Presentation on theme: "With examples from Number Theory"— Presentation transcript:
1With examples from Number Theory Methods of ProofWith examples from Number Theory(Rosen 1.5, 3.1, sections on methods of proving theorems and fallacies)
2Basic 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 assumptionEach statement in the series must be:Basic fact or definitionLogical step (based on rules or basic logic)Previously proved theorem (lemma or corollary)Must end with what you are trying to prove (conclusion).
3Basic Number Theory Definitions from Chapters 1.6, 2 Z = Set of all IntegersZ+ = Set of all Positive IntegersN = Set of Natural Numbers (Z+ and Zero)R = Set of Real NumbersAddition and multiplication on integers produce integers. (a,b Z) [(a+b) Z] [(ab) Z]
4Number Theory Defs (cont.) = “such that”Number Theory Defs (cont.)n is even is defined as k Z n = 2kn is odd is defined as k Z n = 2k+1x is rational is defined as a,b Z x = a/b, b0x is irrational is defined as a,b Z x = a/b, b0 or a,b Z, x a/b, b0p 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.
5Methods 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.
6Direct Proof Prove: nZ, if n is even, then n2 is even. Tabular-style proof:n is even hypothesisn=2k for some kZ definition of evenn2 = 4k2 algebran2 = 2(2k2) which is algebra and mult of2*(an integer) integers gives integersn2 is even definition of even
7Same 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.
8Structure 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.
9Another Direct ProofProve: 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.
10Structure of this 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, then (ad+cb)/bd is rational.AssumedDefBasic facts of arithmeticConclusion from Def
11Example 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
12Discussion 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!
13Example: 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.
14Structure 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.)
152nd 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.
16Is this true? Counterexample? What Proof Approach?(n Z n3+5 is odd) n is evenThe sum of two odd integers is evenProduct of two irrational numbers is irrationalThe sum of two even integers is even2 is irrationalIf n Z and 3n+2 is odd, then n is oddIf a2 is even, then a is evenindirectdirectIs this true? Counterexample?directcontradictionindirectindirect
17Using 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.
18Cases (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
19Even/Odd is a Special Case of Divisibility We say that x is divisible by y if k Z x=ykn 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 = 3kOther casesn = 3k + 1n = 3k + 2n is divisible by 4 if k Z n = 4kThis leads to modulo arithmetic
20Lemmas 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.
21Remainder LemmaLemma: 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. Thena2 = 9k2 + 6k + 1 = 3(3k2+2k) + 1.Since 3(3k2+2k) is divisible by 3, the remainder must be 1.
22Divisibility ExampleProve: 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.
23Divisibility 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 = 3kCase 2: n = (3k+1)Case 3: n= (3k+2); kZ
24n2-2 is never divisible by 3 if n Z Proof:Case 1: n = 3k for kZ thenn2-2 = 9k2 - 2 = 3(3k2) - 2 =3(3k2 - 1) + 1The remainder when dividing by 3 is 1.
25n2-2 is never divisible by 3 if nZ Case 2: n = 3k+1 for kZn2-2 = (3k+1)2 - 2 = 9k2 + 6k +1-2 =3(3k2 + 2k) - 1 = 3(3k2 + 2k -1) + 2Thus the remainder when dividing by 3 is 2.
26n2-2 is never divisible by 3 if nZ Case 3: n = 3k+2 for kZn2-2 = (3k+2)2 - 2 = 9k2 + 12k =3(3k2 + 4k) + 2Thus 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.
27More 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.
28More 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.
29More 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?
30More 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.
31More 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.
32More 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.
33FallaciesIncorrect 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] pFallacy of denying the hypothesis[(p q) p] qFallacy of circular reasoningOne or more steps in the proof are based on the truth of the statement being proved.
34Proof? 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.