1.7 Proof Methods and Strategy

Transcription

1 1.7 Proof Methods and Strategy Introduction We will introduce several other important proof methods: Proofs where we consider different cases separately, Proofs where we prove the existence of objects with desired property. We will provide advice on how to find a proof to a theorem and we will describe some tricks of the trade, including how proofs can be found by working backward and by adapting existing proofs. At the end we will discuss the role of open questions. Exhaustive Proof and Proof by Cases Sometimes we cannot prove a theorem using a single argument that holds for all possible cases. So we can prove it only by considering different cases separately. This method is based on the tautology [(p1 p2 pn) q] [(p1 q) (p2 q) (pn q)]. Instead of proving (p1 p2 pn) q, it is enough to prove each of pi q, i=1, 2,, n, separately. Such an argument called a proof by cases. EXHAUSTIVE PROOF Some theorems can be proved by examining small number of examples. Such proofs are called exhaustive proofs, since these proofs proceed by exhausting all possibilities. Example1. Prove that (n+1)^3>=3^n if n N with n=<4. Solution: We use a proof by exhaustion. That is n=1, (1+1)^3=8>3^1=3 n=2, (2+1)^3=27>3^2=9 n=3, (3+1)^3=64>3^3=27 n=4, (4+1)^3=125>3^4=81. In each of these four cases, we see that (n+1)^3>=3^n. We have used the method of exhaustion to prove the conclusion. PROOF BY CASES A proof by cases must cover all possible cases that arise in a theorem. Example3. Prove that if n Z then n^2>=n. 1

2 Solution: We prove the theorem by considering three cases, when n=0, when n>=1 and when n=<-1. LEVERAGING PROOF BY CASES When to use a proof by cases? What when there is no obvious way to begin a proof? Example 5. Formulate a conjuncture about the decimal digits that occur as the final digit of the square of an integer and prove your result. Solution: By observation one can see that the final decimal digit of a square is 0,1,4,5,6 or 9. We can prove this theorem considering 6 different cases. WITHOUT LOSS OF GENERALITY: When a general condition can be reduced to a specific one, we can start our proof for the specific condition. Example 7. Show that (x+y)^r<x^r+y^r, when x, y, r {x x>0, x R}=R^{+} with 0<r<1. Solution: Without loss of generality we can assume that x+y=1.[generally x+y=t R^{+}. Then (x/t)+(y/t)=1.] Assuming x+y=1, since x, y R^{+}, we have 0<x<1 and 0<y<1. Because 0<r<1, it follows that 0<1-r, so x^{1-r}<1 and y^{1-r}<1. This means x<x^{r} and y<y^{r}. Consequently x^{r}+y^{r}>x+y=1=(x+y)^{r}. Theorem is proven. COMMON ERROS WITH EXHAUSTIVE PROOF AND PROOF BY CASES: not all cases are covered in the proof. Example 8. Is (n+1)^3>=3^n true for all n N? Example 9. What is wrong with this proof? Theorem: If x R, then x^{2} is positive real number. (see the book) Existence Proofs 2

3 A proof of a proposition of the form xp(x) is called an existence proof. By finding an element a such that P(a) is true is called constructive proof. A nonconstructive proof can be given, using proof by contradiction. Example 10. A constructive proof Show that there is a positive integer that can be written as the sum of cubes of positive integers in two different ways. Solution: By computation we find that 1729=10^{3}+9^{3}=12^{3}+1^{3}. So the positive integer 1729 is the integer we are looking for. Existence Proofs Some theorems assert the existence of a unique element with a particular property. First existence and then uniqueness have to be shown. Existence: we show that an element x with the desired property exists Uniqueness: we show that if y x, then y does not have the desired property. Equivalently we can show that if x and y both have the same desired property, then x=y. Remark: I xp(x) is equivalent to x[p(x) y(y x P(y))]. Example 13. Show that if a, b R and a 0, then there is unique r R such that ar+b=0. Solution: 1) Existence: r=-b/a is a solution of ar+b=0. Consequently a real number r exists such that ar+b=0. 2) Uniqueness: suppose that s is a real number which satisfies as+b=0. Then as+b=ar+b. This means as=ar and so s=r. Consequently uniqueness is proven. Proof Strategies Finding proofs can be difficult. Usual process of proving: Replace terms by their definitions Analyze what the hypothesis and the conclusion mean Try to prove the result by using available methods of proof. 3

4 FORWARD AND BACKWARD REASONING Independent of your choice of proof, you need a starting point for your proof. If the proof starts with the premises and constructs steps leading to the conclusion, then this type of reasoning is called forward reasoning. When forward reasoning is not working, then backward reasoning can be used. Example14. Given x, y R^{+} with x y, their arithmetic mean is (x+y)/2 and their geometric mean is (xy)^{1/2}. Can we prove that (x+y)/2>(xy)^{1/2}? Solution: Since it is not obvious how to reach the conclusion, we can work backwards-start working on the conclusion and then see if all the steps can be reverted. If the steps are revertible (equivalent), we can construct a forward reasoning. ADAPTING EXISTING PROOFS Existing proofs can be adapted to prove a new result or some of the ideas used in existing proofs can be helpful. Looking for Counterexamples A conjuncture ends up either with a proof or with a counterexample. In any case looking for counterexamples is extremely important. It provides insights into the problems. Example17. Every positive integer is the sum of the squares of two integers is proven to be false by a counterexample. Now what about Every positive integer is the sum of the squares of three integers? Solution: the first thing we would like to do is to check it for some integers to see if it is going to tend to be true or if we see one counterexample. (Check it until n=7). What about Every positive integer is the sum of the squares of four integers? It is true. Proof Strategy in Action Usual mathematics texts formally present theorems and their proofs. Such presentations do not reveal the discovery process in mathematics. This process begins with exploring concepts and examples, asking questions, formulating conjunctures, and attempting settle these conjunctures either by proof or by counterexample. 4

5 Conjunctures are formulated on the basis of many types of possible evidence. Once a conjuncture is formulated, people will look for a proof or for a counterexample. The Role of Open Problems Famous unsolved problems lead to advances in mathematics. Quote: As long as a branch of science offers an abundance of problems, so long is it alive; a lack of problems foreshadows extinction or the cessation of independent development. Theorem 1 Fermat s Last Theorem The equation x^{n}+ y^{n}=z^{n} has no solutions in x, y, z N with xyz 0 and n N with n>2. Remark: The equation x^{2}+ y^{2}=z^{2} has infinitely many solutions in x, y, z N with xyz 0. These solutions are called Pythagorean triples and correspond to the lengths of the sides of right triangles with integer length. Andrew Wiles has proven the Fermat s Last Theorem using the recently developed theory of elliptic curves of number theory. Example 23. The 3x+1 Conjuncture Let T be the transformation that sends an even integer x to x/2 and an odd integer x to 3x+1. The 3x+1 Conjuncture states that all positive integers x, when we repeatedly apply the transformation T, we will eventually reach the integer 1. For example x=13. This conjuncture has been verified for all integers x up to 5.6x10^{13}. Additional Proof Methods In this chapter we have introduced the basic methods used in proofs. Note that we have not given a procedure that can be used to prove theorems in mathematics. Such a procedure is not existent! 5