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1 Functions (15 P.) Prove or disprove the following statements. a) (4) Given the family of sets N%a = {x 2 N| x mod a = 0} with a 2 N+. There exists a bijection f : N%x ! N%y for any pair x, y 2 N+. The set N%x is the set of all (positive) multiples of x. So the set can also be written as {x, 2x, 3x, 4x, 5x, . . . }. The same goes for N%y = {y, 2y, 3y, 4y, 5y,... }. Now, you only need to define an identity function, which maps x to y, 2x to 2y and so on. I’ll leave it to you as a task to find this identity function and prove its bijectivity. b) (4) The function fn : Nn ! Nn x 7! (xn + 1) mod n is bijective with Nn = {x 2 N| x  n} for every n 2 N+. The set Nn are just all natural numbers less or equal to n. So, N5 = {0, 1, 2, 3, 4, 5}. The function fn has the set Nn as domain and codomain. Since the set Nn is finite for any n 2 N+, the function is either bijective or neither injective nor surjective (but never just one of the two). As a last hint: The function is not bijective. I’ll leave it up to you, to find a respective counterexample. c) (7) There exists a bijection f : N ! N ⇥ N You can imagine N as dots on the number-line and N ⇥ N as a grid of dots in a 2D plane (only covering the upper right). This grid is also indicated in Figure 1. Figure 1: Illustrative idea of how N ⇥ N looks like From here on, you just have to find a plausible way to order these points, like “this is the first one, this is the second one, that’s the third one” and so on. If you can find such a mapping, you also have to show it’s bijective, but that’s again a task left for you. 1

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