@Ayatollah, some remarks:
1. If you want, you can get rid of the subjectivness by not explicitly giving values to each number, but instead saying that s:N->[0,1] is speciality-function, and perhaps giving some more conditions, like s(0)=s(1)=1 etc. If you want this proof to work, you must of course add some rule to ensure it.
2. To give numbers you can also use some nonsubjective method. I think the length of it's description is one that is pretty close to the intuitive one: "The multiplicative unit" has speciality of 1, the least prime has 1/2 /the words "least" and "prime"), and "the least odd prime" has 1/3 (the words "least", "odd" and "prime"). This is relative to the properties we allow on this list of course.
3. Even in fuzzy sets this proof can work, if you think that "not special" means speciality of 0. This would prove that all the numbers have speciality bigger than zero, and that's of course possible. All numbers are special, some of them are just more special than the others.
The additional rule we had to have for function s of remark 1, would then be that s(n)>0 for the least number for which s(n)=0. This is of course the same thing as saying s(n)>0 for all n in N. And this is of course because we are taking off the fuzzines by reducing non-speciality to something that can have ionly two values, 0 and >0.
4. I just understood that I've been trying to figure ot what we must assume of specialness to make the proof work, and you have been thinking is the proof valid. Thus all the comments above may seem to be tangential to the problem.
@Harbinger, what pre-algebra means? If you don't know much about maths, rushing into university/college stuff isn't probably very good idea. You can instead learn about ancient geometry for example. It doesn't require much knowledge, but shows pretty clearly what maths is about. It rarely is covered in universities, so you won't be doing the same thing twice.
You'll have to do exercises also. If you don't like it, it's better to concentrate on something else.
1. If you want, you can get rid of the subjectivness by not explicitly giving values to each number, but instead saying that s:N->[0,1] is speciality-function, and perhaps giving some more conditions, like s(0)=s(1)=1 etc. If you want this proof to work, you must of course add some rule to ensure it.
2. To give numbers you can also use some nonsubjective method. I think the length of it's description is one that is pretty close to the intuitive one: "The multiplicative unit" has speciality of 1, the least prime has 1/2 /the words "least" and "prime"), and "the least odd prime" has 1/3 (the words "least", "odd" and "prime"). This is relative to the properties we allow on this list of course.
3. Even in fuzzy sets this proof can work, if you think that "not special" means speciality of 0. This would prove that all the numbers have speciality bigger than zero, and that's of course possible. All numbers are special, some of them are just more special than the others.
The additional rule we had to have for function s of remark 1, would then be that s(n)>0 for the least number for which s(n)=0. This is of course the same thing as saying s(n)>0 for all n in N. And this is of course because we are taking off the fuzzines by reducing non-speciality to something that can have ionly two values, 0 and >0.
4. I just understood that I've been trying to figure ot what we must assume of specialness to make the proof work, and you have been thinking is the proof valid. Thus all the comments above may seem to be tangential to the problem.
@Harbinger, what pre-algebra means? If you don't know much about maths, rushing into university/college stuff isn't probably very good idea. You can instead learn about ancient geometry for example. It doesn't require much knowledge, but shows pretty clearly what maths is about. It rarely is covered in universities, so you won't be doing the same thing twice.
You'll have to do exercises also. If you don't like it, it's better to concentrate on something else.
) doesn't yield any other solution, but those two, there are no more than 2 solutions.
Uppi has got it.
