Proofs in Finite Math Structures: Direct, Contrapositive, Induction - Prof. Marina A. Epel, Study notes of Systems Engineering

Download Proofs in Finite Math Structures: Direct, Contrapositive, Induction - Prof. Marina A. Epel and more Study notes Systems Engineering in PDF only on Docsity! AMS301 Finite Mathematical Structures A Remarks About Methods of Proof We illustrate some typical proof methods below. For this, we will use the following two definitions: Def. An integer n is even if it can be written as n = 2k for some integer k. Def. An integer n is odd if it can be written as n = 2k + 1 for some integer k. 1. DIRECT PROOF: Start with the premise of the theorem and draw conclusions from it until you arrive at the desired conclusion. Theorem 1 If n is an even integer then n2 is even. Proof. We know that there exists an integer k such that n = 2k. Therefore n2 = (2k)2 = 4k2 = 2(2k2). Note that since k is an integer, so is (2k2), and so we expressed n2 as 2 times an integer, showing that n2 is even. ut 2. PROVING THE CONTRAPOSITIVE: In this method of proof we want to show that if (statement A) then (statement B). Instead we show an equivalent fact: if (not statement B) then (not statement A). Theorem 2 Let n be an integer. If n2 is even then n is even. Proof. In this case statement A is “n2 is even”, and statement B is “n is even”. We show instead that if n is odd (not statement B) then n2 is odd (not A). We know there exists an integer k such that n = 2k + 1, therefore n2 = (2k + 1)2 = 4k2 + 2k + 1 = 2(2k2 + k) + 1 Since k is integer, so is 2k2 + k and therefore we have shown that n2 is odd. ut 3. PROOF BY CONTRADICTION: In this method of proof we want to show that if (statement A) then (statement B). We assume that (statement A) is true and (not statement B) is true, and look for a contradiction. Once we get a contradiction, we conclude that our assumption was false, and therefore the theorme is true! Theorem 3 Let n and m be integers. If n ·m is even then at least one of n or m must be even. Proof. Assume n · m is even (statement A) and neither n nor m are even (not statement B). Therefore we can write: n = 2k + 1 and m = 2c + 1 for some integers k, c. n ·m = (2k + 1)(2c + 1) = 4k · c + 2k + 2c + 1 = 2(2k · c + k + c) + 1 showing that n ·m is odd. Since this is a contraditction, we conclude that the theorme is true. ut 4. PROOF BY INDUCTION: We wish to show that a statement S(n) is true for all integer n ≥ n0 (for some number n0). To do so we can show two things: 1. Base case: S(n0) is true. 2. Inductive hypothesis (IH): Assume that S(n) is true for some n ≥ n0 and show that S(n + 1) is true. Theorem 4 For any n ≥ 0 and x 6= 1, 1 + x + x2 + . . . xn = xn+1 − 1 x− 1 Proof. We first show the base case is true. For n = 0 the statement S(0) claims that 1 = x0+1 − 1 x− 1 , which is clearly true. Now, we make the following induction hypothesis (IH): 1 + x + x2 + . . . xn = xn+1 − 1 x− 1 . We need to show that 1 + x + x2 + . . . + xn + xn+1 = xn+2 − 1 x− 1 . 1 + x + x2 + . . . + xn + xn+1 = xn+1 − 1 x− 1 + xn+1 = xn+1 − 1 + (x− 1)xn+1 x− 1 = xn+1 − 1 + xn+2 − xn+1 x− 1 = xn+2 − 1 x− 1 The first equality follows from the inductive hypothesis (IH), and the rest is simple algebra. ut Remark: It is usually trivial to show that the base case is indeed true; however, it is still very important to do so! For example, consider the following “proof” to an incorrect theorem: Theorem 5 (WRONG) For any integer n ≥ 0 n = n + 5. Proof. “Proof” by induction: Assume the fact is true for some n, namely n = n + 5, we need to show it is true for n+1, namely (n+1) = (n+5)+1. This is “easy” to show: Simply take the (IH) and add 1 to each side of the equality. ut The problem with the alleged “proof” is that we did not check that the base case is true, and clearly, in this case, it is NOT true. Theorem 6 (WRONG) In any set of n students, all students have the same height. Proof. “Proof” by induction on the number of students in a set: We start with the “base” of the induction: For any set of size 1, the claim is clearly true (that all students in that set have the same height). Thus, let us assume that the claim is true for sets of size k (this is the induction hypothesis, IH). Now consider a set, S, of size k + 1. Then, S = S′ ∪ {pk+1}, where S′ = {p1, . . . , pk} is a set of size k, and pi denotes “person #i.” By the IH, all students in S′ have the same height; in particular, p1 has the same height as p2. But also, S = {p1}∪S′′, where S′′ = {p2, . . . , pk+1} is a set of size k. So all students in S′′ have the same height; in particular, pk+1 has the same height as p2, which we already know has the same height as p1 (and all other members of S′). Thus, all k + 1 students in S have the same height, and we are done, by induction. What is wrong with this “proof”? ut 2