I've been assigned to solve an exercise that questions the following:
So far I've found that all non-increasing functions are computable but this isn't an valid argument at all. I've been struggling for much hours now but still don't know how to prove wether we can assure if it's computable or not.
Any idea on which could be the answer? Or which elements should I consider to build it?
Thank you very much!
The set N = {1, 2, 3, ...,}, or perhaps N = {0, 1, 2, 3, ...}, is only infinite in one direction; it has a smallest element. If f is a strictly decreasing function, then f(a) > f(b) whenever a < b. There is a problem though: given these definitions, it's impossible to have a strictly decreasing function from natural numbers to natural numbers. Let's say I make f(0) = k. What choices do I have for f(k+1)? I already must have chosen k values less than k for f(1) through f(k), but there are only k values less than k, and I can't choose one more than once, since the function must be strictly decreasing.
In that sense, supposing we have a function f: N->N is strictly decreasing, we can conclude anything, like 1 = 2, the moon is made of cheese, all cats are dogs, etc. From a contradiction, all things follow, and there cannot be a strictly decreasing function from all natural numbers to all natural numbers. It's like saying, "1 = 2, then 2 = 3". That is a logically true statement by virtue of the hypothesis being false.
