 In the previous video, we learned about what a field is. Now, what I want you to take away from that video is that fields are the number systems that we're used to from previous algebra classes. For example, fields are exactly the environment where linear equations can be solved. So let, for example, F be an arbitrary field. It could be the real numbers or it could be some other field that we haven't discussed yet. It doesn't really matter. All we know about the field F is that we can add, subtract, multiply, divide numbers using the laws of associativity, commutivity, and the distributive laws, excuse me. So let F be a field. And let's solve the given linear equation AX plus B equals C. So what we mean here is that the coefficients A, B, and C, these are numbers that belong to the field and X here is a variable. We want to find a number, a scalar in the field F that we can assign to X to satisfy this equation. But that's all we know here. And so as X is a placeholder, can we find a number X that'll satisfy this equation? That is the solution to this linear equation and an assignment to the number X, not just the assignment to number X here, that'll make the left-hand side equal to the right-hand side when we perform the field operations in this case addition and multiplication. Now, if we don't know what it is, we would just start solving it, right? So how does one solve this? The fact that we don't know what the field is doesn't actually hinder us. The fact that it's a field means we can do the following. We take AX plus B, the number B has an inverse. And so if AX equals C, what we can do is we can add to both sides of the equation the number negative B. We're gonna add to both sides number B here. This is just using the additive inverse property. There exists some inverse to B. We're gonna add it to both sides. And if we add the same number to both sides, the things will still be equal. That's a property of equality here. Now, the next thing is we're gonna re-associate. We're gonna move the parentheses on the left-hand side, excuse me. So by redoing parentheses, we get AX plus B plus negative B. And then on the right-hand side, we simply just get C minus B. So on the left-hand side, we were able to use the additive associativity property. We were able to redo parentheses. Now, on the right-hand side, C minus B at adding negative B is the definition of subtraction. So continue on the left-hand side. If you take B plus negative B, that's gonna equal zero. That happens by construction. And adding zero to AX just gives us AX. Adding zero doesn't change anything. And so we've now used the additive identity property. Oops, identity. So you'll see that we've used all the axioms of addition. We use the inverse axiom associativity identity. And we also use commutivity, as I should mention right here, right? The fact that subtraction is a well-defined operation is a consequence of the inverses, but also commutivity. Cause C minus B, does that mean C plus negative B or does that mean negative B plus C? We can only define C minus B if we're allowed commutivity there. So what we've now done is we've got rid of the plus B. We took all the steps. Now, we usually go much faster than that, right? Usually when you work an algebra question like this, AX plus B equals C, you probably do something like, oh, minus B, minus B. Cancel, you get AX equals C minus B. You do that as like a one-liner. You might even do that in your head. And that's just because you're so accustomed to the field axioms that you can learn how to do it so quickly. What I want you to be aware is that if you can do that for real numbers, we can do that for any field. We can cancel the B on the left-hand side. We have to do the same thing for the A, the coefficient of A now. How do you get rid of the A? Well, if A doesn't equal zero, which if A equals zero, there actually wouldn't be an X and there would be no solution. So if A doesn't equal zero, I would then divide both sides by A. Divide both sides A. But when I say divide both sides by A, what I really mean is I'm going to multiply the left-hand side by A inverse. We're going to do that on the left-hand side as well. So again, we're now using the multiplicative inverse property. There exists this reciprocal element. We then are going to re-associate the left-hand side. We get A inverse A times X equals A inverse, A inverse C minus B right here on. And so we're using the multiplicative and associativity property now, like that, going on. A inverse times A, that's going to equal just one. We get one X and the right-hand side, we get C minus B over A. We can write that as a fractional solution here. And then one times X will just equal X, like so. And so finishing up here, we use the multiplicative identity and the fact that division is well-defined, right? The fact that we're able to go from here to here is because we, in fact, have a multiplicative commutivity. Because if we weren't commutative, dividing on the left by inverse, A inverse can be very different than dividing on the right by inverse. We want this to be well-defined. And so we've now solved for A, or solved for X, excuse me, X will equal C minus B divided by A. And we can solve this linear equation using the axioms. Now you might be like, why didn't you use the distributed property? Why do we want that? Well, that allows us to have more complicated linear expressions. We actually have a very simple linear equation in this example here. And so the properties of a field allow us to solve a linear equation in every field. And so if we look at an example like we've seen before, two X plus one equals six, we can solve this by the usual manner, minus one, minus one, cancel. We end up with two X equals five. And so in order to do that, in order to cancel the one on the left-hand side, or sometimes we say, like we want to move the one to the right-hand side, to move the number one to the right-hand side requires we have, in terms of addition, inverse's identity, associativity, and commutativity. We need all four of those ingredients to move the number from one side to the other side. And to get rid of the two on the left-hand side, we have to divide both sides by two, and we get that X equals five halves. We can very quickly solve this equation, but in order to move the two from the left-hand side to the right-hand side, we need multiplicative inverses, multiplicative identities, associativity, and the commutativity property, right? We didn't actually need distribution here, but it will come up eventually. Again, if these things are a little more complicated. Let's look at a linear equation over a different field with complex numbers. I want you to be aware that the process of solving this is the exact same, although the arithmetic of adding, subtracting, multiplying, dividing complex numbers might be a little bit more involved this time. So to solve this linear equation, because complex numbers are a field, what we have to do is we have to subtract three minus i from both sides of the equation. Now, when it comes to adding two complex numbers, bi, a plus bi, add it to c plus di. In that case, you just add like terms. You add the real parts, a plus c, and then you add the imaginary parts, b plus d. That's how we add them and subtracting is similar, right? So we're gonna get i plus one x equals, we take four minus three, which is one. We take five i plus i, because it's a double negative when you distribute that negative sign there. You get five i plus i, which is a six i, so we get i plus one times x equals one plus six i. That's how we get rid of that part. How do you get rid of the i plus one on the left? Well, you're just gonna divide by i plus one, divide by one plus i right here. This will cancel on the left-hand side, and so we see that x equals one plus six i over one plus i. This is the solution, although we should simplify it. Remember that when one multiplies by complex numbers, a plus bi times c plus di, what you get here is by the distributive property, right, the usual foil, you'll end up with ac plus adi. You get bc times i, and then you get bd times i squared. Now remember i squared is the square, i is the square root of negative one, so i squared is itself negative one, so you get ac minus bd, and then you get ad plus bc times i. And so if you want to, you certainly could just memorize this formula whenever you have to multiply complex numbers. I mean, if you don't, you could re-derive it by foil, but it can be a little bit faster is to kind of memorize what happens here. Division is handled very similarly when you're dividing complex numbers. If you have a plus bi over c plus di, the usual strategy is to multiply the top and bottom by the conjugate of the denominator, switch the sign of the imaginary part. And the reason you do this is when you multiply a complex number by its conjugate, you always end up with c squared plus d squared. It's just gonna be the sum of squares of its real imaginary part. And then you have to foil out the numerator, and when you foil out the numerator, you end up with ac, you're gonna get plus bd, and then that's the real part, and then the imaginary part, you're gonna end up with, I'm just using the formula that I see right here to help me foil out the numerator a little bit quicker here. And so in the numerator, you're gonna get a negative ad, and then we should end up with a bc times i. Let me put that all on the screen. So we divide complex numbers, again, you can just use this formula over here if you want to. Again, you can just re-derive it by multiplying by the conjugate each and every time, but in our example, if we were to go through the details of this, the bottom will look like one square plus one squared, so one plus one. In the top, we're gonna take a and c, which is that one, which is a one. You're gonna take b and d, six times one, which is a six. And then you're gonna start to, for the other one, you're gonna get crosses. So one times one, that's gonna give you a negative one. Remember the negative sign that we saw right here. And then you're gonna take b and c, which gives us a six again, like so. And so this simplifies to be seven plus five i over two. And when one works with complex numbers, it's generally best to write it as a real part and an imaginary part. So you get seven halves is the real part, and then five halves is the imaginary part. And this would give us the solution to this complex linear equation. Now be aware that this is how we could solve one linear equation with complex numbers. But the techniques we used here would apply for any linear equation over any field. The only thing that changes is not the algebra, but it's the specific arithmetic. How do you multiply complex numbers? How do you divide complex numbers? How do you add them? That's what's changed when we went from the rational equation to this complex equation. The algebra did not change. And that's why we're gonna focus on these different types of fields in our course, is that I want you, the student, to recognize that all of the algebra stays the same, but when you change field, it changes arithmetic. It never actually changes the algebra. And so in the next video, we're gonna introduce a new field and see that all of the algebra of linear equation stays the same, but the arithmetic of the field will change.