 Okay, one final screencast for this section here and we're going to take the idea of integer congruence and work a proof out with it. Here's a theorem that doesn't seem like much, but it just gives us a chance to play with this idea. So if m and n are integers and both m and n are congruent to one modulo two, then mn plus seven is congruent to zero mod two. So let's set up a no-show table. Here we go. And this is, again, a conditional statement. So let's begin with the hypothesis. And the second step would be the conclusion and just try to fill in the blank. So we're going to assume the first line in this because it's a direct proof is going to be to assume the hypothesis. And the hypothesis is that m and n are integers. That m is congruent to one mod two and n is congruent also to one mod two. Okay, and that's our first line. That's a hypothesis. And the final line down here, the conclusion is going to be that m times n plus seven, let me parenthesize that for clarity, is going to be congruent to zero mod two. And again, we don't really know exactly how that's going to happen, probably by definition or something like that. So we have our hypothesis stated in our conclusion written out. Let's do either a forward or a backward step. Why don't we try forward this time? So step P one, what we're going to do is just reinterpret what this means here. Okay, they have really two statements that m is congruent to one mod two and n is congruent to one mod two. So what does it mean to say m is congruent to one mod two? Well, that means that two divides the difference between m and one. And likewise, two divides the difference between n and one. Okay, and that's the definition of congruence mod two. Gonna horribly abbreviate that over here in this column here. So by the definition of congruence mod two, we're just going to unpack what these two statements right here mean using divisibility. Now let's unpack that a little farther. This divisibility statement in turn can be written as if two divides m minus one, that means that m minus one is equal to two times k for some integer k. I'll just put that in parentheses that k here is an integer. And also likewise, n minus one is equal to two times, let's call it l, for some l in the integers. Okay, l is an integer. Okay, and that's by the definition of divisibility. Okay, so we've unpacked kind of in two stages here. The congruence can unpack to a divisibility statement. And the divisibility statement unpacks to an equality statement about multiplication. So that's kind of nice. Now, it's kind of hard to know necessarily what to do here unless we look at the bottom. So let's kind of continue to unpack here by making a backward step. What we're eventually going to say is that mn plus seven is congruent to zero mod two. Now what does that mean? Well, that means that two divides the difference between mn plus seven and zero. Normally I would write mn plus seven minus zero. But since we all know that mn plus seven minus zero is just mn plus seven, we don't need to write the zero. Okay, so that's going to be certainly the case. And so the last line would be true by the definition of, again, congruence mod two. It's by what it means to be congruent mod two. Maybe we can make a second backward step here because if two divides mn plus seven, then what that would mean is that mn plus seven is equal to two q for some q in the integers. Okay, that would be correct to say that. So line q one would follow from q two. If I can only get to q two, q one will follow by definition of divisibility, definition of divisibility. Okay, so now how am I going to get from p two to q two? Well, this is where things might look up for us because I'm trying to eventually say something about mn plus seven. Now I have m minus one up here and n minus one here. A little bit of algebra will get us a long way here. Let me just add one to both sides of these two equations. So m is equal to two k plus one and n is equal to two, I called it l plus one. That's just that the s is algebra and the algebra that I did was simply adding one to both sides of the equation. So let's multiply m, let's just compute mn plus seven, right? Let's just go straight for the throat here. What is mn plus seven? Well, let's see, mn, I can make a substitution and write this as two k plus one. That's a plus times two l plus one. That's the m and the n and if I add the seven onto it, that's just kind of hanging out there. That's my substitution and let's do another step here. I might have to scrunch this in to get some room. This is just going to be expanding on the right hand side here. That would be four k l. Okay, using the foil method plus the first outer. Pretty bad when I forget the foil method. Plus two k plus two l plus one plus seven. Okay, so that's algebra, but I'll just say foil method. Now let's just do p six, which would be maybe just adding. Sorry, I said eight, that's supposed to be a seven. My bad, I got ahead of myself because the next thing I want to do is write this as one plus seven. Now, what I'm going to do here is actually combine a couple of steps. I'm going to, you see here I have four k l, a two k, a two l, and the one and the seven come together to be an eight. All of those have a factor of two on them. So I'm just going to factor that out and I'll be left with a two k l plus one k plus an l plus four. Okay, and that would be, this is called an algebra, right? What really would happen is I added the one and the seven together and then pulled out a two. It's factoring. Okay, now there's actually one extra line I need to put in here, p seven. I don't have any room for it, so I'll put it up here. Okay, p seven, it would go right there. And that would be to say that, see this thing in the parentheses? This is going to eventually be my q. Okay, I've got m m plus seven, blah, blah, blah, equals two times a big thing here. That's exactly what I want in line q two here, but I need to just verify again that this thing in parentheses is an integer. So I'm just going to say that. Here's the statement two k l plus k plus l plus four is an integer. Okay, and the reason is by closure. Okay, we already knew from line p, the very beginning, I'm sorry, line p two. I guess that k and l individually are integers. And so if I do a little arithmetic to them, it's going to still have an integer. So that's line p seven. And now finally I can connect. Okay, q two is going to follow because I'm just going to set. And I hope I have room here q equal to two k l plus k plus l plus four. The thing in parentheses, that's my q. Okay, and then my backward steps have prepared the way for me to get to the end here. And so that's a completed proof here. Okay, notice the thing that makes this kind of different is this double unpacking step that we did both in the front end and the back end. I started with a congruent statement. And then the congruent statement using the definition into a divisibility statement. And then the divisibility statement, in turn, through its definition, becomes a statement about multiplication. So now we're kind of back into regular arithmetic land here. So that's it, the completed proof, direct proof using congruence of integers.