My discovery about mathematics

Naoto Meguro Amateur.

                        1

There is a non-standard model of the theory of the formal series in which the integral
functions except the polynomials have no zero points. So Riemann hyposis(RH) is
true in a sense.

                         2

(x-1)ζ(x) is an integral function. You may put (x-1)ζ(x)=∑_i∈N∪{0} r(i)xⁱ
(r(i)∈C for ∀i∈N∪{0})
Let X be the set of the theorems of the complex function theory which doesn't include the
symbols ∑ and #. Put U(k)={n|(n∈U)∧(n≤k)} for U⊂N∪{0} and k∈N.
Let's think the axioms ∑_i∈{a} c(i)=c(a) and ∑_i∈∅ c(i)=0 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 (∑_i∈U c(i))(∑_i∈V d(i))=∑_(i,k)∈U×Vc(i)d(k) and
∀n((n∈N)→∃m((m∈N)∧∀k((k∈N)∧(k≥m)→(|∑_i∈U c(i)-∑_i∈U(k) c(i)|≤n^{- 1}))))
and the elements of X. Set up that ∀x((x∈C)∧(|x|=0)→(x=0)) ∈X and
∀x∀y((x∈C)∧(y∈C)→(||x|-|y||≤|x-y|)) ∈X.
Theorem 1. If X is consistent, the above axioms don't prove ¬ RH.
Proof. Assume that X is consistent and the above axioms prove ¬ RH.
The above axioms prove ∃s((s∈C)∧(ζ(s)=0)∧(Re(s)≥0)∧(Re(s)≠2 ^-1)).
So they prove Q=∃s((s∈C)∧(∑_i∈N∪{0} r(i)sⁱ=0)).
The above axioms are true when you set up that ∑_i∈U c(i)=# for the infinite set U
and #+#=#+d=d+#=##=#d=d#=# for d∈C and |#|=0
and ∑_i∈U c(i) is usual one for the finite set U and c(i)∈C and d(i)∈C
and the axioms in X are true by a suitable model of X.
1+#=# then. So #∉C then. (#∈C)∧(|#|=0)→(#=0) and
(#∈C)∧(d∈C)→( |#|-|d||≤|#-d|) are true then.
So ∑_i∈N∪{0} r(i)sⁱ≠0 is true for ∀s∈C then. Q is false then.
The theorems proved by the above axioms are true then. The above axioms don't prove Q.
This is contradiction.♦
You cannot find counter-example of RH by the above axioms.

                     3

You are to treat only finite sums and finite products in the above non-standard mathematics.
π²6^{- 1}=ζ(2) doesn't exist then as Kronecker said.