 In this final video for Lecture 7, I wanted to demonstrate to you the importance of the group structure and why do we want the axioms of associativity, inverses, and identities? Why are they so important? Well, as one example, the operations of a group are exactly the, excuse me, the axioms of group are exactly the axioms one needs to solve equations. And let me provide you an example by solving the linear equation 2x plus 1 is congruent to 5 mod 7. Let's solve for x. So if one was to solve this equation, how does one typically go about doing it? Well, we want to solve for x, and that means we have to get rid of all the numbers attached to x by various operations. The first operation I notice is the plus, right? We have a 2x plus 1. We want to get rid of the plus 1 part of this equation. So what we're going to do is we're going to first start off by adding negative 1 to both sides of the equations. This is using the inverse axiom. The fact that plus 1 has an inverse, since a 1 has an inverse with respect to addition, that the existence is guaranteed by the inverse axiom. So we're going to add the inverse of 1 to both sides, the added inverse to both sides. But notice here, if we add negative 1 to both sides, we're, you know, the left-hand side, you already have a 2x plus 1 right there. Those are inside of parentheses. On the right-hand side, you just have 5 plus negative 1, no big deal. That would give you 4 mod 7. But on the left-hand side, in order to progress forward, we have to redo the parentheses. We have to re-associate. Redoing the parentheses, we can write this as 2x plus 1 minus negative 1. So this second step right here, use the associativity axiom, the associative axiom, in order to redo it. So that, when we redo the parentheses, now we can add 1 with negative 1 instead of 1 with 2x. 1 plus negative 1, since negative 1 is the inverse, will give us a 0. And the significance of 0 is when you add 0 to anything, you'll get back that thing again. So the 2x plus 0 is just equal to 2x. And so this uses the identity axiom. And so notice, because we have inverses, associativity, and the identity axiom, we were able to, quote-unquote, move the 1. We were able to move the 1 to the other side of the equation. This idea is like, oh, just move the 1 to the other side. Sometimes you say that in a classroom. Sometimes you say that if you're like tutoring a student in a lower algebra class. Oh, just move the 1 to the other side. In order to move the 1 to the other side, you need inverses, associativity, and identity. You need the group axioms to do that. If you do not have a group, then you might not be able to move the 1 to the other side of the equation. All right. Well, that gets rid of the plus 1. How do you get rid of the 2x, right? Well, I should mention that we are able to move the plus 1 to the other side because zn, plus is a group. It is an abelian group, but we didn't actually need commutivity to make that work. You don't need that here. Now, how do we work now with the situation we're in? Now we have to get rid of the plus 1. Now we have this 2x is congruent to 4. Well, now it's tempted to say something like, well, we need to move the 2 to the other side, right? How does 1 do that? Well, it turns out we're going to do the exact same thing. Like we did previously, we're going to use the inverse axiom. We're going to use the inverse axiom here. And so we're going to apply the inverse of 2 to both sides of the equation. Well, what is the inverse of 2? Well, one could go through the Euclidean algorithm here to find it. Notice that 2 does belong to z, or in this case, z7 star, right? We're going to use the fact that z in star with respect to multiplication is a group. So it has inverses. Notice that the GCD of 2 and 7 is equal to 1. I mean, 7 is a prime number. So there's got to be some way of combining 2 and 7 together. So if we go through the Euclidean algorithm here, notice that 7 divided by 2, you're going to get 6, which is 2 times 3 plus 1. And so there we already have it. Is that right? So because we're trying to do, oh, I'm sorry. No, no, we're doing good so far. So notice that 2 times 3 plus 1 is equal to 7. So if we subtract those, we take 2 times 3 minus 7. That's equal to negative 1. That's not what I wanted. So try this again. And if we go the other way around, oh, that's what I wanted. So we want to take 7 minus 3 times 2. This is, of course, equal to 1. And so this tells us that negative 3 is, in this situation, the inverse of 2, multiplicatively, which, of course, if you work mod 7, negative 3 is congruent to 4 if you want a positive representation. So by the Euclidean algorithm, we find this negative 3. Another way of writing this is that 4 times 2 is 8, which is 1 bigger than 7. So you're looking for a number which multiplied by 2 to give you the number 1 bigger than the multiple of 7, which you can probably do that by guess and check the numbers are small enough. But the Euclidean algorithm produces this when we find a linear combination. So use either negative 3 or 4. It doesn't matter. The two numbers are interchangeable when you're working mod 7. So with this inverse in hand, you're going to multiply both sides of the equation by 4. So you're using the inverse action. There exists an inverse. It's going to be 4 right here. Multiply both sides of the equation by 4. And then the next thing is we're going to re-associate, right? So we use the associativity axiom so we can redo the parentheses on the left-hand side. So instead of taking 4 times 2x, we're going to take 4 times 2 times that by x, which, of course, gives us 8x. 8x is the same thing when you're working mod 7 as 1x. That is 8 is the multiplicative identity when you're working mod 7 because 8 and 1 are the same thing, right? So you get 1x is the same thing as 8x. And so the fact that we have an identity, the identity axiom here, gives us that the left-hand side will be an x. And then you have 4 times 4, which is 16, which reduces to 2 mod 7. And so sure enough, 2 is the solution. So we're going to solve this linear equation because zn plus is a group and zn star, with respect to multiplication, is likewise a group. We needed the group axioms to solve this equation. And so hopefully this example illustrates the importance of groups. Groups are some of the most fundamental of all abstract structures. If you one wants to do algebra, you're going to need something that resembles a group in some way. You don't exactly need a group all the time, but fundamentally groups are this essential ingredient for doing algebra. Identities, inverses, and associatives. This is our sort of our native language when it comes to algebra. And so we will then start in the future investigating other group structures that are maybe less traditional than ones we've seen and see that the algebra basically works the same way there as well because we have the axioms of associativity, identities, and inverses.