 So there's a number of cases where we're interested in knowing whether a number can be divided by another number, and we prefer not to actually have to do the division, because division is one of the more difficult of the elementary operations. So this falls under the category of divisibility. So let's consider a few definitions. So suppose N is a product of two numbers, P and Q. Then we say that P divides N, and conversely, if P does divide N, then I'm guaranteed that N is a product of two numbers. Now, this is a fairly important concept, and you can tell how important a concept is in mathematics by how many different ways we have of speaking of it. So divisibility is important enough that we actually have several equivalent ways of expressing the concept of divisibility. So we might say that P divides N, or we might say that N is divisible by P, or we can introduce some notation P vertical bar N, and we can read this as P divides N. And because mathematicians don't like writing a lot of stuff out, this notation here is probably the most common way of expressing this notion of divisibility. Okay, so let's see what we can do with that. Well, so for example we might consider a simple proof. We want to prove that 5 divides 475. So we are using the vertical bar for division, and so we're looking at 5 dividing 475. And we do want to refer back to our definition of divisibility. Again, this is a proof problem. So what do we want? Well, if N is equal to the product of P and Q, then P divides N. And so now I'll compare our definitions. N is 475, P is 5. So the problem is we have to figure out what Q is. And so what I have, N equals P times Q, and this is a division problem. So Q is going to be 475 divided by 5. That works out to be 495. And so I can put these together into a proof statement. Since 475 is 5 times 95, then 5 divides 475. There's a definition of divisibility, which is the same as the statement that we're trying to prove there. Useful idea to keep in mind, commutativity still applies. So 475 is 5 times 95. Then it's also 95 times 5. And I can reverse the two factors and say that 95 is also a divisor of 475. So the fundamental theorem of arithmetic gives us a guaranteed way of determining whether we can write a number as a product of two other numbers. For example, let's consider N to be this mess. And we want to prove or disprove 27 divides N. So I could answer this question by determining whether it's possible to write N as a product 27 times something else. So if we think about the prime factorization of a number as being the recipe for creating that number, the question is, can I find these ingredients, the ingredients that make up 27, in this set of factors? The important thing is that the fundamental theorem of arithmetic only applies if we have the prime factorization. So we want to write 27 as a product of primes. That's 3 to the third power. We also want to write N as a product of primes. Now, 5 and 7 are both prime numbers, so we're good there. 12, on the other hand, is not. So we need to break that up. 12 is 2 to the second times 3. That's 4 times 3 is 12. And expanding this out, that's 2 to the 40th, 3 to the 20th, 5 to the 3rd, 7 to the 4th. So the recipe for making this number N take 42s, 23s, 3 fives, 4 7s, multiply them all together, and that's how you get N. And in this list of ingredients, the question is, can I find this? And if you look, you say, sure. Here we have 23s. I only need 3 to make up 27. So I'll split off the 3 there. I'll have 17 left over, and then all the other prime factors are unchanged. So N is 27 times some other stuff. And so by definition, 3 to the 3rd, 27 divides N. And the proof that 27 divides N really is the portion in green once again. Now, an important variation on this, what if I have a sum? Here's a problem. The fundamental theorem of arithmetic only applies when you have a product of prime numbers. If you don't have a product, all bets are off. So the problem here is I have this sum that's going to mess up my ability to have a product. So let's see what we can do with that. So the question is, can I write N as a product of two things? Again, this is a sum. I want to write N as a product. And so one possibility is I can use the distributive property to simplify that expression so that it goes from being a sum to a product. So I stare at this for a moment, and I notice that everything has a, well, there's a 7 to the 8th in both of these, so I can get rid of that. There's a 5 to the 3rd in both of those, so I can get rid of that. And then I have what's left over this 2 to the 3rd and 3 square. So I can pull out the common factor of 5 to the 3rd. I can pull out the common factor of 7 to the 8th. What's left over 2 to the 3rd, 3 to the 2nd. And now I have 5 to the 3rd times 7 to the 8th times this thing. And since this is still an addition, it doesn't really matter if I figure out what that is. This is 8 plus 9 is 17, which turns out to be a prime number. And so now N is a product of prime numbers. And 11 is prime, and there is no way that I can get an 11 out of this. So 11 is not in the recipe for N, because here it is. So 11 does not divide N. And again, the essential part of the proof is the portion in green.