With examples from Number Theory

1. Methods of Proof With examples from Number Theory (Rosen 1.5, 3.1, sections on methods of proving theorems and fallacies)

2. 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).

3. Basic Number Theory Definitionsfrom 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]

4.  = “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, b0 • x is irrational is defined as a,b  Z  x = a/b, b0 or a,b  Z, x  a/b, b0 • 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.

5. 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 (pq). Show that this leads to a contradiction.

6. Direct Proof Prove: nZ, if n is even, then n2 is even. Tabular-style proof: n is even hypothesis n=2k for some kZ definition of even n2 = 4k2algebra n2 = 2(2k2) which is algebra and mult of 2*(an integer) integers gives integers n2 is even definition of even

7. Same Direct Proof Prove: nZ, 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.

8. Structure of a Direct Proof Prove: nZ, 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.

9. 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.

10. Structure 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. Assumed Def Basic facts of arithmetic Conclusion from Def

11. 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 kZ  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

12. 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!

13. 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.

14. 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.)

15. 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.

16. What Proof Approach? indirect • (n  Z  n3+5 is odd)  n is even • The sum of two odd integers is even • Product of two irrational numbers is irrational • The sum of two even integers is even • 2 is irrational • If n  Z and 3n+2 is odd, then n is odd • If a2 is even, then a is even direct Is this true? Counterexample? direct contradiction indirect indirect

17. 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 kZ  n = 2k. n3+n = 8k3 + 2k = 2(4k3+k) which is even since 4k3+k must be an integer.

18. Cases (cont.) Case 2: n is odd. Then kZ  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

19. This leads to modulo arithmetic 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 • n is divisible by 4 if  k  Z n = 4k

20. 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. Lemmas and Corollaries

21. 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.

22. 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.

23. 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); kZ

24. n2-2 is never divisible by 3 if n  Z Proof: Case 1: n = 3k for kZ then n2-2 = 9k2 - 2 = 3(3k2) - 2 = 3(3k2 - 1) + 1 The remainder when dividing by 3 is 1.

25. n2-2 is never divisible by 3 if nZ Case 2: n = 3k+1 for kZ 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.

26. n2-2 is never divisible by 3 if nZ Case 3: n = 3k+2 for kZ 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.

27. 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.

28. 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, b0. 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, b0 and a and b have no common factors.

29. 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?

30. 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 kZ  a = 2k+1. a2 = (2k+1)2 = 4k2 + 4k + 1= 2(2k2+2k) + 1. Therefore a is odd. So the Lemma must be true.

31. 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 kZ  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.

32. 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, b0. Therefore 2 is irrational.

33. 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.

34. Proof? Prove if n3 is even then n is even. Proof: Assume n3 is even. Then kZ  n3 = 8k3 for some integer k. It follows that n = 38k3 = 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.