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Proofs Zeph Grunschlag Copyright © Zeph Grunschlag, 2001-2002. Agenda Proofs: General Techniques Direct Proof Indirect Proof Proof by Contradiction L14 2 Proof Nuts and Bolts A proof is a logically structured argument which demonstrates that a certain proposition is true. When the proof is complete, the resulting proposition becomes a theorem, or if it is rather simple, a lemma. For example, consider the proposition: If k is any integer such that k 1 (mod 3), then k 3 1 (mod 9). Let’s prove this fact: L14 3 Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) L14 4 Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) 1. k 1(mod 3) L14 5 Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) 1. k 1(mod 3) 2. n k-1 = 3n L14 6 Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) 1. k 1(mod 3) 2. n k-1 = 3n 3. n k = 3n + 1 L14 7 Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) 1. k 1(mod 3) 2. n k-1 = 3n 3. n k = 3n + 1 4. n k 3 = (3n + 1)3 L14 8 Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) 1. k 1(mod 3) 2. n k-1 = 3n 3. n k = 3n + 1 4. n k 3 = (3n + 1)3 5. n k 3 = 27n 3 + 27n 2 + 9n + 1 L14 9 Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) 1. k 1(mod 3) 2. n k-1 = 3n 3. n k = 3n + 1 4. n k 3 = (3n + 1)3 5. n k 3 = 27n 3 + 27n 2 + 9n + 1 6. n k 3-1 = 27n 3 + 27n 2 + 9n L14 10 Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) 1. k 1(mod 3) 2. n k-1 = 3n 3. n k = 3n + 1 4. n k 3 = (3n + 1)3 5. n k 3 = 27n 3 + 27n 2 + 9n + 1 6. n k 3-1 = 27n 3 + 27n 2 + 9n 7. n k 3-1 = (3n 3 + 3n 2 + n)·9 L14 11 Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) 1. k 1(mod 3) 2. n k-1 = 3n 3. n k = 3n + 1 4. n k 3 = (3n + 1)3 5. n k 3 = 27n 3 + 27n 2 + 9n + 1 6. n k 3-1 = 27n 3 + 27n 2 + 9n 7. n k 3-1 = (3n 3 + 3n 2 + n)·9 8. m k 3-1 = m·9 L14 12 Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) 1. k 1(mod 3) 2. n k-1 = 3n 3. n k = 3n + 1 4. n k 3 = (3n + 1)3 5. n k 3 = 27n 3 + 27n 2 + 9n + 1 6. n k 3-1 = 27n 3 + 27n 2 + 9n 7. n k 3-1 = (3n 3 + 3n 2 + n)·9 8. m k 3-1 = m·9 9. k 31(mod 9) L14 13 Direct Proofs Previous was an example of a direct proof. I.e., to prove that a proposition of the form “k P(k) Q(k)” is true, needed to derive that “Q(k) is true” for any k which satisfied “P(k) is true”. Three basic steps in a direct proof: L14 14 Direct Proofs 1) Deconstruct Axioms Take the hypothesis and turn it into a usable form. Usually this amounts to just applying the definition. EG: k 1(mod 3) really means 3|(k-1) which actually means n k-1 = 3n L14 15 Direct Proofs 2) Mathematical Insights Use your human intellect and get at “real reason” behind theorem. EG: looking at what we’re trying to prove, we see that we’d really like to understand k 3. So let’s take the cube of k ! From here, we’ll have to use some algebra to get the formula into a form usable by the final step: L14 16 Direct Proofs 3) Reconstruct Conclusion This is the reverse of step 1. At the end of step 2 we should have a simple form that could be derived by applying the definition of the conclusion. EG. k 31(mod 9) is readily gotten from m k 3-1 = m·9 since the latter is the definition of the former. L14 17 Proofs Indirect In addition to direct proofs, there are two other standard methods for proving k P(k) Q(k) 1. Indirect Proof For any k assume: Q(k) and derive: P(k) Uses the contrapositive logical equivalence: P Q Q P L14 18 Proofs Reductio Ad Absurdum 2. Proof by Contradiction (Reductio Ad Absurdum) For any k assume: P(k) Q(k) and derive: P(k) Q(k) Uses the logical equivalence: P Q P Q P Q (P Q ) (P Q ) (P Q ) (P Q ) (P Q ) (P Q ) P Q Intuitively: Assume claim is false (so P must be true and Q false). Show that assumption was absurd (so P false or Q true) so claim true! L14 19 Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) L14 20 Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) 1. Suppose k 2 is not even. L14 21 Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) 1. Suppose k 2 is not even. 2. So k 2 is odd. L14 22 Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) 1. Suppose k 2 is not even. 2. So k 2 is odd. 3. n k 2 = 2n + 1 L14 23 Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) 1. Suppose k 2 is not even. 2. So k 2 is odd. 3. n k 2 = 2n + 1 4. n k 2 - 1 = 2n L14 24 Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) 1. Suppose k 2 is not even. 2. So k 2 is odd. 3. n k 2 = 2n + 1 4. n k 2 - 1 = 2n 5. n (k - 1)(k + 1) = 2n L14 25 Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) 1. Suppose k 2 is not even. 2. So k 2 is odd. 3. n k 2 = 2n + 1 4. n k 2 - 1 = 2n 5. n (k - 1)(k + 1) = 2n 6. 2 | (k - 1)(k + 1) L14 26 Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) 1. Suppose k 2 is not even. 2. So k 2 is odd. 3. n k 2 = 2n + 1 4. n k 2 - 1 = 2n 5. n (k - 1)(k + 1) = 2n 6. 2 | (k - 1)(k + 1) 7. 2 | (k - 1) 2 | (k + 1) since 2 is prime L14 27 Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) 1. Suppose k 2 is not even. 2. So k 2 is odd. 3. n k 2 = 2n + 1 4. n k 2 - 1 = 2n 5. n (k - 1)(k + 1) = 2n 6. 2 | (k - 1)(k + 1) 7. 2 | (k - 1) 2 | (k + 1) since 2 is prime 8. a k - 1 = 2a b k+1 = 2b L14 28 Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) 1. Suppose k 2 is not even. 2. So k 2 is odd. 3. n k 2 = 2n + 1 4. n k 2 - 1 = 2n 5. n (k - 1)(k + 1) = 2n 6. 2 | (k - 1)(k + 1) 7. 2 | (k - 1) 2 | (k + 1) since 2 is prime 8. a k - 1 = 2a b k+1 = 2b 9. a k = 2a + 1 b k = 2b – 1 L14 29 Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) 1. Suppose k 2 is not even. 2. So k 2 is odd. 3. n k 2 = 2n + 1 4. n k 2 - 1 = 2n 5. n (k - 1)(k + 1) = 2n 6. 2 | (k - 1)(k + 1) 7. 2 | (k - 1) 2 | (k + 1) since 2 is prime 8. a k - 1 = 2a b k+1 = 2b 9. a k = 2a + 1 b k = 2b – 1 10. In both cases k is odd L14 30 Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) 1. Suppose k 2 is not even. 2. So k 2 is odd. 3. n k 2 = 2n + 1 4. n k 2 - 1 = 2n 5. n (k - 1)(k + 1) = 2n 6. 2 | (k - 1)(k + 1) 7. 2 | (k - 1) 2 | (k + 1) since 2 is prime 8. a k - 1 = 2a b k+1 = 2b 9. a k = 2a + 1 b k = 2b – 1 10. In both cases k is odd 11. So k is not even L14 31 Rational Numbers an Easier Characterization Recall the set of rational numbers Q = the set of numbers with decimal expansion which is periodic past some point (I.e. repeatinginginginginging…) Easier characterization Q = { p/q | p,q are integers with q 0 } Prove that the sum of any irrational number with a rational number is irrational: L14 32 Reductio Ad Absurdum Example –English! You don’t have to use a sequence of formulas. Usually an English proof is preferable! EG: Suppose that claim is false. So [x is rational and y irrational] [=P] and [x+y is rational] [= Q]. But y = (x+y ) - x. The difference of rational number is rational since a/b – c/d = (ad-bc)/bd. Therefore [y must be rational] [implies P ]. This contradicts the hypotheses so the assumption that the claim was false was incorrect and the claim must be true. • L14 33 Proofs Disrefutation Disproving claims is often much easier than proving them. Claims are usually of the form k P(k). Thus to disprove, enough to find one k –called a counterexample– which makes P(k) false. L14 34 Disrefutation by Counterexample Disprove: The product of irrational numbers is irrational. 1 1. Let x 2 and y . Both are irrational. 2 1 2. Their product xy 2 1 is rational. 2 L14 35 Blackboard Exercise for 3.1 Proof that the square root of 2 is irrational L14 36