My discovery about mathematics

Naoto Meguro. Amateur. 08A65

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It is easy to prove the consistency of the theory of the formal series including the real numbers.
But this theory has an important defect.

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Let X be the axiom system of the module M(∋x≠0). (0∈M)∧(x∈M)∧(0≠x)∈X.
Think the axioms ∑_i∈{a} c(i)=c(a) and (∑_i∈U c(i) +∑_i∈V d(i)=∑_i∈U∪V e(i))∧∀i(((i∈U-V)→(e(i)=c(i)))∧
((i∈U∩V)→(e(i)=c(i)+d(i)))∧((i∈V-U)→(e(i)=d(i)))) and the elements of X.
Assume that R={∑_i∈U c(i) | (U⊂N∪{0})∧(c(0)∈Z)∧∀i((i∈N)→(c(i)=2 ^{- i}))} is a module by these axioms
and 0(∈Z⊂R) is the zero of R.
These axioms are true when M=R and x=1 and ∑_i∈U c(i)=∞ for the infinite set U and ∞+∞=∞
and ∞+d=d+∞=∞ for d∈M and ∑_i∈U c(i) is usual one for the finite set U and c(i)∈M and d(i)∈M.
If 0=∞ is true then, 1=1+0=1+∞=∞=0 is true. So 0≠∞=∞+d=(∑_i∈N 2 ^{- i})+d is true for ∀d∈M=R then.
∑_i∈N 2 ^{- i}∈R=M. M isn't a module. This is contradiction. So you get the next theorem.
Theorem 1. R isn't a module by the above axioms or 0(∈Z⊂R) isn't the zero of R.
Substitute (M=Z/2Z={0,1} and x=1) for (M=R and x=1). The above axions become true then.
The theorems proved by these axioms are true then. These axioms don't prove the propositions which
never become true like P∧(¬P). So you get the next theorem.
Theorem 2. The above axioms about ∑_i∈U c(i) are consistent.

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Remenber that you can prove every conjecture and its denial in the real number theory which you know.