Re: ** says: Definition: sum{i in N} i = 0

On 20 Jun., 00:52, Franziska Neugebauer <Franziska-
Neugeba...@xxxxxxxxxxxxxxxxxxx> wrote:


The problem is: I see no induction.

Then you should look again.

I have looked again and still see none. A proof by induction requires at
least to statements:

1. The basis: showing a statement which holds when n = 0.
2. The inductive step: showing that if the statement holds for n = m,
then the same statement also holds for n = m + 1.

The problem of your text above is that I cannot identify 1. and 2. Since
you claim to have to have proved (by induction) that ***'s definition
is "wrong" or leads to a contradiction, the burden of proof is yours.

Theorem: The definition 1 + 2 + 3 + ... = r in R yields a
contradiction, namely r =/= r.

Proof: By induction we prove
0 + 1 + 2 + 3 + ... + (n-1) + n + (n+2) + ... < 1 + 2 + 3 + 4 + ... +
n + (n+1) + (n+2) + ...
for all n, and therefore
0 + 1 + 2 + ... < 1 + 2 + 3 + ...
on the other hand 0 + 1 + 2 + ... = 1 + 2 + 3 + ....
Therefore r < r.

For n = 1:
Subtract 1 from the first term of 1 + 2 + 3 + ... = r. Find
0 + 2 + 3 + 4 + ... < 1 + 2 + 3 + 4 + ... = r

Implication on n+1:
Subtract 1 from the first n+1 terms of
1 + 2 + 3 + 4 + ... + n + (n+1) + (n+2) + ... = r.
Find
0 + 1 + 2 + 3 + ... + (n-1) + n + (n+2) ... < r
by
0 + 1 + 2 + 3 + ... + (n-1) + n + (n+2) ... < 0 + 1 + 2 + 3 + ... +
(n-1) + (n+1) + (n+2) + ...
which by inductive assumption is less than r.

Remark: For any fixed n we do not need induction, because we can
calculate the values directly. (This is the same with other proofs by
induction.) Induction is needed in order to show that the whole sum
is covered by the proof.

Regards, WM

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