# ssf - Sequences Series and Foundations These notes by...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 Newton’s binomial formula . For this reason ( n k ) are also called the binomial coefficients. Theorem 4.3 (Newton’s 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)....
View Full Document

{[ snackBarMessage ]}

### Page1 / 21

ssf - Sequences Series and Foundations These notes by...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online