My discovery about mathematics

Naoto Meguro. Amateur.
MSC2010. Primary 03B10;Secondary 03B80.
Key Words and Phrases. The complex number theory, an axiom system.
The abstract. The complex number theory isn't consistent.

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The known complex number theory isn't consistent. Let's see it.

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Let X be an axiom system of the standard mathematics whose object domain is C. Set up that
X⊃{0≠1, ∀x(x0=0.}. Put P=∀x(x=1/(1/x)). 1/0∉C. But x=0 is a removal singular point of 1/(1/x)=x.
P is valid as an axiom of the complex number theory. Set up that P∈X. Let d be a free individual symbol.
X doesn't include d.
Theorem 1. X isn't consistent.
X |- ∀x(x=1/(1/x))→(d=1/(1/d)). X |- d=1/(1/d). X\- ∀x(1≠0=0x). X|- ∀x(1/x≠0)
X |- ∀x(1/x≠0)→(1/(1/d)≠0). X |- 1/(1/d)≠0. X |- d≠0. X doesn't include d. So X |-∀x(x≠0).
X |- 0≠0. X isn't consistent.♦
Theory of the field must treat the function symbol x/y and 1/0. 1/0 isn't an element of the field.
If you admit P, X is an axiom system of the theory of the field like R,Q and Z/pZ.
Theorem 2. The field C doesn't exist in the consistent mathematics.
Proof. If C exists, the standard model of X whose object domain is C exists. (d=0 etc..)
X is consistent then. This is contradiction by theorem 1.♦

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The theory treating C isn't consistent.
Physics isn't consistent too. Every phenomenon like the free energy is possible theoretically.

Naoto Meguro. Amateur.
MSC 2010. Primary 03C65;Secondary 30D30.
Key Words and Phrases. Riemann hypoyhesis,non-standard model.
The abstract. Riemann hypothesis isn't provable.

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I prove that Riemann hypothesis isn't provable.

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Let c₁,c₂,… be the standard zero points of Riemann's zeta function.
Let's think the language in which the individual symbols are 1 and i and c₁,c₂,… and the function symbols are
x+y,x-y,xy,x/y,x^y,Re(x), |x| and f₁(x)=ζ(x),f₂(x)=e^x,f₃(x)=Γ(x)….
Let U be the set of the closed terms which become elements of C when the f_k(x)'s are usual ones.
Put U={u₁,u₂,…}. Let z_k be the complex number which equal u_k when f_k(x)'s are usual ones .
Let's think the language in which the individual symbols are 1,i,d,z₁,z₂,… and c₁,c₂,…,N and C.
and the function symbols are x+y,x-y,xy,x/y,x^y,Re(x),|x|, and f₁(x),f₂(x),…. Let T be the set of the closed terms.
Let X be a countable axiom system of the standard mathematics. N and C become N and C in the standard model
of X. Set up that ∀n((n∈N)→(n∈C)∧(Re(n)≠1/2)∧(Re(n)≥1))∈X and X doesn't include d and the f_k(x)'s.
Theorem 1. If X is consistent, Y=X∪{d∈N,d≥1,d≥2,…(infinitely)} is consistent too.
Proof. X∪{d∈N,d≥1,d≥2,…,d≥m} has a model in which d=m,N=N and C=C and is consistent for ∀m∈N.
So Y is consistent.♦
Let M be an arbitrary model of Y in which the f_k(x)'s are usual ones..
Theorem 2. If M |= 0≠|c|≤1/d, c∉U.
Proof. Assume that (M |= 0≠|c|≤1/d)∧(c∈U). M |= d≤1/|c|=z_k for a k. ∃n((n∈N)∧(M |= d≤n)). But M |= n+1≤d.
This is contradiction.♦
Put C(x)={c|(M |= |c-x|≤1/(3d)}. C(x)∩U={x} if x∈U by theorem 2.
Theorem 3. If x,y∈U and x≠y, C(x)∩C(y)=∅
Proof. Assume that s∈C(x)∩C(y). By theorem 2,1/d< |x-y|≤|x-s|+|s-y|≤1/(3d)+1/(3d). This is contradiction.♦
Y doesn't include the f_k(x).
Z=Y∪W∪W" (W={u₁=z₁,u₂=z₂,…},W"={f_k(u_m)=f_k(u_m+1/(4d)}| k=1,2,3,,m∈N}) is consistent.
Put RH=∀s((s∈C)∧(Re(s)≠1/2)∧(Re(s)≥1)→(ζ(s)≠0)).
You can set up that Re(c₁)=1/2. Z |- (ζ(c₁+1/(4d))=0)∧(Re(c₁+1/(4d))≠1/2)∧(Re(c₁+1/(4d))≥1). Z |- ¬RH.
Let X" be the set of the theorems about ζ(x), Γ(x) and e^x.
Theorem 4. If X is consistent,¬(X"∪W|- RH).
Proof. Z|- ¬RH. X"∪Z |- ¬RH.
¬(X" ∪Z |- RH). ¬(X" ∪W |- RH).♦

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If someone proves RH from X"∪W, standard mathematics isn't consistent.

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

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Let X be a countable axiom system of the standard mathematics. Let d be a free individual symbol.
X doesn't include d. Let f(n.s) be a free function symbol. X doesn't include f(n.s).
Let N and C be the individual symbols which become N and C in the standard model
of X.
Theorem 1. If X is consistent,
Y=X∪{d∈N,d≥1,d≥2,d≥3,…(infinitely)} is consistent too.
Proof. X∪{d∈N, d≥1,d≥2,…,d≥m} has a model in which d=m and N=N and C=C
and is consistent for ∀m∈N. So Y is consistent.♦
Let M be a countable model of Y. Let D be the object domain of M. Put p=∀n∀s(nf(n,s+1)=f(n,s)).
Define M" by M" |= P when M |= P and M" |= f(n.s)=1/n^s when n,s∈C∩D and M" |= f(n,s)=0 when s∈D-C
or n∈D-C.
M" is a model of X∪{p} and Y∪{p}. Put RH"=∀s((s∈C)∧(Re(s)>1)→(∑_n∈Nf(n,s)≠0)). RH⇒RH".
Theorem 2. If X is consistent, ¬(Y∪{p} |- RH")
Proof. If Y∪{p} |- RH", M" |= RH".
M" |= ∑_n∈Nf(n,di+2)=∑_n∈N 0=0, M"|= di+2∈C, M" |=Re(di+2)=2>1. M" |= ¬RH". This is contradiction.♦

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