# With examples from Number Theory

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With examples from Number Theory
Methods of Proof 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]

Number Theory Defs (cont.)
 = “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.

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.

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 = 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: 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.

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.

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. 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

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

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!

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.

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.)

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.

Is this true? Counterexample?
What Proof Approach? (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 indirect direct Is this true? Counterexample? direct contradiction indirect 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 kZ  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 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

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 This leads to modulo arithmetic

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); kZ

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.

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.

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 = 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, 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.

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 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.

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.

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.

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 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.