Math 365 Lecture Notes
... Theorem: If a, b, c, and d are integers and b > 0, d > 0, b d Theorem: Let a/b and c/d be any rational numbers with positive denominators where a c a ac c ...
... Theorem: If a, b, c, and d are integers and b > 0, d > 0, b d Theorem: Let a/b and c/d be any rational numbers with positive denominators where a c a ac c ...
Straight flavor of Binary Number in Decimal Number System
... = 19/1 = 19. As a rule, the bit position that we have called the "first" we may call now the "position 0", and the position on its left will be named "position 1", then "position 2" and so on. These bit positions of the binary will have, respectively, 19/20, 19/21, 19/22 (and so on) integer swops of ...
... = 19/1 = 19. As a rule, the bit position that we have called the "first" we may call now the "position 0", and the position on its left will be named "position 1", then "position 2" and so on. These bit positions of the binary will have, respectively, 19/20, 19/21, 19/22 (and so on) integer swops of ...
+ n
... Suppose that a and b are integers with binary expansions of length 2n. Let a = (a2n−1a2n−2 … a1a0)2 and b = (b2n−1b2n−2 … b1b0)2 . Let a = 2nA1 + A0, b = 2nB1 + B0 , where A1 = (a2n−1 … an+1an)2 , A0 = (an−1 … a1a0)2 , B1 = (b2n−1 … bn+1bn)2 , B0 = (bn−1 … b1b0)2. The algorithm is based on the fact ...
... Suppose that a and b are integers with binary expansions of length 2n. Let a = (a2n−1a2n−2 … a1a0)2 and b = (b2n−1b2n−2 … b1b0)2 . Let a = 2nA1 + A0, b = 2nB1 + B0 , where A1 = (a2n−1 … an+1an)2 , A0 = (an−1 … a1a0)2 , B1 = (b2n−1 … bn+1bn)2 , B0 = (bn−1 … b1b0)2. The algorithm is based on the fact ...
Exam 1 Study Guide MA 111 Spring 2015 It is suggested you review
... (22) Perform the following computations in S5 : (a) [1 2 3] [2 5 4] (b) [1 2] [2 3] [3 5] [5 4] (23) Let g = [1 2 3 4 5 6][7 8 9] in S9 . (a) Let t = [2 5]. Draw convincing pictures to show how the cycle number of g ◦ t differs from the cycle number of g. (b) Let t = [4 8]. Draw convincing pictures ...
... (22) Perform the following computations in S5 : (a) [1 2 3] [2 5 4] (b) [1 2] [2 3] [3 5] [5 4] (23) Let g = [1 2 3 4 5 6][7 8 9] in S9 . (a) Let t = [2 5]. Draw convincing pictures to show how the cycle number of g ◦ t differs from the cycle number of g. (b) Let t = [4 8]. Draw convincing pictures ...
Collatz conjecture
The Collatz conjecture is a conjecture in mathematics named after Lothar Collatz, who first proposed it in 1937. The conjecture is also known as the 3n + 1 conjecture, the Ulam conjecture (after Stanisław Ulam), Kakutani's problem (after Shizuo Kakutani), the Thwaites conjecture (after Sir Bryan Thwaites), Hasse's algorithm (after Helmut Hasse), or the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers (because the values are usually subject to multiple descents and ascents like hailstones in a cloud), or as wondrous numbers.Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process (which has been called ""Half Or Triple Plus One"", or HOTPO) indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. The property has also been called oneness.Paul Erdős said about the Collatz conjecture: ""Mathematics may not be ready for such problems."" He also offered $500 for its solution.