Week 2. Week 1 Recap. Week 2

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1 Week 2 Week 1 Recap In our first session, we outlined our big-picture goals for this course. We are going to spend these first couple weeks looking at mathematical proof, then we will take a couple weeks to look at justification in the sciences, and finally, a couple weeks on justification in the social sciences. The purpose of this overview is to look for similarities and differences across these disciplines and to develop our own theories of argumentation. While we generally shy away from deep epistemological and ontological questions, focusing instead on practice, in our first meeting we did some digging into the nature of mathematics and mathematical knowledge. We went to some interesting places in this discussion a few of us came pretty close to the brink of existential crisis. Some of the main points were: The question of whether math is invented or discovered is a difficult one (to say the least) It may be that the very question is problematic because the only way to ask it is with language, which may just not be up to the task. It seems that doing math requires agreement upon some basic premises called axioms or postulates. We then tried our hands at some basic mathematical proofs. Hopefully, what you found in that process is that mathematical proof falls cleanly within the category of deductive proof scheme. Writing a successful proof required generality (you had to declare a variable for every term), operational thinking (you had to employ mathematical operations mainly algebra), and logical inference (the overall framework for the proof followed a basic logical structure). Week 2 Proofs and Refutations This week we are going to continue looking at proof. Our reading this week is Lakatos Proofs and Refutations. In this play/essay, Lakatos highlights the social nature of mathematics with a dialogue between students and a teacher on Euler s Formula. Euler s formula tells us that for any planar graph (dots connected by non-intersecting lines),

2 [# of vertices] + [# of faces] [# of edges] = 2 V + F E = 2 This formula also applies to convex polyhedra. For example, it applies to shapes like these but not to shapes like these

3 In the conversation that takes place in Lakatos imaginary classroom, we find that the characters conjectures and definitions undergo frequent revision. For example, the teacher asserts that any polyhedron can be flattened out onto a flat surface if you remove one face, but Pupil Alpha questions that assertion. This is, in fact, not true for non-convex polyhedra take a boxy doughnut, for example. There s no way to remove a single face and flatten that shape out onto a flat surface, so, the conjecture has to be refined to exclude polyhedra of this type. As you read through Part I of Proofs and Refutations think about examples of this type of revision process outside of mathematics. Also, consider the characters proof schemes at different times. Do they change? Do they clash? Be mindful of your own proof schemes as you read.

4 Indirect proofs In Week 1, we worked through some examples of direct proof. In a direct proof, you start with a set of assumptions/definitions, and used those definitions to arrive at the desired conclusion in a linear fashion. For example, you can start with two arbitrary odd numbers, x=2m+1 and y=2n+1, and show that their sum is even by adding them up. One example of indirect proof is proof by contraposition. The statement, if p, then q, is equivalent to the statement, if not-q, then not-p. Say you want to prove the statement We could instead prove If x 2 is even, then x must be even. If x is not-even, then x 2 must be not-even. The latter is easier to prove. Try it yourself. Another, very useful, form of indirect proof is proof by contradiction. This is a bit less intuitive. In this case, we say that the negation of the statement if p, then q is the statement not-q and p. A truth-table is helpful to illustrate this relationship. p q If p, then q p not-q p and not-q T T T T F F T F F T T T F T T F F F F F T F T F First let s look at the first three columns. p and q can each be in one of two states: TRUE or FALSE. There are four possible combinations of T and F which we see in the first two columns. Now, if both p and q are true, then it is true that p implies q (row 1). If p is true but q is false, then p does not imply q (row 2). Rows 3 and 4 are where it gets a little funny. If p is false, then regardless of the tuth-value of q, we say that p implies q. Why? Well, consider the following statement: If I am elected, taxes will go down.

5 In this statement, p = I am elected and q = taxes go down. If both statements are true if I am elected and taxes go down - then I was telling the truth. If I am elected and taxes don t go down, then I was lying. BUT, if I am not elected, then whether taxes go down or not, I wasn t lying, so I was telling the truth. Now, you ll notice that the truth-values for p and not-q are exactly the opposite of those for if p then q. The way proof by contradiction works is we assume p and not q, and show that this results in a contradiction. That contradiction implies that p and not-q is false, in which case if p then q is true. Here s a classic example of a proof by contradiction: 2 is irrational An irrational number is a real number that is not rational i.e. it cannot be written as a ratio of two integers. We could write this as an if-then statement in the following way: If x = 2, then x is irrational. So, for the sake of contradiction, we will assume that 2 is rational. This means there exist integers, a, b such that 2 = '. Without any loss of ( generality, we can assume that ' is in reduced form. First, square both sides: ( And rewrite: 2 = a) b ) 2b ) = a ) By definition, this means a 2 is an even number. But, if that s true, we showed earlier that a must also be even. Then, there is an integer n such that a=2n. Now we have 2b ) = (2n) ) 2b ) = 4n ) b ) = 2n ) Well, now this means b is also even. So, we ve arrived at a contradiction. We can t have both a and b be even integers, because we assumed that a/b was in reduced form i.e. a and b have no common factors! We therefore conclude that 2 can t be rational, so it must be irrational.