Mathematical Induction and Cardinality Proofs

1. Lecture 13 3.2,3.3 Sequences & Summations Proof by Induction

2. Sums geometric progression: arithmetic progression: some other useful sums: a=1, n  infinity derivative

3. Sums Example: set notation: note: the order doesn’t matter when sum these elements.

4. Cardinality definition: Two sets have the same cardinality if and only if there is a one-to-one correspondence between them. This is simple for finite sets, but what if a set has infinite elements? definition: A set that is finite or has the same cardinality as the set of positive integers (Z+) is called countable. Example: Consider the sequence {an}, an = n^2, n={1,2,3,4...} Naively speaking, there seem to be much less elements in {an} than in Z+ (since we skip a lot). Infinity is weird! Here is the one-to-one mapping: 1 2 3 4 5 6 7 ... 1 4 9 16 25 36 49 ... infinity (intuitively: you can enumerate them)

5. cardinality Now what about the positive rational numbers: p/q with p,q integer, q not 0 ? skip 1 2 3 4 5 6 7 8 9 10 1 ½ 2 3 1/3 ¼ 2/3 3/2 4 5 set of positive rational numbers are countable, in fact the set of all rational numbers is countable : 4 countable quadrants.

6. cardinality That was amazing? Now check this out: a proof that the real numbers are not countable! When a prove seems impossible, try to prove a contradiction. First this: unions of countable sets are countable intersections of countable sets are countable a subset of a countable set is countable. Prove: assume that the reals are countable,  the reals in (0,1) are countable (since it is a subset),  Thus there is a sequence as follows: r1 = 0.d11 d12 d12 d14 ... r2 = 0.d21 d22 d23 d24 ... r3 = 0.d31 d32 d33 d34 ... etc. construct the number: r = 0.c1 c2 c3 c4 ... with ci=4 if dii not 4 ci = 5 if dii = 4 This is guaranteed to be different than any real in the list, so it isn’t in the list, so the list is not complete. contradiction!  real in [0,1] are uncountable  all reals are uncountable.

7. 3.3 Mathematical Induction If we want to prove propositions P(k) for all positive integers, we may use inductions. First we prove: P(1) is true. Then we prove P(k)  P(k+1). So, is P(100) true? yes, use “modus ponens” 99 times. P(1) P(1)P(2) ____________ P(2) P(2)P(3) ___________ P(3) until P(100). formally:

8. Examples prove that the sum of the first n odd positive integers is n^2. n=1: 1 = 1. assume it’s true for some k. 1+3+5+...+2k-1 = k^2 is true. add 2k+1 on each side: 1+3+5+...+2k-1 + 2k+1= k^2 + 2k + 1 1+3+5+...+2(k+1)-1 = (k+1)^2 

9. Examples Prove that n < 2^n for positive integers n. P(1): 1 < 2 Inductive step: assume P(k) is correct, prove P(k+1) is correct. k < 2^k k+1 < 2^k + 1 < 2^k + 2^k = 2^(k+1)  Note: we could of course also start our induction at another integer b. Example: Prove that 2^n < n! for integers n>= 4 . P(4) = 2^4 = 16 < 4! = 4 3 2 1 = 24. Induction: assume 2^k < k!  2 x 2^k < 2 x k! 2^(k+1) < (k+1) x k! (recall k > = 4). 2^(k+1) < (k+1)! 

10. Fun Example Show that a chessboard with 2^n x 2^n squares where one arbitrary square has been removed can be tiled with L-shapes. (needs drawing) P(1): All 4 possibilities of removing the square for a 2x2 example are precisely covered with 1 L-shape. Assume P(k) is true. Now construct a chessboard with that is twice as large in both directions. Equally divide it into 4 pieces. Remove one piece arbitrarily from one of the 4 pieces. Since P(k) is true that piece can be covered with L-shapes. Next place one L-shape in the middle to remove one square from the remaining 3 pieces. Again due to P(k) these can now be covered as well.

11. Strong Induction induction (I): strong induction (SI): they are equivalent SII: ISI:

12. Strong Induction Example: Consider the game where there are 2 piles of n matches. Each player picks an arbitrary number of matches from one pile. The one who gets the last matches wins. Proposition: The player who starts second can always win. P(1): 2 piles with 1 match each. Second player always wins. Assume player 2 wins when we have 2 piles of k matches. Can player 2 win when we have 2 piles of k+1 matches?  Player 1 will have to take between 1 and k+1 matches from 1 pile. If he picks k+1 matches, player 2 wins by grabbing all matches from the other pile. If he picks between 1 and k matches, player two takes the same amount from the other pile reducing the problem to a smaller equivalent problem. Strong Induction  player 2 always wins 

13. Examples Proposition: every positive integer n>2 can be written as the product of primes. P(2): product of itself (it’s prime). Assume k can be written as a product of primes. Can we prove it for k+1? two cases: k+1 = prime (thus it is a product of one number – itself). k+1 = a x b However since both a and b < k+1 and >= 2, we know that a and b can be written as the product of primes  a x b is a product of primes. 

14. The well ordered property Every non-empty set of nonnegative integers has a least element. This is a trivial statement made explicit so we can give it a name in a proof... Example: Round-Robin tournament: n players play against each other. A cycle is a situation where p1 beats p1, p2 beats p3, pn beats p1. Proposition: If there is a cycle of length greater than 3, then there is also a cycle of 3 among the people in the larger cycle. Prove: (by contradiction) Assume that there is a cycle of length k, where k is the smallest integer > 3 for which a cycle exist and no cycle of length 3 exists. Cycle: p1 p2 p3 ... pk k> 3 Look at p1 p2 p3  if p3 beats p1 we have a cycle of length 3 (contradiction)  if p1 beat p3 we can construct the cycle p1 p3 p4 ... pk which also leads to a contradiction.

15. Infinite Descent A method to prove that a propositional function P(k) is false for all positive integers k. Assume that P(k) is true for at least one k. By the well ordered property there must be a least element s, such that P(s) is true. We show that there is a s’ for which P(s’) is true with s’<s. This contradicts the assumption. In other words, there is no least s’. Example: Prove that sqrt(2) is not rational. Assume it is rational sqrt(2) = M/N where N is assumed the smallest possible positive denominator. We now prove that 2N-M / N-M = N / M with 0 < N-M < N so this leads to a contradiction. We use M^2 = 2N^2: 2N-M / M-N = (2N-M)N / (M-N)N = 2N^2 – MN / (M-N)N = M^2 – MN / (M-N)N = (M-N) M / (M-N) N = M/N Finally ones shows that 0 < (M-N) < N from 1 < sqrt(2) < 2. 