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

WM wrote:

On 20 Jun., 12:05, Franziska Neugebauer <Franziska-
Neugeba...@xxxxxxxxxxxxxxxxxxx> wrote:

Theorem: The definition 1 + 2 + 3 + ... = r in R yields a
contradiction, namely r =/= r. [(T1)]

OK, let's see what theorem your proof proves:

Proof: By induction we prove
0 + 1 + 2 + 3 + ... + (n-1) + n + (n+2) + ... < 1 + 2 + 3 + 4 + ...
+
n + (n+1) + (n+2) + ... [(T2)]
for all n,

So you intend to prove

An (n e N -> E(n)) [(T2)]

with E(n) defined as

E(n) :<-> ( 0 + 1 + 2 + 3 + ... + (n-1) + n + (n+2) + ...
< 1 + 2 + 3 + 4 + ... + n + (n+1) + (n+2) + ...)

This proof was done by induction to show that it is valid for ALL n.

From your original post

,----[ <1182327907.777960.210980@xxxxxxxxxxxxxxxxxxxxxxxxxxx> ]
| 0 + 1 + 2 + 3 + ... + (n-1) + n + (n+2) + ... < 1 + 2 + 3 + 4 + ... +
| n + (n+1) + (n+2) + ... [(*)]
| for all n, and therefore [...]
`----

you suppose that (*) is valid for any n, not only those n e N n > 0.

Do you agree that in orthodox mathematics statement X(n) must be defined
in order to be true or false? So then let me ask whether instanciations
of (*) with any n outside of N \ {0} are defined:

1. Do you agree that in orthodox mathematics 0 (zero) exists and n = 0
is a permitted n? If so can you instanciate (*) for that n?

2. Do you agree that in orthodox mathematics omega exists and n = omega
is a permitted n? If so can you instanciate (*) for that omega?

F. N.
--
xyz
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