 Welcome back to our lecture series, Math 3120, transition to advanced mathematics for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misildine. In lecture 24, we're actually in for a treat. The usual scheme for the lectures inside of this lecture series is that we have three videos from each lecture. The first one focuses on a math topic, the second a logic topic, and the third about some type of communication issue in mathematical writing. Now, we're still going to have three videos, but it turns out both videos. The first two videos in this lecture 24 are going to be focused on logical techniques. We're trying to wrap up our discussion about integers leading to the fundamental theorem of arithmetic. One of the most important theorems involving integers, of course. We're going to prove that in the next video. Our third video, of course, will be about mathematical writing as usual. But I actually want to introduce two new proof techniques in these videos, one in this and one in the next. Honestly, both of them are introduced to give us some very important results about integers. In this video, I want to talk about the idea of proof of a disjunction. Now, by comparison, I want you to note that most of the statements that we've been focusing on have been proofs of conditional statements. If P implies Q, how do you prove that? Well, the main idea that we use is we look at a direct proof, which remember that technique, we assume P to be true, and then we have to argue that Q is also true. That proves the conditional statement. But in addition to direct proof, we, of course, can use the contrapositive, for which the contrapositive, what we do there is that we assume the conclusion is false, and then we argue that the hypothesis is false as well. Those two methods are both valid and essentially logically equivalent to each other. So you can use either one to prove a conditional. I should also mention we've also introduced this method of contradiction, proof by contradiction. A contradiction is not necessarily restricted. I feel like I misspelled the word contradiction there. Contradiction, there we go. A contradiction is the proof by contradiction is not necessarily restricted to conditional proofs. Honestly, to do proof by contradiction, all you have to do is just assume the statement you're trying to prove is false and then derive a contradiction. That works, of course, with conditional statements. You can use that all the time, okay? So what about other types of statements that we have to run across? Not every statement is conditional, even though this is one of the most important ones we deal with. Now, other conjunctions, I should say other connectives that we've run across before is the conjunction, an and statement. How do you prove something like P and Q? Well, it turns out it's not so bad. If you wanna prove P and Q, you first prove P and then you second prove Q, right? So if both of them are true, then their conjunction will be true as well. And so for a conjunction to be true, every term, every statement in the conjunction has to be true. So you can just prove each and every one of them individually and that's fine. So conjunctions deserve no more attention than that to prove because that's all it is. Pretty simple. Now, what I wanna focus now on is the idea of a disjunction. What happens if you wanna prove the disjunction, the or statement P or Q? Now it turns out that the way you can do this is actually to think of it as a conditional. We've seen before that a conditional statement can be rewritten as a disjunction. You can also rewrite a disjunction as a conditional for which this would become not P implies Q, okay? And thus, if you think of the disjunction as a conditional statement, you can prove it by direct proof. How would you do this? You would assume not P and then you would conclude using of course logical implications that Q holds from there. So if P doesn't happen, then Q has to happen as well. And I want you to think in terms of the language here of a disjunction. So there's two options. Let's say I promised my kids like, oh, you've been really good. So we're gonna have pizza tonight or ice cream, okay? And so I make one of those promises. And like we said before, sure. If I give them pizza and ice cream, that's also a true statement, but I gotta give them the least one of those. Otherwise I've lied to them and that'd be horrible. So well, if I come home and I don't have any pizza, it's like, oh, pizza's not true. Pizza's not gonna happen. Then ice cream would have to happen in that situation. This is commonly referred to as like this disjunctive syllogism. This was a logical argument we saw before that oh, P or Q happens, not P happens, therefore Q must happen, this disjunctive syllogism. Now of course in the disjunctive syllogism, you're assuming P or Q is a statement and then you also assume that P is and then Q follows from that. So it's not exactly the same thing, but the idea is if you have an or statement and you know one of the statements is false, then if the or statement is true, the other one has to be true. That's what this implication is saying. If P doesn't happen, then Q has to happen. So basically the way that you handle a disjunctive proof is you actually treat it by cases, okay? So there's basically two cases. The first case is that P is true. Now if P is true, you can then conclude that P or Q is true because again, if one of the statements is true, then the disjunction is true as well. Then the second possibility is that well, P is not true. Now if P is not true, you then want to show that Q is true and then if Q is true, that then would imply that P or Q is true as well. Now this first case is trivial, okay? If P is true, then P or Q is true as well. It's the second case that requires some effort because if P is not true, it's not automatically obvious why Q is true and argument has to be made there. So oftentimes this first case is skipped entirely and like I said, you just start off with assume, assume not P and then therefore you have to conclude Q holds. That's how you prove this thing because if Q holds, then this follows as well. So again, we kind of cut out the obvious parts and we focus just on the non-trivial part. That's what's necessary for the proof. Now some people will include this. You might say something like oh, if P is true, then we're done because it's trivial and therefore assume that not P is the case. So let's see an example of this. Let me get on my screen a little bit, all right? I actually want to use this technique of disjunction to prove Euclid's lima. Euclid of course is a very, very famous mathematician in some regard, one of the first modern mathematicians. Don't get me wrong. He lived around, what was it, 400 BC or so? So I mean, this was thousands of years ago but Euclid's books, the elements, right? Several books there, lots of chapters about number theory, geometry, so many things. It was one of some of the first, well at least in modern history, it was the first known attempts to actually develop rigorous proofs to defend mathematical statements. Now don't get me wrong, Euclid had a lot of mistakes, a lot of unstated assumptions inside of his writings but nonetheless, I mean, again, like 400 BC, this was a phenomenal breakthrough compared to what mathematics was beforehand. Before Euclid, mathematics was often kind of considered really just like arithmetic. It was like for finances, it was for construction and engineering. It was an applied science. But really because of the work of the ancient Greeks, Euclid being very prominent in this conversation here, because of the ancient works of the Greeks, Euclid, the Pythagoreans, et cetera, we actually then started to see that mathematics is more of an abstract science and not necessarily the supplied science. So a lot is attributed to Euclid in that regard. One of which, of course, will be referred to as Euclid's lemma. We've referenced Euclid's lemma several times in this lecture series and we're now at the point where we're gonna prove it. And we're gonna prove it just using this method of disjunction, for which case then it doesn't depend on any of the previous results we've ever done. So there's no circular reasoning here. Let's prove Euclid's lemma using this method. Because after all, Euclid's lemma essentially comes down to proving a disjunction. Let A and B be integers and let P be a prime number. So then what Euclid's lemma says is if A divides the product, sorry, if P divides the product AB, then either P divides A or P divides B, okay? So the structure of this argument is this is a conditional, if, if, then. So if P divides AB then, okay? So we have that initial statement P divides AB, but then the statement, the conclusion, it's itself a conjunction, excuse me, a disjunction, it's an OR statement. And that's why we're proving it here like so. So our, since the statement is fundamentally A, if then statement, we're gonna start with direct proof. We're gonna assume the hypothesis and then argue that the conclusion holds here. So to start off with, suppose that P divides A, that is there exists some integer K such that AB equals PK, like so. Now, so this right here is the hypothesis of the conditional statement. Now what we have to do is we have to now show, given this assumption, we have to show that P divides A or P divides B. So we have to now prove a disjunction, an OR statement. Now, how do you show a disjunction? Well, you'd be like, well, if P divides A, we're done. If not, then we have to argue that P divides B. So that's what's then gonna happen. Suppose next that P doesn't divide A. So this is for our disjunctive statement here. Now, if P doesn't divide A, since P is a prime number, and that means that then the greatest common divisor of P and A has to equal one. Because P only has two positive divisors, one and P. As P doesn't divide A, the only other possibility would be one in that situation. If you take the GCD of a prime and any other number, the GCD is either one, if the prime doesn't divide the number, or it's P, if the prime does divide the number. So we get one in this situation. Now, because of the Euclidean algorithm, again, that's the same Euclid here, because of the Euclidean algorithm, since the GCD of P and A is equal to one, we can actually write one as a linear combination of A and P, where there's two other integers, R and S. I don't really care what they are, but there's two integers, R and S, such that AR plus PS is equal to one. Now, this is an equation of integers. If I multiply both sides of the equation by B, I then get B is equal to ABR plus PBS. It's just like the TV station there, right? Now, remember our original hypothesis, AB was divisible by P. So AB, I can substitute in, oh, I can substitute in for that PK, like so. And then you have P, P. This thing is divisible by P. We can factor it out, and B is then equal to P times KR plus BS, which this shows that since P divides the right-hand side, P has to also divide the left-hand side. So this gives us that P was, P divides B. And that then proves the disjunction we're looking for. If P doesn't divide A, then it must have divided B. So the two cases are then taken care of. P either divided A or divided B. This then proves Euclid's lemma, which will be useful for the theorems we've used it already for, but also be helpful as we prove the fundamental theorem of arithmetic. But also it gives a good example of how one proves a disjunction. You assume one of them is not, and then the one that's surviving you show actually must be the case.