 Now, we've just looked at the left and right distributive properties if we combine two operations. Don't get the left and right distributive mixed up with each other. Now, the way that we structure this course really remember is we're just building the stool set. We're just building definitions, understanding new concepts, and we put them inside of that to review them often so that we know about all of these. When we have our toolbox full, then we're going to start looking at the interesting things. At the moment, I do understand it's a bit difficult. It's all loose concepts. The next concept is actually quite a big one, isomorphism, but we're going to talk about isomorphism, but really we don't have any way to hang this yet. But once we've built our vocabulary, it'll be so much easier to look at the grammar when we learn this language of abstract algebra. Let's work through this next concept called isomorphism. Pack it away in your toolbox and we're going to pull it out later when things become really interesting. Now, I'm going to read to you, if I can, I'm going to put this on the screen as well. Listen very carefully. If we consider an algebraic system, S, that is our algebraic system, if we consider that to be the following, it's a set and all the relations and operations on those sets. So the set, all the relations and operations, that's an algebraic system. Now, if we consider two such algebraic systems, S and T, we consider them isomorphic, isomorphic algebraic systems. If there is a one-to-one correspondence between them, that's number one and number two, any relations and operations defined on the sets are preserved in this correspondence. So read that again. Stop the video here. It's going to continue on for you, but I'm going to write two such systems on the board and we are going to go through these things step by step. Now, let's tackle this as I said, piece by piece. I've got two sets on the board here. I've got my set A sub one and A sub two. A sub one has four elements. A sub two has four elements and they are unordered. In other words, it might have been two, four, one, three, four, three, two, one and the same with these. There's no particular ordering just because it's alphabetical and in order of the natural numbers don't read anything into that. They are unordered. Now, let's just consider these one-to-one correspondence. I'm going to do a mapping of a one-to-one correspondence. Remember, mapping, it's like what we used to say a function of, the f of x equals x squared. So for every x, I now map to an x squared and mapping real numbers to real numbers. That's a function, but now we're calling it mappings because mapping is much more interesting. Now, you can well imagine that there are 24 mappings with a one-to-one correspondence. Now, here's one such mapping, just one of those 24 and I'm going to call it my alpha mapping in functional notation that would be the f of alpha, something like that. But the alpha mapping, it maps one to p in a one-to-one correspondence. So one to p, two to q, three to r and four to s. That'd be one of the 24 valid one-to-one, or mappings at least with one-to-one correspondence. Now, let's bring in a relation on a one and a relation on a two. My relation on a two, it is ordered by, and what I'm going to do with this relation is to order these two unordered sets. So for a two, I'm going to order it by the following relation. R to s, s to p and R to q as well. So it goes to q and s and s goes to p. I've just decided that is the relation and I could call this relation something, you know, that fancy R, two. That's my relation there. On a one, I'm going to do another ordering relation and that is going to be, it's divisible by, so divides. And let's start that ordering. So one, one definitely divides two because one goes into two or two over one. There's no remainder. Two definitely divides four and one also divides three. Three certainly doesn't divide two, two doesn't divide three, three doesn't divide four, four doesn't divide three. Divide means there's no remainder if I do that division. So I've got this ordering here and this is my ordering by this relation, a relation ordering and this is my other relation that gives me order. Now consider my following another one of my two mappings that give me one to one correspondence. So one to R, two to s, three to q and four to p. So you can well see that that is different from this one. Now does this mapping be preserved, the ordering that this relation has brought in? So let's do that. So instead of R, I can put one. Instead of s, I can put two. Instead of p, I can put four. And instead of q, I can put three. I could do it the other way around as well and what do I get? I get exactly that ordering. So this mapping has preserved my ordering. So there's one bit of isomorphism that you understand quite well now. Unordered set, 24 mappings with one to one correspondence. I bring in an ordering relation on them. I take one of my mappings with one to one correspondence and it preserves this ordering. I'm gonna clean the board because I need quite a bit of space. We're going to move on to the next bit of this isomorphism. Good, so here we are with the next part. So again, I'm starting with my two unordered sets, a1 and a sub one and a sub two, or unordered again, remember one, two, three, four, pqrs. And I have these two binary operations on them. These two binary operations and the way these two binary operations are set up is the stable and how do we read the stable? So we would have the following. So two with two, this binary operation with two and two is going to give me four. And on this side, say q and r, q and r, that gives me q and r gives me another q back. I've set up two binary operations, two binary operations, one for each of my set. Remember that I still have this ordering relation. I just want to bring this back. So I have this ordering relation one. This remember was divides one, divides two, two divides four and one divides three. And this one, I'm going to call it my relation prime, r to s to p and r to q. I've set up these two origins. Something, just let's put that in. So if I have u and v as elements of a sub one and I have x and y elements of a sub two with this one to one correspondence, u to x, v to y, which is what we've done with our one to one correspondence mapping, which was on the board previously, then if I have this relation between these two that would imply this relation between these two depending on this one to one correspondence. That's what we saw. So we've dealt with relations. What we are doing now is we're moving on to binary operations. So we set up these binary operations and imagine now I have the following of one of my 24 mappings, which have a one to one correspondence. So one to r, two to s, three to q, four to p. Now we can well say if we do this to this, if I put r in there, I have r, if I put two in there, I'll have s. If I put four in there, I'll have p and if I put three in there, I'll have q. So this holds this mapping that leads to one, this mapping with one to one correspondence, it is going to preserve my ordering relation. What does that do to my binary operations? So let's say for instance, we take on this side, we take two and three. So two and three, what does that going to give us? Well, two and three, that binary operation gives me a one. And let's look if we map this. Two maps to s and three maps to q and that better give us r. And that better give us r. So we've got s on this side, q on that side and that binary operation gives me r. So what have we got here? We've got an isomorphism between my sets, a sub one and a sub two. What has happened? I've set up an ordering relation. I've got this mapping with one to one correspondence. It preserves not only my ordering relations but it also preserves my binary operations. And if we have those two together, we have an isomorphism between sets a sub one and a sub two, not only the sets there, but what goes with the sets. The relations and their operations. So two algebraic systems. So algebraic system a sub one and a sub two are isomorphic. What I can do with the one implies what happens to the other one. I've preserved my relations and my binary operations. So there you have it, an isomorphism between two algebraic systems. As I say, you've got nowhere to go hanging this knowledge. We haven't looked at groups or anything yet, but put this inside of your toolbox, understand this concept of isomorphisms between two algebraic sets and we're going to see it again in the future.