"The idea behind the Galactic Examination is to simulate a situation that can help us to use a self reference constructive criticism, in order to find out if we can be considered as legitimate members of the union of the intelligent civilizations of our Galaxy. The minimal criterion is proof that the examined species can survive technological developments that could destroy it. One of the ways that has to be taken is connected to the language concept in general, and particularly, the language of Mathematics. Through an historical and current research the language of Mathematics is considered as one of the most powerful tools that can help us to implement ideas in a variety of areas, and particularly in scientific research and development."

|P(Z*)| > |Z*|

Let us explain Cantor's proof about |P(Z*)| > |Z*|.

At the first step of his proof, Cantor shows that there is injection between P(Z*) and Z*, so after this step we know
that |P(Z*)| > = |Z*|, or in other words, |P(Z*)| cannot be less than |Z*|.

In order to clearly show that |P(Z*)| > |Z*|, we have to show that |P(Z*)| not= |Z*|.

If we show that |P(Z*)| not= |Z*|, than and only than we have no choice but to conclude that |P(Z*)| > |Z*|.

It can be done only if we can show that there is no bijection (no no_surjection but no bijection) between P(Z*) and Z*.

Cantor tried to show, by using S definition, that any mapping between some arbitrary z in Z* and S, leading us to contradiction, and since this is the case, we can clearly conclude that there is no such z in Z* that is mapped with S (because of S definition) and we prove that |P(Z*)| > |Z*|.

But I show that S is based on a definition that is a self-contradiction, and therefore S does not exist.

My argument goes like this:

1) Any set has unique members, for example: {a,a,b,c,…} = {a,b,c,…}, so S has unique members.

2) Any z in Z* is mapped once and only once with some P(Z*) member.

3) By (1) and (2), when t (which is some arbitrary z in Z*) map is checked against S, we know that t is not in S, but by S definition, t must be in S, but if t is in S than by S definition t must not be in S, … etc.

4) By S definition itself, this set MUST include ALL of z in Z* members that are paired with P(Z*) members that do not contain them (and S is a P(Z*) member; therefore by (1) and (2), it MUST include t within it, but it cannot be done because of the same S definition).

5) If t is in S, then by S definition we can conclude that the term ALL = more_than_ALL.

6) If t is not in S, then by S definition we can conclude that the term ALL = less_than_ALL.

7) By (5) and (6) we can conclude that the term ALL = not_ALL, which is a contradiction in Excluded-Middle logical reasoning.

Since (by (7)) S definition is based on the term ALL = not_ALL, S definition does not exist, and Cantors proof by contradiction does not hold.

Please pay attention that in this post I did not use the Hierarchy of dependency argument (which is another way to show why S does not exist).

In this post I used http://www.geocities.com/complementarytheory/Russell... , which clearly shows that Russell's first paradox is not a paradox in Excluded-Middle logical reasoning.

http://www.physicsforums.com/showthread.php?t=82541&...

Fantastic idea for my friends

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Quote:
Originally Posted by Victor Sorokine
Condition at present

Fantastic idea for my friends

Right contradiction: the number u is infinite

(1°) Let a^n + b^n – c^n = 0,
(2°) where for integers a, b, c the number u = a + b – c > 0, where (a_1b_1c_1)_1 =/ 0, u_(k) = 0, u_{k+1} * 0, k > 0.
(3°) Let's transform the digit u_{k+1} into 1.

(4°) Let's assume that the number u contains only one non-zero digit (u_{k+1}). Then:
(4a°) if ((a_(k) + b_(k) – c_(k))_{k+1} = 0, then a_{k+1} + a_{k+1} – a_{k+1} = 1,
U"_{k+2} = a_{k+1} + a_{k+1} – a_{k+1} = 1 and the number U' contains only one non-zero digit (U'_{k+2} = 1).
Or: u is even, but a^n + b^n – c^n is odd, that is impossible.
(4b°) if ((a_(k) + b_(k) – c_(k))_{k+1} = 1, then a_{k+1} + a_{k+1} – a_{k+1} = 0,
U"_{k+2} = 0 and U'_{k+2} = 1. Or: u is odd, but a^n + b^n – c^n even is, that is impossible.
Therefore there exists second non-zero digit in the number u: u_s.

(5°) Let's assume that the number u contains only one non-zero digit (u_{k+1}). Then:
(5a°) if ((a_(s) + b_(s) – c_(s))_{k+1} is odd, then u is even, but U"_{s+1} (and a^n + b^n – c^n) is odd, that is impossible.
(5b°) if ((a_(s) + b_(s) – c_(s))_{k+1} is even, then u is odd, but U"_{s+1} (and a^n + b^n – c^n) is even, that is impossible.
Therefore there exists third non-zero digit in the number u: u_r.

(6°) Let's assume…
AND SO AD INFINITUM

Victor Sorokine