Re: abundance of irrationals!)
- From: Matt Gutting <tchrmatt@xxxxxxxxx>
- Date: Thu, 21 Apr 2005 11:56:05 -0400
W. Mueckenheim wrote:
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'm confused. Of course 1/9 is not *in* the sequence (0.1,0.11,0.111,...). However, we are not saying that 1/9 is a member of the sequence, but that 1/9 is the *limit* of the sequence; that is, that it is the real number represented by the sequence. Another way to state this idea is to say that 1/9 is simply another way (a short form) of writing the sequence, so that denying that 1/9 is the diagonal is tantamount to saying that the sequence (0.1,0.11,0.111,...) is not the sequence used in looking at the diagonal. But you seem to be asserting that it is the sequence used in looking at the diagonal; thus the diagonal (the sequence) is 1/9.
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.
As implied above, this is not the point. By the way sequences and infinite series are defined, the infinite sum SUM(i=1 to infinity) f(i) could simply be regarded as another way of writing "1", just as "2+2" is another way of writing "4".
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.
I thought you had previously said that you had difficulties accepting the convergence of infinite series but not of infinite sequences.
Matt
Regards, WM
.
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