Ch 1.4: Basic Proof Methods I

1. 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.

2. 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 or x => 0” (P \/ ~P). • Replacement: “f differentiable => f continuous”  “f discontinuous => f not differentiable.”

3. 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.

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

5. 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.

6. 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. 

7. 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. 

8. 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.

9. Example: Direct proof of P => Q 

10. 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

11. 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. 

12. Example: Proof by cases (exhaustion) 

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