Ch 1.4: Basic Proof Methods I A theorem is a proposition, often of special interest. A proof is a logically valid deduction of a theorem, using axioms, premises stated in the theorem, or previously established results. Axioms (or postulates) are initial statements assumed to be true, from which new concepts can be deduced.

Proofs & Tautologies In writing proofs, a working knowledge of tautologies is helpful. For example, De Morgan’s laws, contrapositive, transitivity, modus ponens, etc. See pages 13, 27. These tautologies may be used as justification for proof techniques, or as a replacement. Justification: “Either x 0” (P \/ ~P). Replacement: “f differentiable => f continuous”  “f discontinuous => f not differentiable.”

Modus Ponens Suppose we know that the following are true: P P => Q Then Q is true. Example Calculus result: f differentiable => f continuous. (1) Given: g is differentiable Therefore g is continuous, by modus ponens. (2) Given: g is continouous Cannot conclude g is differentiable by modus ponens.

Direct Proof of P => Q Assume P. …proof details here… Therefore, Q. Thus P => Q.

Number Theory Basics Let a, b be natural numbers. Then a divides b if there exists a natural number k such that b = ka. A prime number is a natural number greater than one that is only divisible by 1 and itself. An integer x is even if there is an integer k such that x = 2k. An integer x is odd if there is an integer j such that x = 2j+1.

Example: Direct proof of P => Q Suppose that a and b are integers. Prove that if a is even and b is odd, then a + b is odd. Proof. Assume that a is an even integer and that b is an odd integer. Then There exists integers k, j such that a = 2k and b = 2j+1. Then a + b = 2k + 2j+1 = 2(k+j) + 1 = 2m + 1, where m=k+j is an integer. Therefore a + b is odd.

Example: Direct proof of P => Q Suppose that a, b and c are natural numbers. Prove that if c divides a and c divides b, then c divides a + b. Proof. Assume that a, b, c are natural numbers, and that c divides a and c divides b. Then There exist natural numbers m, n st mc = a and nc = b. Then a + b = mc + nc = (m+n)c = kc, where k=m+n is a natural number. Therefore c divides a + b.

Strategies for Direct Proof of P => Q Determine precisely the antecedent and the consequent (P and Q). Replace, if necessary, the antecedent with a more usable equivalent. Replace, if necessary, the consequent with something equivalent and more readily shown. Develop a chain of statements, each deducible from its predecessors or other known results, that leads from antecedent to consequent. Every step of proof should express a complete sentence, using important connective words to complete meaning of symbols used. Sometimes it is helpful to work backward to discover a proof, by first determining the end result that needs to be shown, and then seeing what steps are needed to get there.

Example: Direct proof of P => Q

Strategies for Direct Proof of P => Q For P => (Q \/ R), one might prove the equivalent (P /\ ~Q) => R (P /\ ~R) => Q (All three are equivalent to ~P \/ Q \/ R) For (P \/ Q) => R, one might first prove the cases P => R Q => R

Example: Direct proof of P => (Q \/ R) Prove that if n is an odd integer, then n = 4j+1 for some integer j or n = 4i –1 for some integer i. Proof. Assume n is an odd integer. Then n= 2m+1 for some integer m. Case 1: m even => m = 2j for some integer j => n = 2m+1 = 4j+1 Case 1: m odd => m = 2k + 1 for some integer k => n = 2m+1 = 2(2k + 1) + 1 => n = 4k+3 = 4(k+1) – 1 = 4i-1, where i=k+1 is an integer. Thus if n is an odd integer, then n = 4j+1 for some integer j or n= 4i –1 for some integer i.

Example: Proof by cases (exhaustion)

Homework Read Ch 1.4 Do 35(4a-c,5a,b,6a-d,10)