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**Unformatted text preview: **Sequences, Series and Foundations These notes by Mikhail Safonov serve as a supplementary material to the textbook by Weyne Richter Sequences, Series and Foundations. Math 2283 and 3283W Chapter 1. Truth, Falsity and Mathematical Induction 1 Truth Tables The mathematical statements P,Q, etc. can be treated as variables which can take one of two values: T=true and F=false. If we further associate T with 1, and F with 0, then we have the following simple rules: P & Q = PQ, P = 1- P, P Q = (( P ) & ( Q )) = 1- (1- P )(1- Q ) = P + Q- PQ. Exercise 1.1 Simplify the statement P & Q = P Q. Solution. Since P and Q have values 1 and 0 , we always have P = P 2 and Q = Q 2 . Therefore, the given statement can be rewritten as PQ = P 2 + Q 2- PQ, so that ( P- Q ) 2 = 0 , and P = Q. 4 Mathematical Induction Definition 4.1 For n N and k = 0 , 1 ,... ,n, the number of combinations of k objects from a set of n objects is n k = n ! k !( n- k )! , where 0! = 1! = 1 , and n ! = 1 2 ... n for n 2 . The following formula follows either by easy direct calculation, or by considering separately subsets of { 1 , 2 ,... ,n,n + 1) which (i) contain n + 1 , and (ii) do not contain n + 1 . Lemma 4.2 For n N and k = 1 ,... ,n, we have n + 1 k = n k + n k- 1 . In turn, using induction and this formula, one can get the following important Newtons binomial formula . For this reason ( n k ) are also called the binomial coefficients. Theorem 4.3 (Newtons binomial formula) For any n N and (real or complex) a and b, we have ( a + b ) n = n X k =0 n k a n- k b k = a n + n 1 a n- 1 b + n 2 a n- 2 b 2 + + n n- 1 ab n- 1 + b n . 1 In particular, taking a = b = 1 , we get 2 n = n X k =0 n k . This equality has a simple interpretation: the number of all subsets of a set of n objects (including the empty set) is 2 n . Therefore, the notation 2 S =all subsets of a given set S makes sense even for infinite sets S. Definition 4.4 The sets S 1 and S 2 are equivalent if there is a one-to-one function f : S 1 S 2 , i.e. (i) from s ,s 00 S 1 and s 6 = s 00 it follows f ( s ) 6 = f ( s 00 ) ; (ii) for each s 2 S 2 , there is s 1 S 1 such that f ( s 1 ) = s 2 . Theorem 4.5 (Cantor) For an arbitrary set S, the sets S and 2 S are not equivalent. Proof. Suppose otherwise. Then there is a one-to-one function f : S 2 S . Denote S = { s S : s / f ( s ) } 2 S . By our assumption, S = f ( s ) for some s S. Consider two possible cases (i) s S , and (ii) s / S . In the case (i), s f ( s ) , hence by definition of S , we have s / S = f ( s ) , i.e. case (ii)....

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