1. Direct Proof and Counterexample I
Lecture 12
Section 3.1
Mon, Jan 31, 2005

2. Proving 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.

3. 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

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

5. 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.

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

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

8. 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.

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

10. 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. =
= 641

11. 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.

12. 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.

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

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