Title: Direct Proof and Counterexample I
1Direct Proof and Counterexample I
- Lecture 12
- Section 3.1
- Mon, Jan 31, 2005
2Proving Existential Statements
- Proofs of existential statements are often called
existence proofs. - Two types of existence proofs
- Constructive
- Construct the object.
- Prove that it has the necessary properties.
- Non-constructive
- Argue indirectly that the object must exist.
3Example 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.
4Justification
- Argue by SAS that triangles ACM and BCM are
congruent and that AM MB.
5Example 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.
6Example Constructive Proof
- Theorem The equation
- x2 67y2 1.
- has a solution in positive integers.
- Proof ?
7Example Constructive Proof
- Theorem If N is a square-free positive integer,
then the equation - x2 Ny2 1.
- has a solution in positive integers.
8Example Non-Constructive Proof
- Theorem There exists x ? R such that
- x5 3x 1 0.
- Proof
- Let f(x) x5 3x 1.
- f(1) 1 lt 0 and f(2) 27 gt 0.
- f(x) is a continuous function.
- By the Intermediate Value Theorem, there exists x
? 1, 2 such that f(x) 0.
9Disproving Universal Statements
- Construct an instance for which the statement is
false. - Also called proof by counterexample.
10Example 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.
11Example 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.
12Disproving Existential Statements
- These can be among the most difficult of all
proofs. - Fermats Last Theorem is a famous example
- There is no solution in positive integers of the
equation - xn yn zn
- when n gt 2.
13Example 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.
14Example Disproving an Existential Statement
- However, 101010 2 divided by 4 has remainder 2.
- Therefore, x2 y2 ? 101010 2 for any integers
x and y.