This is supposed to be the point where I throw up my hands and angrily tell you to study some math.
For what it’s worth, I’m not sure what exactly your principle means because it hasn’t been stated in math terms, but I can imagine only one math idea that could correspond to your vague statement, and that idea is self-evidently wrong. Namely, if you have a real-valued function defined on the positive integers and that function is strictly decreasing, you cannot reorder the values to make them strictly increasing. If you think that you can, I advise you to try building an example. For example, take f(n) = 1/n, which is strictly decreasing. Try to build a strictly increasing function g(m) such that for each m there’s a unique n (and vice versa) satisfying g(m)=f(n). Both m and n range over the positive integers.
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.
This is supposed to be the point where I throw up my hands and angrily tell you to study some math.
For what it’s worth, I’m not sure what exactly your principle means because it hasn’t been stated in math terms, but I can imagine only one math idea that could correspond to your vague statement, and that idea is self-evidently wrong. Namely, if you have a real-valued function defined on the positive integers and that function is strictly decreasing, you cannot reorder the values to make them strictly increasing. If you think that you can, I advise you to try building an example. For example, take f(n) = 1/n, which is strictly decreasing. Try to build a strictly increasing function g(m) such that for each m there’s a unique n (and vice versa) satisfying g(m)=f(n). Both m and n range over the positive integers.
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.