 An important division in mathematics is between discrete and continuous. Now, watch the spelling, a quantity is discrete, with two consecutive E's, if it's careful in what it says and does. But if we split the E's, a quantity is discrete, if it's made up of indivisible pieces. It's continuous otherwise. Alternatively, we could say that a continuous quantity can take on any value within a range while a discrete quantity can only take on certain values. For example, which of these are discrete quantities and which are continuous? The number of students who run a race, the time it takes each student to complete the race, and the amount of water each student drinks. So the number of students has to be a whole number. You can't have five and three-quarter students running. So the number of students is discrete. The time it takes each student to complete the race, well, that's continuous because time can take on any value. The amount of water is discrete, or maybe it's continuous, and here it depends on how we measure it. If we're measuring by water bottles, then it's a discrete quantity, and bad for the environment. If we're measuring by volume of water, then it's a continuous quantity. In many cases, a discrete resource has to be allocated among several recipients with different worthiness of the resource. This is most often phrased in the problem of assigning congresspersons for voters. The size of a state's congressional delegation should be based on its population, so California, with the largest population, should have more congressmen than a state like Wyoming with the small population. The apportionment problem can show up in other contexts. For example, computers for classrooms, the number of computers should be based on the number of students. Trucks for delivery centers, the number of trucks should be based on the number of deliveries, and so on. And the important question is, how can we apportion a discrete quantity fairly? Now, this might not seem to be a hard problem. After all, there are some easy cases. Well, suppose we have three classrooms. If each classroom has the same number of students, then a fair division of a resource would give each classroom one-third of the resource. But what if the resource can't be divided by three? So if there are four computers, which class gets the extra? Even worse, if there are only two computers, which class gets nothing? And so we see that even in the easiest possible case, there might be no good solution. Well let's make it real. Suppose a school has five computers who distribute among three classrooms. If the classrooms have 21, 41, and 35 students, how should the computers be distributed? Now any reasonable person might proceed as follows. There are a total of 97 students. The standard quota is the number of computers each class should receive if the number was proportional to the class size. So the class with 21 students should receive 2197th of five, or about 1.0825 computers. The class with 41 students should receive 4197th of five, 2.1134 computers. The class with 35 students should receive 1.8041 computers. And that is a perfect distribution of these resources, except you can't give out a fraction of a computer. We can view this problem in another way. Since there are 97 students and five computers, then there are 97 divided by 5, 19.4 students per computer. We can view this result as follows. For every 19.4 students in the class, that class deserves one computer. We can then compute the standard quotas by dividing the class size by 19.4, then round the results. So back to our problem. Again, there are still 97 students and there are still 19.4 students per computer. Using 19.4 as our divisor, we find the quotas, and if we round normally, we assign the classrooms one, two, and two computers. This is obvious and easy, but will it always work? And you should probably suspect that because I've spent more than five minutes of your time on this, there are some cases where we're going to run into problems. For example, suppose students are moved to rounds, so the three classes now have 25, 44, and 28 students. So again, there are still 97 students and there are still 19.4 students per computer. But if we use this as our divisor and round, we only end up assigning four computers. But let's shift the students around some more. So again, there are still 97 students and there are still 19.4 students per computer. But if we use 19.4 as our divisor and round, we have to assign six computers. The problem of rounding gets worse as you get more recipients. And so this leads to two ways we can solve the apportionment problem. First, we can adjust the rounding rules. The other possibility is we can adjust the divisor. So take a look at these solutions next.