I didn’t throw my hands up angrily, I more flicked them out dismissively. I’m not going to learn anything further from this conversation either.
I can construct a real valued functions of the kind you mention easily enough, what I cannot do is construct such a function that is computable in finite time. You want me to draw a conclusion from this that seems obviously absurd.
Assume that g(m) exists. Then there exists an m0 such that g(m0)=f(1)=1. Take any m1>m0. Then g(m1)>1 (because g is strictly increasing). But there also exists an n1 such that g(m1)=f(n1). This cannot be, because f(n) is always less than or equal to 1 when n is a positive integer. Contradiction, QED.
If you have a proof that g(m) exists (computable or uncomputable, no matter), we have a difficult situation on our hands. Could you exhibit your proof now?
Seriously, what you are saying seems to be trivially wrong. You should make it more explicit either way: give an example of what you mean (construct the real valued function, however uncomputable), or engage cousin_it’s discussion.
The only thing I expect to see down that path is more argument on what things are and are not permitted to do with infinity.
If I was to continue discussion on the subject of Occam’s Razor it would be in the form of a new post. Probably involving either a stack of papers and a dart or an extremely large game of spin the bottle. The last one sounds more fun.
I didn’t throw my hands up angrily, I more flicked them out dismissively. I’m not going to learn anything further from this conversation either.
I can construct a real valued functions of the kind you mention easily enough, what I cannot do is construct such a function that is computable in finite time. You want me to draw a conclusion from this that seems obviously absurd.
Such a g(m) cannot exist. Here’s a proof:
Assume that g(m) exists. Then there exists an m0 such that g(m0)=f(1)=1. Take any m1>m0. Then g(m1)>1 (because g is strictly increasing). But there also exists an n1 such that g(m1)=f(n1). This cannot be, because f(n) is always less than or equal to 1 when n is a positive integer. Contradiction, QED.
If you have a proof that g(m) exists (computable or uncomputable, no matter), we have a difficult situation on our hands. Could you exhibit your proof now?
Seriously, what you are saying seems to be trivially wrong. You should make it more explicit either way: give an example of what you mean (construct the real valued function, however uncomputable), or engage cousin_it’s discussion.
The only thing I expect to see down that path is more argument on what things are and are not permitted to do with infinity.
If I was to continue discussion on the subject of Occam’s Razor it would be in the form of a new post. Probably involving either a stack of papers and a dart or an extremely large game of spin the bottle. The last one sounds more fun.
You are not acting reasonably in this thread, which is surprising since elsewhere you usually do.