Re: abundance of irrationals!)

Virgil <ITSnetNOTcom#virgil@xxxxxxxxxxx> wrote in message

> No one is trying to find a limit which IS one of the numbers listed, so
> finding one which is not is a success, not a failure.

The only condition of the diagonal argument is a_nn =/= b_. That has
noting to do with a limit (which is not a diagonal and not the
antidiagonal).

> Who ever said that 0 was one of the terms of sequence {1/n:n in N} ?
> 0 is the limit of that sequence, not one of its terms.
> Only WM seems to be at all confused about which is which.

If 0 is not a term of the sequence then 1/9 is not a diagonal number.
That is no confusin but a clear cut arguing.


> I.e., WM's sequence is precisely {SUM_{1<=k<=n} f(k) : n in N}
>
> and LIM_{n -> oo} f(n) = 0
>
> and LIM_{m -> oo} SUM_{1<=k<=n} f(k) = 1

Right. But neither 0 not 1 are part of the list.
>
> Hence, from the following list,
> >
> > 0.1 = f(1)
> > 0.11 = f(1) + f(2)
> > 0.111 = f(1) + f(2) + f(3)
> > ...
> >
> > we obtain two theorems:
> >
> > A) the nth partial sum of the diagonal can always be found in the nth
> > line.
>
> In fact the nth partial sum of the sequence defining the "dialogonal" is
> exactly equal to the nth line.
>
> > B) The binary representation of 1 is not in a line.
>
> True, but irrelevant, since the limit of a equence is rarely a member of
> that sequence.

> > Conclusion:
> > C) The binary representation of 1 is not in the diagonal.
>
> False. That the limit of a sequence is not a member of that sequence os
> of no significance. In this case thelimit of the sequence exists and is
> the "diagonal".

Now you show your inconsequence. Sequence and series behave alike
beause every series can be expressed as a sequence.

Regards, WM
.

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