 Welcome back to another screencast about functions. So in the last video, we introduced a lot of terminology and definitions that makes rigorous the notion of a function. In the next few videos, we're going to be instantiating those definitions by building examples and non-examples. This is the most important thing you can do when confronted with new terminology and definitions. It's just to build examples and make them concrete. I'll be doing several here in the videos, but it's important that you do several many more than I do on your own. Just to make sure these basic concepts are very, very well understood in your brain. So in these next few videos, we're going to do it like this. I'm going to be presenting several processes here, one for each video. And we're going to return to the same six basic questions here that we ask about these things. First of all, is this process a function at all? And it turns out there are several questions that are sub-questions of this one right here that we'll have to ask. Secondly, if the process is indeed a function, what's its domain? Third, what is its co-domain? Then we're going to pick some points out of the domain and see what their images are. Then we're going to pick some points out of the co-domain and see if they have pre-images, and if so, what are they? And then finally, we're going to ask what's the range? And I encourage you to take these same six questions and apply them to all sorts of different processes of your own making here. So in this video, we're going to work with a real simple sort of non-mathematical example. We're going to take the set A to be the set of all full names that will be first, middle, and last names of all human beings that have ever lived or are currently living. Now, I know that not all human beings have just a first, a middle, or a last name. Some people don't have any middle names. My two daughters have two middle names and so on. But let's just keep it simple and assume that each person has a first, a middle, and a last name. And then we're going to let B be the set of all three-letter combinations of the English alphabet, like R, M, S. And we're going to define a process called initials. That is a process that takes things from A and turns them into things in B. It takes names and turns them into three-letter combinations. You can probably, by the name of the process, tell what it's going to do. The process is, I want to change a person's full name into their initials. So let's talk about the process here and see what we can do with it. First of all, is this really a function? Well, remember those five basic ingredients to being a function. Let's tick through those. Is the input set specified? Well, yeah, it definitely is. That would be the set N of all full names of people. I'm sorry, we call that A. The set of all full names of people who are living or ever live. And that's certainly a well-defined set. I don't think I could list everything in that set, but it's certainly a well-defined set. Is the output set specified? Yes, I said this process is going to send things into the set of all three-letter English alphabet letter combinations. So that's there. Is the process specified? Yes, indeed. I have all the directions I need to carry out the process. There's nothing ambiguous or incomplete about the process that I've described here. Now, the two important properties, we definitely need to check. Does every valid input have an output? I think the answer there is yes. Because if I have someone's three word name, if I have someone's full name, I definitely can create a three-letter combination out of it. There's nobody's full name that would not get an output at all. And secondly, or fifth in the list here, importantly, does every valid input have only one output? And I think the answer to that would be certainly yes. A single person with a first and middle and the last name is not going to have two different sets of initials. So this is really a function because it satisfies all five of these properties and especially the last two. So now that it's a function, we can start talking about things like its domain and co-domain. Well, the domain of this function is definitely A, the set of all first, middle, last names of people. And the co-domain is just the set B, the set of all three-letter combinations. So we don't have to think too hard about domain and co-domain here. In fact, if we page back and just look at the way that the process was set up, the domain and the co-domain are essentially given to us in the definition of the process. Now let's think about some images of things. So let's take a point in the domain. A point here would just be someone's name. So let's say we had John Joseph Alexander, for example. That person's name, what would their image be? Well, that would be the output of the function. If I take this person's name and evaluate it into my initials function, I'm going to get the result J, J, A. And we can play this with your own name or your friend's name or your mother's name or whatever. And we can see that it's very easy to take a point in the domain and find its image. So J, J, A is the image of this input here. Now what about pre-image as a point in the co-domain? And so pick an element out of the co-domain, like let's say RNT. And ask yourself, could I put something into this initials function? To get that as an output. Does RNT actually have a pre-image at all? If it does, then I should be able to fill into this blank here the name that produces that three letter combination. It may have more than one co-domain. You might, John Joseph Alexander, for example, would be a pre-image of J, J, A. But there could be another person like Jessica Jenna Austin, for example, I don't know, that has the same three letter initials. You see that quite often in real life. Now in fact, RNT does have one particular point in the co-domain. It's my name. My name is Robert Nathan Talbert. And that would be so I would be a pre-image of this particular three letter combination. And I'm sure there are a million people out there at least with the same three letter combination. So this is another example of where you could have many different inputs getting sent to the same output, for example. But if I just use my name alone, I'm not going to get sent to three, two or three different sets of initials. So that's pre-images of a single point in the co-domain. Now let's think about the range here. What is the set of all actual outputs of the function? Now that we've played with inputs and outputs, what's the set of all actual outputs of the function? Well, let's first of all keep in mind what the co-domain is, okay? The co-domain B is the set of all three letter combinations from the English alphabet, three letter combinations. Now the set of all actual outputs certainly seems like it's going to be a subset of this co-domain here. And this is something we're going to prove a little bit later on. That the range of my function, the range of the initials function will definitely be a subset of B, okay? But it may not be equal. We don't know if those two sets are equal or not. So the question here is, are there any three letter combinations in the co-domain that do not appear as actual outputs of the function? I don't really know this for a fact, but I think not. I think that every three letter combination that you could possibly make would be somebody's initials. And so I would have to go through and, I don't know how I would check that, ask the linguists about this. Are there any three letter combinations like Z, Z, Z, or something like that, who were nobody ever lived that had those three initials? It seems unlikely. It seems like there has to be somebody out there whose initials are these. So I think the range here is going to be B. I think the range is going to be the entire co-domain in this particular situation. That's not always the case. Functions, ranges do not necessarily have to equal their co-domains. But I think they do in this case. I think that every point in the co-domain does actually appear as the output of this initials function at some point. Okay, so there's our first instantiation of all the definitions here. And again, you'll see more in future videos, but don't wait for me. Go out and do some of this on your own and you will achieve a very high level of understanding. Thanks for watching.