Proofs

Information about Proofs

Published on December 10, 2007

Author: avsar

Source: authorstream.com

Content

Basic Proof Methods:  Basic Proof Methods Rosen 6th ed., §1.5-1.7 Nature & Importance of Proofs:  Nature & Importance of Proofs In mathematics, a proof is: A sequence of statements that form an argument. Must be correct (well-reasoned, logically valid) and complete (clear, detailed) that rigorously & undeniably establishes the truth of a mathematical statement. Why must the argument be correct & complete? Correctness prevents us from fooling ourselves. Completeness allows anyone to verify the result. Rules of Inference:  Rules of Inference Rules of inference are patterns of logically valid deductions from hypotheses to conclusions. We will review “inference rules” (i.e., correct & fallacious), and “proof methods”. Visualization of Proofs:  Visualization of Proofs … Various Theorems The Axioms of the Theory A Particular Theory Rules Of Inference Inference Rules - General Form:  Inference Rules - General Form Inference Rule – Pattern establishing that if we know that a set of hypotheses are all true, then a certain related conclusion statement is true. Hypothesis 1 Hypothesis 2 …  conclusion “” means “therefore” Inference Rules & Implications:  Inference Rules & Implications Each logical inference rule corresponds to an implication that is a tautology. Hypothesis 1 Inference rule Hypothesis 2 …  conclusion Corresponding tautology: ((Hypoth. 1)  (Hypoth. 2)  …)  conclusion Some Inference Rules:  Some Inference Rules p Rule of Addition  pq “It is below freezing now. Therefore, it is either below freezing or raining now.” pq Rule of Simplification  p “It is below freezing and raining now. Therefore, it is below freezing now. Some Inference Rules:  Some Inference Rules p q  pq Rule of Conjunction “It is below freezing. It is raining now. Therefore, it is below freezing and it is raining now. Modus Ponens & Tollens:  Modus Ponens & Tollens p Rule of modus ponens pq (a.k.a. law of detachment) q “If it is snows today, then we will go skiing” and “It is snowing today” imply “We will go skiing” q pq Rule of modus tollens p “the mode of affirming” “the mode of denying” Syllogism Inference Rules:  Syllogism Inference Rules pq Rule of hypothetical qr syllogism pr p  q Rule of disjunctive p syllogism  q Formal Proofs:  Formal Proofs A formal proof of a conclusion C, given premises p1, p2,…,pn consists of a sequence of steps, each of which applies some inference rule to premises or to previously-proven statements (as hypotheses) to yield a new true statement (the conclusion). A proof demonstrates that if the premises are true, then the conclusion is true (i.e., valid argument). Formal Proof - Example:  Formal Proof - Example Suppose we have the following premises: “It is not sunny and it is cold.” “if it is not sunny, we will not swim” “If we do not swim, then we will canoe.” “If we canoe, then we will be home early.” Given these premises, prove the theorem “We will be home early” using inference rules. Proof Example cont.:  Proof Example cont. Let us adopt the following abbreviations: sunny = “It is sunny”; cold = “It is cold”; swim = “We will swim”; canoe = “We will canoe”; early = “We will be home early”. Then, the premises can be written as: (1) sunny  cold (2) sunny  swim (3) swim  canoe (4) canoe  early Proof Example cont.:  Proof Example cont. Step Proved by 1. sunny  cold Premise #1. 2. sunny Simplification of 1. 3. sunny  swim Premise #2. 4. swim Modus tollens on 2,3. 5. swimcanoe Premise #3. 6. canoe Modus ponens on 4,5. 7. canoeearly Premise #4. 8. early Modus ponens on 6,7. Common Fallacies:  Common Fallacies A fallacy is an inference rule or other proof method that is not logically valid. May yield a false conclusion! Fallacy of affirming the conclusion: “pq is true, and q is true, so p must be true.” (No, because FT is true.) Fallacy of denying the hypothesis: “pq is true, and p is false, so q must be false.” (No, again because FT is true.) Common Fallacies - Examples:  Common Fallacies - Examples “If you do every problem in this book, then you will learn discrete mathematics. You learned discrete mathematics.” p: “You did every problem in this book” q: “You learned discrete mathematics” Fallacy of affirming the conclusion: pq and q does not imply p Fallacy of denying the hypothesis: pq and  p does not imply  q Inference Rules for Quantifiers:  Inference Rules for Quantifiers x P(x) P(o) (substitute any object o) P(g) (for g a general element of u.d.) x P(x) x P(x) P(c) (substitute a new constant c) P(o) (substitute any extant object o) x P(x) Example:  Example “Everyone in this discrete math class has taken a course in computer science” and “Marla is a student in this class” imply “Marla has taken a course in computer science” D(x): “x is in discrete math class” C(x): “x has taken a course in computer science” x (D(x)  C(x)) D(Marla)  C(Marla) Example – cont.:  Example – cont. Step Proved by 1. x (D(x)  C(x)) Premise #1. 2. D(Marla)  C(Marla) Univ. instantiation. 3. D(Marla) Premise #2. 4. C(Marla) Modus ponens on 2,3. Another Example:  Another Example “A student in this class has not read the book” and “Everyone in this class passed the first exam” imply “Someone who passed the first exam has not read the book” C(x): “x is in this class” B(x): “x has read the book” P(x): “x passed the first exam” x(C(x)  B(x)) x (C(x)  P(x))  x(P(x)  B(x)) Another Example – cont.:  Another Example – cont. Step Proved by 1. x(C(x)  B(x)) Premise #1. 2. C(a)  B(a) Exist. instantiation. 3. C(a) Simplification on 2. 4. x (C(x)  P(x)) Premise #2. 5. C(a)  P(a) Univ. instantiation. 6. P(a) Modus ponens on 3,5 7. B(a) Simplification on 2 8. P(a)  B(a) Conjunction on 6,7 9. x(P(x)  B(x)) Exist. generalization More Examples...:  More Examples... Is this argument correct or incorrect? “All TAs compose easy quizzes. Ramesh is a TA. Therefore, Ramesh composes easy quizzes.” First, separate the premises from conclusions: Premise #1: All TAs compose easy quizzes. Premise #2: Ramesh is a TA. Conclusion: Ramesh composes easy quizzes. Answer:  Answer Next, re-render the example in logic notation. Premise #1: All TAs compose easy quizzes. Let U.D. = all people Let T(x) :≡ “x is a TA” Let E(x) :≡ “x composes easy quizzes” Then Premise #1 says: x, T(x)→E(x) Answer cont…:  Answer cont… Premise #2: Ramesh is a TA. Let R :≡ Ramesh Then Premise #2 says: T(R) Conclusion says: E(R) The argument is correct, because it can be reduced to a sequence of applications of valid inference rules, as follows: The Proof in Detail:  The Proof in Detail Statement How obtained x, T(x) → E(x) (Premise #1) T(Ramesh) → E(Ramesh) (Universal instantiation) T(Ramesh) (Premise #2) E(Ramesh) (Modus Ponens from statements #2 and #3) Another example:  Another example Correct or incorrect? At least one of the 280 students in the class is intelligent. Y is a student of this class. Therefore, Y is intelligent. First: Separate premises/conclusion, & translate to logic: Premises: (1) x InClass(x)  Intelligent(x) (2) InClass(Y) Conclusion: Intelligent(Y) Answer:  Answer No, the argument is invalid; we can disprove it with a counter-example, as follows: Consider a case where there is only one intelligent student X in the class, and X≠Y. Then the premise x InClass(x)  Intelligent(x) is true, by existential generalization of InClass(X)  Intelligent(X) But the conclusion Intelligent(Y) is false, since X is the only intelligent student in the class, and Y≠X. Therefore, the premises do not imply the conclusion. Proof Methods:  Proof Methods Proving pq Direct proof: Assume p is true, and prove q. Indirect proof: Assume q, and prove p. Trivial proof: Prove q true. Vacuous proof: Prove p is true. Proving p Proof by contradiction: Prove p (r  r) (r  r is a contradiction); therefore p must be false. Prove (a  b)  p Proof by cases: prove (a p) and (b p). More … Direct Proof Example:  Direct Proof Example Definition: An integer n is called odd iff n=2k+1 for some integer k; n is even iff n=2k for some k. Axiom: Every integer is either odd or even. Theorem: (For all numbers n) If n is an odd integer, then n2 is an odd integer. Proof: If n is odd, then n = 2k+1 for some integer k. Thus, n2 = (2k+1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Therefore n2 is of the form 2j + 1 (with j the integer 2k2 + 2k), thus n2 is odd. □ Another Example:  Another Example Definition: A real number r is rational if there exist integers p and q ≠ 0, with no common factors other than 1 (i.e., gcd(p,q)=1), such that r=p/q. A real number that is not rational is called irrational. Theorem: Prove that the sum of two rational numbers is rational. Indirect Proof:  Indirect Proof Proving pq Indirect proof: Assume q, and prove p. Indirect Proof Example:  Indirect Proof Example Theorem: (For all integers n) If 3n+2 is odd, then n is odd. Proof: Suppose that the conclusion is false, i.e., that n is even. Then n=2k for some integer k. Then 3n+2 = 3(2k)+2 = 6k+2 = 2(3k+1). Thus 3n+2 is even, because it equals 2j for integer j = 3k+1. So 3n+2 is not odd. We have shown that ¬(n is odd)→¬(3n+2 is odd), thus its contra-positive (3n+2 is odd) → (n is odd) is also true. □ Another Example:  Another Example Theorem: Prove that if n is an integer and n2 is odd, then n is odd. Trivial Proof:  Trivial Proof Proving pq Trivial proof: Prove q true. Trivial Proof Example:  Trivial Proof Example Theorem: (For integers n) If n is the sum of two prime numbers, then either n is odd or n is even. Proof: Any integer n is either odd or even. So the conclusion of the implication is true regardless of the truth of the hypothesis. Thus the implication is true trivially. □ Vacuous Proof:  Vacuous Proof Proving pq Vacuous proof: Prove p is true. Vacuous Proof Example:  Vacuous Proof Example Theorem: (For all n) If n is both odd and even, then n2 = n + n. Proof: The statement “n is both odd and even” is necessarily false, since no number can be both odd and even. So, the theorem is vacuously true. □ Proof by Contradiction:  Proof by Contradiction Proving p Assume p, and prove that p (r  r) (r  r) is a trivial contradiction, equal to F Thus pF is true only if p=F Contradiction Proof Example:  Contradiction Proof Example Theorem: Prove that is irrational. Another Example:  Another Example Prove that the sum of a rational number and an irrational number is always irrational. First, you have to understand exactly what the question is asking you to prove: “For all real numbers x,y, if x is rational and y is irrational, then x+y is irrational.” x,y: Rational(x)  Irrational(y) → Irrational(x+y) Answer:  Answer Next, think back to the definitions of the terms used in the statement of the theorem:  reals r: Rational(r) ↔  Integer(i)  Integer(j): r = i/j.  reals r: Irrational(r) ↔ ¬Rational(r) You almost always need the definitions of the terms in order to prove the theorem! Next, let’s go through one valid proof: What you might write:  What you might write Theorem: x,y: Rational(x)  Irrational(y) → Irrational(x+y) Proof: Let x, y be any rational and irrational numbers, respectively. … (universal generalization) Now, just from this, what do we know about x and y? You should think back to the definition of rational: … Since x is rational, we know (from the very definition of rational) that there must be some integers i and j such that x = i/j. So, let ix,jx be such integers … We give them unique names so we can refer to them later. What next?:  What next? What do we know about y? Only that y is irrational: ¬ integers i,j: y = i/j. But, it’s difficult to see how to use a direct proof in this case. We could try indirect proof also, but in this case, it is a little simpler to just use proof by contradiction (very similar to indirect). So, what are we trying to show? Just that x+y is irrational. That is, ¬i,j: (x + y) = i/j. What happens if we hypothesize the negation of this statement? More writing…:  More writing… Suppose that x+y were not irrational. Then x+y would be rational, so  integers i,j: x+y = i/j. So, let is and js be any such integers where x+y = is/ js. Now, with all these things named, we can start seeing what happens when we put them together. So, we have that (ix/jx) + y = (is/js). Observe! We have enough information now that we can conclude something useful about y, by solving this equation for it. Finishing the proof.:  Finishing the proof. Solving that equation for y, we have: y = (is/js) – (ix/jx) = (isjx – ixjs)/(jsjx) Now, since the numerator and denominator of this expression are both integers, y is (by definition) rational. This contradicts the assumption that y was irrational. Therefore, our hypothesis that x+y is rational must be false, and so the theorem is proved. Proof by Cases:  Proof by Cases To prove we need to prove Example: Show that |xy|=|x| |y|, where x,y are real numbers. Proof of Equivalences:  Proof of Equivalences To prove we need to prove Example: Prove that n is odd iff n2 is odd. Equivalence of a group of propositions:  Equivalence of a group of propositions To prove we need to prove Example:  Example Show that the statements below are equivalent: p1: n is even p2: n-1 is odd p3: n2 is even Counterexamples:  Counterexamples When we are presented with a statement of the form xP(x) and we believe that it is false, then we look for a counterexample. Example Is it true that “every positive integer is the sum of the squares of three integers?” Proving Existentials:  Proving Existentials A proof of a statement of the form x P(x) is called an existence proof. If the proof demonstrates how to actually find or construct a specific element a such that P(a) is true, then it is called a constructive proof. Otherwise, it is called a non-constructive proof. Constructive Existence Proof:  Constructive Existence Proof Theorem: There exists a positive integer n that is the sum of two perfect cubes in two different ways: equal to j3 + k3 and l3 + m3 where j, k, l, m are positive integers, and {j,k} ≠ {l,m} Proof: Consider n = 1729, j = 9, k = 10, l = 1, m = 12. Now just check that the equalities hold. Another Constructive Existence Proof:  Another Constructive Existence Proof Definition: A composite is an integer which is not prime. Theorem: For any integer n>0, there exists a sequence of n consecutive composite integers. Same statement in predicate logic: n>0 x i (1in)(x+i is composite) The proof...:  The proof... Given n>0, let x = (n + 1)! + 1. Let i  1 and i  n, and consider x+i. Note x+i = (n + 1)! + (i + 1). Note (i+1)|(n+1)!, since 2  i+1  n+1. Also (i+1)|(i+1). So, (i+1)|(x+i).  x+i is composite.  n x 1in : x+i is composite. Q.E.D. Non-constructive Existence Proof:  Non-constructive Existence Proof Theorem: “There are infinitely many prime numbers.” Any finite set of numbers must contain a maximal element, so we can prove the theorem if we can just show that there is no largest prime number. i.e., show that for any prime number, there is a larger number that is also prime. More generally: For any number,  a larger prime. Formally: Show n p>n : p is prime. The proof, using proof by cases...:  The proof, using proof by cases... Given n>0, prove there is a prime p>n. Consider x = n!+1. Since x>1, we know (x is prime)(x is composite). Case 1: x is prime. Obviously x>n, so let p=x and we’re done. Case 2: x has a prime factor p. But if pn, then p mod x = 1. So p>n, and we’re done. Limits on Proofs:  Limits on Proofs Some very simple statements of number theory haven’t been proved or disproved! E.g. Goldbach’s conjecture: Every integer n≥2 is exactly the average of some two primes. n≥2  primes p,q: n=(p+q)/2. There are true statements of number theory (or any sufficiently powerful system) that can never be proved (or disproved) (Gödel). Circular Reasoning:  Circular Reasoning The fallacy of (explicitly or implicitly) assuming the very statement you are trying to prove in the course of its proof. Example: Prove that an integer n is even, if n2 is even. Attempted proof: “Assume n2 is even. Then n2=2k for some integer k. Dividing both sides by n gives n = (2k)/n = 2(k/n). So there is an integer j (namely k/n) such that n=2j. Therefore n is even.” Begs the question: How do you show that j=k/n=n/2 is an integer, without first assuming n is even?

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