My discovery about mathematics

Naoto Meguro. Amateur.
MSC 2010. Primary 03C65 ; Secondary 30D30.
Key Words and Phrases. Non-standard model, Riemann hypothesis.
The abstract. Riemann hypothesis isn't a theorem of the standard mathematics.

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I solved Riemann hypothesis negatively though a troublesome by-product appeared.

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Let X be the countable set of the logical formulas which are the theorems
of the standard mathematics treating the next s. (s=(1/2)(1/2)(1/2)…(infinitely)) ∈X.
Set up that ∀x((t(x)={x,x+1,x+2,…(infinitely)})∧(t"(x)={x,x-1,x-2,…(infinitely)})) ∈X and
∀x((s(x)=∏_m∈t(x)(1/2))∧(s"(x)=∏_m∈t"(x)(1/2))∧(s(x)=s"(x)=s))∈X.
(∀x∀y∀z(((x∩y=∅)→(∏_m∈x∪y1/2=(∏_m∈x1/2)(∏_m∈y1/2)))∧(∏_m∈{z}1/2=1/2)) ∈X.)
These formulas are written by countable symbols. (1/2)s=s. So s∈R⇒s=0.
Let's see X as an axiom system. Let d be a free individual symbol. X doesn't include d.
Let N,R,C be the individual symbols which become N,R,C in the standard model of X.
Theorem 1. If X is consistent,
Y=X∪{s=0,d∈N, d≥1,d≥2,d≥3,…(infinitely)} is consistent too.
Proof. X∪{s=0,d∈N,d≥1,d≥2,…,d≥m} has a model in which d=m,N=N,R=R,C=C and is consistent
for ∀m∈N. So Y is consistent. ♦
You may set up that ∀n((n∈N)→(1≥1/2ⁿ≥0)∧(0(1/2ⁿ)=0)) ∈X. Assume that X is consistent.
The logical formulas treating the terms s,t(x),t"(x),s(x),s"(x) are countable. You can make a countable model of Y.
For the closed logical formula P, P is true.⇔ P∈Q=Y∪{P₁(e_n)→∀x_nP₁(x_n),P₂(e_n")→∀x_n"P₂(x_n"),…}∪{Q₁,Q₂,…}.
Q is consistent. You can set up that the elements of Y are true in a consistent mathematics.
Let Z be the set of logical formulas which are true in the mathematics in which the elements of Y are true.
Put U(n)={m|(m∈N)∧(1≤m≤n). Define 1/2ⁿ=∏_m∈U(n) 1/2 for n∈N.
Theorem 2. If X is consistent, (1/2^d=0)∈Z.
Proof. Let's think the mathematics in which the elements of Y are true.
m∈U(d)⇔(m∈N)∨∃k((k∈N)∧(m=d-k+1))∨(t(m)∪t"(m)⊂U(d)).
1/2^d=s(1)s"(d)V. V=∏_{m∈U(d)-t(1)-t"(d)} 1/2. 0≤V=(1/2)(1/2)…≤1.
1/2^d=ssV=00V=0. So (1/2^d=0)∈Z. ♦
You may set up that ∀s((s∈C)→(ζ(s)=(2π)^sζ(1-s)/(2Γ(s)cos(πs/2))))∈X
and ∀s((s∈C)→(0/Γ(s)=0)) ∈X and; ∀s((s∈C)∧(s≠1)→(0ζ(s)=0)) ∈X and
∀s((s∈C)∧(0≤Re(s)≤1)→(0(2π)^s=0)) ∈X for these are true in the standard mathematics in which C=C.
Put RH=∀s((s∈C)∧(ζ(s)=0)∧(0≤Re(s)≤1)→(Re(s)=1/2)).
Theorem 3. If X is consistent, RH∉X.
Proof. Let's think the mathematics in which the elements of Y are true.
If r≥1, 0≤1/e^rd≤1/2^d=0. 1/cos(rdi)=2/(e^rd+1/e^rd)=2/(e^rd+0)=0 (i²=-1).
ζ(di)=0, Re(di)=0. This is a non-standard counterexample of RH. ¬RH∈Z. X⊂Y⊂Z. If RH∈X, RH∈Z.
This is contradiction.♦
Put FC=∀x∀y∀z∀n((x∈N)∧(y∈N)∧(z∈N)∧(n-2∈N)→(1/(yz)ⁿ+1/(zx)ⁿ≠1/(xy)ⁿ)).
Theorem 4. If X is consistent,FC∉X.
Proof. Let's think the mathmatics in which the elements of Y are true.
0≤1/6^d≤1/3^d≤1/2^d=0. 1/3^d+1/6^d=0+0=0=1/2^d. (d-2∈N). So ¬FC∈Z. FC∉X.♦
If abc conjecture is a theorem of the standard mathematics,FC∈X.
So abc conjecture isn't a theorem of the standard mathematics if X is consistent. Similarly,Beal conjecture isn't
a theorem of the standard mathematics if X is consistent.
RH etc. are solved negatively. The troublesome by-product is the next theorem.
Theorem 5. If X includes Peano's axioms,X isn't consistent.
Proof. Assume that (1∈N)∧∀x((x∈N)→(x+1∈N))∧∀y((1∈y)∧∀z((z∈y)→(z+1∈y))→(N⊂y)) ∈X.
Put E={n|(n∈N)∧1/2ⁿ≠0}. 1∈E. n∈E⇒1/2ⁿ≠0⇒
1/2ⁿ⁺¹=(1/2ⁿ)1/2≠0⇒n+1∈E. So E=N. In the mathematics in which the elements of Y are true,
d∈N=E. 1/2^d≠0. (1/2^d≠0)∈Z. X isn't consistent by theorem 2.♦
E may not be a set.

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You can prove any conjectures and the denials of them in the standard mathematics if X isn't consistent.
Actually,FC is proved.