 Direct proof and proof-by-contradictory positive are two methods we've introduced so far in our lecture series for proving conditional statements like P implies Q, okay? We've also introduced a method of proof known as proof-by-contradiction for which you assume the negation of your conclusion and then you derive a contradiction that there's some statements true but also is false. Thus, managing the negation you assumed actually the opposite. It was actually the other way around. And so I wanted to mention specifically how does one use proof-by-contradiction specifically when you're trying to prove a conditional statement? Because it turns out you can squish this method of contradiction into the direct proof method because if you're trying to do a direct proof, if you're trying to prove it in this statement right here, you would assume in your proof here, whoops, you would assume that P is true and then you would make some arguments and then you would conclude that Q is true. That proves then the conditional statement. But what if you're struggling to prove that Q is true? What you could then do is you could assume P is true and then you could also assume by way of contradiction, you could assume that not Q is true. Then you're like, okay, so now you have two assumptions instead of just one. You can then be like, ah, there's a statement R, we proved to be true. Some more arguments, you get not R and that gives us of course a contradiction and therefore that means that Q is actually true, thus proving the conditional statement as well. So one can prove a conditional statement by assuming the hypothesis and assuming the negation of the conclusion. You derive a contradiction, that means the conclusion was actually true, which means the conditional statement P implies Q is true as well. So let me show you this variation of proof by contradiction on the case of a conditional statement. So for example, let's take this statement. If A and B are both integers such that A is greater than or equal to two, then it must be the case that A does not divide B or A does not divide B plus one, okay? That is to say that an integer greater than two can't divide consecutive integers. It can't divide both of them. So this isn't if then statement, right, if then. So if A and B are integers such that A is greater than or equal to two, then we have some conditions about divisibility. So if I want to prove this by direct proof, then what I would do is I would assume the hypothesis there. Oh, let A and B be integers such that A is greater than or equal to two. Now, when you look at the statement here, I'm actually trying to prove a negation, right? I'm trying to prove that A doesn't divide B or A doesn't divide B plus one. You have some negations here. You have an or statement. It actually might be best to look at the negation of this statement. So proof by contradiction might be very helpful. So it's useful to declare this in your proof. Suppose for the sake of contradiction that A divides B and A divides B plus one. So notice I took this statement right here, applying the De Morgan laws. If you negate an or statement, it becomes an and statement. And then of course, if you negate a statement like A doesn't divide B, that would become A doesn't not divide B. That is A divides B, like so. And then A divides B plus one. So okay, I'm thinking that because of how this is phrased, I'm trying to prove that an integer can't divide two consecutive integers, at least not if you're bigger than two or equal to two. But it feels a little awkward because of the negations of the or. And so I think the negation would be easier to work with here. So we're gonna assume for the sake of contradiction that A does divide B and A divides B plus one. Okay, well, how's that gonna go then? Well, if A divides B, that means there's some integer in such that B is equal to An. And likewise, because A divides B plus one, that means there's some integer M such that B plus one is equal to Am. And so we have then these factorizations of B and B plus one. Let's play around with those for a moment. For example, if I take these two equations and subtract them from one another, I'll get the following B plus one minus B on the left-hand side, and then you're gonna get Am minus An on the right-hand side. Now B plus one minus B is gonna give you two, excuse me, where did two come from? It's gonna give you one. B plus one minus B is one. Then you're gonna get Am minus An. You can factor out the common divisor of A right there. This is gonna equal A times N minus N. Now, if you look at that equation there, this tells us that A divides one. But now this is where the problem's gonna come into play. The only divisors of one are plus or minus one. And by assumption, we have that A is greater than or equal to two. If A divides one, that means A equals plus or minus one, which means A is less than two. We get a contradiction. A cannot be less than two while also being greater than or equal to two. So we get this contradiction. Once you arrive upon your contradiction, you can actually end the proof right there. I mean, we said that earlier, for the sake of contradiction, we're gonna assume the opposite of what we want. Once you get a contradiction, that means you've now proven it. You can stop at that moment. Contradiction's been found into the proof. If you wanna put some words in there and say things like, well, because we have a contradiction, this means the other statement holds. You can do that if you want to. But as the statement of the proof is still on the, you know, it's listed there on the page, once you get that contradiction and you've told the reader you're proving my contradiction, there is no actual need to go any further. There's no clarity that's necessary. It should be clear to the reader. Oh, you're proving something by contradiction. You found the contradiction. Therefore, the proof then follows. This is a valid proof argument. You can stop at that moment there. Let's see another example of this, where again, we're gonna prove the conditional using contradiction here. This time, let the statement be, suppose that A is an integer. If A is even, excuse me, if A squared is even, then A is even as well, okay? So conditional statement, if, then. So we're gonna assume that A squared is even, but for the sake of contradiction, we're gonna suppose that A is not even. That is, A is odd. And so that's how we start our proof here. For the sake of contradiction, it's generally a good idea when you do a proof by contradiction to tell the reader that. Because if you don't preface that you're doing a proof by contradiction, they might be like, why are you getting the negation of the thing you're trying to prove? Oh, proof by contradiction. For the sake of clarity, just tell the audience that you're gonna prove by contradiction here. So for the sake of contradiction, suppose that A squared is even, but A is not even. A squared is even, but A is not even, which of course, if A is not even, then it's odd. Well, what does it mean to be even? If A squared is even, that means there exists some integer n such that A squared equals two n. If A is odd, what does that mean? It means there's an integer m such that A equals two m plus one. We have to derive a contradiction somehow. How are we gonna do that? Well, we could take this statement for A and plug it in for A squared and see what happens. So if you take A squared and replace the A with a two m plus one, you now have two m plus one square. You could foil it out. You'd end up with four m squared plus four m plus one. Combining terms there, like, because after all four m squared and four m are both even numbers, you could factor out a two. You end up with two times two m squared plus two m plus one. This right here is an integer. Times it by two, of course, is an even integer plus one makes it a non-integer. This is a non-integer. So wait a second. A squared is an even number and it's a non-number. That's not a possibility. This gives us the contradiction we were looking for. Now, I do wanna make mention here, particularly with this last proof. While this proof does, in fact, prove the statement. We now know this is a true statement. This is a valid proof. One could make the argument that the proof that we just provided using the contradiction is perhaps a little bit more awkward than how we could have done it, like, say, proof by contrapositive. It actually would be maybe more straightforward. And in fact, I'll talk more about this in the next video for, the next and last video for lecture 23. How do you decide between direct proof, proof by contradiction or proof by contrapositive? There are strengths and weaknesses to all of those methods. And so, like I said, I'll say some more about that in the next video. What I wanna mention right now as we finish this video is that with these different methods we've introduced. So like I said, we have direct proof. We have contrapositive. We have proof by contradiction, which I'll write that one here as well. These are all techniques that we can use to prove conditional statements like P implies Q. But these three techniques are just one of many methods you can use to prove a conditional. But there's other things we're able to prove like, we can prove things about integers using induction, strong induction, the well-ordering principle, proof by smallest counter example, also should be included in that. We've proven things from commonatorics using commonatorial proof. We can prove and statements, conjunctions. We can prove or statements, disjunctions. There's a lots and lots and lots of proofs out there. We do proofs by cases. Sometimes you have to break it up into cases. My point is that as our proofs get longer and more complex, it becomes more important and actually commonplace to start combining all of these different proof techniques, the three I have listed on the screen, some of the other ones I listed but didn't write down, and then so many other proof techniques that we haven't even discussed so far. You might have to start combining lots of different proof techniques into a single proof. Like we said beforehand, if you're trying to prove a conditional statement, you might begin with direct proof. So you start off with like, okay, in our proof we're going to assume, whoops, assume that P is a thing. Now we then want to prove that Q is true as well, but if Q, it's itself like a compounded statement, like instead of Q, if you have something like, if itself is a conditional statement, Q implies R, well then you're like, well then, okay, to prove that I'm then gonna assume, again just using by direct proof, we're gonna assume Q in that manner, but what if R is itself like, oh, it's R or S? Well then how do you prove an or statement? You're gonna assume like, well okay, assume not R, and then this is just, and then you have to keep on going, eventually you wanna conclude S is a thing. There's a lot of things you could do there, and so when you start writing, you start writing proofs inside of proofs, inside of proofs, and this can get really complicated and long, and so sometimes when one writes a proof, you have to break it up because it just starts getting too big. This is the notion of a lemma. One might introduce a lemma because the proofs start to get long that you're gonna sort of encapsulate a small part of the proof in this one lemma, and then the next lemma has the next part of the proof, and then after you have a bunch of lemmas combined, then you prove a theorem where you're gonna start citing lemmas and propositions you've proven previously. This is what you have to sometimes do. Now, fortunately for the students of this course, our proofs are probably not gonna get that long, that you have to start breaking up into smaller lemmas and lemmas and lemmas, but nonetheless, this idea of writing a proof inside of a proof is a necessary thing to do, and so I do wanna illustrate this as we finish this video on combining techniques here. Consider the proposition that every non-zero ration number can be expressed as a product of two irrational numbers. How are we gonna do that? Well, it is essentially just a if then statement. If you have a non-zero ration number, then it can be factored as a product of two irrationals. So we start with our non-zero irrational and then argue that it has such a factorization. So that's how we begin. Suppose R is a rational number, which is non-zero. R belongs to the set Q, which makes it rational, and then we specifically say it's non-zero. Well, because it's a rational number, that means there exist integers A and B such that R equals A over B. And B, of course, we can assume is also a non-zero number. So that's what it means to be a non-zero rational number. We could also assume that A is non-zero. I guess I should throw that in there as well. A is non-zero because it's a rational number, B can't be zero, the denominator can't, but because it's non-zero, we also have that A is non-zero. That'll be necessary later on in the proof. Now, previously in our lecture series, specifically in theorem 610-2 there, we proved that the square root of two is in fact an irrational number. We've argued that this cannot be rational. This is just following Euclid's classic argument there. And be aware, if you're confused about the numbers, we're not necessarily going the order that the book is in, so just to see you're aware, no big deal there. It's just siding in here. So we've already previously seen in our lecture series that the square root of two is not a rational number. So what we're gonna do is we're gonna rewrite R as, well, we know it's A times B. I'm gonna rewrite it as the square root of two times A over B times the square root of two, okay? Now, we know that the square root of two is an irrational number. If I can show that A divided by B root two is likewise irrational, then this would be the factorization we desire. And so this is where we get to that idea of a proof inside of a proof, right? We need to show that this number is likewise irrational. We've already used a proof inside of a proof because we cited a theorem, we've already proven. That's a more subtle way, but even still, we have to then prove something we haven't proven already. And so this sentence right here serves to just tell the audience what we're gonna do. We have a claim. I claim that this number is irrational. We are now gonna show it, and if we show it, that finishes the proof that we wanna do. And now how are we gonna show that A over B root two is irrational? Basically the same way that we showed the square root of two is irrational, proof by contradiction. So suppose to the contrary that A over B root two is a rational number. Now if it's a rational number, there exist integers CD such that A over B root two is equal to C over D, which of course be aware that the square root of two is not zero, R itself is not zero. So this product is not zero. This is helpful because when we work with this thing we wanna make sure we don't divide by zero. Since the square root of two is not zero, it's reciprocal is not zero. And so in particular, when you look at this CD here, this is R times one over the square root of two. R is not zero, one over two is not zero. So this product is also not zero. The reason this is relevant is that CD has a reciprocal and particularly it's D over C. This is that relevance why I said A earlier was non-zero so that when you flip this thing upside down, this is a well-defined rational number. We wanna make sure, because CD could be a rational number, but DC might not be. Particularly if you take like zero over one, this is a rational number, but one over zero is not. Oh shoot, I just said zero is not a rational number. That's an official JK there. One over zero is not rational because it's not even a real number. So you have to watch out for that, but we don't have to worry about that because we've taken care of things. R is non-zero so the reciprocal is gonna be non-zero here. So if you take this equation and you work things around, you can actually show that the square root of two is equal to R times D over C, okay? And which remember, R is a rational number. We also have that DC is a rational number because it's the reciprocal of CD, which is a rational number, non-zero. So you get that the square root of two is a product of two rational numbers, which would imply that it's a rational number. But we also know by the theorem we stated earlier that the square root of two is not a rational number. Therefore we get a contradiction. The contradiction then gives us the opposite of what we assumed. Now, in this case, because the original expression is not what we're negating right now, we actually should probably mention what it is that we negated. What do we do contradiction on? Just to remind the reader here, we assumed, we're trying to prove that A over B root two is irrational. So we assumed it was rational. We got a contradiction. Therefore that means A over B root two is irrational, which then finishes the proof for the reasons we had mentioned before. So again, as your proofs get longer and more complicated, it becomes more necessary to write proofs inside of proofs. And you can use language to help it make clear to the audience what you're doing here. You can tell them the proof pattern you're using. Like we're gonna use proof by contradiction, or we can prove by the contrapositive. If you're doing direct proof, don't say that. I mean, that's sort of like the default. That's why we call it direct proof. You only need to specify to the reader what technique you're using. If it's an indirect proof, like contrapositive or contradictions. But then you can also make claims in the middle of the proof like we did before. We have the claim, this claim that we needed, that we need this fact to finish the proof. And then you can remind the reader after you've proven that, that finishes it. Giving your reader some hints, some guideposts, some mile markers on what you're doing in the proof, can be very helpful the more complicated your proofs get. Again, you can use any of the methods we've learned, transposition, contrapositives. That first one was none of the things. Sorry, contrapositives, contradictions, and direct proof induction. You can start to get all of these different proof techniques come together to form more complicated proofs. Give your audience some help and guide them through the proof together. Much in the same way that I narrate these proofs, we wanna put that into writing to help your reader understand these proofs as you start combining techniques together.