More Examples of Proofs

Size: px

Start display at page:

Download "More Examples of Proofs"

Rosaline McBride
6 years ago
Views:

1 More Examples of Proofs

2 Contradiction Proofs Definition: A prime number is an integer greater than 1 which is divisible only by 1 and itself. Ex: 2, 5, 11 are primes; 6, 15, 100 are not primes. There are an infinite number of primes. Pf: BWOC suppose that there are only a finite number of primes. Let p 1, p 2,..., p k be the complete list of all prime numbers. Consider the integer N = p 1 p 2...p k + 1. N is not a prime because it is larger than every element in the list. Suppose the integer d, with 1 < d < N, divides N. Let p be a prime divisor of d. p divides N since d divides N. Thus, p divides N p 1 p 2...p k since p is in the list of primes. Therefore, p divides 1. Thus, there are an infinite number of primes.

3 Quiz #2 If all the integers of A are either even or squares, then either there is an even integer in A or all the integers of A are squares. ( x A)(P(x) Q(x)) ( x A)P(x) ( x A)Q(x) ~ (A B) A ~B ( x A)(P(x) Q(x)) ( x A) ~P(x) ( x A) ~Q(x) All the integers of A are either even or squares, and (but) none of the integers in A are even and there is a non-square in A.

4 unless When used in an English sentence the meaning can be ambiguous. The usual interpretation of A unless B is : ~B A ~(~B ~A) B A. Thus, Tuition will increase unless Referenda C & D pass does not usually imply that tuition will not go up if C & D pass, i.e., B ~A ~(B ~~A) ~B ~A. However, I will fix your car unless you don't pay me. Would have the stronger interpretation of ~B A.

5 Contradiction Proofs There do not exist three consecutive natural numbers such that the cube of the largest is the sum of the cubes of the smaller numbers. Pf. Look at (n+2) 3 = (n+1) 3 + n 3 n 3 + 6n n + 8 = n 3 + 3n 2 + 3n n 3 0 = n 3-3n 2-9n - 7 = f(n) f'(n) = 3n 2-6n - 9 = 3(n-3)(n+1) f(-1) = -2, f(3) = f(5) = -2, f(6) = 47 the only root of f(n) is non-integral (between 5 and 6).

6 Contradiction Proofs There are no prime numbers a,b,c such that c 3 = a 3 + b 3. Pf. At least one of a, b is even. May assume b = 2. 8 = c 3 - a 3 = (c - a)(c 2 + ca + a 2 ) but each term in the sum is greater than or equal to 4.

7 Contradiction Proofs If a,b,c are integers such that a 2 + b 2 = c 2, show that at least one of a or b is even. Pf: Suppose that both a and b are odd. There exist integers k and m so that a = 2k + 1 and b = 2m + 1. a 2 + b 2 = (2k+1) 2 + (2m+1) 2 = 4(k 2 +m 2 ) + 4(k+m) + 2. hmmmmmmmm??????? Try again. Pf: Suppose that both a and b are odd. a 2 and b 2 are odd, so c 2 is even. Since c 2 is even, c is even. There is an integer n so that c = 2n and so c 2 = 4n 2. hmmmmmmmmmmm???????

8 Contradiction Proofs If a,b,c are integers such that a 2 + b 2 = c 2, show that at least one of a or b is even. Pf: Suppose that both a and b are odd. There exist integers k and m so that a = 2k + 1 and b = 2m + 1. a 2 + b 2 = (2k+1) 2 + (2m+1) 2 = 4(k 2 +m 2 ) + 4(k+m) + 2. Thus, c 2 = 2[2(k 2 +m 2 + k+m) + 1], which, since k 2 +m 2 + k+m is an integer, is two times an odd number. Since a 2 and b 2 are odd, c 2 is even. Since c 2 is even, c is even. There is an integer n so that c = 2n and so c 2 = 4n 2. So, c 2 = 2(2n 2 ) which is two times an even number.

9 Case Method The case method is based on the tautology: [(P Q) R ] [(P R) (Q R)] If the hypothesis of an implication can be broken up into special cases (which cover all possibilities) and each special case implies the conclusion, then the implication is true. The special cases may arise naturally from the form of the hypothesis or a definition used in the hypothesis. Other times, the special cases may be more "forced" imposed because you see how to easily prove the statement under additional conditions.

10 Case Method Prove that every natural number n is either a prime, a perfect square or divides (n-1)! Pf: If n = 1, then it is a perfect square, so we may assume n > 1. Any natural number is either a prime or not a prime. If n > 1 is not a prime, then there are integers a and b with n = ab and 1 < a,b < n. Case I: a = b If a = b then n = a 2, so n is a perfect square. Case II: a b Since a < n, a (n-1)!. Since b < n and b a, b (n-1)!/a So, ab (n-1)! Thus, n is either a prime, a perfect square or divides (n-1)!

11 Pigeon-Hole Principle The Pigeon-Hole Principle is the formal statement of a common sense idea that we are all aware of. If there are 7 pigeons and 6 pigeon holes and all the pigeons are in a pigeon hole, then some pigeon hole must have at least 2 pigeons in it! - or - If there are 13 pieces of paper all of which are in one of 3 drawers in a desk, then some drawer must have at least 5 pieces of paper.

12 Pigeon-Hole Principle Pigeon-Hole Principle : If kn + 1 objects are distributed amongst n sets, one of the sets must contain at least k + 1 objects. Pf: If no set contains at least k+1 objects, then each of the n sets has at most k objects. Thus, the total number of objects is at most nk.

13 Problem Let there be 9 points in 3-space with integer coordinates. Show that there is a pair of these points whose line segment contains an interior point whose coordinates are integers. Outline of Proof: Points in 3-space have 3 coordinates, (a,b,c). Integer coordinates are either odd or even. There are 8 odd-even patterns of integer coordinates in 3-space. Since there are 9 points, at least two must have the same pattern by the Pigeon-Hole Principle. The midpoint of the line segment joining two points with the same pattern has integer coordinates. Special Assignment 1 : Fill in the missing details of this proof.

CSCE 222 Discrete Structures for Computing. Proofs. Dr. Hyunyoung Lee. !!!!! Based on slides by Andreas Klappenecker

CSCE 222 Discrete Structures for Computing Proofs Dr. Hyunyoung Lee Based on slides by Andreas Klappenecker 1 What is a Proof? A proof is a sequence of statements, each of which is either assumed, or follows