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Division into Cases - Discrete Mathematical Structures - Lecture Slides, Slides of Discrete Mathematics

During the study of discrete mathematics, I found this course very informative and applicable.The main points in these lecture slides are:Division Into Cases, Elementary Number Theory, Methods of Proof, Quotient-Remainder Theorem, Unique Integers, Nonnegative Integer, Integer Remainder, Integer Quotient, Parity Property, Square of Odd Integer

Typology: Slides

2012/2013

Uploaded on 04/27/2013

ashwini
ashwini 🇮🇳

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Download Division into Cases - Discrete Mathematical Structures - Lecture Slides and more Slides Discrete Mathematics in PDF only on Docsity! Chapter 3 Elementary Number Theory and Methods of Proof Docsity.com 3.4 Direct Proof and Counterexample 4 Division into Cases and the Quotient- Remainder Theorem Docsity.com Div & Mod • Definition – Given a nonnegative integer n and a positive integer d • n div d = q, the integer quotient obtained when n is divided by d, and • n mod d = r, the integer remainder obtained when n is divided by d. • where q and r are integers and 0 ≤ r < d. Docsity.com Example • Computing the Day of the Week • Suppose today is Tuesday, and neither this year nor next year is a leap year. What day of the week will it be 1 year from today? – 365 days per year (non-leap year) and 7 days in a week. – 365 div 7 = 52 and 365 mod 7 = 1 – 365 = 52 * 7 + 1 (365 days will be 1 year), – Answer: Wednesday (the day after the start day) • General form: – DayN = (DayT +N) mod 7, DayT day of the week today, DayN is the day of the week in N days. – where Sunday = 0, Monday = 1, … , Saturday = 6. Docsity.com Integers • Parity Property – The parity of an integer refers to whether the integer is even or odd. • Theorem 3.4.2: Any two consecutive integers have opposite parity. – Proof: • Suppose that two consecutive integers are given, m and m+1. By the parity property either m is even or m is odd. – Case 1 (m is even): In the this case m = 2k for some integer k, and so m + 1 = 2k + 1, which is odd by definition. Hence in this case one of m and m+1 is even while the other is odd. – Case 2 (m is odd): In this case, m =2k+1 for some integer k, and so m+1 = (2k+1) + 1 = 2k + 2 = 2(k+1). k+1 is an integer (sum of integers), hence k+1 = r. Therefore m+1 = 2r which is even by definition. • It follows that regardless of which case actually occurs for the particular m and m+1 that are chosen, one of m and m+1 is even and the other is odd. Docsity.com
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