Naoto Meguro Amateur.
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.
(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.
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.
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.