 In many proofs, it's necessary to consider different cases. For example, suppose we need to prove that the set A union B is a subset of C, where in this situation, A is all those numbers divisible by six, that is multiples of six. B is all those integers divisible by nine, that is multiples of nine. And then C is all those integers divisible by three, AKA those multiples of three, okay? So how might we go about proving such a thing here? Well, honestly, the way we would start is, because we're trying to prove a set as a subset of another, what we're gonna do is we're gonna start with an arbitrary element of the set, that's supposed to be a subset. So we're gonna say X belongs to the union of A and B. Now, then the next thing we would do is we'd unravel the definition here. We know what A and B stand for. We also know what the union means, so that's what we apply next. Since X belongs to the union of A and B, we would then be like, oh, then X equals A, or X equals B, all right? But then at that moment, we kinda get stuck because by nature of the or, one of these things is true, and they could both be true, but we don't know that. At least one of them is true. So how do we proceed forward? And the idea is we can proceed forward using cases. One of those two things is true, and therefore, let's consider if the first one was true, what does that mean? And then once we're done with that, then we'll consider the second one, see if that one's true, and then go from there. So we're gonna break it up into cases. There's the case where X equals A, and then there's the case where X equals B. One of those two things must happen, so we will consider the two possibilities separately. And this gives us a huge advantage because when you assume that X equals A, then you inherit the properties of B inside of A. And later on, when you assume that X is inside of B, then you inherit the property of B and in B. So by putting it into cases, you're gaining assumptions which you didn't have beforehand. But you don't necessarily get both assumptions. You're gonna have to prove for X in the side of A, you'll have to prove that X belongs to C. And then if X belongs to B, you have to then argue it belongs to C. And you're gonna use different arguments there, and so you treat these things individually. And so how you might proceed by proofs would be the following. The same proposition as we had before. We wanna show that A union B is a subset of C. So we start off just like we did. X is a member of A union B. Therefore, X belongs to A or X belongs to B. We're gonna consider the first case first. If X belongs to A, what does that mean? Well, then there exists an integer K such that X equals 6K. We couldn't say that when we thought of X as belonging to A union B, but since we're assuming now that X belongs to A, I mean, it could belong to B. I don't know that. It could be a common multiple of six and nine, but we don't need to consider that. We don't have to consider the intersection. We just have to consider right now X is in A, and therefore X is a multiple of six. But six, of course, can be factored as three times two. So if X is a multiple of six, then we can write X as a multiple of three. It's three times 2K, whereas since 2K is an integer, this is three times an integer, so this shows that X belongs to C. Now, on the other hand, so I'm now indicating to the reader here that I'm considering a different case, and there's only two cases, so you can think of one hand versus the other hand. If on the other hand, X belongs to B, well, to be inside of B suggests that you're a multiple of nine. Therefore, there exists an integer L, such that X equals nine L. And if you wanted to, you could overload the symbol K because it's just an arbitrary integer. I chose to use a different integer here just so there was no confusion whatsoever. There's no connection between this L and that K whatsoever. So X equals nine L. But just like we did with the multiples of six, if X equals nine L, you can factor nine as three times three and as three times an integer is an integer, this shows that X is equal to three times an integer. That means that three divides X, so X belongs to C, like it did up here as well. And so in either case, notice what happened. In either case, we ended up with X belongs to C. And so since X belongs to C, that then shows that A union B is a subset of C. So when you're trying to prove something about a union of sets, you often have to do it by cases. You have to treat like, okay, if I'm in the union, I'm in one of these sets. Well, if I'm in that set, or if I'm in that set, and you can do that, that's okay. The cases are nice because they give us extra assumptions. B and an A that empowers us to say that X is a multiple of six. B and B says that X is a, it empowers us to say that X is a multiple of nine. And we're able to use that to show that in either case, X is a multiple of three. Now, an important thing to also know about cases is that if we're gonna consider cases, then the cases have to be exhaustive. That is, we have to try every possibility. For a union, there's only two possibilities, X is in A or X is in B. Which of course, X could simultaneously be in X and B, like if X was 18, it would be divisible as six and by nine, that's okay. We're not saying that these cases are mutually exclusive, but we do consider every possible case we were exhaustive. If we miss even one case, we then fall into the danger of the false dilemma, the limited choice, which was this logical fallacy we had talked about before, we have to make sure if we're doing cases that we include all of the cases. Which oftentimes, like in this one, there was only two cases, so it wasn't that big of a deal. We handled both of them. Let's look at another example. Suppose that N is a natural number, so it could be zero, one, two, three, any positive integer. Then the quantity one plus negative one to the N times two and minus one is a multiple of four. Now that's sort of, when you first look at that, that might seem like a strange result. Is that even true? Oftentimes before starting a proof, we should stop to strategize and consider how the proof is gonna come out. Cause there are some situations, the proof almost writes itself. Like there's only one way to write it, you know, up to stylistic changes here. And if such a thing's gonna happen, we would wanna know. Or maybe the proof would be more difficult and still we wanna investigate that. Now, for this one and for many situations, sometimes it's useful to collect examples of the phenomenon in question. And you know, you're collecting data, seeing that it's supported by examples. And now, no number of examples will prove the theorem, but it's possible that if we have a lot of examples, perhaps we can extrapolate a general pattern from these examples. And then that general pattern is the proof. Like what happens in general here? And so what I want us to do is consider the following. Let's actually make a table to consider these quantities. So on the very top, we'll have N right here. And then this will be our quantity, one plus negative one to the N times two N minus one. And so let's just do a couple of natural numbers here. Zero, one, two, three, four, five, six, seven. We'll go that far. And so notice what happens when you plug zero into this formula here. You're gonna get one plus negative one to the zero, which negative one to the zero is one. Then you're gonna get two times zero minus one. That's gonna be a negative one. You're gonna end up with zero, which of course, zero is the same thing as four times zero. So that is a multiple of four. When you plug in one, plug in one for the Ns, well, in this case, when you plug in a one, you're gonna get a negative one here. You get two times one, which is two minus one, which is one, so you get one minus one. You're also gonna get zero, okay? Now for reasons that'll be making much more sense in a moment, I'm gonna say that zero this time is four times negative zero. You're like negative zero, isn't that just zero? Yes it is, but to indicate the pattern that's forthcoming, I want you to think of it as a negative zero, bum, bum, bum. When you plug in two, when you simplify it, pause the video and do these calculations on your own if you would like to, I'm gonna kind of speed through some of these. When you plug in two, you're gonna end up with four, which four is four times one. When you plug in three, you're gonna end up with negative four, which is equal to four times negative one. And when you plug in four, you're gonna end up with eight, which then is the same thing as four times two. When you plug in five, you end up with negative eight, which is four times negative two. When you plug in six, the quantity will simplify to be 12, which is four times three. And then when you plug in seven, it'll pop out a negative 12, which is four times negative three. So maybe we should also include this as a negative zero right there. So now the negative start to become apparent here. You are getting multiples of four here, zero, four, eight, 12, and also negative ones as well. But when you look at it, it seems like there's a difference of sign based upon the parity of your input. If you look at the positive, excuse me, if you look at the even natural numbers, you always get a positive multiple of four. And then when you look at the odd natural numbers, one, three, five, seven, you seem to be getting negative multiples of four. And so because of this data we've collected, it sort of begs to reason that we should separate our proof into cases. There's gonna be the case where n is an even number and n is an odd number because they seem to be behaving differently. Now, as every natural number is either even or odd, if we consider the evens and consider the odds, this will be an exhaustive list of possibilities. And so that's how we're gonna proceed forward here. Let n be a natural number. We're gonna first consider the possibility where n is even. If n is even, that means there exists some integers such that n equals two to the k. And so then consider the quantity one plus negative one to the n times two n minus one. I'm gonna substitute each of those ends now with my two k like so. Now, some things to note, if you take an even power of negative one, that actually is a positive, which then it just disappears there. Maybe that's why the even ones are actually turning out to be positive because the negative sign disappeared there. Of course, two times two k is the same thing as four k. And those should get one minus one, those will cancel out. And so this just becomes a four k, which clearly that's a multiple of four. And therefore we see four, in fact does divide the quantity when n was an even number. Now let's consider the possibility where n is an odd number. This time, that means there's some integer k such that n equals two k plus one. And so let's start with our quantity again. One plus negative one to the n times two n minus one. Let's replace each of the ends with a two k plus one. Now, one thing to note here when you take an odd power of negative one, that actually becomes negative one. So the negative sign sticks around. Distribute this two here, you're gonna get four k plus two. And so simplifying things, what happens is you're going to get two minus one, which is one, one minus one, which is canceled out to give you zero. And so you're left with, this will simplify just to be negative four k, which is also a multiple of four because you could write this as four times negative k like we were doing over here, okay? And so therefore when n is odd, you get four, it divides the quantity. And so since every natural number is even or odd, it then follows that this quantity is always divisible by four. And so we completed the proof because we have exhausted the possibility you're either even or odd. And as we consider both, we then completed the proof. Now I wanna provide one more proof inside of this lecture video here and we're gonna go the other direction. So we came up with this quantity and showed that this quantity is always a multiple of four. What I wanna do is actually go backwards to kind of flip the direction here. Can every multiple of four be written as this quantity right here? And again, we're gonna break this thing up into cases. This time I'm gonna break it up into three cases for which as we take a number k, is it always possible that four times k can be expressed like this? Can every multiple of four be expressed that way? I'm gonna break it up into three cases because it turns out that k is either zero, it's positive or it's negative. Every integer is either zero positive or negative. That's true for real numbers as well, but we only have to consider integers. And so I'm gonna break it up into three cases. k is zero, k is positive, k is negative. It turned out like we saw, there was something different about positive and negatives that they have to do with even and odds. And then zero is neither positive or negative, so maybe I just treat it separately. But from the data I've already collected, I already know how to deal with zero, that's easy. So if k is zero, then four k is equal to zero, which is equal to one plus negative one to the zero times two zero minus one. And therefore n would equal zero in that situation. There is a number n which produces four times zero. That's the number zero. One also did it, but I don't have to demonstrate every natural number that doesn't. It just says there exists an existential statement. There is a natural number, there's actually two that do it, but I only have to demonstrate one, okay? So then my second case, I'm gonna then assume that k is a positive number. So if k is positive, choose n to be two k, which is not all of four k, but that's what we're gonna choose for n. And so notice here, if you take one plus negative one to the n times two n minus one, if n here is two k, since k is positive and you have a two k, those are things we're gonna consider here. Negative one to an even power is gonna be a positive one like we saw before. You're gonna get two times two k, which is four k, like so. And then you get negative one plus one cancels out. This then produces the four k that we're looking for. And so this then proved the result. When k is positive, you can choose n to be two k. One thing I should note here, what did the positivity have to do anything? Our theorem says that given any integer k, there exists a natural number n. A natural number is either zero or positive, not negative. Now, since k is positive, it's a positive integer, it's actually a natural number. Two times a natural number is likewise a natural number. So I've produced a natural number that satisfies this condition. And then let's consider the last possibility. What if k is negative? Well, if k is negative, then consider the number one minus two times k. Now, because k is negative, negative two times k is a positive integer, which makes it a natural number. Adding one is still a natural number. So this is a natural number like we need. That's where the negativity comes into play. And then consider the quantity one plus negative one to the n times two n minus one. We're gonna get one plus negative one to an odd power this time. One minus two k is an odd number. So that's gonna make this quantity be a negative one. But then we also distribute the two here. We have a one minus two k, distribute the two. You're gonna get two minus four k minus one. Much like we saw before, two minus one is one. One minus one goes away. So you're left with a negative negative four k. This simplifies just to be a four k. Now be aware that I don't claim that four k is a positive number because I actually know the k is negative. Four k is a negative, that's okay. The point is this is equal to four k even though it's a negative number. But the number we use for the exponent here, the n was in fact a natural number. So we did have to use a different argument for negative multiples of four as opposed to positive multiples of four. But like we saw, we're just gonna use odd numbers for the negatives and we used evens for the positives. That was exactly what we saw just now in the reverse direction. As our list is exhaustive, we then know that we've considered every possibility and as every possibility led to the same conclusion, we have then a complete proof. But that's the thing is we have to make sure that our list is exhaustive. Proof by cases is a powerful technique and I recommend we use it because each time you look at an individual case, you get extra assumptions which then strengthens your ability to prove something but you have to be exhaustive. If you miss even one case in a proof, then your proof actually is then invalid because it's at the false dilemma, fallacy and as such the result might not even be true because of one case. You have to consider all of those cases when you prove by case.