Problem Set 2 Solutions
... as input. So we conclude that P cannot say yes to itself. Now consider the possibility that P says no to itself. Then, by how P was defined, we can conclude that P is not self-hating. But if P is not self-hating, then it must say yes when given itself as input. So P cannot say no to itself either. W ...
... as input. So we conclude that P cannot say yes to itself. Now consider the possibility that P says no to itself. Then, by how P was defined, we can conclude that P is not self-hating. But if P is not self-hating, then it must say yes when given itself as input. So P cannot say no to itself either. W ...
computer applications - IndiaStudyChannel.com
... Use the following functions: inputs(): Inputs ten numbers from the user using InputStreamReader. factorial(): Returns the factorial of a number. iskrishnamutry(): Checks if a number is krishnamutry. mainmenu(): Main function in the program. [15] 6. Initialise a String name to "Mohan Das Karam Chand ...
... Use the following functions: inputs(): Inputs ten numbers from the user using InputStreamReader. factorial(): Returns the factorial of a number. iskrishnamutry(): Checks if a number is krishnamutry. mainmenu(): Main function in the program. [15] 6. Initialise a String name to "Mohan Das Karam Chand ...
Exam 1 solutions
... (a) This function is injective. The best way to see this is if we let x 6= y, then 2x 6= 2y, and therefore f (x) 6= f (y). This function is not surjective. The only possible outputs are even numbers, and thus the range will not be all of Z. For example, 2x = 3 has no solution for x ∈ Z. Since the fu ...
... (a) This function is injective. The best way to see this is if we let x 6= y, then 2x 6= 2y, and therefore f (x) 6= f (y). This function is not surjective. The only possible outputs are even numbers, and thus the range will not be all of Z. For example, 2x = 3 has no solution for x ∈ Z. Since the fu ...
![[Part 2]](http://s1.studyres.com/store/data/008795912_1-134f24134532661a161532d09dceadfe-300x300.png)
1 Solutions to assignment 3, due May 31
... (b) This is false. Consider A = {0}, and B = C = {a, b}. Let f (0) = a, and let g : B → C be the identity function. Then g ◦ f is not surjective, even though g is. (c) This is false. Consider A = {a, b}, and consider B = C = {0}. If we let f : A → B be given by f (a) = 0, f (b) = 0, and if we let g ...
... (b) This is false. Consider A = {0}, and B = C = {a, b}. Let f (0) = a, and let g : B → C be the identity function. Then g ◦ f is not surjective, even though g is. (c) This is false. Consider A = {a, b}, and consider B = C = {0}. If we let f : A → B be given by f (a) = 0, f (b) = 0, and if we let g ...
