Mathematical Writing and Methods of Proof

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1 Mathematical Writing and Methods of Proof January 6, 2015 The bulk of the work for this course will consist of homework problems to be handed in for grading. I cannot emphasize enough that I view homework as a writing assignment. In particular you should view each homework problem you turn in as a small essay. Among other things this means that 1. First drafts are not acceptable. Just as in an essay you would write in any other course you should write what you want to say about a problem then read and revise what you have written. Obviously you will have to solve the problem you are writing about before you can say much about it so it is important to begin working on the assignments early. If you wait until the night before an assignment is due to solve the problems you will certainly not have the time to write anything but a first draft. It will show. 2. All mathematical writing has two overarching goals. In order of importance these are (1) Correctness and (2) Clarity. These are the only criteria upon which your homework will be graded. In general a correct but hard to read solution will fare better than a clear and eloquent incorrect solution. 3. This is a mathematics course. You are allowed to and expected to use mathematical notation in the course of your writing. In fact it will be impossible not to. Be sure you use it correctly. For instance, the statement x 2 4 lim x 2 x 2 = lim x + 2 = 4 x 2 is correct while x 2 4 lim x 2 x 2 = x + 2 = 4 1

2 is not. (Why?) The remainder of this handout defines and gives examples of the four methods of proof. Notice in particular that (1) I begin each proof by stating which method of proof I will be using and (2) that at the end of each proof I write down the conclusion I have reached. You should also begin and end every proof you write in this fashion. This is not a creative writing course. You get no points for originality of presentation. 1 Direct Proof A direct proof is a sequence of statements, each of which is either known to be true or is implied by the preceeding statements. This is the easiest kind of proof to understand though not always the easiest to construct. A direct proof is also called a Proof by Deduction. Here is an example. Theorem 1. If an integer is even then it s square is also even. Proof. Our proof will be direct. Let k be any even integer. Then k = 2n for some integer n. 1 Thus k 2 = (2n) 2 = 4n 2. Since 4 is divisible by 2 (is even) it follows that k 2 is also even. 2 Proof by Mathematical Induction Proof by Mathematical Induction can be very confusing at first. The first time you encounter this method of proof it can seem as if you are getting something for nothing, or as if you are assuming what you are trying to prove. Neither of these is true but the logic here can get a little slippery if you aren t paying close attention. Mathematical Induction is a way of proving arbitrarily many statements simultaneously. Suppose that you have a class of statements, say {S 1, S 2,...} that you want to prove inductively. In principle you could prove each of them independently but that would take, well, it could take the rest of your life (and then some) since the list of statements is potentially infinite. Instead we proceed as follows. 1 Specifically, n = k/2. 2

3 1. Prove the first statement directly. (In the kinds of problems that lend themselves to an inductive proof this is often extremely easy.) 2. Next prove that if statement S n is true, then statement S n+1 must also be true. (Notice that we have assumed that statement S n is true. We do not necessarily know this at the moment.) These two steps are sufficient to prove all of the statements in the given class. Here s how. We know by step 1 that S 1 is true. Therefore, by step 2 S 2 is true. Therefore, by step 2 S 3 is true. Therefore, by step 2 S 4 is true. And so on. Once the two steps given above are carried out the statements form an ascending ladder of implications: Here s an example. S 1 S 2 S 3 S 4... Theorem 2. If n is an arbitrary integer then the sum of the first n odd integers is equal to n 2. You might (quite reasonably) ask, Where are the statements mentioned above? Before proceeding with the proof let s stop and consider that question for a moment. Since n is taken to be arbitrary it might be equal to 5. Then the theorem says that = 25 which is true. We might also take n to be 8. Then the theorem says that = 64 which is also true. In fact, for any value of n we care to choose this theorem can be interpreted for that value. Clearly this theorem gives us a whole class of statements, one for each different possible value of n. Here are the statements in ascending order: S 1 : 1 = 1 2 = 1 S 2 : = 2 2 = 4 S 3 : = 3 2 = 9 S 4 : = 4 2 = 16. 3

4 Proof. Our proof will be by induction. Step 1: Let n = 1. Then 1 = 1 2 part.) = 1. (I said this was usually the easy Step 2: Induction Assumption: Suppose that (n 1) 1 = (n 1) 2 for some n. 2 Then (n 1) 1 + 2n 1 = ( (n 1) 1) + 2n 1 = (n 1) 2 + 2n 1 = n 2 2n n 1 = n 2 and the proof is complete. 3 Proof By Contradiction (Reductio Ad Absurdum) Proof by contradiction or Reductio Ad Absurdum relies on what is known as the Law of the Excluded Middle. That is, given a statement and it s negation one will be true and the other will be false. They cannot both be true and they cannot both be false. To prove a statement by contradiction you show that the negation of the statement is false and conclude, by the Law of the Excluded Middle, that the statement itself is true. Generally, you prove that a statement is false by showing that it implies some other statement and its negation. Thus you contradict the Law of the Excluded middle or equivalently you reduce (reductio) the statement to and absurdity (ad absurdum). Here is an example. Theorem 3. There are infinitely many prime numbers. 2 Note that the kth odd integer is given by 2k 1. 4

5 Proof. We will prove this theorem by contradiction 3. RAA Assumption: There are finitely many prime numbers. Since we are assuming that there are finitely many primes we can enumerate them. Let {p 1, p 2,..., p n } be all of the prime numbers. Next, form the number, P, obtained by multiplying all of the primes and adding one. That is, let P = p 1 p 2 p 3 p n + 1 Clearly P is not divisible by any of the primes on our list. (Why?) Therefore either there is a prime number between the largest prime on our list and P or P itself is prime. In either case we are unable to enumerate the primes. This contradicts our assumption that there are finitely many primes. Therefore our RAA assumption is false. Therefore the negation of our RAA assumption is true. Conclusion: There are infinitely many primes. 4 Proving the Contrapositive Proving the contrapositive is in some ways the subtlest form of proof 4. It relies on the following point of logic. If Statement A implies Statement B then the negation of Statement B implies the negation of Statement A and conversely. More succinctly and using the notation of symbolic logic we have: (A B) ( B Ã) Read the preceeding very carefully. Notice that we are not saying that a statement and its negation are equivalent 5. In fact they are not. A statement and its contrapositive are equivalent however 6. Theorem 4. Let a and b be positive real numbers. If a b then b a b. 3 This proof is due to Euclid. 4 So much so that I have been unable to come up with a good example of it for this handout. The theorem I prove using the contrapositive could just as easily have been proved directly. 5 For instance the negation of A implies B (A B) is not A implies not B (Ã B). 6 The contrapositive of A implies B (A B) is not B implies not A ( B A). 5

6 Before proceeding with the proof notice that if we get the statement of the contrapositive wrong, everything that follows will be wasted (though not necessarily incorrect.). It is therefore very important to get be certain we have stated the contrapositive correctly. Proof. We will prove the contrapositive of this statement. Contrapositive: Let a and b be positive real numbers. If b < b then a a > b. Since we are proving the contrapositive we begin with the assumption b < a b. Dividing both sides by b gives b a gives b < a which completes the proof. < 1 and multiplying both sides by a 6