Re: the irrational number system

On Jul 20, 2:01 pm, magi...@xxxxxxxxxxxxxxxxx (Arturo Magidin) wrote:
In article <1184953814.946734.249...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,

Calvin <cri...@xxxxxxxxxxxxxx> wrote:
How the irrational numbers were discovered does not
concern us here. We know that the set V exists, and that it
is uncountable. (Cantor's diagonal proof works just as well for
the irrationals as for the reals.)

No, the diagonal argument by itself does not help you with
irrationals; you would need to modify it in order to guarantee that
the diagonal number that you produce is not a rational, and you have
no warrant to assume it is not.

By restricting the diagonal argument to irrationals all we are
doing is removing the repeating decimals from the list.
The operation on diagonal elements to produce a new number
not in the list can be simply to replace them with the 'next'
digits. Thus diagonal element 1 becomes 2, 2 becomes 3, ...,
9 becomes 0, 0 becomes 1. The new number is also a non-
repeating decimal, but is not in the list.

Instead, you use the fact that the reals are uncountable, the
rationals are countable, and that the reals are the disjoint union of
the rationals and the irrationals. If the irrationals were countable,
then the reals would be the union of two countable sets, and therefore
countable.

I'm specifically resisting this kind of argument. I know
it's true, but I would like to show that the irrationsals
are uncountable *within* the irrational number system (if
it had been a reasonable system, which I now know it isn't.)

.

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