Presentation on theme: "1 In this lecture Number Theory Quotient and Remainder Floor and Ceiling Proofs Direct proofs and Counterexamples (cont.) Division into cases."— Presentation transcript:
1 In this lecture Number Theory Quotient and Remainder Floor and Ceiling Proofs Direct proofs and Counterexamples (cont.) Division into cases
2 Quotient-Remainder Theorem Theorem: For n Z and d Z + ! q,r Z such that n=d·q+r and 0r
3 Example of using div and mod Last year Halloween was on Sunday. Q.: What day is Halloween this year? Solution: There are 365 days between 10/31/10 and 10/31/ mod 7 = 1. Thus, if 10/31/10 is Sunday then 10/31/11 is Monday.
4 Proof Technique: Division into Cases Suppose at some stage of a proof we know that A 1 or A 2 or A 3 or … or A n is true; want to deduce a conclusion C. Use division into cases: Show A 1 C, A 2 C, …, A n C. Conclude that C is true.
5 Division into Cases: Example Proposition: If n Z s.t. neither of 2 or 3 divide n, (1) then n 2 mod 12 = 1. (2) Proof: Suppose n Z s.t. neither of 2 or 3 divide n. By quotient-remainder theorem, exactly one of the following is true: a) n=6k, b) n=6k+1, c) n=6k+2, d) n=6k+3, e) n=6k+4, f) n=6k+5 for some integer k. (3) n cant be 6k, 6k+2 or 6k+4 because in that case 2 | n (which contradicts (1) ). (4) n cant be 6k+3 because in that case 3 | n (which contradicts (1) ). (5)
Division into Cases: Example(cont.) Proof(cont.): Based on (3), (4) and (5), either n=6k+1 or n=6k+5. Lets show (2) for each of these two cases. Case 1: Suppose n=6k+1. Then n 2 = (6k+1) 2 =36k 2 +12k+1 (by basic algebra) = 12(3k 2 +k)+1 (6) Let p=3k 2 +k. Then p is an integer. n 2 = 12p+1. ( by substitution in (6) ) Hence n 2 mod 12 = 1 by quotient-remainder th-m. Case 2: Suppose n=6k+5. (exercise)
7 Floor and Ceiling Definition: For any real number x, the floor of x: = the unique integer n s.t. n x < n+1; the ceiling of x: = the unique integer n s.t. n-1 < x n. Examples:
8 Properties of Floor and Ceiling For x R and m Z,. is false. Counterexample: For x=1.7, y=2.8, Note: If x,y>0 and the sum of their fractional parts is <1 then
9 Properties of Floor and Ceiling Theorem: For n Z, Proof: Case 1: Suppose n is odd. Then n=2k+1 for some integer k. (1)
10 Properties of Floor and Ceiling Proof (cont.): By substitution from (1), (2) because k Z and k k+1/2 < k+1. On the other hand, n=2k+1 k=(n-1)/2. (by basic algebra) (3) Based on (2) and (3), Case 2: n is even (left as exercise).