My discovery about mathematics

Naoto Meguro. Amateur.
MSC 2010. Primary 03C62; Secondary 11M26.
Key Words and Phrases. Riemann hypothesis, a non-standard model.
The abstract. A proof of Riemann hypothesis.

                         1

I prove Riemann hypothesis(RH) by making a non-standard model for which Riemann's zeta function ζ(s) has no
non-trivial zero point.

                         2

Let X be a countable and consistent axiom system of the mathematics like the following.
Let C",R",N" be the individual symbols which become C,R,N for the standard model of X.
(1∈N")∧∀n((n∈N")→(n+1∈N")) ∈X and ∀x((1∈x)∧∀y((y∈x)→(y+1∈x))→(N"⊂x)) ∈X.
Let d be a free individual symbol. X doesn't include d. Let H be the individual symbol for which
∀s((s∈H)↔(s∈C")∧(¬(Re s≤0))). You may set up that
∀x∀y∀s((x∈N")∧(y∈N")∧(x≤y)∧(s∈C")→(1/x^s-1/y^s=s∫_x≤t≤y 1/t^s+1dt)∧(|∫_x≤t≤y 1/t^s+1dt|≤∫_x≤t≤y |1/t^s+1|dt)) ∈X.
Theorem 1. X has s countable model M for which M |= (d∈N")∧(d≥n) for ∀n∈N.
Proof. X∪{d∈N",d≥1,d≥2,…,d≥m} has a model M_m for which
M_m |= (d=m+1)∧(N"=N)∧(R"=R)∧(C"=C) and is consistent for ∀m∈N.
So X∪{d∈N",d≥1,d≥2,…(infinitely)} is consistent and has a countable model M. M is a model of X
and M |= (d∈N")∧(d≥n) for ∀n∈N.♦
Let M be that in theorem 1. Let O be the object domain of M. Define M" and M(x) and z(x) by
M" |= M(n)= Mertens function when n∈N and M" |= M(n)=0 when n∈O-N and μ(n)=M(n)-M(n-1) and
M" |= z(s)=u when (M |= (s∈C")∧(u∈C"))∧P(s,u) and M" |= z(s)=? when (M |= s∉C")∨(¬∃u((M |=u∈C")∧P(s,u))
(P(s,u)=∀r((M |= (r∈R")∧(¬(r≤0))→∃m"((m"∈N)∧∀m((m∈N)∧(m≥m")→(M |= |1/(∑_1≤n≤m μ(n)/n^s)-u|≤r)))).)
and M |= p(c₁,…,c_m) ⇒ M" |= p(c₁,…,c_m).
p(x₁,…,x_m) is an arbitrary predicate symbol and c_i∈O. M" |= ∀c(|M(c)|≤d) then.
μ(n)=M(n)-M(n-1)=0-0=0 when n∉N for M". The object domain of M" is O. So M" is a model of X.
Let Y be the axiom system gotten by adding the above definitions of M(x) and z(x) to X. M" is a model of Y.
Theorem 2. M" |= ∀s((s∈C")∧(¬(Re(s)≤0))→(z(s)≠0))
Proof. Assume that M" |=(¬(Re(s)≤0))∧(z(s)=0). M" |= 0≠?. So (M" |= s∈C")∧P(s,0). By Abel's theorem,
M" |= |∑_1≤n≤m μ(n)/n^s|=|M(m)/(m+1)^s-∑_1≤n≤m M(n)(1/(n+1)^s-1/n^s)|
≤d+d|s||∑_1≤n≤m∫_n≤t≤n+11/t^s+1dt|≤d+d|s|∫_1≤t≤m+1|1/t^s+1|dt≤d+d|s|/Re(s)∈R" for ∀m∈N. By P(s,0),
∃m, M" |= 1/(1+d+d|s|/Re(s))≥|1/(∑_1≤n≤mμ(n)/n^s)-0|. This is contradiction.♦
If |1/f(c)|≥n for ∀n∈N,f(c)=0 in the standard mathematics. But it may be that M" |= d≥|1/f(c)| and M" |= f(c)≠0
though it isn't verified by concrete calculation. The non-trivial zero points of ζ(s) are such ones.
Let P be the set of the prime numbers in N. M" |= z(s)=1/∑_1≤n≤∞ μ(n)/n^s=1/∏_p∈P(1-1/p^s)
So RH=∀s((s∈C")∧(z(s)=0)∧(¬(Re(s)≤0))→(Re(s)=1/2)) for M". M" |= RH by theorem 2. ¬(Y |- ¬RH)
You never find a counterexample of RH in the mathematics by Y.

                         3

ζ(s) has no non-trivial zero point. It may be impossible to treat ζ(s) in the formalism.