 Hello and welcome to this first screencast on proof using cases. Let's talk first about what this technique of proof by cases involves, and we'll do this by looking at an example. We're going to prove the proposition for all integers n, the integer 3n squared plus n plus 14 is even. Let's try a direct proof of the statement first. Now, this is a universally quantified statement, so we can convert it into a conditional statement as such. If n is any integer, then 3n squared plus n plus 14 is even. So with a direct proof, the one assumption we could make to begin here is to assume that n is an integer, and then we need to work forward and try to show the conclusion that 3n squared plus n plus 14 is even. The problem right away with the direct proof of this statement is that the assumption doesn't give you much to work with, so we're assuming that n is an integer and that's all. Well, what could you do with that? Not much. If we knew that n was an even integer, then we could rewrite n into a different form and perhaps work forward, or maybe if we knew n were a multiple of four or something about n other than that is merely an integer, something that would allow us to rewrite n in a different form would be really useful, but we don't have that in a direct proof. But what we do know is that if n is an integer, then one of two things must be true. Either n is even or n is odd. Of course, every integer falls into one of these two categories and no integer falls into both of them. So let's try this strategy. First, we'll assume that n is even, and then see if we can prove the result that 3n squared plus n plus 14 is even. Once we have a complete proof of this result with the additional assumption that n is even, go back to the beginning and assume that n is odd, then prove the result again. So we will have two complete proofs. One proof is for the case that n is even, and the other proof is for the case that n is odd. Since every integer is one or the other, and no integer is both even and odd, these two proofs cover all possible integers n, and so together they fit to form a complete proof. This strategy is known as proof by cases. It's useful in situations where you need to prove a result, and your assumptions don't provide you with enough information. But when the objects you're working with can be split into a finite number of cases that don't overlap. For example, if you're proving something about integers, then every integer is even or odd, so you can do two cases, one for when the integer is even, and one for when it's odd. We'll see more examples of cases in later videos. If you can construct a correct proof for each of the cases, and if all possibilities are covered by the cases you've considered and no cases left out, then all those many proofs fit together to constitute a correct proof of the main result. So let's proceed with our proof here. For case one, we're going to assume that n is even. By itself, this would be an unwarranted assumption. But remember, we're going to go back once we're done here and re-prove the result in the other case where n is odd. Since n is even, there exists an integer k such that n is equal to 2 times k. Now we want to show that 3n squared plus n plus 14 is even. So substitute n equals 2k in to this expression and compute. And after a few steps, we will have our result. But this is just one case, so the proof is not completely done yet. There's one other case we need to loop back and consider. So case two, let's assume that n is odd. Then there exists an integer l such that n is equal to 2l plus 1. Substitute this into 3n squared plus n plus 14 and compute. And after a few lines, we will have our result. Since these two cases cover all possibilities of integers and the proofs are correct in both cases, then we've proven the result in its entirety. So to recap the strategy of a proof by cases, proof by cases is often helpful when the assumptions of reproof don't give enough structure to work with. In this situation, think about a finite number of non-overlapping cases into which the objects you're working with could fall such that all possibilities are covered. For example, if n is an integer, you can look at two cases, one for n even and the other for n odd. If you're working with real numbers, then you might have three cases. One for x less than zero, one for x bigger than zero, and a third for x equal to zero. There are all kinds of different ways of splitting something into cases. Go through each case one by one and make an additional assumption based on the case. Then construct a correct proof of the main result for each case. When you are done proving the result for every case, and as long as all the cases represent all possibilities and nothing is left out, then the result is entirely proven. In the upcoming videos, we're going to look at some different proofs using cases where the cases are built along different possibilities other than simply even and odd integers. So thanks for watching.