 Welcome to this screencast one to two, which we're going to discuss direct proofs of conditional statements using no show tables. This is part one of two examples that we're going to see here. No show tables are described in the Sunstream textbook. So if you haven't read those examples in that text, please go ahead and do so first. So the problem we're going to be working with in this screencast is the conjecture that says if n is odd, which implies that n is an integer, then n cubed is also odd. So what we're going to hide in this video is why we would care or even believe that this conjecture is true. But let's just try to prove it. If you do some instances and practice with this conjecture, play with it, you'll see that it looks to be true, but we need proof. So first of all, we have to understand what the word odd means because being able to move forward in a proof presupposes we know what the terms mean. So looking up the definition of the word odd in our textbook that we've developed, it says that an integer is said to be odd if there exists another integer k such that n equals 2k plus 1. For example, 7 is odd because I can find another integer, namely the integer 3, such that 7 is equal to 2 times 3 plus 1. So we're clear on the terms. We know what it means to cube something. So now we're going to go and set up a no show table and attempt to prove this conditional statement. So here's our blank no show table for this proof. Again, we're proving a conditional statement that if n is odd, then n cubed is odd. In a conditional statement, there's a hypothesis and a conclusion. Okay, so the hypothesis here is that n is odd. The conclusion is that n cubed is odd. And so, remember conditional statements are always true, except for one situation that could happen. That's when the hypothesis is true and the conclusion is false. So our goal in a direct proof is to rule that possibility out that if you assume that the hypothesis is true, then the conclusion must follow. The conclusion is never false. It always happens. So the first step in this proof, and we're going to label that p, which is our usual label for the hypothesis, is to assume the hypothesis. So if I assume the hypothesis, what I know is that n is odd. Okay. And why is that? Because that's the hypothesis. Assuming the hypothesis is always the first step in a direct proof of a conditional statement. The next thing we're going to do in the no show table is kind of jump to the end and think, okay, where do I want to end up? Having assumed that the hypothesis is true, what do I want to end? Well, I want to end, of course, in the conclusion. And that would be to say that n cubed is odd. This is sort of a happy state that's off somewhere in the future. And so I don't really know what the reason is for this yet. I would like for this to finally be true, but I don't know why yet. So there's no reason for this yet. Although the reason we're using the no show table is because a mathematical proof that's well constructed, every statement that you make, every claim that you make is backed up by some legit mathematical reason, okay? And a no show table forces us to be explicit about what that reason is. And so that's why it's a good sort of scaffolding to get started. Okay, so it's the stuff in the middle, the steps in the middle that are mysterious to us. So let's work on this. So we could either do one or two things. I could start with what I know and kind of move forward, or I could start with and back up a step. I think we're going to start here and go forward a step, which would mean, let's go, let's call this line P1. And maybe we could say, alright, we know that n is odd. What does that mean? Well, what it means is if you look up the definition of odd, is that n is equal to 2k plus 1 for some integer k. I'm going to abbreviate a lot here. Okay, so n means integer. So n is equal to 2k plus 1. Now what allows me to say that? That's the definition of odd. Okay, so you see, I've made a claim that there does exist a k such that n is 2k plus 1. What gives me the right to say that? That's just the definition of the term that I'm using. So unless you have some composing definition, then you have to accept line P1. Okay, so now what? Well, let's see what we're trying to do here. I eventually want to say something about n cubed. What seems like a logical thing to do next is I've got something that expresses n. Why don't I think about cubing that? Okay, so n cubed would be 2k plus 1, the quantity cubed. Alright, the reason that you should accept line P2 is because that's just algebra or arithmetic, really. It's just cubing both sides of an equation. Okay, so so far, all the claims that I've made in this proof at the front end of it are legit and you have to believe them in different definitions or you're just crazy. Alright, now what? Well, why don't we do what comes natural? Let's expand that right side out and see what we got. So n cubed is, using some algebra, that's going to give us 8k cubed plus 12k squared plus 6k plus 1. Okay, and that's of course just by algebra, just expanding that cube out. Okay, so I think we can buy that. But now what? Where do we go from here? Well, it's kind of hard to say, so let's try not, let's don't even try to. And let's go down here to the end. Okay, so I want to prove that n cubed is odd. I want all this stuff to eventually land me at knowing that n cubed is odd. But what does that mean? So let's rephrase the thing we're trying to prove. Okay, so n cubed being odd would mean that n cubed is equal to 2 times another integer plus 1. That's what odd means. Now I don't want to use k for the name of my integer because that's already taken. So k is kind of off limits here. So let's just choose another letter of the alphabet. Let's call it l. So 2l plus 1 for some integer l. Okay, that's what I really want to prove. I want to prove that n cubed is odd, sure, because that's what the problem says. But what that means is I need to show that n cubed is equal to 2 times a whole number plus 1. Okay, and we don't know why that's going to be true yet either. So we have some more work to do still. So let's look up at where we are. Well, you know, I have n cubed, of course I could copy that down here, is equal to a bunch of stuff plus 1. So I wonder if I could take the stuff that is out here that I'm underlining in the wavy line and write it as 2l. Well, it does appear that I can. There's a common factor of 2 on all those things. So n cubed, if I just factor out the 2 would give me 4k cubed plus 6k squared plus 3k. And that's all plus 1. And that's from Algebra 2. I'll just be a little more descriptive and say it's because I'm factoring. Okay, so now look at what I have. I have n cubed equal to 2 times something plus 1. That sounds good, but remember, it can't just be 2 times anything. It has to be 2 times an integer. Okay, so I need to make one more step here and explain why I believe that this quantity right there is an integer. So I'm going to make that claim right here that 4k cubed plus 6k squared plus 3k is an integer. An integer. Now why is that the case? It's going to be because of the closure properties of the integers that you learned from your reading. Closure properties, we know that the integers are closed under addition and multiplication, which means that if you take two integers and add them, you get another integer, you take two integers and multiply them, you get another integer. And that's really all we're doing here. We're starting with an integer. That's why that term is important. And what am I doing to it? A bunch of stuff that boils down to adding and multiplying. So this is going to be an integer as well. Okay, so I'm going to jump right to here, in fact, and say that n cubed is 2l plus 1 for some integer l. Why? Because I can go over here and now set that l equal to 4k cubed plus 6k squared plus 3k. That's an integer because of line p5. And now I have n cubed equals 2 times an integer plus 1. And by definition that's the final reason, final nail in the coffin here. That proves that n cubed is odd. So if you read this from top to bottom, you get a complete flow of an argument. But notice you don't construct it top to bottom. You construct it nonlinearly by bouncing around, either pushing forward from what you know or rephrasing what you want to prove and trying to get them to connect up in the middle. In the next video, we're going to look at another example of a direct proof of a conditional statement also using a no-show table. And there are many, many more to come throughout the course. So thanks for watching.