Out of curiosity, if natural numbers don't continue to infinity, there must be a maximal natural number, right? Assuming we call it "m", what is the result of "m+1"?
Let's define a few numbers: google= 10^100 googleplex= 10^google googleplexplex= 10^googleplex googleplexplexplex= 10^googleplexplex Now, how much is googleplexplexplex+1? I think it's about the same as googleplexplexplex. I can't make a difference between them. Let's call it googleplexplexplex, too. Now, you will probably yell I'm inconsistent. How can a number plus one be equal to itself? Well, they are not equal, they are about the same. Nothing is equal or not equal in my logic. Nothing is definite. Are googleplexplexplex and googleplexplexplex equal? Sometimes they're equal, sometimes not. Now, you can ask me whether googleplexplexplex can be divided by 2, whether it's a prime number, and so many questions. The real answer is - I don't know. But I can define big numbers and ask you questions which you will not know, either. Your guess is as good as mine. By the way, there is no maximal natural number. I can define bigger numbers. You get the point.
> We > can define numbers so big, that 2n and 2n+1 is almost the same. Almost, yes. > In > any representation, whether in bits or in turing machines, if we > devide both numbers in 4 we will not necessarily get two different > results. See, not "any representation". The fact that you, or your computer, cannot solve a given problem does not impossible to solve make it. In mathematics, the numbers exist whether you can represent them in a finite space or not. If you have an infinite number of natural numbers, it is obvious you will need an unbounded number of bits to represent an unknown natural number. That does not make that number not exist. As a side note, a Turing machine has a semi-infinite storage, and would therefor have no problem to represent any natural number precisely.
I know, but all of those models are inconsistent, even in theory. Each of them can be defined to contradict itself. The existence of an unbounded sequence of integers, countably infinite, all well defined, without contradictions, is an illusion.
> I can't define such a specific number, since you will be > able to contradict me. That's where you prove yourself wrong. If a specific number you name turns out not to be the largest natural number, we have proven nothing. If, however, we are in agreement that ANY specific number you will name will not be maximal, or, in other words, that it is impossible for you to name the maximal number, then THERE IS NO MAXIMAL NUMBER.
OK, I agree with you, there is no maximal number. But there is not an infinite number of integers as well. An integer number plus one will not always be an integer. They are integers by definition, but this definition contradicts itself. And therefore, I don't accept it. For example, take googleplexplexplex (the first one I defined). And define these numbers: googleplexplexplexplex= 10^googleplexplexplex googleplexplexplexplexprime= the smallest prime number who is larger than googleplexplexplexplex googleplexplexplexplexprimeplex= 10^googleplexplexplexplexprime googleplexplexplexplexprimeplexnumberofprimes= the number of prime numbers who are smaller than googleplexplexplexplexprimeplex Now, consider googleplexplexplexplexprimeplexnumberofprimes as the numeric representation of an algorithm who solves any specific decision problem for any specific input in O(1). All you have to do is call it, and you will get a result in no time. Now, can you prove whether the result you will get from this algorithm is correct or not? Not only you can't prove it, but even in theory I don't think there is an answer to such a question. It contains contradictions (for example, it's unknown whether it halts on any input or not) and therefore the answer is not defined even in theory. By the way, when I said O(1) I meant there is a constant number c for which this algorithm will return an answer in less than c steps. I don't know the constant. You will have to prove my assumption for *any* constant. But I can make it even more complicated. For any input with less than googleplex bits, the result will be calculated the old way, by brute forcing - by checking all the possible results. Only for more than googleplex bits, googleplexplexplexplexprimeplexnumberofprimes will be used. Can you prove it will be incorrect even once? 2^googleplex is still a constant. If any problem can be solved in less than 2^googleplex steps, it's still O(1). There is no limit on how things can get complicated. I agree with that. I'm just trying to show that the ordinary logic is inconsistent. If any question can be defined, and there is an answer, proving that any computer (whether deterministic or not) can't give the answer in a short time is always inconsistent, and this can never be proved. Uri. ================================================================= To unsubscribe, send mail to [EMAIL PROTECTED] with the word "unsubscribe" in the message body, e.g., run the command echo unsubscribe | mail [EMAIL PROTECTED]
