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

Proofs Zeph Grunschlag Copyright © Zeph Grunschlag,

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**Agenda Proofs: General Techniques Direct Proof Indirect Proof**

Proof by Contradiction L14

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**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

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Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) L14

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Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) k 1(mod 3) L14

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**Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) k 1(mod 3)**

n k-1 = 3n L14

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**Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) k 1(mod 3)**

n k-1 = 3n n k = 3n + 1 L14

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**Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) k 1(mod 3)**

n k-1 = 3n n k = 3n + 1 n k 3 = (3n + 1)3 L14

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**Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) k 1(mod 3)**

n k-1 = 3n n k = 3n + 1 n k 3 = (3n + 1)3 n k 3 = 27n n 2 + 9n + 1 L14

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**Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) k 1(mod 3)**

n k-1 = 3n n k = 3n + 1 n k 3 = (3n + 1)3 n k 3 = 27n n 2 + 9n + 1 n k 3-1 = 27n n 2 + 9n L14

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**Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) k 1(mod 3)**

n k-1 = 3n n k = 3n + 1 n k 3 = (3n + 1)3 n k 3 = 27n n 2 + 9n + 1 n k 3-1 = 27n n 2 + 9n n k 3-1 = (3n 3 + 3n 2 + n)·9 L14

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**Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) k 1(mod 3)**

n k-1 = 3n n k = 3n + 1 n k 3 = (3n + 1)3 n k 3 = 27n n 2 + 9n + 1 n k 3-1 = 27n n 2 + 9n n k 3-1 = (3n 3 + 3n 2 + n)·9 m k 3-1 = m·9 L14

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**Proofs Example PROVE: kZ k 1(mod 3) k 31(mod 9) k 1(mod 3)**

n k-1 = 3n n k = 3n + 1 n k 3 = (3n + 1)3 n k 3 = 27n n 2 + 9n + 1 n k 3-1 = 27n n 2 + 9n n k 3-1 = (3n 3 + 3n 2 + n)·9 m k 3-1 = m·9 k 31(mod 9) L14

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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

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**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

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**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

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**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 31(mod 9) is readily gotten from m k 3-1 = m·9 since the latter is the definition of the former. L14

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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

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**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

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**Indirect Proof Example**

PROVE: The square of an even number is even. (Ex b) L14

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**Indirect Proof Example**

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

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**Indirect Proof Example**

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

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**Indirect Proof Example**

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

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**Indirect Proof Example**

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

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**Indirect Proof Example**

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

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**Indirect Proof Example**

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

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**Indirect Proof Example**

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

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**Indirect Proof Example**

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

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**Indirect Proof Example**

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

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**Indirect Proof Example**

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

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**Indirect Proof Example**

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

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**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

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**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

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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

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**Disrefutation by Counterexample**

Disprove: The product of irrational numbers is irrational. L14

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**Blackboard Exercise for 3.1**

Proof that the square root of 2 is irrational L14

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