Mr Dildo Baggins

hat P(N): N-> N is countable b) This means that Pi(N) is enumerable P0, P1, P2, such(prenominal) that Pi(N) = a subset of N c) straightaway submit a expire g where g(n) = Pi(n) + 1 d) g must(prenominal) be virtually subset Pi(n) for nigh i. But this is impossible considering that Pi(i) = g(i) = Pi(n) + 1 e) After erectceling, 0 = 1 which is impossible. f) Therefore P(N): N -> N is uncountable. 2) direct that there be only countably m any(prenominal) texts, where text is a finitely immense string of emblems, whose symbols are chosen from a finite alphabet. deem of facts: Jason Seemann a) Given a finite continuance, apiece blank quadruplet of the length can be mapped to some x lively in N. b) Given a finite alphabet, severally symbol in the alphabet can be mapped to some y existing in N. c) Given these 2 sup poses, any finite series of symbols can be represented by concatenating the symbols x position and the symbols y value. d) e.g. Consider: Symbols: a , b, c role a->1, b->2, c->3 Now consider the text abc. This is mapped to 112233. Consider: Symbols: a, a, a Mapping a->1, a->1, a->1 Now consider the text aaa. This is mapped to 111111. e) Because each symbol and position is represented by a 1-1 birth with N, we can undertake that each possible text ordain be unique. f) Since each possible text is unique, it can be represented by an element in N with a 1-1 correspondence. g) As shown above, we energize created a function that for every presumption text, it has a 1-1 routine with the set N. f: N-> S | (f(i) = s(i)) 3) Show that the following job is undecidable using the undecidability of the gimpy worry. For any function F: A->B tick off whether or non F is total. In other words, launch that the function: Total(F) = 1 if F is total 0 differently is computed by no effective procedure. Given: halt problem is undecidable. a) For the sake of contradiction, suppose that total(F) exists. 1) Read p. (p | p is a chopine and |p|