Lecture 3 21 127 Concepts of Math 08 31 2012 Prime Time The following result and proof are due to Euclid way back in Anicent Greece Definition A natural number p that is greater than 1 is called a prime number if the only positive divisors of p are 1 and p A non prime positive integer is called a composite number Theorem There are infinitely many prime numbers Proof Assume instead that there are only finitely many prime numbers Let s say there are k primes total where k is some natural number List those primes in ascending order p1 p2 p3 pk so that pk is the largest of these prime numbers Define the new number N p1 p2 p3 pk 1 It must be true that N is a prime number We can show this by pointing out that N does not have any prime factors because it is not divisible by any of the prime numbers in the list above Specifically notice that pNi leaves a remainder of 1 for every prime pi Remember that in this hypothetical world p1 pk are all of the primes Also N pk so pk is not actually the largest prime number Therefore our original assumption was invalid there must be infinitely many prime numbers More on Gauss Sum Recall that last time we discovered the following formula S n n X k 1 2 3 n k 1 n n 1 2 We found a way to pair off the terms of the sum and write a product instead of a sum However we had two separate cases depending on whether n is odd or even Maybe we can avoid having cases because we ended up with the same formula for both cases Notice that we can write the sum one way and then write it backwards 1 n n 1 2 n 1 n 1 n 1 n S n 2 1 S n n 1 n 1 2S n This yields n 1 n 2 S n and therefore S n n n 1 2 1 There is even a geometric interpretation of this problem Draw an n n square of boxes and express its area in two ways We find n2 S n S n n so then 2 S n n2 n and therefore n n 1 2 What if we looked at this differently We might notice that S n n2 S n S n 1 is another valid way of expressing the area of the square We might even write this as just S n S n 1 n Does this help us Only if we knew what S n 1 was Is there any sum that we definitely know Sure S 1 1 That s easy This helps us find S 2 S 1 2 1 2 3 This helps us find S 3 S 2 3 3 3 6 This helps us find S 4 S 3 4 6 4 10 And so on We are eliminating some of the work right Rather than evaluate S 5 as a sum of five terms we only have to add two terms How can we use this to tell us a general formula for any natural number n Let s be more efficient and develop a proof machine that generalizes our method Essentially we want to use the basic fact S 1 1 and the general fact that S n S n 1 n to prove something about all natural numbers for an arbitrary n Proof We want to show that S n n n 1 2 for every natural number n Observe that this fact holds for n 1 because S 1 1 and 1 2 2 1 Also observe that for any n we have S n S n 1 n Let s suppose that there is a natural number k for which the formula works That is suppose k is some arbitrary and fixed natural number and that S k k k 1 Why can we do this 2 Let s now try to prove something about S k 1 We know that S k 1 S k k 1 2 Substituting in what we supposed about S k and simplifying algebraically we find S k 1 S k k 1 k k 1 k 1 2 k k 1 2 k 1 2 k 1 k 2 2 By our assumption Making a common denominator Factoring Voila We have a formula for S k 1 and it perfectly matches what we thought it would k 1 k 1 1 k 1 k 2 S k 1 2 2 We have shown that our proposed formula holds true for the value n 1 We have also shown that whenever our formula holds true for the value n k it must also hold true for the value n k 1 Therefore our proposed formula holds true for every natural number n Whoa does that work We just claimed that it does but how do we know for sure If you have seen mathematical induction before this should be familiar because this is precisely what we are doing here We are making an inductive argument Every instance of our formula i e for every value of n depends somehow on previous instances of the formula and we can use some knowledge about those previous instances to show something about a later instance If we also show that the first instance or first few as the case may be is valid then our inductive argument gets off the ground and our proof machine takes over covering all possible cases for us There are other popular analogies for how inductive arguments work the domino analogy which I use in the textbook and Mojo the Magical Mathematical Monkey ask Prof John Mackey about that one Does this seem convincing It might make it believable However this is not a proof Induction is a great strategy but we don t know yet that it works What I have just provided is an intuitive description of how the method kinda works I did not give you a rigorous mathematical proof of the method That is something we will need to do We will get to that in due time a few weeks perhaps and it will require us to talk about a bunch of mathematical topics along the way Inductive Arguments For now we can look at a bunch of inductive arguments to get used to playing around with facts like this that depend on other instances of the fact That way when we eventually prove that induction is a valid method we will already be great at writing out these kinds of arguments An example is the game Takeaway you played in recitation on Thursday Some Other Sums We claim that T n 1 3 5 2n 1 n X 2k 1 n2 k 1 There is a neat geometric picture to describe this Just like before this is a helpful way to come up with a proposed formula for T n but it does not prove anything 3 Another method is to use previous results …