CSE 215 Lecture Notes - Lecture 12: Boolean Function, Qix, Contraposition
7 Mar 2017
School
Department
Course
Professor
Document Summary
If a = {1,2,3}, then 1 is in the set of a. This is valid since r is a set. Above x s1, and f(x) s2. N = dq + r and 0 <= r <= d. 54 = 52+2 = 54 4*13+2 ; hence q = 13 and r = 2. 54 = -56 + 2 = 4*-14 + 2; hence q = -14 and r = 2. To show q(x), consider each possibility of the world. Given any integer n and positive integer d, there exists unique integers q and r such that. This is invalid since it is stating that all x belongs to b(x), where b(x) is a boolean function and not a set. For example, we can draw a graph to represent what values n can be when d = 3: Square of any odd integer has the form 8m+1 for some integer m. Forall n in z, (n is either odd or even)
