneously so let's see how it's true. So to prove this formula I first want to introduce the notion of a ferris diagram now a ferris diagram You can think of as a matrix array. It's a two-dimensional ray consisting of dots and omissions So if you want to think in terms of a matrix, you could think of like a zero one matrix where a one Represents sorry a dot represents a one and no mission represents a zero something like that But the rule is that the dots are always to the left of the blinks So if you are if you think of a zero one matrix the ones always appear to the left of the zeros inside of it And that makes this ferris diagram and typically it's not required But typically if they have different sizes, you're gonna put larger sizes downward like so So the following diagram the following ferris diagram you see right here This is the type of object. We're interested in right now So you have one row two rows three rows four rows five rows So if you could count the size of a ferris diagram with five rows That would then be the thing you're looking for now I'm gonna slightly change my fight ferris diagram here I'm not gonna look at this diagram But keep this in mind because we're gonna find it again in a moment what I'm gonna do Instead as I'm gonna consider the ferris diagram, which is an in plus one in plus one square That is each row has in plus one mini dots and each column Likewise has in plus one mini dots in it as well And what I want to do is I want to count the area of this ferris diagram Well with the first attempt this is a square with in plus one rows and in plus one columns So the standard area formula comes into play here that a squares area is just going to be The side length squared so the number of dots is going to be in plus one squared in plus one times in plus one So that's the that's our first account of how many dots are in this square now I'm gonna take the same square. I'm gonna count it slightly differently Notice this time. What if I just look at the diagonal elements of my square How many are those going to be well? There's one exactly one diagonal element for every row of the square There's in plus one rows, so I'm gonna get in plus one mini dots along the diagonal So just consider the diagonal for a moment now consider the elements which are above the diagonal You're gonna look at this object right here now. This is no coincidence this set We have right here above the diagonal is exactly the sum of one plus two plus three plus four all the way up to Plus in that's why we had a square of in plus one So when we remove them we actually get the thing we're counting before and likewise look at this ferris diagram this sub diagram I should say this is the thing we were interested in counting Okay, let's call this number s for the moment that is s is the number of dots in the Triangular figuration this these are sometimes called triangle numbers Triangular numbers because they count the number of dots in one of these ferris triangles So if you then re Redistribute the things here. We have in plus one mini diagonal entries So let's mark that there. We have in in plus one mini diagonals We have one of our triangle diagrams right here. There's s many of those We don't know if that is yet, but there's s many of them and then over here you have an identical triangle I'm sure it's been reflected, but still there's s of those So when you put these together, you're gonna have two s plus in plus one mini dots In this in plus one square, but like we saw before there's in plus one Squared many dots in this as well. This is where the combinatorial proof came into play here I counted the square once using just in plus one times in plus one But then I counted the square with a different a different scheme And I got a different formula two s plus in plus one notice s is the number I'm interested in this equation is valid because the counted the same set twice So I can solve this equation for for s right here to do that. I'm actually going to Foil out the right-hand side. You're gonna get an n squared plus two n plus one So then we can subtract the n and the minus one from both sides minus n minus one So we're then gonna get something like this right here. You're gonna get two s is equal to n squared plus n which of course you can factor the n squared plus n as n times n plus one Divide by two and you end up with s equals n times n plus one over two That was the formula we were trying to prove right there And so this is our first example of a combinatorial proof We counted the number of dots in this in plus one square in two different ways That gave us then a formula this formula right here As as the number I actually cared about was embedded inside of the formula We're able to manipulate that equation to get the formula we were looking for. It's a really really nice method It's very clear to understand But of course, how did I know I should be counting n plus one times n plus one size squares? That's intuition that is a little bit difficult to just Tell to you one as you get better and better with combinatorics You start to get better intuition a better gut feeling on the things you like to count But ferrous diagrams are a good place to go to try to count many of these combinatorial formulas Let's provide another example a slightly harder formula one that's far less intuitive This time let n be a positive integer Then if we take one plus three plus five plus seven plus nine all the way up to two n plus two n minus one So this is this on the left hand side is the sum of consecutive odd integers In the previous one we were adding consecutive integers one plus two plus three plus four plus five Now we're only adding together consecutive odd integers one plus three plus five plus seven etc I claim that is equal to n squared. That is if you take the sum Of consecutive odd integers. This always gives you a perfect square and you'll get the next perfect square by adding the next odd Integer it's fantastic. It's it's not very intuitive at all that that would be the case But it is it is the case. It's a pretty cool thing Now when I look at this formula right here If I was trying to prove this using combinatorial proof um, the previous example might have been a little mysterious how we did it, but this one has a has a super super Obvious blues clues that we need to be looking for that tells us what we're going to count when I look at the right hand side I see n squared and much like because I'm thinking of the last proof Squares count the area of a square. That's why we call it square We don't typically say n to the second power We say n squared because of its relationship to the area of a square So when I look at the right hand side, I'm thinking I should be counting dots of a square So that you want to grab the low hanging fruit? Oftentimes the formula has a low hanging fruit and that's what's going to motivate the combinatorial proof So in this situation, that's exactly what I'm going to do I'm going to prove the formula using combinatorial proof And I'm going to do that by counting the dots in the n by n ferrous diagram Which you can see illustrated right here now like we saw in the previous proof It has n rows and n columns So the the number of dots in that square is going to be n squared Okay, that gives us the right hand side We often start with the low hanging fruit in that regard So what I have to now do in order to count The the the left hand side that is in order to prove the left hand side is equal to the right hand side I have to consider this square again But then in such a way that I can produce these odd integers And this takes a little bit of guessing and checking as one tries to write the proof But notice the following idea here. I have one dot right here. I have three dots right here I have five dots right here I have seven dots right here and then nine dots and then 11 dots Like so so if I look at these l shape configurations You're always getting these odd numbers and that's the relationship that we're going to be looking for in the second attempt I've now color coded to make it a little bit easier to see here On the other hand, let's partition the cells in the in the cell based upon their address So we're going to go back to linear algebra here Notice that this is this right here is the 1 1 position 1 2 position 1 3 position 1 4 So the first index is the row that you live in the second index is the column This would be the 2 1 position the 2 2 position 2 3 2 4 2 5 This will here be the 4 1 position just just you're aware that each of these dots It is a matrix after all each of these dots has an address I'm going to partition the cells I'm going to partition the dots in this matrix based upon their address But we're going to follow the rule that you belong to the group Where you have the larger of the indices, right? So the 1 1 position is all by itself because the max of your indices 1 and 1 is itself 1 And that's the only position where the max of the two indices will be 1 On the other hand, if you take the positions 1 2 2 2 and 2 1 The largest coordinate in those addresses is 2 and so they're all going to put together in the 2 group So you get the one group which the one group was this one right here You then get the 2 group, which is this one right here The next cell in our partition would be the 3 group We're going to grab all of those addresses for which the maximum coordinate is 3 That would include 1 3 2 3 3 3 3 2 3 1 which gives you these ones right here Then the next group would be the 4 group. You're going to get 4 1 4 2 4 3 4 4 You'll get 1 4 2 4 3 4 and then 4 4 again And let's just do one more example for completeness here if we look at the 5 group You're going to get 5 1 5 2 5 3 5 4 5 5 You'd also get 1 5 2 5 3 5 4 5 and 5 5 So each of these groups that if you think of the kth group I claim that the kth group is going to contain 2 k minus 1 many elements And why is that because of those elements you're going to end up with 1 k 2 k 3 k All the way up to k k And that coincides with considering these right here this column, but you're also going to end up with k 1 k 2 k 3 All the way up to k k again Okay, that grabs the things in the row. All right, so notice how many things are in this list The first this first consideration There's k objects here because the first coordinate is allowed to range from 1 to k with this second list You again get k objects because the second coordinate is now allowed to range from 1 to k But you also should notice that you counted the element k k twice So by the inclusion exclusion principle, there's k things in the first row There's k things in the second row, but there is an overlap of 1 the intersections 1 So each of these l shape objects is going to have 2 k minus 1 many elements in it And that's that itself is just the arbitrary odd number. So when k Is 1 you get 1 when k is 2 you get 3 when k is 3 you get 5 when k is 4 you get 7 etc etc etc up until 2 k 2 in minus 1 Okay, so what this now shows us is that we can count the objects Um as of the square by n squared We can also count the objects by taking the sum of things of the form 2 k minus 1 As k ranges from 1 up to n which the left hand side is then the is the sum of all of the odd integers And so by the method of combinatorial proof these things are equal to each other and thus proves the formula We're going to do some more examples of combinatorial proof in the next lecture lecture 18 So if you want some more practice look for those videos as well