In the last video, we calculated some distances between points in the xy-plane and even ventured into distances in space using the theorem of Pythagoras applied to certain right-angled triangles. In this video, I'd like to step back a little and just consider distances in the real line. In the last video, all of the diagrams used positive numbers or zero implicitly. Now, I'd like to introduce an apparatus for dealing carefully with all numbers, whether they are positive, negative, zero or mixtures of positive, negative, and zero. We'll look at the notion of absolute value or magnitude of a real number and some important related techniques. Absolute value or magnitude captures precisely the notion of distance along the real line. Let x be a real number. To find the absolute value or magnitude of x to be, this symbolism with x inside two vertical lines, and that's equal to x if x is greater than or equal to zero, or minus x if x is less than zero. This is just the distance between x and zero on the real line. Well, how could that be? Here's the real line with zero in the middle. If x is greater than zero, then the distance between x and zero is just x itself, which matches the formula for the absolute value of x. If x is less than zero, then the distance is minus x. Remember, x is negative, so minus x becomes positive, again, matching the formula for the absolute value of x. If x happens to be zero, then the distance of course is zero, which is by definition the absolute value of zero. For example, 1 and minus 1 above one unit distant from zero, so the absolute value is 1. 2 and minus 2 are both distant two from zero, so the absolute value is 2. 1 and a half and minus 1 and half are both three and two units from zero, so the absolute value is 3 on 2. Of course, as with have observed, the absolute value is zero, so minus zero is zero. It's an important and useful fact that the distance between two numbers is the absolute value of the difference, and it doesn't matter which order you take the difference. For example, if x is greater than y, so x appears to the right of y on the number line, then the distance between them is just x minus y. That's the absolute value of x minus y because it's positive. If x is less than y, then the distance between them now is y minus x, which is the negative of x minus y, and that's by definition the absolute value of x minus y. Then the final case, if x equals y, then the distance between x and y is zero, and that's just the absolute value of x minus y. For example, the distance from 2 to 4 is just the absolute value of their difference, which is 2. You can see this on the real line. If you draw the point 2 and 4 on the real line, you can say that the distance really is two units. On the other hand, if we look at the distance between say minus 2 and 4, that's the absolute value of the difference, which turns out to be 6. You can say that directly on the real line, the distance between the two numbers turns out to be 6. Here's some practice for you. I want you to solve the following equation, that is, once you've defined all x distance exactly three units from 2, okay. So, here's the solution. Let's draw the line and locate some points. In particular, let's locate the number 2 and look for all the points that are distant exactly three units from 2. In fact, there's two possibilities when you move to the right or the left. If you move to the right, you find the number 5, and if you move to the left, you find the number minus 1. So, the solution set just consists of the two numbers, minus 1 and 5. Let's theory the previous exercise by turning the equation into an inequality. Let's replace equals by greater than. Okay. Let's solve it. Here's the diagram from the previous solution where we had equals, but now because we've got greater than, neither minus 1 nor 5 are solutions. So, we want to exclude them. So, we'll draw little circles around them. Now, we want the distance to 2 to be more than 3, so move further away from 2 either to the right or to the left. Then we can read off the solution set. It's just the union of two intervals. One interval extends indefinitely to the left from minus 1, but not including minus 1, and the other interval extends indefinitely to the right, from 5 but not including 5, and we put those two integrals together with the union symbol. Wherever you can, it's best to use the geometric approach to solving such problems. Sometimes it's not easy or even possible to interpret the information geometrically. Then it's important to have strong technique and be able to fall back on algebraic methods. Several illustrations given in the notes that accompany this video. Please read and digest the notes carefully. When you're ready, please attempt the exercises. I look forward to seeing you again soon.
