Direct Proof and Counterexample I Lecture 12 Section 3.1 Wed, Feb 8, 2006

Proving Existential Statements Proofs of existential statements are often called existence proofs. Two types of existence proofs Constructive (a la Euclid) Construct the object. Prove that it has the necessary properties. Non-constructive Argue indirectly that the object must exist.

Example: Constructive Proof Theorem: Given a segment AB, there is a midpoint M of AB. Proof: Draw circle A. Draw circle B. Form ABC. Bisect ACB, producing M. C A B M

Justification Argue by SAS that triangles ACM and BCM are congruent and that AM = MB. C A B M

Example: Constructive Proof Theorem: The equation x2 – 7y2 = 1. has a solution in positive integers. Proof: Let x = 8 and y = 3. Then 82 – 732 = 64 – 63 = 1.

Example: Constructive Proof Theorem: The equation x2 – 67y2 = 1. has a solution in positive integers. Proof: ?

Example: Constructive Proof Theorem: If N is a square-free positive integer, then the equation x2 – Ny2 = 1. has a solution in positive integers.

Example: Non-Constructive Proof Theorem: There exists x  R such that x5 – 3x + 1 = 0. Proof: Let f(x) = x5 – 3x + 1. f(1) = –1 < 0 and f(2) = 27 > 0. f(x) is a continuous function. By the Intermediate Value Theorem, there exists x  [1, 2] such that f(x) = 0.

Disproving Universal Statements Construct an instance for which the statement is false. Also called proof by counterexample.

Example: Proof by Counterexample Disprove the conjecture (Fermat): All integers of the form 22n + 1, for n  1, are prime. (Dis)proof: Let n = 5. 225 + 1 = 4294967297. 4294967297 = 6416700417.

Example: Proof by Counterexample Disprove the statement: If a function is continuous at a point, then it is differentiable at that point. (Dis)proof: Let f(x) = |x| and consider the point x = 0. f(x) is continuous at 0. f(x) is not differentiable at 0.

Disproving Existential Statements These can be among the most difficult of all proofs. Fermat’s Last Theorem is a famous example: There is no solution in positive integers of the equation xn + yn = zn when n > 2.

Example: Disproving an Existential Statement Theorem: There is no solution in integers to the equation x2 – y2 = 101010 + 2. Proof: A perfect square divided by 4 has remainder 0 or 1. Therefore, x2 – y2 divided by 4 has remainder 0, 1, or 3.

Example: Disproving an Existential Statement However, 101010 + 2 divided by 4 has remainder 2. Therefore, x2 – y2  101010 + 2 for any integers x and y.

Example: Disproving an Existential Statement The standard 8 x 8 checkerboard, with two opposite corners removed, cannot be covered with 1 x 2 and 2 x 1 rectangles.

Example: Disproving an Existential Statement The standard 8 x 8 checkerboard, with two opposite corners removed, cannot be covered with 1 x 2 and 2 x 1 rectangles.