 Welcome back to another screencast in which we're going to use what we learned about integer divisibility in the previous screencast and prove something with it. That's the main heart of this course is proof and communication, and now we're going to prove things involving divisibility. Namely this conjecture here, which is really a fact. It's going to be a true statement, a theorem. It's a property of divisibility that is similar to many of the ones that you see in the section in the Sunstrom textbook. Also something you probably picked up along the way in your school math career. It says that A divides C and B divides D. Remember that's how you read that symbol. This isn't a fraction. This isn't the fraction A over C or any such thing. It's a sentence. A divides C is a statement. And B divides D. If all that takes place, then A times B divides C times D. Now before we begin to prove, we need to just take a moment to check and make sure that the conjecture is true and that we believe that it's true. So let's just pick some A's and B's here. A and B can, we're free to choose what we want. Let's just pick three and let's say five for fun. Let's keep it simple. And then C and D can't be any integers. They need to be such that A divides C and B divides D. So remember what this means is that C is an integer multiple of A. There is some integer Q such that C is equal to A times Q. So let's pick an integer multiple of three. Let's say it's 27, for example, 9 times 3. And let's pick an integer multiple of 5. Let's say 10, 2 times 5. Now does the conclusion follow? Does A B divide C D? Well A B is obviously 15 and C D is obviously 270. So the question is does A B divide C D? And of course the answer is yes and you'll find that. And how do we know we could do long division or we could use sort of our definition. Can you fill in the blank here with an integer? And the answer is yeah, use 18 for example. You might notice that the 18 is 9 times 2 and we use 9 to get from A to C and 2 to get from B to D. So maybe it has something to do with what integer goes in the blank here. Okay, so maybe you need a couple more run-throughs of testing. Use some different kinds of integers like negative integers and so forth. But let's assume that we believe this conjecture is true and now we're psychologically capable of moving on to a proof. Now I've got a blank no-show table set up over here. And again this is a conditional statement. Okay, it's an if-then statement. The only way we know so far how to prove such a thing is directly. So what we're going to do is step through this no-show table starting with by assuming the hypothesis and ending with the conclusion and then the proof, the argument is going to fill in the middle here. So what we know here is by assuming the hypothesis, I know that A, B, C, and D are integers. And I'm just going to shorthand that like so using our set notation. A, B, C, and D are integers. And A divides C and B divides D. And again the reason for that is that's the hypothesis. We always start a direct proof by assuming the hypothesis. The conclusion at the end of this proof is going to say that A, B divides C, D. We don't know the reason for that yet. We will fill that in as we go. Okay, so we've got a bracket around this proof here. Let's move either forward or backward. And perhaps it's easier to move forward first. What does it mean to say that A divides C and that B divides D? That's the essence of a forward step. Take what you already know and rephrase it by just interpreting what it means. Well, to say that A divides C means that there exists an integer, let's call it Q1, okay, using some shorthand. Again, this means there exists an integer Q1. It's in the integers. Such that, I'll abbreviate such that with an ST. There exists an integer Q1 such that C is equal to A times Q1. That is just using the definition of divides. Okay, just like we did in the test. How did I know that 3 divides 27? Well, 27 is equal to some integer multiple of 3. And so the definition gives us that. And we can also say the same thing for B. And there exists an integer. Now we need it to be something different. We don't know, just like in the test, the integers, the quotients for the test cases we use were different. So I don't want to use the same letter here. Let's call it Q2 in the integers. Such that D is equal to B times Q2. And again, that is by both of these or by the definition of divides. That's by definition, okay? The only way you could disagree with that step is to say that I think divides means something different than what you think it means. If we agree on the definitions, there's no way you can deny this step. So there exists, to say that A divides C means there exists an integer Q1 such that C equals A Q1. And to say B divides D is something similar. But I want to make sure I'm using different Q's here. I don't really mean for them to be the same quotient necessarily. Okay? Now what can we do? Well, it's kind of hard to know. Let's move down to the bottom end of the proof here. Maybe work backwards. Because it's a similar case when I say that A B divides C D, which I don't know yet, but what does this statement mean? And that's the essence of a backwards step too. It would mean that there would have to exist an integer, let's call it Q3 such that C D, the thing being divided, is equal to A B Q3. Okay? So the very, very last line is going to be true by definition as well. How does it abbreviate a definition of divides? And this is what it means. Now maybe this gives us an idea how to get from here to here. I want to eventually end up saying something about C times D. So why don't I go up here to P1 and actually calculate C times D? This gives me an in road to making a forward step. I'm going to say what is C times D anyway? Well, C is equal to A Q1. So let me just substitute A Q1 in for C. And then substitute B Q2 in for D. Okay? And what I just did there was I substituted, did two substitutions. This for C and this for D. And that was established in line P1. Now what I can do next is a little bit of arithmetic here or algebra if you like, and that is to shuffle some things around. Remember down here in line Q1 where I eventually want to end is on the right side of that equation is A times B times something. So I'm just going to manipulate the ordering of multiplication here and commute things around. So A times B times the quantity Q1 times Q2. And that is by algebra or what specifically I did here algebra was the commutative property to change the ordering of multiplication and that's okay. And the associative property to group off the Q1 and Q2. Now why did I group off the Q1 and the Q2? Well just notice that I'm very close to the end here. I have CD equals AB times something. And that's exactly where I wanted to end up. I need to end up saying that CD equals a multiple of AB. And that's exactly how I've written it just in case this is somehow not clear. This is a C right here and an A right here. Now what's left is to say that the thing that I'm multiplying by the Q3 really is an integer, right? This can't just be AB times anything. It's got to be AB times an integer, right? It has to be in the integer. So what I need to do next is it's all important but easy to miss step that the thing I have right here is truly an integer. So I'm just going to make that assertion. Q1 and Q, not and, Q1 times Q2 is an integer. Now why is that? It's by closure. It's by the closure property of the integers under multiplication because Q1 is an integer. We saw that up here. And so was Q2. And I'm just saying that when I take two integers and multiply them together, I get a third integer. Okay, so now the only thing left in this no show table is to fill in the reason right here. So what's that going to be? Well, how do I get to the definition from the previous lines? Basically, I needed to say what the Q3 is. What is the Q3? A Q3 is the simply Q1 times Q2. That's my Q3. That's my quotient that I'm going to be using to set up the definition of divisibility which gets me to the end. So there's a completed no show table using the definition of divisibility. Notice that in a no show table, any given point, especially at the beginning, at the top end of the proof and at the bottom end, involves interpreting, rephrasing things. And many times that's using definitions of things. So instantiating that definition of divisibility super important. Thank you for watching.