 So welcome to the screencast where we're going to take a look at another simple direct proof of a conditional statement. Actually in this video we're going to do slightly more than this. We're going to put some ideas together that you've seen so far by developing a conjecture and then trying to prove it if we believe it's true. So in an earlier video we defined the concept of an even number as follows. An integer n is said to be even if n is equal to 2 times k for some integer k. So take a look at the expression n cubed plus n squared plus n plus 1. The simple powers of n up through the third power. I'm wondering if we know that n is even. What could we say about this expression in terms of whether it's even or odd? Is this expression going to come out always even? Will it flip and become always odd? Or will it be either even or odd? Could be both depending on what n is. So let's play with the problem and form a conjecture. And if we can do that form a conjecture then we can try and prove for that conjecture. The conjecturing part I'm going to leave to you as a concept check. So I want you to take several even values of n. Skip the odd ones because this situation only calls for even integers for n. And compute n cubed plus n squared plus n plus 1 and see what happens each time. Then I want you to pause the video and identify the answer that seems most correct. If n is an even integer, any even integer then is n cubed plus n squared plus n plus 1 even? Is it odd or could it be either even or odd depending on the value of n? Well once you play with the problem just a little bit you discover that when you try an even n, no matter what it comes out, no matter what it is, the expression n cubed plus n squared plus n plus 1 seems to come out odd every time. So the best conjecture we can make is to answer b. If n is an even integer then at least it appears that n cubed plus n squared plus n plus 1 is odd. This is a conjecture or an educated guess based on experimentation in play and based on the outcomes of our experiments we should totally believe that this conjecture is true. But we do not have a proof of it yet. A proof of a conditional statement must cover all the possibilities given the hypothesis. We haven't seen all even integers. We've only tried a few. We haven't tried all of them. So we don't know about any yet. We believe the statement is true but we don't know it's true until we have a proof. So for that we're going to go over and build a no show table and try to prove that if n is an even integer then n cubed plus n squared plus n plus 1 is odd. Let's go do that. Okay so here's our blank no show table and we're going to try to prove this statement again if n is even then n cubed plus n squared plus n plus 1 is odd. So I'm going to give you a question here as a concept check. What is the first thing I should do in this in this proof here? Should I assume that n is even? Should I assume that n is odd? Should I assume that n cubed plus n squared plus n plus 1 is even? Should I assume that n cubed plus n squared plus n plus 1 is odd or none of the above? Well the answer of course is I'm going to assume the hypothesis. This is always the first step in a proof of a conditional statement is to assume the hypothesis and in that case again we'll label it p and that's to assume that n is even. Okay so we can always make the first step in a proof of a conditional statement. Okay and the reason I'm doing this of course again is by the hypothesis. Okay now what we're going to end up the very last statement in this proof is to conclude that n cubed plus n squared plus n plus 1 is odd and again I don't, I'm not there yet so I don't know what the reasoning is. All ready? So let's move forward. So what does it mean? Maybe a forward step is good here to show that n is even. Well not to show. I'm assuming that n is even. So what that means is that n is equal to 2k for some integer k. Why is that true? Why can I make that statement? Because that's what even means. That's the definition, definition of the term even. That we took so much pain to develop. Now what? Well let's see. I want to prove something about n cubed plus n squared plus n plus 1. Maybe I can make a backward statement here. Okay it's similar to the forward statement. What I want to prove really in the end here is that this whole shebang here is equal to 2 times some other integer plus 1. That's an L there. So n cubed plus n squared plus n plus 1 is equal to 2L plus 1 for some integer, drop down a line here, L. This is what I, this is, if I prove q1 then q follows because that would be the definition of odd. Okay. So all this seems to gravitate around n cubed plus n squared plus n plus 1. So why don't I take p2 and compute that quantity? Okay. My next forward step. So n cubed plus n squared plus n plus 1 is equal to, I'll just slap these in parentheses here. Since n is equal to 2k I can just substitute it. Probably not supposed to be a 3. Plus 2k squared plus 2k plus 1. And this is just a substitution I guess you could say. Substitution. And now maybe the next thing should be what seems to be natural is to expand all these powers on the right hand side. So that would be 8k cubed plus 4k squared easier algebra than the previous problem. Plus 1. Okay so that's just algebra if it can even rise to that level. And now keep in mind where we want to end up. I want to end up with all this stuff. Okay here it is. Here it is. I want this to be 2 times an integer plus 1. So I got the plus 1 part going for me. And I think I see the 2 times part going for me as well. Let me factor out the 2 here. And I got 4k cubed plus 2k squared plus k plus 1. Okay and that's again algebra. You could say algebra. You could say factoring. Either way that nails that step. That's why that's true. Now I have, this is very similar to the last problem now. I have the thing that I'm working with, n cubed plus n squared plus n plus 1 is equal to 2 times something plus 1. I just need to make sure that something is an integer. Okay so again this is going to follow from closure. k is an integer. We already knew that from line p1 definition of even. And so k cubed is also an integer because k cubed is just repeated multiplication. And so 4k cubed is also an integer. And so if I take all these integers and add them together I get another humongous integer. Okay so 4k, I'm just going to say it 4k cubed plus 2k squared plus k is an integer because of closure. If I had more time and better handwriting I would maybe stretch this out to say because of the closure of the set of integers under multiplication and addition. That would be a more precise way to say that. And in a written version of this which we're going to look at in the next video I will say that. Okay so anyhow this is an integer and so I'm really done actually. I've got n cubed plus n squared plus n plus 1 equal to 2 times an integer plus 1. What gives me the right to say this line? It's because by setting l equal to the thing that I just proved was an integer. Okay? I'm just going to, this is just a, I can set anything I want equal to anything else as long as they're like variables. Okay so l, if I set l equal to this then I've written n cubed plus n squared plus n plus 1 equal to 2 times an integer plus 1. That means that this expression is odd by definition. Okay so there we have a completed framework for a proof that starts with the hypothesis, uses logic, uses definitions, uses previously known results like all this algebra stuff here, axioms of closure and so forth to establish the conclusion must be true in that case. Okay that's all for now. We have much more practice to do with constructing direct proofs but the next video is going to talk about how to take this stuff and turn it into a nice looking nicely flowing English paragraph so stay tuned.