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Some Proof Techniques • Induction • Pigeonhole Principle • Diagonalization Induction Let P (n) be a proposition involving integer n. The Principle of Mathematical Induction states that P (n) is true for n0 ≤ n, if the following are true: • P (n0), and • for n ≥ n0, P (k), n0 ≤ k ≤ n implies P (n + 1) Problems 1. The number of subsets of a set of size n is 2n. 2. n X i = n(n + 1)/2 i=0 3. n X i=0 i2 = n(n + 1)(2n + 1)/6 4. n X 3 i =( i=0 5. n X n X i)2 i=0 2i = 2n+1 − 1 i=0 6. n X 1/(i(i + 1)) = n/(n + 1) i=1 7. For n ≥ 4, n! > 2n Pigeonhole Principle If A and B are finite sets and |A| > |B|, then there is no one-to-one function from A to B. Example: In any group of at least two people there are at least two persons that have the same number of acquaintances within the group. Example: How many shoes must be drawn from a box containing 10 pairs to ensure a match? Example: In NYC there are at least two people with the same number of hairs on their heads. Example: If A is a set of 10 numbers between 1 and 100, there are two distinct disjoint subsets of A whose elements sum to the same number. Diagonalization Proofs • Theorem (Georg Cantor). The set of all subsets of natural numbers is uncountable. • Proof by contradiction. – Suppose there is a one-to-one function f from N onto pow(N). – pow(N) = {S0,S1,S2,...} where Si = f(i) – D = {n in N | n is not in Sn}. D is the diagonal set for N. – D is a set of natural numbers, hence D = Sk for some k in N – if k is in Sk, then k is not in D. But Sk = D. – if k is not in Sk, then k is in D. But Sk = D. – So pow(N) is not countable.