Copyright © Zeph Grunschlag, 2001-2002. Proofs Zeph Grunschlag Copyright © Zeph Grunschlag, 2001-2002.

Agenda Proofs: General Techniques Direct Proof Indirect Proof Proof by Contradiction L14

If k is any integer such that k  1 (mod 3), 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

Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) L14

Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3) L14

Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3) n k-1 = 3n L14

Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3) n k-1 = 3n n k = 3n + 1 L14

Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3) n k-1 = 3n n k = 3n + 1 n k 3 = (3n + 1)3 L14

Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3) n k-1 = 3n n k = 3n + 1 n k 3 = (3n + 1)3 n k 3 = 27n 3 + 27n 2 + 9n + 1 L14

Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3) n k-1 = 3n n k = 3n + 1 n k 3 = (3n + 1)3 n k 3 = 27n 3 + 27n 2 + 9n + 1 n k 3-1 = 27n 3 + 27n 2 + 9n L14

Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3) n k-1 = 3n n k = 3n + 1 n k 3 = (3n + 1)3 n k 3 = 27n 3 + 27n 2 + 9n + 1 n k 3-1 = 27n 3 + 27n 2 + 9n n k 3-1 = (3n 3 + 3n 2 + n)·9 L14

Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3) n k-1 = 3n n k = 3n + 1 n k 3 = (3n + 1)3 n k 3 = 27n 3 + 27n 2 + 9n + 1 n k 3-1 = 27n 3 + 27n 2 + 9n n k 3-1 = (3n 3 + 3n 2 + n)·9 m k 3-1 = m·9 L14

Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3) n k-1 = 3n n k = 3n + 1 n k 3 = (3n + 1)3 n k 3 = 27n 3 + 27n 2 + 9n + 1 n k 3-1 = 27n 3 + 27n 2 + 9n n k 3-1 = (3n 3 + 3n 2 + n)·9 m k 3-1 = m·9 k 31(mod 9) L14

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

Direct Proofs 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

Direct Proofs 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

Direct Proofs 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 31(mod 9) is readily gotten from m k 3-1 = m·9 since the latter is the definition of the former. L14

Proofs Indirect In addition to direct proofs, there are two other standard methods for proving k P(k)  Q(k) Indirect Proof For any k assume: Q(k) and derive: P(k) Uses the contrapositive logical equivalence: P Q  Q  P L14

Proofs Reductio Ad Absurdum 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

Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) L14

Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) Suppose k 2 is not even. L14

Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) Suppose k 2 is not even. So k 2 is odd. L14

Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) Suppose k 2 is not even. So k 2 is odd. n k 2 = 2n + 1 L14

Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) Suppose k 2 is not even. So k 2 is odd. n k 2 = 2n + 1 n k 2 - 1 = 2n L14

Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) Suppose k 2 is not even. So k 2 is odd. n k 2 = 2n + 1 n k 2 - 1 = 2n n (k - 1)(k + 1) = 2n L14

Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) Suppose k 2 is not even. So k 2 is odd. n k 2 = 2n + 1 n k 2 - 1 = 2n n (k - 1)(k + 1) = 2n 2 | (k - 1)(k + 1) L14

Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) Suppose k 2 is not even. So k 2 is odd. n k 2 = 2n + 1 n k 2 - 1 = 2n n (k - 1)(k + 1) = 2n 2 | (k - 1)(k + 1) 2 | (k - 1)  2 | (k + 1) since 2 is prime L14

Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) Suppose k 2 is not even. So k 2 is odd. n k 2 = 2n + 1 n k 2 - 1 = 2n n (k - 1)(k + 1) = 2n 2 | (k - 1)(k + 1) 2 | (k - 1)  2 | (k + 1) since 2 is prime a k - 1 = 2a  b k+1 = 2b L14

Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) Suppose k 2 is not even. So k 2 is odd. n k 2 = 2n + 1 n k 2 - 1 = 2n n (k - 1)(k + 1) = 2n 2 | (k - 1)(k + 1) 2 | (k - 1)  2 | (k + 1) since 2 is prime a k - 1 = 2a  b k+1 = 2b a k = 2a + 1  b k = 2b – 1 L14

Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) Suppose k 2 is not even. So k 2 is odd. n k 2 = 2n + 1 n k 2 - 1 = 2n n (k - 1)(k + 1) = 2n 2 | (k - 1)(k + 1) 2 | (k - 1)  2 | (k + 1) since 2 is prime a k - 1 = 2a  b k+1 = 2b a k = 2a + 1  b k = 2b – 1 In both cases k is odd L14

Indirect Proof Example PROVE: The square of an even number is even. (Ex. 1.16.b) Suppose k 2 is not even. So k 2 is odd. n k 2 = 2n + 1 n k 2 - 1 = 2n n (k - 1)(k + 1) = 2n 2 | (k - 1)(k + 1) 2 | (k - 1)  2 | (k + 1) since 2 is prime a k - 1 = 2a  b k+1 = 2b a k = 2a + 1  b k = 2b – 1 In both cases k is odd So k is not even L14

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

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.  You don’t need the bracketed symbols: [=P] [= Q] [ P  P  Q ] In the actual proof. These are added to point out where the symbolic version of Reductio Ad Absurdum is really being used. L14

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

Disrefutation by Counterexample Disprove: The product of irrational numbers is irrational. L14

Blackboard Exercise for 3.1 Proof that the square root of 2 is irrational L14