 Okay. Okay. So I guess, I guess we can, I guess we can go ahead and go ahead and start and probably the, probably the easiest, probably the easiest way to work this. I guess we can start going through. We think. Okay. Maybe like the last final or something. Yeah. Okay. So, yeah, so let's take a look at probably, let me see if I can pull that up. Okay, give me a second here too. The final exams that are posted on the department website, they are going to be a little bit different in structure and format from the, from the actual final because those were written under the assumption that you'd have, we'd have in-person classes, and the structure was a little bit different. So some of the questions are similar and some of them, some of them are not. Let me, let me pull out a couple of questions that'll probably be, probably be, let's see. Alright, let me just go ahead and copy this whole thing. That's probably okay. Yes, let me pull a couple of somewhat representative questions. Okay, so I'll pull a couple of questions out of here. I have a question for Professor Jean-Bassiste. Like how do we know if the accommodation for the different time is going to be like granted for us? I'm going to grant everybody's time. I have all your emails. So just pop up during the time we told you and. Yeah, yeah, it's fine. Don't worry. I still have a question. I just wanted to ask, since the exam is going to be taking on the MyOpenMath website, I noticed that on some of the homeworks, like if you didn't put like the correct parentheses or something like that, you don't get any points. Is that going to be the same thing for the exam? Like if we get the correct answer but don't, how do I explain? Like don't put the right, put it in the right sequence as it requires us to, are we going to lose? Is it going to be a complete zero or are we going to have some points? So part of the reason that you'll be able to add work on to the exam is that is that your professor will be able to see that, okay, you did the actual work on this question and be able to give you credit for the question. So it's sort of this backup. I mean, in addition to showing the work that you did, it's sort of a backup to that. I don't know if that really answers your question, but yeah, the idea is that you should be able to, you should be able to get credit for it. If there's like a syntax error or something like that, it will, as long as you have your work, as long as you attach your work to the exam, then your professor will be able to take that into account. All right, thank you. Okay, so. All right, so let's let's take a look at some of the some of the somewhat representative questions from the from one of the previous exams. So I think this is the last paper exam that we have up there. And so, so here we have a situation where we have a Venn diagram. And we know something about our universe of sets and what the sets T and C and a all represent. And so so so the idea here is that remember the Venn diagram. So if you are inside a circle that means that you're a member of a particular set. So, for example, something like this. The we want to describe the set notation, let me, let me submit a little bit more on that. Okay, so so again the idea here is that so so if I'm in a set so if I'm say at this point, then, well let's think about that so if you're in that point. Then you're in the set T. So that's the set of tennis players. Well, let me ask so so let's ask this question. Is this a person who plays chess. No, yeah it's not a person who plays chess if they play chess they should be in the in in the in the set C. So how do you put a person who is a tennis player and also somebody who plays chess. So our regions are numbered there so where would they end up. And to in to probably yeah, potentially they may be in five as well we haven't committed ourselves to whether or not they're artists, but certainly somebody who plays tennis and is also a chess player they should end up in this region to. And so let's talk about this so we so okay so let's try to translate this so we want all everybody who's an artist, but they're not people they do not play tennis. So, first of all, so the artists are the artists are all in this set a, and they don't play tennis so. So let's think about this. So there's four regions that are there four five six and seven. So, who so so which regions should we include in artists who don't play tennis. Seven and six and seven right just these two regions we don't want these regions four and five. So this is region six and seven. And now let's think about how we describe that. So remember there's there's three. There's three set symbols for really if you count. There's a three or four set set operations. And so we have the Union of two sets. We have the Union of two sets. We have the intersection of two sets. We have the set compliment, and we don't really like to use it but it does exist as a notation we also have the set difference. So remember that the Union is either or you're either in a or you're in be the intersect is both and you have to be both in a and also in B. The compliment is not. So the compliment is we're not in a and the set differences is kind of the weird one this is in a but not in B. Now part of the reason that we don't use the set compliment sorry the set difference very much is that we can actually reword this we don't really need set difference it's not a it's not a critical operation. Because if you take the statement apart in a but not in B. So what's another way we can phrase in a but not in B. Well, so here's a starting point is that an either something or something, or is it a both and either. Is it either in a or not in B. Is that what, but translates as. Wait, can you repeat the question. So the question is this word, but the word. But when we use that in in our set notation. It's, it's what the question is how do we interpret, but is it an either or, or is it a both and so give me a sentence. That uses the word but something but something. How about green, but how about not alive. Describes most of my plants when we both and it's got to be a both and so if I say something but something I'm really meaning green and also not alive. So here's a useful idea when you run into that word but in a in a in a statement it really translates as a me to do that it really translates as an intersection. So this set difference, another way we can look at it so in a so you have to be in a, and also not, well there's our negation, not in B. So remember there's our four set our for four set operations again this this set difference is not really is not really a separate operation it's really a special case of a combination of an intersection and a compliment. Okay, so let's think about this. How do you want to phrase the set of artists who do not play tennis. So first of all, are we talking about an either or. Or are we talking about a both and challenge with conjunctions by the way is that usually if we have an either or statement will actually say something or something. It's hard to phrase a union without using the word or. But as you can see, it's really easy to phrase a conjunction and not use the word and we could use the word, but this but something else. So do we have a is this statement is that a both of these things, or is it one of these things or the other of these things. It's a both right so if you think about this let's reword this a little bit. So, these are people who. Well to both and so are able to phrase it this way both artists and they don't play tennis. And so because this is a both and I just think I heard somebody just show up. Okay, so, so because this is a both and well we know what the artists are the artists are this set a so that's a it's a both and and they don't play tennis. So that's tennis and will negate that. So we would. So we describe the this group of people, both artists they're both in the set a and also intersect the gate tennis they do not play tennis. And so there's one way of describing our region. And the other way we could describe this region. Remember that this particular construction also corresponds to a set difference. So we could also, we could also phrase it this way. So this is going to be a remove anybody who's in tea. Now, if you actually have the Venn diagram in front of you if you actually have the Venn diagram in front of you. Then this set difference notation is maybe a little bit more intuitive. There's a useful approach to many problems in mathematics, but particularly any problem where you're given a diagram or a picture or a figure. I call it scotch tape and scissors. The idea is that you want to construct something and you can do two things. You could either tape two things together, or you can use scissors and remove a section from something else. So again, if you have the words, if you have the words, it's probably easiest to go through this idea that union is neither or intersect is a both and and then that'll get you to this form. If you have the picture. So actually, so let's even take a look at this describe set six in set notation. If you have the picture, it might be easiest to think about this as either gluing together regions or gluing together regions or cutting out the regions that you don't want. So, so let's say so so this so the second form artists who don't play tennis as soon as you identify that region six and seven. One way you might look at that is, this is also this region. And we're going to remove that region that region there. And if you look at where that region comes from that region is actually part of one. So the other way we could have gotten this once we identify that it's this region six and seven. It's really the set a, but we're going to take away we're going to remove everything that's in this in this set T. That's where this a remove T notation comes from. Again, from the verbal description, it's probably easier to figure out whether we have an either or or a both and and then get the notation. If you have the visual picture if you have a picture, you might find it easier to write it as a set difference. Specifically, in this particular case we do have the second problem here, we want to describe this region six. Let me reset my diagram. So if we want to describe that region six in set notation. Let's build six. So scotch tape and scissors. If you want to build six. So to get six. So, what do we want to start with. If I want to get this region six. So I'll pull up my scotch tape and scissors. What do we start with. What you mean. Yeah, we want to get this region six so we could either glue it together from smaller pieces, or we could cut it from a larger piece. So, well, let's cut it from a larger piece. What is it part of. We have three sets T, C and a so what's a part of what. Okay, let's start with a so let's start with a. So that's this set. So here's the set a. Okay, so now if I want to get if I want to get six. What I need to do that's a really terrible drawing. If I want to get six. If I just want to get six. What do I have to do what do I have to cut out. Do you have to cut out four and five. I have to cut out four, five, and I also cut out seven. Okay, so here's here's where here's where we might take here's where a little bit of artistry might might be useful. Let's. How do you want to cut out four and five. So, cut out. Okay, so that means I'm going to use set difference. Sorry about this one second. So I'm going to remove I want to take out. And here's the question. How do you describe four and five. The intersections of T and a. Yeah, this is if you look at where four and five are that's the intersection of sets T and a. So I'm going to remove that. So that gets rid of four, five. Okay. So, now, how do I get rid of seven. Wouldn't it be minus all integers and a. Well, the problem is I can't write minus a because that would take out the entire thing I don't want to get rid of. I don't want to get rid of six. So, let's. What's that. Is it I forgot the symbol that it would be the a adjacent. So what, so what do you mean, what do you mean adjacent, like all all sets in a that are not in C. Ah, so we want everything well actually actually actually. Okay, so you want to see you want to remove. Sorry. Okay, I see what you're saying. Okay, so you want to remove. So let's get rid of everything that's in a, but not in C. Okay, that's actually a good description. So how do I. So, so what is that in set notation. What is everything in a, but not in C. Is it a compliment. Well, a compliment is everything not in a C compliment. A compliment is everything not in C. Okay, so, so how do I get how do I limit ourselves to just the things that are in a so we want everything in a that's not in C. So, we have. Well, let me ask the same question that we asked that I asked a little bit earlier. If you want everything in a that is not in C. Either or, or is it a both and both and it's a both and and it's it's again one of these. So we want to remove everything in a so it has to be in a and also not in C. So we should actually throw parentheses around that to, we should actually throw parentheses around that to, to make it clear that we're going to remove this particular set a intersect C compliment. So, so, so here's one way that we can get six. Just as, just as an observation here. There's actually several different ways we can isolate that set. And there's not a unique answer to what that expression is. There's several different ways of describing the same set. So here's, so here's a different way of looking at that. So let's reset. No pun intended. But if I reset. So, again, I want this region six. So here's a, here's a different way I might look at that. If you wanted to cut six down from a larger region so back here we took six as part of this big set a. So we started here, we started with a and then we've removed a whole bunch of things. If I want to remove six, if I want to cut six down from a small from another set. What's another set that I might or what's another thing that I might cut six from from C. I might cut it from C. Actually, let's actually what we might do. Look at this region right here. So one of the problems we have if we start with a is if I want to get six I have to get rid of three pieces. So if I wanted to get six this way, I could start with a and then I give remove the three pieces that I need there. But if I start with this bit here, then I only need to remove one piece. So the question is, how would you describe this bit right there. How would you describe those things, five and six together. Would it be both a and C, but not a and see yeah so let's start with that it's both a and see so we'll start with that remember both and that's an intersect. So it's both a and see. So that gives us five and six together. Now what do I want to I don't want all of it I need to remove. I remove T. So, so again here's one way we can write that region. Here's another way we can write the region again you should probably include the parentheses there just to just to just to keep everything keep everything straight. So there's two answers right so there's at least two answers. If you're creative you could probably come up with several more answers to this. So yes there's at least two answers here there's actually an infinite there's not an infinite number but there's there's quite a few more answers that you could come up with, really depending on how you look at the region. And again the thing I would I would strongly emphasize here is or the thing that makes this easier is think about the scotch tape and scissors. If I want to make this region six or really any region. I cut it down from a larger region or how do I glue it together from a from a from a from smaller pieces. By the way, if we had to glue regions together. How would we just how would we, what's our glue. So, so removing regions cutting regions out that corresponds to a set difference. What is gluing them together, taping them together mean in terms of our set operations. Union, that's a union right if I want to regions together. I'm going to scotch tape them together and that's going to be a union. Okay, so now a related question here so I'm going to zoom out just to be able to pick up the. So here's a related question related type of question. By the way, if you want to get a screenshot of this let me know and I could I could zoom in and focus in on it. We can do that afterwards. So here's a related question that gives us a bunch of information about sets. Okay, so here we have 60 students and let's see 41 like basketball hockey and like neither. And again, the reason that Ben diagrams are so useful is they're really good ways of organizing a whole bunch of information. So here, we have two sets of the universal set all the students surveyed the B set they like base basketball the eight set they like hockey so there's two sets be an age and then there's a universal set to draw a little mastercard diagram. Okay, so let's think about this so there's 60 students so that means that there's 60 students. So I would just mind zooming in a little bit the words are very tiny. Let me let me know actually do that. Yeah, so there's 60 students in this universal set. So there's 60 students in our universal set. So 41 students 41 students like basketball. Now now here's here's here's a strategy for thinking about this. I call this the $20 strategy. Before you write a definitive thing down so so yes scrap, scratch, scratch work is scratch work and nobody pays attention to it. But before you write something down are you willing to put $20 on the table to guarantee that you're correct. So here 41 students like basketball. So, am I willing to say there's 41 students in this region, are you willing to put $20 that says there's 41 students in this particular space of our Venn diagram. No, why not. I mean we know 41 students like basketball. So the thing is that when we write this 41 here, we are committing ourselves to there's 41 students who like basketball, and none of them are in this region I don't know if you can see that none of them are in this region between the two sets. But can I be a razor. Can we have a student there. Is that a student who likes basketball. No. Well they are in the set be aren't they. Yeah. So they are a student who likes basketball, but they're not that they're not in this group of 41. So the idea is we don't know that there's 41 students in just this region we don't know that there's 41 students just here that 41 students is going to be distributed. So 41 students is going to somehow be distributed among the group that is in this region. That's in this region, but also in this region in between the two sets. And similarly, there's 28 students who like hockey, and we know that those students are going to be in this set, or they in this part of the diagram or maybe they're going to be over here. They're going to be outside. What can you tell me about the students outside. They don't like either. They don't like either sport. Are you willing to put $20 on the table and tell me how many people are outside. Well you just told me they don't like either sport. We know there's eight right and and here, here we know there's eight students who like neither sport the only place that those eight students can be is they have to be outside. So they cannot be in the intersection. They cannot be in the intersection because if they're in the intersection, they either like hockey, or they like basketball, or they like both. Okay, so we know. So we can't really say yet how many students we have, how many students we have in each of these circles but we do know those eight students who are entirely outside the circles. So we know there's eight students at the outside. What does that tell you about the students inside the two circles together. So of the students that are in here that they like either basketball or hockey, and how many of them are there. So you have to add up the total and then minus eight right. Well, it's actually easier than that. What do we know there's 60 students altogether. 60 minus eight is 60 minus eight. There's 60 minus eight there's 52 inside our two circles. I get it. It makes sense. So now, well we know there's 41 inside this circle. We know there's 28 inside that circle. So here's the part we couldn't figure out how many are in the middle. Let's look at those numbers 41 and 28. We know there's 52 that are inside the two circles together. So we know there's 41 inside this circle. There's 28 inside that circle. What does that tell you about the group in the middle. So here's a useful strategy for for for approaching many problems in life and also mathematics. Take an extreme position and see where it takes you. So in this case, if there's no one in the middle. There's only in this middle region within what happens when we have 41 people 41 students in the one set 28 and the other. So that's 41 plus 28 and that's 69. If there's no one in the middle, then 41 students are in that first circle 28 students are in the second, and then we have to have 69 students altogether. Now, what does that tell you 17 students are in the middle. We got to have 17 in the middle so there's 17 in the middle. That's the only way that we can basically I mean you can look at it a couple of different ways. One way you can look at it is these 17 students are counted twice. They're counted in the 41 students who like basketball. They're also counted in the 28 students that like hockey. And because they're counted twice, our numbers add up to too much, but the way we can reconcile that is that we drop out those 17 that are that are in the middle. Let's let's go ahead and redraw that diagram a little bit. I'm sorry how did you get 17. So, so if there is nobody in the middle, then we had to have a total of 69 students altogether, because there were 41 who like basketball. There's 28 who liked hockey. And so if there is nobody in the middle, then that for that group of 41 students is off here to one side, the 28 students is off here to the other side and there's nobody in the middle. So this MasterCard diagram would represent 69 students altogether. Now what we know, though, is that there can only be 52 students inside. And so that means there has to be some number of students in the middle, and where that 17 came from, which is because we knew there's 52 inside, this is 17 too many. So that's where that 17 came from. Right, so we know there's 17 students here. And let's go ahead and piece the rest of it together. I know there's 41 students inside this left circle. I know where 17 of them are. So that means there has to be 41 minus 17, what 2424 have to be here. There's 17 students in the middle. I know that in the entire circle, there has to be 28 altogether. So there has to be 28 minus 17, there has to be 11 in that region. And so now, once we have the picture we can answer all of our questions how many students liked exactly one of them. Well, we know that 24 liked basketball and didn't weren't hockey fans. We know that 11 liked hockey and were not basketball fans. And so, so we know that 24 plus 11 so that's going to be what 35 liked exactly one sport. And then again here, once we have the diagram this is the number that are not in the middle that are not in the intersect age. So, if we know there's 17 students in the middle, if we know there's 60 altogether, then outside has to be the remainder. That'll be. So we have 17 the middle so we have 60 minus 17 outside of the intersect. So that says the cartonality of be intersect age vacation. It's gonna be 17. Okay, questions on that one. I think you mean 60 minus 17 right. Yeah, 60, yes, sorry, yeah, sorry, 60 minus 1743. Thank you, I think this is clear. What's that. I said this is clear. Thank you. Okay. Okay, so let's take a look at what's another one here. Okay. All right, so base conversions let's talk a little bit about base conversions. Okay, so. Okay, so I'll make one modification on this one so I can tell you right out that that question there is not a type of question we've we've we were asking the question in a different format now. But, well let's say so so this one is. Okay, so let's, so let's think about how these numbers actually let me pull, let me pull a question from. We get a better question for this one. The base conversion questions are probably one of the ones that have the biggest difference between what's the printed sample exams and what the, what the final will look like so Okay, let me pull some, let me pull some somewhat more representative questions if the. Okay, so, so, so here's a couple of a couple of somewhat more representative questions and then I'll and then I'll do the, the other question, but this is sort of to set the stage for this. If we, if we want to convert so remember the idea behind writing a number in any base is that what we're going to do is we're going to take our number, and we're going to group by whatever our base is. So if I'm working in base three, for example, then I'm going to form sets of three, and every time I get a set of three, it's a unit. So one of the big ideas here is that a unit is, sorry, one of the big ideas here is that arithmetic is bookkeeping you're keeping track of how many of which units. So if I want to express a number in base three. So let's take a look at this here's 12312. So I can look at this. So a unit is either an individual object, or it's a set of, in this case, three objects that I'm going to treat as a single thing. So here's a set of three. And so I'm going to record what I have. So I have one of these things. And two leftovers. So, so you can think about this as an intermediate stage for writing units. I have one of these, and two of these. And the way that I get my final notation is I leave out all of the, all of that information. So just to record, I have one of one type of unit, I have two of the next type of unit, and then because it's in base three I'll actually spell it out that way. So there's my there's my answer in base three. If I have more objects, the bigger. So, so let's take more objects. So again, I am working in base three. So I'm going to collect sets of three. So here's a set of three. Here's a set of three. This morning I'll add in. Here's another set of three. Here's another set of three. And then there's one thing left over. Okay, so now we might think about this as 12345 we might think about this as five of these. And one of these. So that's the way we might start. But the thing to recognize is that the, the finding sets of three extends to any set of three we can find. And so the thing you should notice here is here. We have another set of three. That's a set of three of these things. Wait, how do you have a remainder. Well, it's it's whatever's left over so so here we have here we have a bunch of things and so I formed my set of three but then I had two things I couldn't I didn't have a set of three left over here. So I just had two of these things left over. So so so the idea is that is that is that we're getting you're starting with a bunch of objects however many objects you have. So the idea is you form sets of you form your sets, however you're going to form them. Again, because we're working in base three we're forming sets of three, and we may run out we may we may, we might have exactly the right number to form enough sets, or there may be one or two things that a couple of things left over that we can't fit into a full set. So that's the basic idea we just we just start by forming our sets and then figuring out how many of those sets we have. Oh, and I apologize. I don't know if you saw but someone in the group just said that they're trying to get into the review but they can't. Okay. Okay, so I Okay. Yeah, I think she I think she just got in. Yeah, I did I did see her try to try to join in so I think I added her. So, so here's the important thing we do have this set of three of these objects. So again arithmetic is bookkeeping. We have one of these things and I'm just going to draw it that way. We have so that takes care of all of these that takes care of all these. We have two of these things. Those two, and one last leftover, and that's this last one over here. So again, our first step is we indicate we can draw out our units we don't have to draw out our units but first few times you do this it's useful to do that. So we have so now we can record how many of which units is one of the big things to the medium. One of the small. And again this is base three. And when we're writing numbers this is all we're really doing. So when we work in base 10 are sets that we're forming our sets of 10 so I take 10 of something and that gets me a single thing. Okay, so that's writing numbers. Now let's talk a little bit about doing arithmetic. So actually so so here's something else we might do so once I have a number written in a particular base, I can go back to a base 10 you could think about these as so to speak ordinary numbers. So here, this number one, one, four, four, and this is base five. So, so in base five. I'll move this down. So if I want to figure out 1144 in base five. So my units in base five. So the units in base five the smallest is always going to be a one. So the smallest unit this is always a one, because it's base five, then I, my next size unit has five. So this is a five. How big is the next size unit. Well it's five of these. So how much is five fives 2525. And so our number here we have our smallest unit, one size up one size bigger, we have to go one size bigger. What's our next size unit. That's a that's an even bigger one that I'm not going to try to draw. And again, it's five of, it's five of these. So every unit is five of the preceding unit. So five fives that gets 25. 525 how big is that how much is five quarters one 25 it's $25 right so so these are the size of the units. So again, arithmetic is bookkeeping. When you write a number like this, we're saying we have one big one. That's this one. We have one large. We have four of these. And then we have four of our smallest units. So, well think about this is making change base five is kind of convenient for us because we actually have coins penny nickel quarter, we don't have a dollar 25 coin but but that's that would be the extension to the next place up. And so how much do we have well this is 125. We have 25. We have four nickels and four pennies. So that's 174. So there's our number 174. The, the, the idea you want to keep coming back to is arithmetic is bookkeeping we're keeping track of how many of which units that we have. And the only real difference here is that our base tells us how big each of the units are relative to the next. So again, whatever our base is each unit is that many times larger than the unit before it. Okay, so let's extend that to doing arithmetic. But the answer would be 175 base 10 right 174174 base 10 yes. Yeah, yes, yes, so this is the so this would be our final answer. If I wanted to convert this into a number in base 10. Yeah, this is what I would want to do. Just as a quick note, there's not any good reason you want to convert this to a number in base 10. What you really want to do and that actually kind of goes to this type of question. And let me pull up. Let me pull an addition question as well. What you really want to do is is you want to be able to work with the with the amount in in whatever base they have. So, we pull an addition question first. And the least efficient way of trying to do many of these problems is converting back into base into into base 10 and then doing the arithmetic in the usual way. Because part of the part of the problems you do want to get the final answer in terms of whatever the base is. So, here's a, here's a useful way we might represent the units. Again, this is base five. So we have our smallest unit. We have five of these and one way we often represent this is we represent as a rectangular block it doesn't really matter what we represented at, but the idea is that five of these make up one of those. And then oftentimes we represent as you saw I draw the we we might represent a bigger unit this way so this is five of these is one of those. So again because we're working base five, five of any unit is the next larger unit. So, this 302 base five, again visually you can think about it as we have three of these, none of the rectangles and two of the little squares, and then 201 base five, we have two squares, one of these. And so if I add them together. So if I add them together. What happens well I have three of the squares, and then I have five. I'll do this in two ways by the way. So, so if I add them together here's where they end up with, and the most important thing to recognize here is again because we're working in base five, anytime you have a set of five objects. And treat that as a single thing that term the term that we use for this is we can bundle and trade. We're going to bundle this set of five and trade it for the next bigger thing I'll represent that way. So this set of five we can bundle that up we trade it and we get the next bigger thing. And so again arithmetic is bookkeeping. We have one big thing we have none of these. We have none of these, and we have 123 we have three of these. And so our final answer 1003 base five. So, so so again arithmetic is bookkeeping we just want to keep track of how many of which units. Now, you don't have to draw everything out. So there's a couple ways you might organize this one way you could organize this is sometimes called a place value chart. And the important idea here is that we're going to separate out the different amounts. And again, you but if you want to you can think about these are small, these are medium, these are large. And so how many of the small so I have a three of the smalls. And again that's what we got there. I have one of the mediums, and I have five of the large. But now, because I have five, because I have five of the large what I can do is I can bundle these up and trade them for one more of the combo size. So I can bundle the five here and trade them for one more of the jumbo size. And so that's where I get my final answer 1003 again in base five. This is a yes so so again either way you want to approach this problem. I like the visual the first couple of times that that I do this just because there are some tricky features about doing this. The most common mistake, by the way, if you do this, the most common mistake is bundling when you trade remember that what you get ends up in the next column over. So the most common mistake here is bundling this five into one, but then leaving the one in the in the column that you started with. So if you trade it it's gone and then it's one more the next place over. So, so so questions questions on questions on that one on the on the addition there. Okay, now subtraction subtraction is is follow something very similar. And again we could think about subtraction as removal. So, again, again I'll do this in two ways we can. We're working in base four. So our units are one. We have four of these. And four of these. So what I have is I have 2214 I want to remove 234. So I have two big to medium one small. And I want to remove to medium and three small. Okay, so I can see that I can remove to medium because I do have to medium to remove. I want to remove three small, but I only have one small that I can remove. So what was what can we do. Well we can break one of these up remember we have our exchange rate. If exchange rate one of these one big is one big is for medium, and one medium is for small. So what I can do is I can the term is on bundle. You might have you might use the word borrow borrow is a terrible term borrow implies that you're going to get it back. You're never going to get it back. The idea is I'm going to unbundles I'm going to take this one apart I'm going to take this apart it's gone. It's it's no longer there it's never going to come back so we were not going to call it borrowing. But I can trade it for four. One, two, three, four of the smaller pieces. And now I have plenty of these medium I can remove I still want to get some of the smalls I can remove. So again I'm going to unbundle. I'm going to take one of these I'll take this one and I'll break it apart and my exchange rate says that I can break that into four of the smalls. So I can break that in so if I get rid of it if I trade it out. I get this amount. Now this is still two to one base four. The only difference here is that instead of having two of these and two of the medium and one of the small. I've basically made change I've taken my $10 bill and I've broken it up into two fives. Do I still have the same amount is still $10, but it's two fives instead of 110. So the amount I have here hasn't changed at all. So I still have the same amount. And so now I want to remove two and three. So at that point I can and maybe I'll switch colors here just to see if I can see that. So I'll remove one, two, and one, two, three. So there's my removal and arithmetic is bookkeeping. So what do we have left. We have one big. We have 123 medium. And we have one, two, we have two small. And so the amount is one, three, two base four. So again, here's the here's the those of you who've taken up one of the education pedagogy courses might might refer to this this is a this is how we do this with tangibles. And these are quesadilla quesadilla rods, I think are the things that these are. And and again, what we're really doing is we're just saying I could take this big thing and I could break it apart. I can trade it for some number of the smaller pieces, and then I can remove things as I need to. And so I can get my computation. This is 132 base four. If you want to do that using a place value chart. So again, we have our two to one, we want to remove two, three. So what we did with our play what we did here is we took one of these, and we broke it apart. We still have one left. But now we have four in the for more of the mediums. We took one of our mediums and broke it apart and we got for in the next place over. And so now, again, arithmetic is bookkeeping. We have one, we didn't remove any. We had four, five, we removed two. We had five. We remove three. And so we get our final answer same answers before 132 base four. Again, you can either write it out tangibly using you or you can you basically use the manipulative use the quiz in your rods to to to perform that subtraction, or we can use the place value chart. Again, I can again if you again the the drawing I think the drawing helps the first couple of times to do this it does take a little bit of time to get used to the place value chart. And again, very the one thing to be careful with the place value chart is arithmetic is bookkeeping. The most important thing to remember here is that when you trade things you still have, you still probably have things left in the place that you traded from. So here we started with to in this largest place, we broke one of them apart, but we still had one left. So we had one left, and we got for more in the next place over. Okay, so questions on the subtraction. No, thank you. Okay, so now, let me zoom out to this one. And okay, so now. Okay, so now let's take a look at this without good. So let's take a look at the first question. So again, here's a, here's the, again, that problem. Don't worry about that one. But again, here's a subtraction the difference years are using base 12. But here. Okay, so let's think about the next two numbers in the sequence. So one way you might look at this again there's there's two ways you could approach this. This one I would think is almost easier if you actually draw the pictures out. So one one base for that's one. Not that. That's one medium. Small 22 base for that's two mediums, and two small 110 that's one large, one medium, not and no small. And then let's see to zero three base for that's too big. All right, so let's think about this. Let's think about a rule that converts that takes us through each of these steps. Okay, so what is one possibility. If I have this amount. And I do something to get that about. What's one thing I could have done to go from one to the next. You added one small one medium. I could have added one small and one medium. Are you able to group them as well. Yeah, I am able to group the most. That's interesting I I was just thinking like what about grouping the medium size. Well, we can't quite group the medium size because remember this is base four so we don't have quite enough to do that. I think there may be a typo in that problem. I think about that. I think there's actually a typo on that problem because I don't know what I have no idea what that is supposed to get to let me let me pull up a let me pull up a different version of that one because that one seems to actually be wrong. It seems to be unanswerable within the context of what you've done. Here's a better one. Let's do a better one. Problem is that sequence is not an arithmetic sequence is not a geometric sequence is not any sequence that that we would have done. Okay, so here's so here we actually have an arithmetic sequence. We have an arithmetic sequence we form by adding the same thing every time. So our first two terms 325 and 445. So again, if you draw it out. You might say okay well if I draw it out then I can see what's been added. Here. What did we add to go from the first one to the next one. So one medium size and two small. One medium and two small. So again, we can see that we have three and two. We have one more medium we have two more small. And because it's an arithmetic sequence, then the next term. We're going to add again one medium. To small. Now what I'll do is I'll draw these out first and then we'll figure out what they are going to be so we have four. Four. And then I'm going to add one more. And two more. Okay, so that's what I would get. Now, if I wanted to write this as a number. That we're working in in this case we're working base five. So every time I see a set of five, I can group them as a single thing. So it's easiest if we start grouping with the smallest sets. So here, I have this set of five here. That's a set of five. So again, we're bundling and trading them. So when we do that this is now gone. And now I have another medium. And just so it doesn't get lost, we do have this one leftover right there so don't forget about that one. Do I have any more sets of five. Yes, you have the five medium. Yeah, I have the five mediums one two three four five. So I can group all five of these mediums. And again, when I group them, they're gone and they've been replaced with a large. And so arithmetic is bookkeeping. This is one large, one medium. And again, don't forget we I drew this badly, but we have that one leftover piece in there the 111 base five. So there's our next term is going to be 111 base five. And then I want the term after that. So again, I'm going to add again, it's an arithmetic sequence. So I'm going to keep adding the same amount every time one large to small. And so what does that get me. Again, what I start with is this, I'm going to add one large to small, and I don't have fives that I can group. So this is just going to be one of these one big to medium three small. And there's my next term. And I can keep going from there. I can I can keep adding terms if I wanted to. All right. So questions on the arithmetic. Those are basically the type of base and the base base and arithmetic questions you should be able to do. Any anybody want to talk a little bit more about those or do we want to move on to the next next topic. Can we try using subtraction possible. Sure. Let's okay yeah let's do another subtraction problem how about. Let's see how about two one base five. Okay, so here's a so here's another subtraction. Two one three base five minus four four base five. So how do you want to do this to use place value chart. Well, we'll do both. So if I do this with a with the actual object so again remember this is two of these two big one medium. Three small, and they want to remove for these and one. So I don't have enough medium or small to be able to remove them. So I have to break them up. So again base five. So I'm going to break this one apart, and it gets me five. One, two, three of those. And let's see I need four smalls. So I'll break one of these apart. And that gets me five smalls 123. So that's really just prepping the table as it were for the things I need to get that stuff so I now actually am in a position where I could remove for medium for small. So that's going to be so remove 123 for the medium and 123 for the small. And arithmetic is bookkeeping. So what do I have left, I have one big, one medium. And three small, and that's 113 base five. Um, I have a question. Question that you did before with the arithmetic sequence. This one where professor taught it like a totally different way so I just wanted to see if we could go over it that way. It's not using pictures at all so it was like 44 minus 32 but there was no pictures being used we kind of did it like a regular math problem so it would be like. Yeah, yeah, so yeah, so I mean, yeah, so what you're really doing here is so the so what so again it's an arithmetic sequence. So remember that when you're doing an arithmetic sequence, the the you're always adding the same thing. So what are we adding to this to get this so if you translate this into a number and base five this is one of the medium to the small. Yeah, so what we're doing is we're always adding one to base five. So, yeah, so our arithmetic sequence. So it's 32 base five, and then we add one to base five, and that gets us our 44 base five. Okay, and then my next term I'm going to add one to base five. So remember that having that place value chart can actually be useful. So again, that six of these five of those, and I now can bundle and trade. So I have six in this, I'm going to take five of them that leaves me with one that leaves you with one and then there's going to be one more the next place. Now I have a set of five so I'm going to take five of these. And that leaves me with one more the next place. So there's my 111 base five. Thank you for going over it because we were learning that way. So, yeah, yeah, again, I mean, I mean, this is more efficient once you get used to it. But yeah, I like pictures. Yeah, I like drawing things. So, so I mean this also this will you know if you can if you can do it this way this is actually this will be more efficient. And, and certainly you do want to move towards towards being able to do it this way. The pictures help the first couple of times you do it so that they said that you don't have to the trickiest part about the pictures remembering that when you trade something. But it goes away and then you have, and then what you have ends up in the next column over. And so that's one of the, that's one of the things to be careful with if you're using the place value chart. Yeah. But again, certainly you want to get ideally you want to be able to move to not having to draw the picture every time because certainly once the numbers get large or once the number of places gets like you really don't want to have to draw these things out. Yeah. I'd like to do it on the exam, like with the place value chart because I feel like I. Yeah, that's fine. Yeah, that's fine. Yeah. Yeah. I'm nervous. I wouldn't have time to draw the pictures, but I'm confused. I'm wondering if you could explain it again as to why the, like the one goes stays and then moves over. Okay, so okay so okay so probably the easiest way of doing that is to so so remember this line came from back here so here we had four of these and for these. And then we added one, and we added two more. So if we just combine them. So if we just combine them. So there's our six. This is where that six came from. So and that was our we have four we have to we put them together we have six so that's where that six is coming from this five that's from our four and one so here's four there's one more we get a total of five. Five came from. Okay, so that's where we started with our five six here. Okay, so now the next thing we did is we bundled them and and because this is based five, we look for sets of five. And so here we have the set of five. Right. So we combine that. Now here's the important thing. When we combine that that gets us one of the medium sized pieces. If you draw them out, it really doesn't matter where it is. If you draw them out you can leave the medium sized piece in here it's somewhat disorganized but there's nothing you know it's still clear what we have. But the that where it gets tricky to transition to the table is this isn't this medium thing is not one of the small things. So it's actually it's it should be over in this next place. So, so let's take a look at what we did there. So we had six. We bundled five, and that got us one more in the next place over. That's where this one came from. But when we bundled the five, there was still one of the small units left over. And so that's why we still had a one in that last in that last place. Yes. And so now, so now we had so again we had the five that we had before we had the five that we had before and then the one more that we just created. And we said, Oh, well here's a set of five. I can bundle these together and to make one big. And if you're drawing them out it doesn't really matter where we draw these, but on the place value chart this these big pieces have to be in the next column over. So that was we took this five that we had, we bundled it and then we got one more. And that's where that's coming from. That's where that one there is coming from. And so we have one more in this in this third place over this one here that was from the one that we created. So so that was so that one is still there. And so that that's where we get our final answer one one one. I get it now. Thank you. Yeah, yeah, the tricky part about using the using the using the chart is remember always remember that when you bundle and trade what you bundle has to end up in the next column over. And the other thing is it's all bookkeeping so sometimes we are not able to bundle everything. We may have leftovers that stay behind because they don't get they don't get bundled into a single package. Thank you. Okay. All right, so yeah okay so there's our subtraction. Let's see. Okay. All right, so other other questions you want to go over on base and arithmetic. Yeah, is it possible to do like an arithmetic pattern similar to that question but with subtraction, instead of addition. You can there there there won't be a question where you have to do the subtraction but it would really it would really be the same thing. And at that point, yeah, the arithmetic the arithmetic sequence and the and the the arithmetic sequence is really a question of just repeatedly adding the same thing. Yes, I don't would say I don't worry don't worry about having to do a arithmetic sequence where the numbers decrease. Okay, thank you. I mean if you had to it would be the same thing but you but you won't have to do something like that. Okay, so let me get where we are way over here. All right, so let's take a look at some of these other questions. Okay, so now some of these questions are really based around the idea of when we when we when we do an arithmetic, when we do an arithmetic computation what we're actually doing is where there is a there is a significance to what we're doing. So, so if we want to so we have here we have seven and three quarters land for a for a building project. And so, okay, so let's see seven and three quarters. So, so we have two problems and part and the reason these are asked is they actually are very closely connected seven and three quarters. So as you can probably guess I'd like drawing pictures so let's think about so so here's how. So here's, here's the, here's what we might do here. So this is, I'm going to set aside one third of the land I want to figure out how much is left. Let's think about that I have a plot of land. So if I want to set aside one third of it. So what I'm going to do. Well, what does what does one third look like, how do I get one third if this is the land that we bought. How do you want to represent one third of that. What's the picture that you have in your mind for what one third looks like. In threes and she didn't want to cut it in three pieces. And I'm going to set aside one third and I really care about how much is left so I'm going to basically get rid of that piece. Okay, so the idea here is that I'm looking at at this plot of land. I've cut it into three equal pieces, and I get rid of this piece. Okay, so and then the question is I want to know how much is left. So, so here's a question. Again this question is asking how much is left. So how much of what I have, what I started with is left as a fraction, don't worry about the computation yet but just think about it in terms of how much we have. So if I got rid of one third, then what's the leftover part to third. It's going to be two thirds. So it's two thirds of seven and three quarters. And let's think about this as an arithmetic operation how do we translate two thirds of an amount. What arithmetic operation does that correspond to your metric. Well so arithmetic operation so add subtract multiply divide. When I say of, what do I mean. So I have five of three things. So how many do I have. So you subtract. Well is five of three. So if I have five of three. You multiply right so so usually when we talk about of that's a multiply I have five packs of three candy bars so I have five times 15 candy bars that's a multiply this is two thirds times seven and three quarters. And so we have this computation that will make whatever that numerical value is. So this first problem of in order to answer that the question we need to answer is what is two thirds times seven and three quarters. All right now let's take a look at the second question we have three quarters and be the berries filled two thirds of the jar. So, well I have my jar. What is two thirds look like two and a three shaded. Yeah, I'm going to take I'm going to again I'll do the same sort of thing I'll divide this up into three quarters. Divide this up into three pieces jar has been top and two of those two out of three are going to be shaded and there's my seven and three quarters. So seven and three quarters is going to be that amount. So let's think about that. So if I wanted to figure out the question is how much does the how much is the entire jar. So, give me a statement. So there are three terms, the jar, seven and three quarters, and two thirds. Give me a statement that uses these three terms in some combination plus with some other words, possibly that describes this situation here, seven and three quarters. Okay, start with seven and three quarters. What is that. What about the other two so now incorporate the other two terms. So you want to give me a sentence yeah. So so here we said, you know, what's left is two thirds of seven and three quarters that was our statement that represented the situation. Now we want a statement that represents this situation. And again the terms we want to use jar seven and three quarters, two thirds. What's the jar. So what's the relationship between seven and three quarters. So remember this is our seven and three quarters. What's our relationship to the jar. Can you ask the question again. So so what's what's our seven and three quarters. What is what is the relationship of seven and three quarters to the jar. How much is full. Well, it's, it doesn't fill up the jar but it's part of, I mean it's partly filling the jar. So give me give me a sentence that relates seven and three quarters and the jar and potentially even somehow involves this two thirds. What could you do seven and three fourths jar of two thirds. I'm trying jar. Well, so. So, so here's a jar. The idea is here's a jar. Here's a seven and three quarters. The verb is seven and three quarters the jar the seven and three quarters is the jar. Is that true. Isn't it part of the jar. It's part how much seven and three fourths no. Well that's that's that's the amount but it's you just told me it's part of the jar, how much of the jar is it. Look at your picture, or again so that so you know it's part of the jar. So you know it's part of the jar. It's two thirds of the jar. It's two thirds. I get it. Okay. A little more space there. So we have the picture. And if we if we if we know if we know that there are seven and three quarters well that's that fills up our jar to two thirds. Seven and three quarters is two thirds of the jar. And well there's my of again. So how do I translate that into mathematics to a mathematical statement. Well that seven and three quarters. Do you translate is what is the mathematical symbol we associate with is equal. That's it equals right yeah if I say five is two plus three say oh well five is equal to two plus three so we say is okay so seven to three quarters two thirds of well there's our multiplication, whatever our jar is. And I'll remember our question how much of the whole jar. So I have this statement seven and three quarters well that's two thirds of the jar. So if I wanted to know how much is in the jar. What do I have to do with that two thirds figure out what one third of the jar is. Well I went well I want the whole jar right I just want I want the jar I want the whole jar. So how do I get rid of this two thirds what do I need to do. Add one third. I can't add remember that's a times. So what happens so if I have a if I have an equation, five equals two times x, if I want to solve for x, what do I need to do with the two. You move it to the other side. How do you do that subtract it. Or not. Oh you multiply both sides by that. I don't know. Divide. I have to divide so this is a times. So if I want to get rid of a times I have to divide. So if I want to find the amount of the jar, because it's a times I need to divide by the amount. And that's going to give me the amount that's in the jar. And so if I want to find the amount in the jar, I have to divide by two thirds. So, so again there's a computational process that we use to compute what these are. But the more important idea here is that is that if I want to if I want to find these amounts I either need to multiply or divide and that's really going to depend on the information that I have. And let's see. Some things to keep in mind here. Probably the probably the most useful thing to keep in mind here is that if you tell if you have the pictures if you have the pictures that represent the amounts that you have and what their relationship is. And then what you need to do becomes a lot easier to becomes a lot easier to figure out, because it allows us we can go from the picture to writing a statement. And often it's easier for us to translate that statement into a mathematical expression, then it is to go from the go from the question itself. And then into a mathematical statement. So the, the idea here is that is that having those pictures are really is a really useful stepping stone to figuring out what needs to be, what needs to be done to solve a particular problem. Appreciate it as just you know we've we really never worked on this before like this so it's like you know, a little bit of, you know, a lot going on. Okay, let me touch. Let's take a look at another one. So here's. So part of the idea here is the other idea that you really want to think about is all of our basic arithmetic operations, they correspond to doing something they correspond to some sort of physical action you can take. So if I'm adding, if I'm adding a plus be three plus four, what I'm doing is I'm taking two piles, and I'm running them together. Right, so there's addition if I'm subtracting for minus one what I'm doing is I'm taking something, and then I'm going to remove an amount, and I have something left. What happens when I multiply, what am I doing when I multiply you have to turn it into an improper fraction. Well, that's that's what we're that's the that's the, that's the algorithm but what is the quite the important question here is, what is what picture. Do we have when we multiply things when I say three times to what type of thing am I looking at what am I doing. You're multiplying two three times. Adding I'm taking yeah I'm taking two. But I'm taking three sets of two. So there's my well actually that really is my three times to how about division. What am I doing when I'm taking 12 divided by four. What I find 12 divided by four what am I doing. Well, here's 12 let me draw 12. So there's 12. What am I doing when I find 12 divided by four. There's many times. I'm there's actually there's actually two things we do. Part of the recent division is complicated as there's two ways we can look at any division. Addition subtraction multiplication there really aren't too many variations, but one of the ways I can look at division is if I want to find 12 divided by four. What I'm going to do is I'm going to split off sets of four. So there's a four. 123, there's a four. And so I have 123. There's my three. So one of the things we do when we divide is we say okay, I'm going to take pieces that are this size I'm going to take pieces of size four, and I'm going to figure out how many sets of four we can make. Now the other way we can also look at this. So, again, part of the reason that division is the most difficult of the arithmetic operations is that there's two ways we can look at a division. And the way we usually introduce division is completely different from the way we usually do division. This, by the way, this is how we do division, right, because what do you do you say well 12 divided by four, how many times, how many fours can I make from a 12. And so I can make 123. At this point I usually make the observation are symbols for the arithmetic operations are no accident. There's a symbol for times. There's a symbol for plus. What do they look alike part. Well, why do they look they look alike because we decided to make them look alike but the reason that they, they, the reason that this particular notation is common is that when you write times this way. It's a reminder that every multiplication is really just a repeated addition. I take a bunch of twos, and I add them together. And that's why we have our symbol for plus and a symbol for times, looking alike. Here's a symbol for subtraction. Here's a symbol for division. Here's our other symbol for division. There's a reason these symbols look alike. When we divide, what do we often do we say well how many times can I subtract this divisor from the amount that I have. And so here I can subtract for 123 times, and there's my, and there's my quotient. Now, we do introduce we do talk about division in another way. What's another way that we can look at 12 divided by four as a fraction. Well, as so so what does it look like in our picture. So how do I show 12 divided by four in our picture divide 12 four times and then you get three. So so when you say divide divide 12 four times what do you mean so so again or so so I so I have a thing right I have. So that's a 12. Okay, call that a 12. So how should I show how can I show 12 divided by four, create four sections. I create the four sections I break this I go one and maybe I do something like that and so there's 12 divided by four. My whole thing is 12. I have divided it into four pieces. This is in fact usually how we introduce division we say okay I have a pie I have a cake, and I have five people. I don't want to find I have six people coming to the party and I want to serve everybody a slice of cake. So I'm going to divide the cake. I'm going to divide the cake into six pieces. And so maybe I my division looks something like that and so I divide the cake into 123456 pieces and this is how we usually talk about division. So here we divide into four pieces and each piece is three is three units. So, so we can see how we can think about division in two different ways, either the divisor tells us how big each piece is, and the quotient is the number of pieces, or the divisor tells us how many pieces, and the quotient is how big each piece is. And again we usually introduced division this way, we usually introduced division by saying well here I've divided this into four, I've divided it into six I've divided it into three or whatever. We usually introduced division by talking about dividing something into equal pieces. The problem is that when we actually compute divisions. This is what we do. When we actually compute divisions this is what we do we, we pull out pieces of size whatever the divisor is. So let's think about this problem. You have 55 eggs. So if I have 55 eggs and each cake requires eight eggs. Well let's think about that how do you figure out how would you figure out how many. How would you figure out how many cakes you could make. I have 55 divided by eight or vice versa. Well it's one or the other. Right. So is it 55 divided by eight, or is it eight divided by 55. Well what's the significance of eight. So I have a whole bunch of eggs. I'm not going to try to draw 55 but eight to the divisor eight to the divisor because each cake requires eight eggs. So I have this whole pile of eggs and I want to make a cake so what do I do well I peel off eight eggs, and that's one cake. And then I peel off another eight eggs, and that's another cake. And so what am I doing. Well I'm, I'm doing this. I'm saying I have a whole bunch of things, and I'm going to sort them I'm going to I'm going to group sets of eight, and each eight is going to make one cake. And so that's going to tell me how many cakes I could make it's going to be 55 divided by eight. Now, we can write that as a division but just as a note. We don't have to do the division, because what are we really doing I have 55 eggs so I bake a cake what does that do well that takes up eight eggs. Can I make another cake. Do I have enough eggs left to make a cake. Yes. Yeah sure I'll make another cake. I'll have leftovers yeah and the and the idea is that the idea is that I'll keep going until I until I don't have until I can't make another cake right so I can I can do this division so I'm really doing this division, but you can do that by just subtracting eight over and over again. And eventually we get down to what happens I think we have eventually we get down to a point where we have a number of eggs we peel off eight eggs and we have seven left over and that's not enough. So that's not enough for a cake. Let me just let me just do that really quickly so here 55 minus eight that leaves me 47 I bake another cake. I bake another cake. I bake another cake and another and another and that's where my seven is how many cakes did I make. Let's count them. Here's one, two, three, four, five, six, so I made six cakes, and I had seven eggs left over at the end. So for this answer would you just have to write the expression itself and not solve it. The question is I mean I mean as the question is worded as the question is worded the the answer is 55 divided by eight. The question may actually say something like you know like you know how many cakes does, you know, how many cakes does is she able to make. So let's go ahead and finish that off there so what six cakes. What is that seven tell you what's left over what's another word for what's left over remainder that's our remainder. There's a reason that we use the words that we do for division. So, and the symbols again the idea here is that when we do our division, we're really doing a repeated subtraction. And so that's why our symbol for division looks a lot like a subtraction symbol, because this is really saying if you want to do this the basic way you can do this. Again, you don't really get this is this is slow, but but this is really this is really what we are doing. We're subtracting eight, as many times as we can, and it turns out we could subtract eight, six times. And then we had some things left over. And so what we do this division 55 divided by eight. Well it's six, and we have this remainder, what's left over is going to be seven. We're going to have to use, like what would be a question they that would be asked that we would need to write a remainder of seven, or not really just information to know. Well, I mean it might be I mean the question might just be to let me think how, how many cakes can Amy bake and, and how many eggs are having eggs that you have remaining or something like that, or you might even just say perform the computation and express your answer as a quotient and remainder. That would probably be the most directly of saying it so express your so so so the the computation we need is this one, and as a quotient and remainder, our quotient is six our remainder is seven. So would you write six remainder seven as the answer. Yeah, if the question if the question were to actually compute the amount. So again the way that this question is worded the expression is really 55 divided by eight is the expression that that you want to evaluate. If you actually wanted to find out how many cakes you would need to go ahead and do that division. Yeah. Hey, and again it depends on it depends on how that how the question how the question is specifically worded. Okay, we're not going to lose points like if we write remainder obviously not six like God forbid we don't write remainder or something we might lose points I feel like. Yeah, well, again it depends on how the question is worded so if the question does ask you to perform to find the quotient and to do the division and give your answer in terms of a quotient and remainder. So you actually need to write out, you know, your quotient is six or remainder is seven. Just as just as a note. Yeah, yeah so just to make one more connection here. So the other way we can do this. So, so the other connection that you want to make here. So, so this is six with remainder seven. And many of you want to also express this as a fraction. So here's an important idea here. So if I'm expressing this as a fraction the idea that is that's important here is that when I do a division. Because a frag a division is equivalent to not that a division is equivalent to a fraction where my dividend is the numerator and my divisor is the denominator. So this 55 divided by eight I can express that as a fraction 55 eights. By the way, how you speak influences how you think always read fractions as numerator denominators. So, don't read this as 55 over eight. The reason that you don't want to read it as 55 over eight is you lose information. This is 55 eights. And if you say 55 eights. The reason that it's better to say that is this tells you you have a whole bunch of these things. So this is the arithmetic is bookkeeping how much of how many units. When I write this as a fraction, when I read this as 55 eights that tells me I have a whole bunch of eights. And I have a whole bunch of these things and part of the reason that's useful is, what's that equal to one. Why is one full piece right. Because read what it says this is eight. So that also be eight eighths. It's eight eighths right it's eight eights. And the idea is that if I again if you think about what our eighths look like and I can actually draw those. So eight eighths equals one whole right eight eight equals one whole because the idea at eight is a piece that if I take eight of them. I get one whole. And so eight eight that's going to be one. And so that's going to be true in general. Wow that's interesting I've never learned this part like this like visualizing it wise. Yeah the visual the visual the visualization of this is really is really is really very very important. How about 16 eights. Well again I have a whole bunch of these eights. Well what do I know about the eights. Well I know eight of them make one. This is eight eights. And I have 16 I have eight more eights. So this is two. So so when we when we write fractions again the thing to remember is that the numerator tells you how many of whatever the pieces are. And so if you know how many in this case, if you know how many make up one. Then when you have an improper fraction, you could say okay I can split off the amounts that I have. So if I have something like 13 eights. Well that's eight and five. So if I have something like 13 eights that's well 13 well that's eight and five. And these are eights. And I know that eight eights. Well that's one and five. And I can write that as a mixed number one and five eights. So there's your transition between improper fractions and mixed numbers. And it all goes back to the same basic idea that when we, when we do arithmetic we are keeping track of how much of what units and all of arithmetic goes back to this bookkeeping. We're keeping track of how many of which units and sometimes we can trade them. So here, eight eights we can trade eight eights for one. And you know if we don't have enough we can't trade them because you know we don't want to shortchange somebody. And so we have to express that as a fraction. So just to kind of complete the complete that idea here. This goes into well remember what this said is we were able to get six full eights. Sorry. Yeah, six, six full sets of eight so that's six. And that remainder seven. Well, if you think about that that seven eights is what we have left. So this improper fraction becomes the mixed number six and seven eights. And that mixed number the easiest way of getting that mixed number is actually going to be from the quotient of remainder form of the division. Mr Suzuki you'll be uploading this recording. Yeah, yeah. I see there's something else that would be good. We'd be like having a problem that like involves LCM and GCF. Potentially. Yeah, yeah, yeah, I could pull one of those. Let's see if I can find a good one for that. And also, Mr Suzuki, we need I think most of the class needs help also with the Gauss method if possible. Okay. Professor, if I could also request. I have problems with the repeating. Excuse me, the repeating decimal ones where it's like, for example, like point seven to nine to nine to nine to nine. And then you have to like convert that. Okay, yeah, while you were pulling questions to be easier to tell you now. Okay, yeah, yeah, I could do that when that one's actually pretty easy for me to this problem right now. Okay, I'll get over that in a second. I just want to pull a couple of other questions out of here that. All right. Okay, I seem to have lost that. Okay. All right, so, all right, so I'll just make up and make one up. Okay, so let's take a look at this one. So, Okay, so, so develop recreational league, blah, blah, blah. Okay, so we want to make sure that everyone's on the team, the teams of the same size, and each team has the same number of males and the same number of females. So let's say we have 42 men. Okay, so, so if we want to make so if. So let's think about that so so we have our requirements of each team has the same number of men. So if so so let's start off with this if each team has the same number of men, then how many men could be on a team. 21. Okay, 21. Any other numbers. Six or eight. Six, sure. Or seven or seven. Any others. To choose three. And there's two that people always forget. We could put all 42 men on one team, and then you could have one man on each team. 14. Okay, are there any other possibilities. I don't think so those are the. Yeah, so again you may want to think about that prime factorization 42. Six times seven seven is prime. And so to three. So the ways that the divisors of 42 things that are formed by multiplying two three and seven. And these are the only possibility. There's also 14. 14. Yes, good. Good catch. So let's take each of these and steps if I put 42 men on a team. How many teams would there be one team. If I put 21 people on a team, how many teams would there be two and 14 would be three, six, seven, two. Notice that that's just these numbers in reverse order. So so so I could so bit so I can make this number of teams. So with 60 women. So how many teams could there be with 60 women. Okay, that could be 15. What are the possibilities. Two teams. Three. 1010. There's five 12. 15. Yeah, okay. Yeah, so I hear somebody thinking about this in in probably the easy way of thinking about this once you know one of the divisors, then you get the other one so let's let's let's try and let's try and listen out so I could have one or 60. I could have two or 30. I could have three or 20. I could have four or 15. Five or 12, six, or 10. I need that that's kind of this nice information. Okay, so I can so so if I if the if the teams have the same number of men, I can get any number of tea, I can get any of these numbers of teams I can get any of these numbers of teams. So, so here's a question. We want to make sure that the teams are the same size and we want to make sure that everybody gets on a team so let's think about that. If we make 10 teams. So let's think about this if we make 10 teams. So they're 60 women. So with 10 teams that's going to be six women on each team and again that fits our requirement. So there's one team and there's the same number of women on each team. What about 42 men. If I make 10 teams what happens. There's people left over. There's people left over. I can't, I can't, I can't make 10 teams and distribute the 42 men equally among all those teams. If I want to make sure everybody plays, then some of the teams have more men than others. If I want to make sure that all the teams have the same number of men, then some men don't get to play. So I can't make 10 teams. So, so what seems to be the requirement here then if I want to make sure everybody plays, I want to make sure that all the teams have the same number of men and also have the same number of women. So what, what am I looking for. I'm looking for a common actually a common divisor. I'm looking for a number that's in both lists. 10. I can divide 60 women among 10 teams, but I can't make 10 teams with the 42 men. I can make seven teams with the 42 men, but seven isn't on this list, which means if I try to do seven teams with the 60 women again we've run into the same situation. Either some of them don't get to play, or some teams have more women than other teams. So, so let's break this into two pieces here. We can. So let's think about that. So these are the number of teams I can make that will meet the requirements all men play they're the same number of men on each team. These are the number of teams I can make with the requirement that all women play and they're the same number of women on each team. If a number is on both. So one is on both. If I make one team, all men play, there's the same number of men on each team. And because it's on this list, all women play, and there's an equal number on each team. Okay, can I make more teams. Is there another situation where the requirements are met. If I make two teams, can with two teams, will all women play, and will they will all teams have the same number of women. Well, it's not the list of possible teams. Well, if I have two teams, then it is one of our possibilities with two teams, all women can play. All teams have the same number of women. If I make two teams, will all the men be able to play and have all teams have the same number of men. Yeah, yeah, because two is on this list as well. So I could do two teams. Can I do another number of teams. Can I do three teams. Yeah, because it's on the list. Yeah, three is on well three is on both lists and that's the important thing three is on both lists. So I know that I could get all women to play and all teams have the same number of women. It's on this list. So I know that I can get all men to play and all teams have the same number of men. How about four teams. Can I make four teams. No, no, it's on this list. So I could get all the women to play and have all teams have the same number of women. But it's not on this list. If I try to make four teams, either some men won't be able to play, or some teams will have different numbers of men. What's the next larger team I could make next, next larger number of teams I can make six, six, four. And is there any larger number. No, no. So, in fact, and there's nothing, there's no larger number. So six is largest. So if there's six teams. So let's now tie everything together. If there's six teams. So how many women aren't each team. All together. So there's going to be 10 and each women. And how many men there's 42 men, seven, seven men. And so an answer to so the first question is the largest number of teams we can make at most six teams that meet those requirements. And so they're 17 people each. So Suzuki, these are multiples, considered multiples, right? These are actually divisors. So where we get these numbers from these numbers here, these numbers are all divisors of 42. These are all divisors of 60. And what is six. It's a divisor of both. Right. It's a common divisor. And so, so one, two and three, one, two, three and six are all divisors. Sorry, one, two, three and six are all common divisors. What's special about six? It's the largest one. It's the largest of the common divisors. This is actually the greatest common divisor. Okay, this makes sense. This is this is the idea behind the GCD. I have. So the idea in general is, I have a bunch of divisors. I have a bunch of divisors. And there are situations where I want to look at the, I would look at the divisor, the largest divisor that shows up in both lists of in both sets. So I want to find the greatest common divisor. Okay. Now, the, so let's see. Okay, so, so there's, so there's our greatest common divisor. Now, the difficult thing here is that when we, when we look for the greatest common divisor of two numbers. This is actually, well, this is an easy problem, but not the way we usually do it. So if I pick two numbers, finding the greatest common divisor, frequently we talk about factoring factoring is actually the hardest easy problem in mathematics. It's easy in the sense of it's easy to describe how to what we're trying to do we're trying to find two numbers that multiply to a number. It's hard because it's actually very hard to do. So if I try to factor something like that. That's actually a difficult problem. But, but we can find the greatest common divisor using something called the Euclidean algorithm. And I'll say a few things about this, this, this, if you haven't, this one, this Euclidean algorithm takes a little bit more, a little bit more time to go into but, but here's a quick version of the Euclidean algorithm. If I have a number that divides both A and B, it also divides A plus or minus B. And what this means is that, so here's the short version of the Euclidean algorithm. And the idea here is you can find candidates for the greatest common divisor this way. So here's an example if I want to find the greatest common divisor of 75 and 85. The thought process here is whatever it is, it must divide the sum and difference. Now in practical terms, we want to look at the difference because that gives us a smaller number to deal with it has to divide. So what are things that divide 10 to and five to five and don't don't forget one as well to five and then 10 itself. So so so here is your so here is the here is the here is where we can. Here is where we can use the Euclidean algorithm effectively. Whatever the greatest common divisor is it must be one of these numbers. So what's the largest of these numbers that divide both of these numbers. And there's your GCD. So let me let me let me write up one that's going to come up with a number here. So for example, let's take a look at something like this one. You do not want to try and factor these numbers and you can factor them eventually but it'll take a lot of effort. But whatever it is must divide. So whatever the greatest common divisor is it must divide 510. So what we could do is we could go and say, well, things that divide. So we could find a list of things that divide 510. And there and we could find a list without too much difficulty. But so there's a strategy in life and in mathematics that I call rather Vince repeat. Once we do something we can do this as many times as we need to. If I want to find the greatest common divisor of two numbers, it has to divide their difference. But now whatever that greatest common divisor is it's got to divide 391, 901 and 510. So it has to divide any difference that I want. And the difference that I want to take advantage of is this one 510 minus 391. So my greatest common divisor of these two numbers also has to divide 510 also has to divide 119. So any number on the board, any number on the board. Pick two of them and subtract. Which two do we want to subtract 901 minus 510. Well 901 minus 510 just shuttles us back to 391. So we don't want to subtract those two because that just takes us back to a number that we have. Our goal really is we want to get small numbers. It is easier to find things that divide a small number than it is to find things that divide a large number. So which numbers will give us the smallest difference. 391 and 119. Yeah, I could do 391 and 119. I made a mistake there someplace. What's that? Is it 272? I might be wrong. Hold on. No, you're right. 272. Yeah. Yeah. Okay, that's okay. I was I was worried there because I was not getting. Yeah. I was not getting an answer that I expected. Yep. 272. There we go. That has a lot of factors. So subtract two numbers. Give me another number. Any number on the board. Any number on the board. Which two do you want to subtract? 272, 119. Yeah, I would probably go 272 minus 119. Again, the goal here is we want to get as small a number as possible. So what is that? That's going to be, can we get smaller numbers? 153 and 119. Okay. And that's small enough that we can actually find the factors. What divides 34? Well, one, don't forget one. What is it? What is it? 244. What else? Two and 17. So our greatest common divisor. And so let's go back to, go back to our original question. Our greatest common divisor 391 and 901. So again, the idea that all this comes back to this idea that if I have, if I have a number. I have to divide this summer difference and again, typically the useful thing to look at is their difference because that will get us to smaller. That'll get us to smaller numbers. So if I want to find the greatest common divisor of these two, I find their difference. The insight that makes this work is that now I have three numbers. That my divisor has to work on. And so any two. I can subtract and find their difference. I just keep subtracting numbers that I get until I get something that I feel like factoring. In this case, that's 34. So here's my list of possibilities. Which of these, which of these can you know, immediately are not going to be divisors? 17. Well, the 17 is not a divisor. Two. Yeah. You know, two is not a divisor because these are odd numbers. Those are odd numbers. You know, two isn't going to work. And if, and same reason 34 can't work either, because again, if a number has 34 as a divisor, it's got to be even. 17 may work. Now here's the, here is the one important thing that we do have to verify. We do have to check to make sure that, that this does work. If this doesn't work, the only choice is the only, the only choice is going to be one. And so we check. And we find. That in fact, that does work 391 is in fact 17 times, I think it's 23. And I know one. Is 17 times 53. So, so it turns out that 17 will be our greatest common divisor. And again, the, the, the, the key idea here. Again, the key idea here is that what I can do is I can keep doing subtractions. I can keep subtracting the numbers that I get until I get to a number that's small enough where I feel comfortable factoring it. The other possibility is I may get to a number that's equal to one and to which point I know that the greatest common divisor has to be something that divides one. And in fact that. So greatest common divisor, these two numbers. Again, whatever it is, it must divide. Has to divide two. So what are the things that divide two? One always. Can it be two? No. Nope. Those are odd numbers. Can't be two. So greatest common divisors one. And we got that without having to, without having to try to factor these two numbers. And generally speaking, if you want to find the greatest common divisor of two randomly chosen numbers, this Euclidean algorithm is actually the only really effective way of the only really efficient way of solving it. Okay. So. Okay. So somebody asked about the, about Gauss's method. So let me, let me pull up. Let me do a problem with Gauss's method. Thank you. Mr. Okay. And so Gauss's method. Oh, and then a, and then somebody was asking about converting repeating decimal. So, yeah. Thank you so much. Okay. So a little bit of background on this Gauss's method is based on, and Gauss's method and repeating decimals actually have a certain similarity that I'll point out. So Gauss's method is based on the following idea. So here's the problem. Let's say I want to add. So here's the basic idea behind Gauss's method. If I want to add the numbers from one to a hundred, I'm going to do two things. The first thing I do is actually an important connection to make with the repeating decimal. What I can do is I can say, well, this is, I don't know what this is, but I can at least give it a name. I'll call it S. And Gauss's method is based on the following idea. If I reverse the order of the sum. First of all, does reversing the order change the value? No. No, it's still going to be S. So the value is still going to be S. And here's the important insight here. If I add these two things together, if I add vertically, what do I get? This is 101 plus 101 plus 101. And the important insight here is that if I add vertically, all of those things add to 101, what's on the right hand side? Arithmetic is bookkeeping. I have an S plus an S. I have two Ss. So you can't do S to the second power, right? No, because remember S to the second, well, that really means S times S. This is really S plus S. So that's a, yeah, so that's two S. So, well, that doesn't do us any good. We have to add 101 a whole bunch of times. That isn't significantly easier, but what does make it easier? What do I call a bunch of things added, a bunch of the same things added together? What is that? Let's take a smaller example. What's five plus five plus five. What's another name for that? What happens when I add the same number to itself over and over again? What am I doing? What arithmetic operation am I doing? Remember that is a, where did we put that? Remember all of our arithmetic operations correspond to doing something. So when I'm adding, I'm putting combining sets. When I'm subtracting, I'm removing sets. When I'm multiplying, what am I doing? I'm adding sets of the same size. So if I have three twos, that's the same as three times two. So here, so here I'm adding together a whole bunch of 101s. So there's a multiplication here. And so I'm adding together 101s. Now here's the important question. How many 101s do we have? Six. Well, I've written six, but I have not really, we've got to add all the numbers from one to 101. So how many 101s do we actually have here? 100. There's 100. The easy way of seeing that is every column gets you 101. So here's the first 101. Here's the second 101. Here's the third. And we don't write a whole bunch of the middle. Here's the 98th, the 99th, and the 100th, 101. So importantly, when we have a repeated sum, that is a multiplication. So the sum of a whole bunch of 101s, well, that's really 100 times 101. And that's equal to 2s. And now I want to solve for s. So what do I need to do to solve for s? Well, it's two times. So how do I get rid of a times? Remember, if you want to get rid of multiplication, we can divide by 2. Divide by 2. And so s is 100 times 101 divided by 2. And that's equal to whatever number we get. That's 5,050. Okay. So this is the basic idea behind Gauss's method. So the two things you really want to take away from this, I'm going to give the sum a name. I'm going to call it something so I can work with it algebraically. And if I reverse, if I reverse the sum ends, then when I add them vertically, they all add to the same thing. And this sum is some number of whatever they are. And then I just have to figure out what they are. So let's take a look at this problem then. Okay. So let's see what are we doing. We're doing 60 in the first month. Plus six more. That's 66. Plus 72. Plus. Wow, that's a long time. 15 years. So each month, so that's 15 times 12. That's 180 months. So let's see. This is the first month. Here's our second. Here's our third. So I need that 180th month. So how much do I make in the 180th month? Well, let's think about that. What are we doing? We're adding six each month. Here's a kind of a useful thing to, to keep in mind in many problems. It's sometimes useful not to do the arithmetic. So here in my first month at 60, my second month, I add six. My third month, I add six. My next month. I add six. The next month. I add six and so on. What am I doing here? I'm adding the same thing over and over again. So what is that? That's a multiplication, right? So this is really 60 plus four sixes. This is really 60 plus three sixes. This is really 60 plus two sixes. 60 plus one six. And that's just going to be 60. Now, one thing that'll also help, let's try to be consistent. Every other one of these is 60 plus some number of sixes. Well, this first one is 60 plus how many sixes? Zero sixes. And the reason that that's useful is we go back to, if we go back to here, this is 60. So remember how this worked out. This was 60 plus zero sixes. This is 60 plus one six. This is 60 plus two sixes. What's going to be in our 180th month? Well, it's 60 plus how many sixes? 170, yeah, somebody said 179, right? Because look at our third month, we add two sixes. Our second month, we add one six. It's always one less than the month number. So it's 179 sixes. And that works out to be what, that's 1134. And there's our sum. So this is what we're trying. So, so actually that answers that first question, how much in the last month? So that's going to be 1134. And then if I want to look at the total savings again, Gauss's method works because we can reverse the order. Remember the month before we have one less six. So that's one, one, two, eight plus one, one, two, two plus. That's all the way down to 60. Still the same sum still to us, to us. That's still going to give us the same sum. And again, this is a whole bunch of 1194 is added together. So that's a multiplication. And how many do we have? Well, here's our first one. Here's our second one. Here's our third one. So which one is that one? Someone put it in the chat 180. Oh, sorry. Yeah, I'm not seeing the chat. Okay. So 180 there. Yeah, there's 180 of them. Oh, there it is. Okay. So there's a hundred and 80 of them. So there's our 180. And that's going to be two s. And so s, I'm going to divide everything by two. And so there's our total amount that we are able to save over that. So I am, I did the formal Gauss's method. Then I sort of gave them the formula. So I said the beginning term is B, plus E. And then it's going to be B plus E times the number of terms divided by two. Yeah. So that's how I introduced it after I did the formals, formal. Exposition of Gauss formula. Okay. Yeah. Yeah. So yeah. So that's, yeah. So that's really, yeah. So if you look at what, if you take this, if you take this apart so that, yeah. So the, so the other way you can look at Gauss's method, it's really the first plus the last times the number of terms divided by two. So if you take this equation, sorry, if you take this expression, so we have 180 terms divided by two and we're going to multiply first plus the last 1134. So, so, so yeah. So there's where our 1194 is coming from and then our 180 divided by two is coming from that, that quotient. Okay. So with repeating decimals, we almost have the same sort. We have something very similar going on with converting a repeating decimal into a fraction. So if I want to convert the, well, let's take a fraction like zero point, I don't know, one, two, one, two, one, two repeating. So the idea here is that I'm going to say, I'm going to call S to be 0.121212 repeating. And my recommendation on this is that if you're dealing with repeating decimal, it's convenient to write down at least three repetitions of it just to just to sort of get a sense of what, of how it works. Excellent. Let me try. Let me do two different ones here. How about let's try zero point. Okay. So the key idea here is what happens if I multiply S by 10. So if I multiply something by 10, that shifts the decimal over one place. So if I multiply S by 10, 10 S, that decimal shifts over one place. And again, because it's a repeating decimal, all that happens is those twos fill in. Now, now, what you can think about as the, what I think about as the troublesome part of the, repeating decimal is all this stuff. It's the, it's the repeating part of the repeating decimal. That's the, that's the problem. I want to get rid of that stuff. And the thing to recognize here is if I subtract over on the right hand side, because the repeating decimals line up, they'll go away. I'm left with just the integer portion. What do I have on the left hand side? Arithmetic is bookkeeping algebra, algebra is something that we haven't talked about, but algebra is generalized arithmetic. The way you want to think about any sort of algebraic expression is that is that algebra is just a generalized, is a generalized arithmetic. So what do I have? I have 10 S's. I'm going to remove one S. So what do I have left? Well, I have 10 S. I took away, I removed one of them. So what do I have left? Nine S. I have nine S's, right? So algebra is generalized arithmetic. So I had 10. I took away one. I have one, I have nine left over. There's still S's. They're still whatever the S is. And now I want to solve for S. So now I can divide both sides by nine. And I guess we S is equal to two nights. And there's my expression. So what about something like something like this one? Now the important thing to keep in mind here is don't get caught up in the steps. Focus on the goal. So the obvious thing to try here is I multiply by 10. That shifts to decimal one place. And they subtract. They get nine S is equal to S. The problem here is that the digits don't line up. And while we can do the subtraction, it's kind of a mess. It's, it's not a non-repeating decimal. So I don't want to multiply by 10. I don't want to do that. What do I want to do instead? Again, what makes this work is that we're able to line up the digits. What do I want to multiply by? 100. I want to multiply by 100. If I multiply by 100, that shifts to decimal two places. And I get 22. 12, 12, 12. And so on. And so again, we can do the same thing now. And so I have 100 S's. I subtract one of them. And what do I have left? 99. I have my repeating decibels line up. So they're all gone. I left with 12. And so S is 12. 99. And we probably want to reduce this. 12 is three times four. 99 is three times 33. So I can remove that common factor and get a fine Lancer. For 30 thirds. Okay. Professor. Yeah. What if it's like, um, seven, one, two, one, two, one, two, like where you have to get rid of that first number? Ah, okay. Good question. Uh, so let's see. Seven. One, two. Well, let's try it out. See what happens. Uh, so it's zero point seven. One, two. One, two. So again, we'll call this S. Okay. So. Okay. So that's what we're going to do. We're going to do this by, so that our repeating part will line up. Well, do you want to remove this seven first thing? You just do 10, or do you want to do all three numbers and do a thousand? That's what I'm confused on. Well, okay. So let's, let's look at that. If we both play by a thousand, let's try that. So if we both play by a thousand, that shifts our decimal over three places. So that gets seven. One, two. Point. One, two. One, two. One, two. And the thing that you'll notice here is that our repeating portion doesn't line up. We have twos and ones that are offset. And if we multiply by 10, we've run into the same problem. What if I multiply by a hundred? So if I multiply by a hundred. Well, that work. Well, it won't completely eliminate it, but it almost works because look at the last portion there. All of those line up. So if I'd actually do that. Everything past that point. Drops out. And while I do have, I still have a decimal here, but the nice thing is it's not a repeating decimal. That. And I could still solve for us. Ended with that. Now here's a question. Two questions. One is that. And answer. Does this express our. Original repeating decimal as a fraction. Well, yeah. Sort of. Yeah. I mean, we, we typically, I mean, typically we'd like our fractions to be quotients of whole numbers. But certainly we've gotten to a point, we've gotten to a point where at the very least, at the very least, we no longer have this repeating decimal. So at least we have taken a step in the right direction. It's no longer a repeating decimal. Now we like our fraction typically to be whole numbers. So here's a question. I don't really like having a decimal in my numerator or denominator for that matter. So is there something we can do to change it to get rid of that, of that decimal. Michael, my ultimate goal here is I would really kind of like to have a quotient of whole numbers. So is there something I can do that will allow me to no longer have a decimal any place. Well, one of the things we can do to fractions, if I have a fraction, I can remove a common factor. But anything I do in mathematics, I can generally reverse, I can also include a common factor. We usually say that we can multiply numerator and denominator by the same number. So can I, is there something I can multiply this by that will get rid of the decimal. What's the easiest thing to multiply by that will get rid of a decimal. So I have a number like 5.3 times something. And what I want is something that I don't, I no longer have a decimal place. I can multiply by what? You say 10. Yeah, I can multiply by 10. Sure. If I multiply a decimal by 10, I can get rid of the decimal over one place. And depending on how far I need to move it, that might get rid of the decimal. So this is also the same thing as I can multiply the numerator by 10, as long as I also multiply the denominator by 10. 705 divided by 990. And then you just get the simplest form if it asks. Yeah, at this point, we see that numerator and denominator both have a factor of 5 so we can remove that. That's 140. In any case, with a repeating decimals, would you always have to multiply by 100 to get the S away or no? Well, yeah, it's, it's a power of 10. So sometimes we multiply by 10. So in that first thing we, in the first case we did, we multiply by 10. Because our rep attend the digits that were repeating, it was a single digit that was repeating. So here in the, in the second example, in the second example, sorry. In our second example, the rep attend the digits that were repeating, there was a two digit rep attend. And so we multiplied by 100. And here, again, we had the, this leading seven, which was different. But then the repeating portion was a two digit number. So it was convenient to multiply by, by 100. And so, so the, what we multiply by is governed by how many digits we have in the rep attend. So, yeah, so here if we have that, if we have a non repeating portion at the beginning, you can, I mean, there's two ways of handling this. You can split it off at the beginning and then deal with a deal with, deal with a, just the repeating digits or you can just include it and then, and then, and then you will end up with a fraction where we have a terminating decimal as the numerator. And at that point, we can get rid of the decimal by multiplying by a power of 10. So either way works. Personally, I like doing it this way better because otherwise the, the alternative there is that this decimal portion, this is going to be, it's, you have to convert that there will be seven tenths plus an amount. And then at some point, at that some point you, you would need to deal with that seven tenths. Why can't you submit it with a decimal on top? It depends on the question, but typically this type of problem, you want to write it as a fraction reduced to lowest terms. So what that means is that we do need to have it, we do need to write it as a fraction where both numerator and denominator are whole numbers. And so we're not, because of the form that we're supposed to write our answer in, we do need those to be, to be whole numbers. All right. Let's see. Okay. So does that, does that, does that help out on, on terminating decimals and so on? Let me see if there's anything else. Yeah. So, so it's, so we've been here about what three hours now. So, so are there other questions that said, so I'd like to finish off soon, but, but if there are other questions that people want to go over, if there's any other topics that you'd like to talk about, I can, I can, I can, I can spend a little bit more time here. Okay, surface area volumes. Okay. Yeah. All right. All right. So, surface area. Professor, can we also do a problem where we have to find the end term? Okay. Yeah. So the end term. Yeah. So that was basically like this, like the, of the Gauss's method. So here, so this one, we were actually finding the end term of the, of this arithmetic sequence. And so, so again, the idea here is the most important thing here is identifying that what we're actually doing when we add the same thing repeatedly does correspond to a multiplication expression. So finding the end term of the sequence really is, is, is going to be, is going to be something like this. How would the equation look like? Like, like, um, for example, there was a problem and it said, yeah, I wanted me to write the equation. Do you know what that one was from? I think she means when you have right the equation for the end term in an arithmetic sum or geometric sequence where n is just a variable. Ah, okay. Okay. Well, let's, let's take a look at this one. So yeah. So that, that one's actually pretty easy. So, so do you see, do you understand where, do you understand where this formula is coming from? Yeah. Okay. So let's, let's take a look at that. So let me just take down a couple of notes here. So our third term is 60 plus two times six and are 80th. So, so, so here's, you know, again, I'm a big fan of writing things out. So if you, if you look at our third term, if you look at our 180th term, now we want to generalize it. We want to write the nth term. So let's look at this. We always have a 60. We always have a plus. We always have a times six. The thing that changes is the multiplier. So how do you get the multiplier? So for our third term, we multiply by two. For our 180th term, we multiply by 179. So what's the relationship between the, between the term number and the thing we're multiplying by? The term minus one. It's the term minus one. So if I'm dealing with the nth term, then the, then what I want to multiply by is n minus one. So that's going to be n. So, yeah. So again, that's that. So, so that's making this connection that once you know the term number, you're multiplying by one less than the term. So if I want to multiply, if I want to find the nth term, then that's going to be the term n minus one. Okay. I get it now seeing it this way. Thank you. Okay. All right. So let's look at those surface area and surface area and volume problems. Okay. So we want to find the area of the figure. So remember area is, area is the number of, again, arithmetic is bookkeeping algebras generalized with particular geometry is bookkeeping area is the number of unit squares. So, so here our distance between adjacent rows and columns is one unit. So if I take a look at a figure that looks like this, then the area of this figure is going to be, well, there's one square. There's two squares. So the area of that figure is going to be two square units. So, so we can figure out the figure out the area of a, of a region. Rectangles are pretty easy. We just add up the, out of the figures. The other thing that's the other thing that we want to be able to compute is if I have a triangle. So if I have a triangle and here's the thing. If you know the area of a rectangle, if you know the area of a right triangle, you can compute any area that's bounded by straight lines. What's the area of a triangle? Well, the way to remember that triangle is if I fill it out so that it's a rectangle. So if I fill out a rectangle, that triangle is half of the enclosing rectangle. So if I want to find the area of a triangle, it's half the area of the rectangle that would include it. So, so those are your basic area formulas. Now let's see if we can put them together. So earlier I mentioned the idea of scotch tape and scissors. So I have a figure. If I want to find the area of this figure, one of the things I can try to do is I can try and see how I would make this from a, from a, from a larger figure. So I'm going to cut it out from, I'm going to cut this out from this figure. I'm going to enclose it. I'm going to put it into a box. And if I want to get the area of the figure, I need to get rid of a bunch of pieces. So let's see if we can do that. So the area. So let's start off with a box. The whole thing. How big is it? How many squares do I have inside that box? And we can do this in two ways. We can either count them, which, you know, again, counting is easy enough. The other thing to remember is that, well, if I count them, here's one, two, three, four. Here's another four. Here's another four. And another, and another, and another. So how many fours are there? One, two, three, four, five, six, fours. And so my area is six times four. You might remember that the area of a rectangle is product of length and width. Well, that's where it comes from. Here's my width, four. Here's my height in this case, and the height is six. And so I'm adding together six, fours. And that corresponds to that multiplication. So that's the box that came in. So now let's start removing pieces. Let's start with this triangle. That's a terrible drawing. Let's start off with this bit here. How big is that triangle? Well, it's half the rectangle it came in. So how big is a rectangle it came in? Is it two? The rectangle that triangle came in that has one, two, it has three. It has three squares in it. So when I remove that triangle, the triangle is half of three. It's three halves. So that's gone now. Let's get rid of this triangle. So how big is this triangle? Well, it's half the box that came in. So how big is the box? Well, here's the box. How big is that? Six. Six. So the triangle is half of that. So the triangle is going to be three. So I'm going to get rid of another three. How about this one? Is it one? Okay, so how big is the box? So be careful on that. So this is probably not helped by my drawing here. It's easier to see what that triangle is. Yeah, there we go. So there's the box the triangle came in. So how big is that box? Three. That's three. So that triangle that we're getting rid of this amount here is going to be another three halves. Okay. Now this piece is going to be a little bit complicated. So the key idea, the thing to remember here is that we know how to find the areas of rectangles and we know how to find the areas of right triangles. So the trick here is, well, again, Scotch tape and scissors. If I want to cut, well, I want to cut this piece out, but then I do need to figure out what the area of that piece is. If I wanted to make this piece, what do I tape together? I would probably tape together this square. And then these two triangles. So how big is the square? That's going to be four, right? There's one, two, three, four square units there. And then the two triangles, this triangle here, this one in the, this triangle in the upper right, what's the area of that one? Well, it's half the box that comes in. So fill out the box. How big is the box? Two square, so it's one. That's going to be one, right? So yeah, so that's going to be one. Subtract that. And then we have this one, this triangle. So it's two? It's going to be two, right? So again, if you look at the box that that triangle would come in, that box has area four. So the triangle has half that area. And so that triangle is going to have area two. And so that's going to be 24 minus. Then we have to do a bunch of computations here. That is what is that? That's three, six, 10, 11, 12, 13. That's minus 13. And so that says that our area is going to be 11. And so, so really finding any, any area of any figure that's bounded with straight lines. It really comes down to. That take your figure and start removing or sometimes gluing together some, maybe the thing to do and start adding or subtracting triangles and rectangles because we know how to find the area of a triangle or a rectangle. And in the worst case scenario, we just count boxes and that's our area for a rectangle. Again, the thing to remember is that a triangle is half of the box it comes in. And so we can start with our bigger figure. And again, Scotch tape and scissors, mostly scissors in this case, we cut out the pieces that we don't want. So volumes. So volumes and surface areas are going to be. The cube. Cubes, right? So volume is how many cubes do we have? So, so for example, we'll pretend these are cubes. Come on. Okay. So, you know, I'll try to move it. I'll try to move it off the. So, so if we want to find volume, so again, volume is a number of cubes. And surface area. Is squares. On the outside. So let's think about this. So again, the thing that you really want to, the thing you really want to, I'm having some computer difficulties here. I'm not going to try to move it. Okay. So, so the, so the volume is just the number of cubes and so we can go ahead and count them. So one of the difficulties here is to remember, remember that our picture is shown in perspective. So we can't actually see all of the cubes. But it looks like we have one, two, three, four, five, six, seven. That's 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24. So we have 25. Cubes in the first layer. How many cubes behind it? 25. Yeah, we'd expect it to be 25. So I could write 25 plus 25. But here's a nice little review. What's another way of saying 25 plus 25. 25 times two is 25 times two. So if you know the volume formula for a rectangular prism, what that comes from is again, the idea is that I'm doing a repeated addition. Of a, of an amount. So the volume here is going to be 50 cubic units. Okay. Now. So now let's take a look at our surface area. All right. So let me erase those markers. So the volume is just the front and back technically. Well, it's, so, so this had three layers. So if this had three layers, then we'd have 25 times three. So, so it's basically, because the, because the way that the figure is, once we know how many cubes are in this front. Outermost layer. It's just a question of how many layers that we have, how many layers we have behind it. So if this had a third layer behind it, then we'd have three times 25. But how would you know? How would you know? Yeah, yeah, yeah. Well, in this, we only, we only see that there's two layers. So, so we have our first layer here. Oops. Yeah, I mean, it would depend on the drawing. So here we have our first layer here and our second layer there. So if we had another one, we would have, yeah, so I try to draw this and anything resembling perspective. But yeah, so, so, so much. Yeah, that's horrible trying. Yeah. So here. So if you had something like that, then we'd have three layers and then we'd multiply it by three. But. Yeah. Yeah. Yeah. You would go by what the drawing looks like. Okay. So how about surface area? So again, these are the squares that are showing on the outside. So this one is, so let's think about this. So let's count the ones that are facing us. So here's one. Well, actually. So, so here is one, two, three, four, five, six, seven. And if you think about it, we should actually get 25 because we are just, we're still counting the cubes. It's just recounting the cubic faces in front of us. Now, here's where you have to think about this as a three dimensional object. So there's 25 squares facing us. What else does that tell you? There's 25 on the other side. Yeah. There's an opposite side to this figure that we can't see it. We can't see it, but on the opposite side of this figure, if you think about what that figure has to look like, there has to be another 25 squares on the opposite face as well that we can't see. So we can think about this as front and back. So our front face, 25 squares, our back face also 25 squares. How about some, let's look at what we might think about as our right hand side. So here, so these are the, these are the squares we're going to see on the right hand side. So it's these two, these are not on our right hand side. We might call these on the top. So we have these. 12. These we have 12, right? So there's going to be 12. Now this one takes a little bit more visualization. How about our left side? 12. It's going to be 12 as well. Here's kind of how, here's how I see it. Just, yeah, just to give you some insight into, into how you might think about this, the way I see it is if I take these 12 squares that are facing to the right, if I just slide them over, if I slide these over to the left hand side, they cover up the left hand side. So there's 12 squares at the left. And so there's going to be 12 squares on the right. How about the top? If I look down from the top, what do I see? 10. I see 10, right? So it's the six here. It's the six here. But then I also see these two. And then I also see those two. So there's 10 on the top face. And then on the bottom, 10. There's also going to be another 10. And again, again, the way that I think about this is if I take these, these 12 squares that are on the top face, if I just drop them straight down to the bottom, they cover up the bottom. So our surface area is we're just going to add up all of the, all of the, all of the, all of the squares that we have. So what that's 50 plus 24 plus 20. And that's going to be what 94. So a total of 94 squares altogether. All right. So let me close with, let me close with one last idea on this. So, so what happens if we, so, so here's a, here's an interesting thing that actually happens that is a relevant in a lot of areas outside of mathematics. So basically any of the sciences, any of the, any of the sciences actually have to contend with this. This figure has a volume of 50 cubic units. It's if we break this apart. So ask this question. If we break this part into individual cubes, what's the total surface area? So I know there's 50 cubes. So if I take one of those cubes, what's the surface area of a single cube? Wait, say that again. I'm sorry. So, so, so if I take a single cube, so I take one of these cubes, doesn't matter if we join, I take one of these cubes. And if it's separated from all of the others, so I'm going to break this figure apart into its individual cubes. What's the surface area of a single cube? How many squares do I have on the outside? Six. So it's how many? Six. Six, right? So, yes, there's a front back, left, right, top, bottom. So there's six. Each individual cube would have a surface area of six. And so my total surface area, that's going to be 50. That's going to be 50. Sixes. 50 times six. And that's going to be a total surface area of 300. And so, so the, the interesting thing here is that if I, I have a volume and it has a surface area, the figure has a surface area of 94. But if I break this volume up into a lot more pieces, that surface area goes up by a considerable amount. And this is actually a fairly important thing in, again, many of the sciences because all chemical reactions, all changes, all interactions take place at the surface. And so what happens is if I take a substance, it might not be very reactive if I keep it all together. But if I break it apart into a whole bunch of pieces, if I powder it, the surface area goes way up and the reactivity increases significantly. And that's a nice little consequence of this relationship between surface area and volume. Thank you. So to be clear, the volume is the front, like you count the front face and add it and do like a repeated addition of the amount of columns that are on your right that you see on your right. Right. Yeah. Again, basically, I mean, what you're really doing is you're just counting the total number of cubes that you have. And the easiest way of doing that is, again, in this particular case, we can count the number of cubes we have that are facing us. And if all of our layers are going to be identical, then it's just going to be the number of cubes on that layer times however many layers we have. Got it. Thank you so much. Okay. So, all right. So any other questions over any, any, anything else? Hello, I have a question, but it's about the exam. Sure. For the final, when once we're finished, do we have to do, how do we, do we upload or work after we submit or we have to, or are we given time? How does that work? I don't have time afterwards. So, so I mean, probably the easiest thing to do is as you finish it, you can upload it then. That's, but, but after the exam finishes, you'll have, you know, you'll have about, yeah, call it about a half an hour at least to upload the work that you've written up for each individual question. So we have to press submit and then upload or work or we have to. So, so the way that it, so, so typically the way a problem will be, let me see if I can pull something up from. Let me get, let me see. So, so the problems that you'll get, they'll look like, yeah, so, so they'll typically look something like this. So again, this is calculus. So don't worry about the questions, but there'll be like a question here and then there will be a place where you can upload your work. So here, if you, if you've done your work and it's easy enough, you can upload it here at this point into, into this answer box that's right next to the question. Otherwise, after you finish the exam, let me see if I can, let me see if I can show you what this looks like. After you finish the exam, you'll also have a chance to do that. I have to switch to a student few, right? So, yeah, so after you, so after you finish the exam, then what you'll be able to do is to, so, and after you finish the exam, there will be a screen that says do you want to upload work and it'll give you the option, but what that'll look like is something like this. So you'll see the exam questions. You'll see the exam questions. You won't be able to change any answers at that point, but then you'll have, again, this type of answer box where you can upload or add, add any work that's relevant to the, to the problem and, and add that as well, or add that to the, to the, to what you're submitting. Thank you. Okay. Okay. All right. Anything else, any, any other questions any? Let's see.