 So, welcome to the screencast on proving subset inclusion. In this screencast, in the ones that follow it, we're going to be proving things about sets. And this one we're going to be proving that one set is a subset of another. How do you do that? How do you show that one set is a subset of another? And then later screencasts we're going to look at how do you prove that two sets are equal and some other ideas. So to get a handle on proofs about sets, let's go back to something we picked up in an earlier screencast about what does it mean for a set to be a subset of something else. So what does it mean to say that A is a subset of B? Well, we could translate that into a conditional statement and we know how to prove conditional statements. We have lots of tools for doing that. So this is a helpful way to leverage what we know. When I say that A is a subset of B, what it means is that every element of A is also an element of B. And in logical language, this would say, for every X that belongs to A, if X is in A, then X is in B. This is really an if, then statement. And that's how we're going to think about proving subset inclusion. So to prove that A is a subset of B, whether it's a proper subset or a regular subset, we're going to assume the hypothesis. We're going to assume that we have an X that belongs to A. In other words, we're choosing an unspecified element of, of A. And then we're going to say, what does that mean? If X belongs to A, what do I know about it? What's a forward step I could do from there? And if A is given in set builder notation, that we have something we can do. A set builder notation is a way of writing sets that describe elements of sets by a property that they all have in common. So if I know that X belongs to A, and A's got some description for all of its elements, I can rewrite X using that description. And the goal of the proof here is to prove that X belongs to B. And to do that, we're going to show that X not only has the property that it takes to belong to A, it also has the property that it takes to belong to B as well. So that's how we're going to proceed just by proving a simple conditional statement. So let's look at an example here. Here is, here are two sets. S is the set of all integers that are congruent to 9 mod 12. And T is the set of all integers that are congruent to 1 mod 4. And what we're going to prove here is that S is a subset of T. So again, the, the strategy here is I'm going to prove that if X belongs to S, then X belongs to T. That's the conditional statement. We want to prove to get subset inclusion established. Now just remember, S means, or the set of all things that are congruent to 9 mod 12. And T is the set of all things that are congruent to 1 mod 4. So let's proceed with the proof here. And we're going to begin by assuming the hypothesis. So let's just let X belong to S. Okay, in your book this is referred to as the choose an element method for solving or for proving a theorem. And that's what we're doing. We're going to assume that we have an element that belongs to S, the set of all integers that are congruent to 9 mod 12. So just pick one out. We don't know what it is, but we know that it just, it's simply an element of that set. So what does that mean? If X belongs to S, then I'm going to use the property that we used to define S in set builder notation to rewrite it. X belongs to S, that means that X is congruent to 9 mod 12. All elements of S have that thing in common. So now let's just kind of work forward from there. Now what does it mean to say that X is congruent to 9 mod 12? Well, let's just use the definition of integer congruence and I would say then there exists some integer q such that, such that X is equal to not congruent to but actually equal to 12 times q plus 9. So being congruent to 9 mod 12 is equivalent to saying that if you divide X by 12, it goes in a certain number of times with a remainder of 9. Now it'll be helpful at this point to remind ourselves what we want to show. So I'm going to put this in red because we don't want to assume this. We're going to show that X belongs to T. That's the ultimate goal here. And what it means for X to belong to T is to say that in other words, we want to show that X is congruent to 1 mod 4. That's what it means to show that X belongs to T, that it's congruent to 1 mod 4. So let's work toward that goal here. So I know that X is equal to 12 times q plus 9 for some integer q. Now what I want to do is write X equal to 4 times an integer plus 1. That's what it means for X to be congruent to 1 mod 4. And I think I can see that pretty clearly here. If I write this 12q plus 9 is 12q, let's peel off an 8 and add a 1. That's certainly a true equation here that 12q plus 9 is actually equal to this stuff now here. But what I can do now is factor a 4 out of this block right here and get this. And then we can see what happens from here. By closure, this is an integer right here. And so what this is telling me is that X is not only congruent to 9 mod 12, it's also congruent to 1 mod 4 because I've written X is equal to 4 times an integer plus 1. And so that's going to be my conclusion. Therefore, we're just going to wind it up here. Therefore, X is congruent not equal to but congruent to, sorry, 1 mod 4. And that's the end of the proof. So what I've done here, maybe I should say one other thing here. Let's not end the proof just yet. Since X is congruent to 1 mod 4, we can then say therefore X belongs to T. And so now I know that S is a subset of T because I've proven that if I take an arbitrary value of X out of S, let's trace the proof out. Here's an arbitrary value of X out of S. Then I can rewrite it and use the criterion to rewrite it as something that belongs to T. And this is exactly what it means to belong to T here. So I picked a random element out of S and show it belongs to T. That makes S a subset of T. And that's the end of that particular proof. Now, the converse of this is not true. T is not a subset of S. And why is that? Well, it's because there exists an integer Y that is in T but Y is not in S. And you can almost take your pick of very, of just about anything. But for example, Y equals 5 will be a great counter example because 5 is congruent to 1 mod 4. So therefore 5 belongs to T. But 5 is not congruent to 9 mod 12. Okay? Because 12 does not divide 5 minus 9. So 5 does not belong to S. Okay? So those two sets are not equal to each other. So what we could come in here and conclude then is that S is actually a proper subset of T. There's something in T that doesn't actually belong to S. So that's how we prove that one set is a subset of another. We choose an element out of, let's back up and see our proof here. We choose an element out of the set that I'm saying is being contained. And then I prove that it belongs to the set that's containing it. And that's enough to get the proof done. Thanks for watching.