Well for question one on problem set four, we have to choose which one of the following is equivalent to to this expression. Okay, we're negating a universally quantified statement in which there's conditional. And we're told that there's only one of these, one of these answers is correct. The correct answer, let me just jump to the correct answer and then see what's wrong with some of the others. Is, is part C, the notforall becomes an exist and the negation goes into this part. When you negate a conditional, you end up with the antecedent together with the negation of the consequent. And when you negate the disjunction, the negations filter in, and the disjunction becomes a conjunction. So if you follow the rules about, well they're not rules, they're, but they sort of our rules. But I, I recommend you not to think of them in terms of rules because that's, that's not really getting at what this course is about. but there's certainly patterns of activity that you can get to get to recognize, and then what happens is universals become exists. You have the truth of the antecedence, in place of an implication, of a conditional you have a conjunction. And then when you negate these guys the negation applies to each one and that becomes a conjunction. Okay, but that, but as I used to indicate or I just referred to, really what I want you to do is concentrate always on why it is that you get a behavior. Why does this give you px and not that. Why does negating this give you a conjunction there, so it's all about understanding. and if you, if you simply learn to apply the rules, you really don't have a useful skill, okay? Computers are good at applying rules. That's what they do, that's all they can do. People can go much beyond that. Okay, what's wrong with some of the others. Well, the first one is just hopeless. I mean, there's just nothing remotely like that. If you got that as your answer, then either you were having either a temporary aberration, or you really, really, really haven't got the issue with this. The others, there were reasons why people could go wrong. And the others were put in, because there were mistakes that people frequently make. Okay, in the case of this one, the negation is in the wrong place. when you negate a conditional, the negation doesn't come together with this one. The negation should come inference of here and inference of there. So it's just ,um,mixing up where the negation comes in. Okay, but, otherwise, zero place. So, so it's like applying your bis, you're applying the right sort of, let me call it rule, you're doing the right thing but you're, you're, you've got one step out. It's like getting a negative sign, a minus sign in the wrong place in an equation. looking at this one, well everything went fine except you forgot to change disjunction to conjunction, but everything else was fine. And in the case of this one, you forgot to, then that should be conjunction, and everything else was fine. Okay, so in cases b, d, and e, there was just one thing wrong. It's even possible you just did that by a slip. I mean, heaven only knows you've seen me make slips often enough in the, in the lectures and the tutorials where I write the wrong thing down or I whatever. It's uhh, you have to keep a lot in our minds when we're doing these things. And frequently what a hand does isn't what we're thinking it's doing and sometimes what we say, isn't what we think we're doing. Mathematics is like that, when you're really focusing on the mathematical concepts, in the heart of it. You can make slips with the writing and with the wedge you. I do it all the time, and we all do. That's, that's just part of thinking mathematics. It takes a lot of concentration to focus on the mathematics. And the, the every day things like writing and using words, tend to miss out on that because our mind is focused on the content. Okay, let's go on and look at number two now. So for number two, let me just begin by reading this formula out in everyday English, or at least, stilted everyday English. It says for every person p, there's a person q, and a game t such that p beats q, in game t. So more colloquially, everybody beats somebody in some game. Everybody beats somebody in some game. Everybody beats someone in some game. Well, which one says the same as that? That one certainly does. That means the same as everybody beats somebody in some game. doesn't mean the same as that, so it's not that one. That's not a possibility. for every player there is another player they beat all the time. That's not quite the same. If there was a for all there, for all t, then whenever they play p would beat that person, that person q. So that's, that's not quite, you would need a different quantifier. That's just some game. If there was a, for all there you would have it in all the games. There is a player who loses every game. No, no that's all about losing. So hopefully, well, actually, while, while we're on this, what would this look like? if I wanted to put that in a formula, I would start with the following. I would say, there is a q, such that for all p, and for all t, W, p, q, t. In a minute I'll say why I said start with. Because I was looking ahead to what would be wrong with this. But, let's just read it, it says, there is a player, such that for all players and for all times, p beats q in that game. So it sort of says the player loses every game. The problem is, when you take for all p here, one of the p's that are going to pop up is q, him or herself, and a player can't beat himself. So this doesn't quite work. Because the p could include the q itself. In fact one of the p's will be the q, okay? And it's also, well it depends this is, that players going to, yeah, I think that would be okay. No, you've got to be a little bit careful actually. It depends how you interpret the thing. for all t, t just goes over games of tennis. Whereas we're just talking about games of tennis in which people, people win. So you, you, you'd have to sort of think about it and decide whether you need to put clauses in here, to make it clear that you're only looking a t's where they play together. There'll be similar issue here. Again, you could start with, if you wanted to do this one, you would start with saying there isn't a p, let's see, for all q, for all t, w, p, q, t. The only difference is here we had qp and here we've got pq. But you'd have the same issue., among the qs is a possible p. And you'd have to be careful to take account of the fact that the t itself ranges over all possible games. And you only want to be making a statement about the games that we did play together. and you could have, amend this, you could put clauses in. There are various ways out of that. if I thought this was a big deal I'd have thought about it before I started to record this little piece. But, I think we've, we've solved the problem in any case. That one came up the first time, that was the one that that has that meaning. None of the others do, so put those as no's, and we've solved that one, okay? Let's go on to number three. Now normally when I'm going through examples in classes, I don't make deliberate mistakes. I don't need to because I make plenty of real mistakes, when I'm going through mathematics in any case. But this time, I'm going to make a deliberate mistake and I'm going to do it to emphasize the issue that's going to come up here. Okay, so what I'm going to do is go through these three and give you answers to all of them. but one of them's actually going to be a wrong answer, and then we'll correct it. So this one, actually if you remember the previous question, this is the one that we saw. That means everyone wins the game, okay? For all the players p, there's a player q and a time t to a p beats q and that time. What does this say, everyone beats, well it's almost the same, right? Except instead of playing every player beats one player, you say everyone beats everyone else, at some time, in some game. Okay, because it's all, it's all pairs of p and q. And this one, it's essentially well, actually, what's the difference? That one is for all p, there exists a q. This one is for all q, there exists a p. So, this isn't, doesn't say everyone wins a game. This one says, everyone loses again, okay? Seem plausible? Well, one of these is actually is actually not possibly true. Let's just see. Could that be true? We, we're talking about, whoops, there's a typo there. That should be possibly. Okay, let's correct the typo. I got it right up there. Okay, let's see which one cannot possibly be true. Okay, that's what the issue is here. Well, that could possibly be true. Okay, that's okay. That could be true, everyone beats everyone else all the time. Not sometime. Well, everyone loses a game. Well, that's certainly possible, that's okay. Coming to this one, can this, is it possible for this to be true? It's certainly possible for that to be true. It's certainly possible for this to be true. We're talking about the real world here when people are playing tennis. Is this one, the one that can't possible be true? No, there doesn't have to be one. You could have said none of these are the case. In fact if you got this far you probably did say they're all okay, but actually this one is not okay. Because, when you say for all p and for all q that includes the case of q and p being equal. That would mean everybody has to beat themselves at some time amongst everything else. So a player cannot beat every other player. Because the problem is, no player, can beat herself, or himself, whichever you want, okay? No player can beat herself. So that can't possibly be true. So this is the guy, that can't possibly be true. And it can't possibly be true because the q and the p have to be equal. What you would have to say if you wanted to make that something that could be true, was you'd have to say, if for all p, for all q. if p and q are different then there is a t so it should w, p, q, t, then you'd be okay. Because you'd say for all pairs p and q providing they're different. Then at some time, at some game, p does beat q in that game, so that would be a way of making it possible. But as it stands, that's cannot possibly be true because players cannot beat themselves. Okay, so the issue was whether you can take the formulas and correctly tie them into what they say about the real world. It wasn't mathematics that was deciding these, it was the real world. And in the real world, players cannot beat themselves. Well, actually, in a figurative way, players beat themselves all the time. But in the sense we're talking about here, that's not the case. Okay, now let's move on to question four. Okay, now this expression is, is, is a colloquial expression. And so, when I set this question up, I, I realize that for non native speaking English students this would be a, this would be somewhat challenging. So what I did is I only gave you options that you should be able to distinguish between, by, by doing the logical structure. the reason I like to give these kind of examples is because they capture an awful lot of social and cultural knowledge and, and there's an interesting challenge. In capturing that kind of thing in, in formalisms. but to help you along, I, I gave you three options, that you should be able to sort out just on the basis of logic. If you know what, what being a lover means in this case. So being a lover means you're in a mutual relationship which means you've got, you love someone and that person loves you. But if you look at this one this says for this person x, so these are all about person x, they all talk about some person x. This doesn't say that person's a lover, this part says that person is in a love, in a relationship with everybody else. So this part, because it's a universal quantifier, that says L love, x loves z and z loves x. So, this person x is in a loving relationship with everybody. But, well that's nonsensical, so we can forget that one. So, it comes down to these two. Because in each of these, it says the person is in a loving relationship. X is in a relationship with some z and it's mutual. X loves z and z loves x. The same clause here. So the choice is between a and c. Well let's look at what a, what a says. That says, for all x and for all y, if the x is in a loving relationship, then y loves x. Now the y doesn't come in here so the, the for all y has to do with this part. So it says take any person x, if that person is, is a lover, then every person loves them, if they know them, so that actually is the correct one. Okay, all people love a lover, of these three, that's the, that's the one. You know, you could argue about whether that's the absolute best interpretation, but out of these three it's certainly a correct one. And so it's, it's the one here, let's look at this one. This one is is, is a little bit different, because it's got for all y in here. So, it says for all x, if x is a lover, then it doesn't say then, it says and for all y, L, y, x. So the for all actually applies to this part as well. So what this really, what follows from this is that for all x and for all y, L, y, x. In other words, everybody loves everybody. Well that's not the case, because this isn't conditional on being a lover. This just really says that's the case and that's the case. So part of this is saying that for all x and for all y, L, y, x. Oh, that's not the case, I mean it's not the case that everybody loves everybody. The world would be a nice place I guess if that was true, but it's not true. So it can't be that one. So we can, we've eliminated this one, because it's, it doesn't capture, it doesn't use the fact that, being a lover. And we've eliminated this one. Because it basically just boils down to saying everybody loves everybody. And that leaves this one. And, and this is definitely one, one good interpretation of everybody loves a lover. Okay, so we, we will be able to reason that one out, and I would hope that even without the detailed understanding of what this means in English. there's only one of these that will stand up to analysis, and after all the idea is to sort of look at how the formulas capture relationships from the real world. Well for question five, we have to find which statements are false. Okay, so let's just see what they say. for all x, for all y, for all z, if x is less than or equal to y and y is less than or equal to z, then x is less than or equal to z. That's true. It's actually known as the transitivity of the of the order relationship. If x is to the left of y, and y is to the left of z, then x is to the left of z. Okay, so, that one's true. What does this one say. For all x, for all y, if x is less than or equal to y and y is less than or equal to x, then x is equal to. Well, it is true, okay? the only way you can have x is less than or equal to y or y is less than or equal to x is that they're actually equal. What about part c? For all x is equal to y, [SOUND]. Well, that's certainly the case, because given the x, you can take y equals x and then you have x less than or equal to y and y is, so that one was true. It's beginning to look like they're all true. Let's look at the last part. there is an x such that for all y, y is less than x or x is less than y. Well you might be tempted to say, yup, given any x it's the case that every [UNKNOWN] y is either less than x or bigger than x. But wait a minute, among the y's, governed by a universal quantifier is the x you start with. So if there was an x with that property, how could this happen? Because when you look at all the y's, among those y's would be x itself. So you would have x less than x, which is impossible, so that one's false. And it's false because the universal quantifier includes the x itself. Given an x, any x that you find you, when you universally quantify over y's, you include that x, and then that fails, and that fails. So there's the one that's false. And there is a false one. And it, it fails because universal quantifiers go over everything. And that means the, the, the, the y will, among the y's you will be looking at, is the x that you start with, okay? Well for the last question on problem set four, which is question six, we have to look at this piece of reasoning that a student gave. This is actually when I first gave this course in the Fall of 2012 and a student was trying to understand the, Euclid's proof that the prime numbers are infinite in number. And there was a key step where you from that product and you add one. And a student wanted to verify that in fact N plus 1 was not divisible by p. And came up with this argument. So I thought this would be a good one to look at. It raises a number of interesting issues. So, first of all let's just see what a student does. Suppose N is divisible by p. Well, arguably the student, you could say the student doesn't begin by saying what N is, and doesn't say what p is. But in this context there's no need to. It, it's already been stated that N is an integer. It doesn't matter whether it's positive or negative by the way. And it's already said that p is a prime. So I'll just, in this context, I don't think there's any need to demand that the student repeat it here. What's key is that the student is, is, is writing down the assumption on which the argument is going to be based. Okay, so I'm going to say that the, oh I'll come to the various other issues in a minute, the opening I think is fine. Okay, that's good. What does the student do then? then as an integer q to the a equals pq. Yes, that's the definition of divisibility. I'm not going to demand that the student spells it out because it's, it's fairly clear what's going on. So N plus 1 is P [INAUDIBLE], that's okay. Then dividing through by by q by p, you're going to have N plus one over p. Now the student's now written a lower case q there and that's obviously meant to be an upper case q, I'm not going to worry about that. That's just a typo in writing it, ditto here. so I'm not going to deduct anything for that those are just typos. I'm just using the wrong lowercase and uppercase confusion, so NH is not divisible by p. Okay, logical correctness. But in a sense this is okay, right? I mean there's everything steps fine and the arithmetic's true. So, I'm going to give 3, why aren't I giving four? I'll, I'll come to that in a minute. What about clarity? I'm going to give 4. This is absolutely clear what's going on. State the conclusion, yep. Conclusion stated, that's 4 for that one, a reasons given, absolutely reasons are given. yeah I'm going to sort of observe at the time the student sort of doesn't say, this is by definition of divisibility, and so forth. But you know, there's a limit to how much you can write down within the context of this class, and the intended audience, which is the students in this class. Then this is fine you know, it's fine, so we put more details in the reasons if you, if you want to weigh on the side of caution. But given the fact this is absolutely clear I think there's enough reasons given, okay. And that brings me to the overall evaluation. And this is where I'm going to put a zero down. And this is also why I didn't give a full four for this, because this is logically correct in terms of doing mathematics, but this is an argument about integers. And the integers is a number system in which you can add, subtract, and multiply. What you can't do, is divide. Division is not an operation on the integers. It's an operation on the natural, on the rational numbers and on the real numbers. But, it's not an operation on the integers. So, in, in going into the equation. This argument goes beyond the realms that we're permitted to use. So, I'm giving it, I'm being generous here actually in saying I'll give you three because it's, this is mathematically correct in terms of the rational numbers. I mean I could have been tougher. I could have reduced it to two, I could have even put a zero down. But as I've said before I'm looking for reasons to give people marks, not to take them away. We're trying to make people better thinkers, not to make them feel bad about themselves, okay, so I'm going to give credit where it's due. But strictly speaking it's not right. Here's what the student should have done, okay, so we've got the first park we've written N=PQ. Okay, that was okay. Okay, now what I'm going to do is I'm going to argue by contradiction. I'm going to say suppose N plus 1 were divisible by P. Then, there is an integer called R such that N plus 1 equals PR. Then, we have N plus 1 minus N, equals PR minus PQ, which is P into R minus Q. But N plus 1 minus N is 1. So what I've shown is that P times R minus 1 equals 1. That means that P, no, that means that 1, is divisible by P. But, that's a contradiction, P is a prime number. So, it's at least equal to 2. So, it can't divide into 1. One isn't divisbly by any, by two or three or anything so there's a contridiction. Hence, the original assumption here was false. Hence N plus 1 is not divisible by P. Notice that this is almost the same, in a sense this is equivalent to this. It's not the same as this, but it's equivalent, because the, the R here, is N plus 1 over P. So this is the R. So I've said, and I'm sort of doing the same thing except here, I'm not using division. I'm doing everything in terms of divisibility, which means I'm doing it in terms of addition, subtraction and multiplication. All I'm using here is addition, subtraction and multiplication. I'm not using division. I'm getting around it, by introducing this R, if you like. And that's a significant difference, because we simply don't have division as an operation in the integers. We have divisibility, which is a property that may or may not hold between two, two integers. But we can't divide one integer by another because, division's not an operation in the integers. Okay, so I, I gave a reasonable amount of credit for this. I think I was generous with this one because I tend to be generous. And that means the total grade is 19, okay? Well that was the end of problem set four.