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With examples from Number Theory (Rosen 1.5, 3.1, sections on methods of proving theorems and fallacies) Basic Definitions Theorem - A statement that can be shown to be true. Proof - A series of statements that form a valid argument. • Start with your hypothesis or assumption • Each statement in the series must be: – Basic fact or definition – Logical step (based on rules or basic logic) – Previously proved theorem (lemma or corollary) • Must end with what you are trying to prove (conclusion). Basic Number Theory Definitions from Chapters 1.6, 2 • • • • • Z = Set of all Integers Z+ = Set of all Positive Integers N = Set of Natural Numbers (Z+ and Zero) R = Set of Real Numbers Addition and multiplication on integers produce integers. (a,b Z) [(a+b) Z] [(ab) Z] = “such that” Number Theory Defs (cont.) • n is even is defined as k Z n = 2k • n is odd is defined as k Z n = 2k+1 • x is rational is defined as a,b Z x = a/b, b0 • x is irrational is defined as a,b Z x = a/b, b0 or a,b Z, x a/b, b0 • p Z+ is prime means that the only positive factors of p are p and 1. If p is not prime we say it is composite. Methods of Proof p q (Example: if n is even, then n2 is even) • Direct proof: Assume p is true and use a series of previously proven statements to show that q is true. • Indirect proof: Show q p is true (contrapositive), using any proof technique (usually direct proof). • Proof by contradiction: Assume negation of what you are trying to prove (pq). Show that this leads to a contradiction. Direct Proof Prove: nZ, if n is even, then n2 is even. Tabular-style proof: n is even hypothesis n=2k for some kZ definition of even n2 = 4k2 algebra n2 = 2(2k2) which is algebra and mult of 2*(an integer) integers gives integers n2 is even definition of even Same Direct Proof Prove: nZ, if n is even, then n2 is even. Sentence-style proof: Assume that n is even. Thus, we know that n = 2k for some integer k. It follows that n2 = 4k2 = 2(2k2). Therefore n2 is even since it is 2 times 2k2, which is an integer. Structure of a Direct Proof Prove: nZ, if n is even, then n2 is even. Proof: Assume that n is even. Thus, we know that n = 2k for some integer k. It follows that n2 = 4k2 = 2(2k2). Therefore n2 is even since it is 2 times 2k2 which is an integer. Another Direct Proof Prove: The sum of two rational numbers is a rational number. Proof: Let s and t be rational numbers. Then s = a/b and t = c/d where a,b,c,d Z, b,d 0. Then s+t = a/b + c/d = (ad+cb)/bd . But since (ad+cb) Z and bd Z 0 (why?), then (ad+cb)/bd is rational. Structure of this Direct Proof Prove: The sum of two rational numbers is a rational number. Proof: Let s and t be rational numbers. Assumed Then s = a/b and t = c/d where a,b,c,d Z , b,d 0. Def Then s+t = a/b + c/d = (ad+cb)/bd . But since (ad+cb) Z and bd Z 0, then (ad+cb)/bd is rational. Basic facts of arithmetic Conclusion from Def Example of an Indirect Proof Prove: If n3 is even, then n is even. Proof: The contrapositive of “If n3 is even, then n is even” is “If n is odd, then n3 is odd.” If the contrapositive is true then the original statement must be true. Assume n is odd. Then kZ n = 2k+1. It follows that n3 = (2k+1)3 = 8k3+8k2+4k+1 = 2(4k3+4k2+2k)+1. (4k3+4k2+2k) is an integer. Therefore n3 is 1 plus an even integer. Therefore n3 is odd. Assumption, Definition, Arithmetic, Conclusion Discussion of Indirect Proof Could we do a direct proof of If n3 is even, then n is even? Assume n3 is even . . . then what? We don’t have a rule about how to take n3 apart! Example: Proof by Contradiction Prove: The sum of an irrational number and a rational number is irrational. Proof: Let q be an irrational number and r be a rational number. Assume that their sum is rational, i.e., q+r=s where s is a rational number. Then q = s-r. But by our previous proof the sum of two rational numbers must be rational, so we have an irrational number on the left equal to a rational number on the right. This is a contradiction. Therefore q+r can’t be rational and must be irrational. Structure of Proof by Contradiction • Basic idea is to assume that the opposite of what you are trying to prove is true and show that it results in a violation of one of your initial assumptions. • In the previous proof we showed that assuming that the sum of a rational number and an irrational number is rational and showed that it resulted in the impossible conclusion that a number could be rational and irrational at the same time. (It can be put in a form that implies n n is true, which is a contradiction.) 2nd Proof by Contradiction Prove: If 3n+2 is odd, then n is odd. Proof: Assume 3n+2 is odd and n is even. Since n is even, then n=2k for some integer k. It follows that 3n+2 = 6k+2 = 2(3k+1). Thus, 3n+2 is even. This contradicts the assumption that 3n+2 is odd. What Proof Approach? • (n Z n3+5 is odd) n is even indirect • The sum of two odd integers is even direct • Product of two irrational numbers is irrational Is this true? Counterexample? • The sum of two even integers is even direct contradiction • 2 is irrational • If n Z and 3n+2 is odd, then n is odd indirect • If a2 is even, then a is even indirect Using Cases Prove: n Z, n3 + n is even. Separate into cases based on whether n is even or odd. Prove each separately using direct proof. Proof: We can divide this problem into two cases. n can be even or n can be odd. Case 1: n is even. Then kZ n = 2k. n3+n = 8k3 + 2k = 2(4k3+k) which is even since 4k3+k must be an integer. Cases (cont.) Case 2: n is odd. Then kZ n = 2k+1. n3 + n = (8k3 +12k2 + 6k + 1) + (2k + 1) = 2(4k3 + 6k2 + 4k + 1) which is even since 4k3 + 6k2 + 4k + 1 must be an integer. Therefore n Z, n3 + n is even Even/Odd is a Special Case of Divisibility We say that x is divisible by y if k Z x=yk • n is divisible by 2 if k Z n = 2k (even) • The other case is n = 2k+1(odd,remainder of 1) • n is divisible by 3 if k Z n = 3k Other cases • n = 3k + 1 • n = 3k + 2 This leads to modulo arithmetic • n is divisible by 4 if k Z n = 4k Lemmas and Corollaries • A lemma is a simple theorem used in the proof of other theorems. • A corollary is a proposition that can be established directly from a theorem that has already been proved. Remainder Lemma Lemma: Let a=3k+1 where k is an integer. Then the remainder when a2 is divided by 3 is 1. Proof: Assume a =3k+1. Then a2 = 9k2 + 6k + 1 = 3(3k2+2k) + 1. Since 3(3k2+2k) is divisible by 3, the remainder must be 1. Divisibility Example Prove: n2 - 2 is never divisible by 3 if n is an integer. Discussion: What does it mean for a number to be divisible by 3? If a is divisible by 3 then b Z a = 3b. Remainder when n is divided by 3 is 0. Other options are a remainder of 1 and 2. So we need to show that the remainder when n2 - 2 is divided by 3 is always 1 or 2 but never 0. Divisibility Example (cont.) Prove: n2 - 2 is never divisible by 3 if n is an integer. Let’s use cases! There are three possible cases: • Case 1: n = 3k • Case 2: n = (3k+1) • Case 3: n= (3k+2); kZ n2-2 is never divisible by 3 if n Z Proof: Case 1: n = 3k for kZ then n2-2 = 9k2 - 2 = 3(3k2) - 2 = 3(3k2 - 1) + 1 The remainder when dividing by 3 is 1. n2-2 is never divisible by 3 if nZ Case 2: n = 3k+1 for kZ n2-2 = (3k+1)2 - 2 = 9k2 + 6k +1-2 = 3(3k2 + 2k) - 1 = 3(3k2 + 2k -1) + 2 Thus the remainder when dividing by 3 is 2. n2-2 is never divisible by 3 if nZ Case 3: n = 3k+2 for kZ n2-2 = (3k+2)2 - 2 = 9k2 + 12k +4 -2 = 3(3k2 + 4k) + 2 Thus the remainder when dividing by 3 is 2. In each case the remainder when dividing n2-2 by 3 is nonzero. This proves the theorem. More Complex Proof Prove: 2 is irrational. Direct proof is difficult. Must show that there are no a,b, Z, b≠0 such that a/b = 2 . Try proof by contradiction. More Complex Proof (cont.) Proof by Contradiction of 2 is irrational: Assume 2 is rational, i.e., 2 = a/b for some a,b Z, b0. Since any fraction can be reduced until there are no common factors in the numerator and denominator, we can further assume that: 2 = a/b for some a,b Z, b0 and a and b have no common factors. More Complex Proof (cont.) (2)2 = (a/b)2 = a2/b2 = 2. Now what do we want to do? Let’s show that a2/b2 = 2 implies that both a and b are even! Since a and b have no common factors, this is a contradiction since both a and b even implies that 2 is a common factor. Clearly a2 is even (why?). Does that mean a is even? More Complex Proof (cont.) Lemma 1: If a2 is even, then a is even. Proof (indirect): If a is odd, then a2 is odd. Assume a is odd. Then kZ a = 2k+1. a2 = (2k+1)2 = 4k2 + 4k + 1= 2(2k2+2k) + 1. Therefore a is odd. So the Lemma must be true. More Complex Proof (cont.) Back to the example! So far we have shown that a2 is even. Then by Lemma 1, a is even. Thus kZ a = 2k. Now, we will show that b is even. From before, a2/b2 = 2 2b2 = a2 = (2k)2. Dividing by 2 gives b2 = 2k2. Therefore b2 is even and from Lemma 1, b is even. More Complex Proof (cont.) But, if a is even and b is even then they have a common factor of 2. This contradicts our assumption that our a/b has been reduced to have no common factors. Therefore 2 a/b for some a,b Z, b0. Therefore 2 is irrational. Fallacies Incorrect reasoning occurs in the following cases when the propositions are assumed to be tautologies (since they are not). • Fallacy of affirming the conclusion • [(p q) q] p • Fallacy of denying the hypothesis • [(p q) p] q • Fallacy of circular reasoning • One or more steps in the proof are based on the truth of the statement being proved. Proof? Prove if n3 is even then n is even. Proof: Assume n3 is even. Then kZ n3 = 8k3 for some integer k. It follows that n = 38k3 = 2k. Therefore n is even. Statement is true but argument is false. Argument assumes that n is even in making the claim n3=8k3, rather than n3 = 2k. This is circular reasoning.