Gerry Myerson wrote: > In article <email@example.com>, > "Peter Webb" <firstname.lastname@example.org> wrote: > [...] > > He then goes on: > > ------------------------- > > Think for a few more days, and you will be able to see how to make these > > statements without any reference to `infinite sets', and that this suffices > > for Cantor's proof that not all irrational numbers are algebraic. > > ------------------------- > > > > Sorry, I am having a lot of trouble understanding how I can think about > > "all" irrational numbers without thinking about an infinite set. I have even > > more trouble comparing the cardinality of "all computable numbers" and "all > > algebriac numbers" without using set theoretic constructions. > > > > True, crank or deliberately misleading? > > You can give a rule to construct an irrational number which you can > then prove to be unequal to any algebraic number someone else hands you.
"Any algebraic number that someone else hands you"? What is that supposed to mean? Looks like you are talking of the class of all algebraic numbers here. If someone were to "hand you" algebraic numbers one by one, the above statement cannot be said to have been verified unless all algebraic numbers have been "handed to you".
> You can do this without thinking about the set of all irrational > numbers, just as you can prove the square root of two is irrational > without thinking about the set of all rational numbers (heck, Euclid > pulled that one off). >
No, he didn't. You are talking about Euclid's proof by contradiction where he starts with the hypothesis "Let the square root of two be a rational number r = m/n, where m and n are integers that are mutually prime" and then arrived at a contradiction. Note that r is an *arbitrary* rational number (which can equal any member of the infinite class of rationals). What has been demonstrated is that in a given enumeration of the rationals, the square root of two is not equal to the first rational, and if it is not equal to the n'th rational, it is not equal to the n+1 th rational. More generally, the square root of two does not belong to the class of all rationals.
> You can (Norm says) work with functions by knowing what kind of thing > is an input, what kind of thing is an output, and what the rule is, > without ever thinking of the set of all inputs. If you can show that > any rule that takes an integer as an input and gives a computable > number as an output must omit some computable number, then you have > a proof that computable numbers are not countable, and you've done it > without using set-theoretic constructions. [...]
"What kind of thing is an input", "what kind of thing is an output", "takes an integer as input" all have to be precisely defined. For example, what do you mean by "takes an integer as input"? Takes some specific integer? Or do you mean "zero is taken as input and if the n'th integer is taken as input (in some enumeration of the integers) then the n+1 th integer is also taken as input"? The latter clearly defines an infinite class of integers even if we don't explicity name this class.