Chapter 4 (Part 1): Induction & Recursion Proof Strategy (4.1)

Proof Strategy (4.1) Introduction Provide advice on how to find a proof of a theorem How proofs are constructed Method of proof by cases CS 210 Discrete Structures, Chapter 4 (Part 1)

Proof Strategy (4.1) (cont.) Proof Strategies Proof of a statement Replace terms by their definition Analyze what the hypotheses and conclusions mean CS 210 Discrete Structures, Chapter 4 (Part 1)

Proof Strategy (4.1) (cont.) Prove the result using one of the available methods of proof Example: p  q Go from p and derive q Assume q false and p true and derive a contradiction Prove the contrapositive q  p. CS 210 Discrete Structures, Chapter 4 (Part 1)

Proof Strategy (4.1) (cont.) Strategy of proof of an implication The forward reasoning Start with hypotheses Use axioms and known theorem with hypotheses Use a sequence of logical steps to derive the conclusion  This is the most common type of reasoning to prove relatively simple results. CS 210 Discrete Structures, Chapter 4 (Part 1)

Proof Strategy (4.1) (cont.) CSE 504 Discrete Structures & Foundations of Computer Science Proof Strategy (4.1) (cont.) Dr. Djamel Bouchaffra The backward reasoning Start with the negation of the conclusion Use a sequence of logical steps to derive the negation of the hypotheses Unfortunately, forward reasoning is often difficult to use to prove more complicated results. We often use backward reasoning. CS 210 Discrete Structures, Chapter 4 (Part 1) Ch. 3 (Part 1): Sections 3.1 & 3.2

Proof Strategy (4.1) (cont.) Example: Prove that the arithmetic mean is always greater than the geometric mean for positive real numbers. Proof: (a + b)/2 > ab (a + b)2/4 > ab (a + b)2 > 4ab a2 + 2ab + b2 > 4ab a2 – 2ab + b2 > 0 (a – b)2 > 0 since this last inequality is always true if a  b and since all these inequalities are equivalent, it follows that (a + b)/2 > ab (a  b). CS 210 Discrete Structures, Chapter 4 (Part 1)

Proof Strategy (4.1) (cont.) Example: Show that there are no solutions in integers x and y of x2 + 3y2 = 8. Proof: when |x|  3  x2 > 8 when |y|  2  3y2 > 8 The remaining cases are those when x takes one of the values –2, -1, 0, 1 or 2 and y takes one of the values –1, 0 or 1. By computing x2 and 3y2 with all these possible values and by combining the sum x2 + 3y2, we find that the largest sum is equal to 7. Therefore, it is impossible for x2 + 3y2 = 8 to hold when x and y are integers. CS 210 Discrete Structures, Chapter 4 (Part 1)

Proof Strategy (4.1) (cont.) Conjecture and Proof Conjecture: Statement whose truth value is unknown No matter how a conjecture was made, once it has been formulated, the goal is to prove or disprove it When mathematicians believe a conjecture may be true, they try to find a proof If proofs cannot be found by mathematicians, they may look for a counterexample Few conjecture resist attack for hundred of years and lead to the development of new parts of mathematics CS 210 Discrete Structures, Chapter 4 (Part 1)

Proof Strategy (4.1) (cont.) Example: Conjecture: Do other primes of the special form an –1, where a and n are positive integers, exist? 26 –1 = 63, 28 –1 = 255 34 –1 = 80, 45 –1 = 1023 These numbers are not prime! We cannot find any primes beside the Mersenne primes (2p-1) We can conjecture that an-1 is composite when a > 2 or when a = 2 and n is composite CS 210 Discrete Structures, Chapter 4 (Part 1)

Proof Strategy (4.1) (cont.) Proof: We need to find a factor of an-1 when a > 2 or a = 2 and n is composite. an-1 = (a –1) (an-1 + an-2 + … + a + 1) when a =2  a –1 = 2 –1 = 1 we cannot conclude! when a > 2  a –1 is a factor of an –1 with 1 < a –1 < an –1  an –1 is not prime. when a = 2 and n is composite  r and s with 1 < r < n and 1 < s < n such that n = rs an-1 = ars –1 = (ar –1) (ar(s-1) + …+ ar +1)  ar –1 is a factor of an –1. CS 210 Discrete Structures, Chapter 4 (Part 1)

Proof Strategy (4.1) (cont.) Theorem 1: The integer an-1 is composite when a > 2 or when a = 2 and n is composite CS 210 Discrete Structures, Chapter 4 (Part 1)

Proof Strategy (4.1) (cont.) Conjectures & Counterexamples The role of open problems One famous problem unsolved for approximately three hundred years led to the development of an entire branch of number theory This problem asked whether the statement known as Fermat’s last theorem is true CS 210 Discrete Structures, Chapter 4 (Part 1)

Proof Strategy (4.1) (cont.) Theorem 2: Fermat’s Last Theorem The equation xn + yn = zn has no solutions in integers x, y, and z with xyz  0 whenever n is an integer with n > 2. Proof: Refer to Andrew Wiles (Theory of Elliptic Curves) CS 210 Discrete Structures, Chapter 4 (Part 1)

Proof Strategy (4.1) (cont.) Example: The 3x + 1 conjecture When we repeatedly apply the function f, we will eventually reach the integer 1. CS 210 Discrete Structures, Chapter 4 (Part 1)

Proof Strategy (4.1) (cont.) x = 13  f(13) = 3 * 13 + 1 = 40 x = 40  f(40) = 40/2 = 20 x = 20  f(20) = 20/2 = 10 x = 10  f(10) = 10/2 = 5 x = 5  f(5) = 3 * 5 + 1 = 16 x = 16  f(16) =16/2 = 8 x = 8  f(8) = 8/2 = 4 x = 4  f(4) = 4/2 =2 x = 2  f(2) = 1 This conjecture has been verified for all integers x up to 5 * 6 * 1013. CS 210 Discrete Structures, Chapter 4 (Part 1)

Exercises Section 3.1: #6, 8, 10 and 12. CS 210 Discrete Structures, Chapter 4 (Part 1)