Bruno Marchal skrev:
Le 15-nov.-07, à 14:45, Torgny Tholerus a écrit :
But m+1 is not a number.
This means that you believe there is a finite sequence of "s" of the
type
A =
s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(
....s(0)))))))))))))))))))))...)
where "..." here represents a finite sequence, and which is such that
s(A) is not a number.
Yes, exactly. When you construct the set of ALL natural numbers N, you
have to define ALL these numbers. And you can only define a finite
number of numbers. See more explanations below.
BTW, do you agree that 100^(100^(100^(100^(100^(100^(100^(100^100)],
and 100^(100^(100^(100^(100^(100^(100^(100^100)] +1 are numbers? I am
just curious,
Yes, I agree. All explicitly given numbers are numbers. The biggest
number is bigger than all by human beeings explicitly given numbers.
If you define the set of all natural numbers N, then you can pull out
the biggest number m from that set. But this number m has a different
"type" than the ordinary numbers. (You see that I have some sort of
"type theory" for the numbers.) The ordinary deduction rules do not
hold for numbers of this new type. For all ordinary numbers you can
draw the conclusion that the successor of the number is included in N.
But for numbers of this new type, you can not draw this conclusion.
You can say that all ordinary natural numbers are of type 0. And the
biggest natural number m, and all numbers you construct from that
number, such that m+1, 2*m, m/2, and so on, are of type 1. And you can
construct a set N1 consisting of all numbers of type 1. In this set
there exists a biggest number. You can call it m1. But this new
number is a number of type 2.
There is some sort of "temporal" distinction between the numbers of
different type. You have to "first" have all numbers of type 0,
"before" you can construct the numbers of type 1. And you must have
all numbers of type 1 "before" you can construct any number of type 2,
and so on.
The construction of numbers of type 1 presupposes that the set of all
numbers of type 0 is fixed. When the set N of all numbers of type 0 is
fixed, then you can construct new numbers of type 1.
It may look like a contradiction to say that m is included in N, and to
say that all numbers in N have a successor in N, and to say that m have
no successor in N. But it is not a constrdiction because the rule "all
numbers in N have a successor in N" can be expanded to "all numbers of
type 0 in N have a successor in N". And because m is a number of type
1, then that rule is not applicable to m.
--
Torgny
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Received on Fri Nov 16 2007 - 12:14:47 PST