Four Fundamental Proof Techniques

1. Four Fundamental Proof Techniques Section 1.5 Fri, Sep 9, 2005

2. Four Proof Techniques • The Principle of Mathematical Induction • The Pigeonhole Principle • The Diagonalization Principle • Proof by Contradiction

3. The Principle of Mathematical Induction • See Discrete Math.

4. Proof by Contradiction • See Discrete Math.

5. The Pigeonhole Principle • Let A and B be finite sets with |A| > |B|. Then there does not exist a one-to-one function f : A B. • Thus, if |A| > |B|, then for any function f : A  B, it must be the case that f(a1) = f(a2) for some a1, a2  A. • That is, if you have more pigeons than pigeonholes, then at least two pigeons must occupy the same pigeonhole.

6. Example of the Pigeonhole Principle • Theorem: Let G be a graph with n vertices. If there is a path from vertex A to vertex B (A  B), then the shortest path from A to B has length at most n – 1. • (The length of a path is 1 less than the number of vertices along that path.) • Proof: • Let the path be (a0 , a1 , a2, …, am), where a0 = A and am = B, be the shortest path from A to B. • This path has length m.

7. Example of the Pigeonhole Principle • Suppose that m > n – 1. • Then some vertex ai must be repeated along the path as aj. • That means that the path contains a loop from ai to aj . • This loop may be excised, resulting in a shorter path (a0, …, ai – 1, ai, aj + 1, …, am) from A to B. • That is a contradiction. Therefore, the shortest path must have length no more than n – 1.

8. The Pigeonhole Principle • The Pigeonhole Principle applies to sets of infinite cardinality as well. • Let A and B be infinite sets with |A| > |B|. Then there does not exist a one-to-one function f : A B. • Similarly, there does not exist an onto function g : B  A.

9. Example of the Diagonalization Principle • Theorem: The set 2N is uncountable. • Proof: • Suppose that 2N is countable. • Then there is a one-to-one function f : 2N N. • That is, the elements of 2N may be listed A0, A1, A2, … • Define the diagonal set D to be D = {i | i  Ai}. • Then D Ai for all i. (Why?) • But D is a subset of N, so it must equal some Ai. • That is a contradiction. Therefore, 2N is not countable.

10. Another Example • Theorem: The set of all functions from N to N is uncountable. • Proof: • Suppose that the set is countable. • Then its elements can be listed f0, f1, f2, … • Define a function g as g(i) = 0 if fi(i)  0, and g(i) = 1 if fi(i) = 0. • Then g  fi for all i. • But g is a function from N to N, so it must equal some fi. • That is a contradiction. Therefore, the set is not countable.

11. Another Example • Theorem: The set of real numbers in the interval [0, 1) is uncountable. • Proof: • Suppose that the set is countable. • Then its elements can be listed x0, x1, x2, … • Write each xiin its decimal expansion: xi= 0.di0di1di2… • Define the i-th digit of a number x to be 0 if dii  0 and 1 if dii = 0. • Then x  xi for all i.

12. Another Example • But x is a real number in [0, 1), so it must equal some xi. • That is a contradiction. Therefore, the set is not countable.

13. The Diagonalization Principle • What do these three examples have in common?

14. The Diagonalization Principle • Let R be a binary relation on a set A. • Let D be the diagonal set for R: D = {aA | (a, a)  R}. • For each a in A, define a set Ra to be Ra = {b A | (a, b) R}. • Then D is distinct from each Ra.

15. The Diagonalization Principle • To see that this is true, note that for every a in A, • if aD, then a Ra, and • if a D, then aRa.

16. The Diagonalization Principle • Adapt this general framework to each of the preceding examples.