My discovery about mathematics

Let's make a non-standard model of the complex number theory.
Put C"=C^N, N"=N^N,Z"=Z^N,R"=R^N,{0,1}"={0,1}^N.
Define f((a₁,a₂,…), (b₁,b₂,…)…,(p₁,p₂,…))=(f(a₁,…,p₁), f(a₂,…p₂),…) for f(a,b,,…,p).
(a₁,a₂,…)+(a"₁,a"₂,…)=(a₁+a"₁,a₂+a"₂,…) etc. then.
Let's write a=(a,a,a,…). 2^k=(2,2,…)^(k,k,…)=(2^k,2^k,…)
(a₁,a₂,…)/2^k=(a₁/2^k,a₂/2^k,…)
Define ∑_k∈N (a_{k 1},a_{k 2},…)/2^k=(∑_k∈Na_{k 1}/2^k,∑_k∈N a_{k 2}/2^k,…).
R"={m+∑_k∈N b_k/2^k| m∈Z",b_k∈{0,1}"}.
You get a non-standard model of the complex number theory by the above.
Put {c₁, c₂,…}={c |(c∈C)∧(ζ(c)=0)∧(0≤Re(c)≤1)} and d=(c₁,-2,c₂, -4,…).
ζ(d)=(ζ(c₁),ζ(-2),…)=(0,0,…)=0, Re(d)=(Re(c₁),-2, Re(c₂),-4,…)≠(1/2,1/2,…)=1/2,
d/(-2)=(c₁/(-2),1,c₂/(-2),2,…)∉N". Put RH=∀s((ζ(s)=0)∧(Re(s)≠1/2)→(s/(-2)∈N")).
RH is false in the above model. The complex number theory doesn't prove RH.
The above model doesn't treat the predicate symbol x≤y. Z" isn't a domain. The above isn't final solution.

Naoto Meguro. Amateur.
MSC2020. Primary 03C62; Secondary 03H05.
Key Words and Phrases. Peano's axioms, Goldbach conjecture,Riemann hypothesis.
Abstract. Peano's axioms have a model in which the prime numbers don't exist.
So they don't prove Many conjectures.

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Peano's axioms(PA) don't prove many conjectures. Let's see it.

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Put N={1,2,…}, N"=N^N/~ ((n₁,n₂,…)~(n"₁,n"₂,…) ⇔ ∃m∀k((k≥m)→(n_k=n"_k)).)
[n₁,n₂,…]={x|(x∈N^N)∧(x~(n₁,n₂,…)}.
[M₁,M₂,…]={[n₁,n₂,…]| n_k∈M_k for ∀k∈N.} for M_k⊂N for ∀k∈N.
Set up that "the subsets of N" " are these.
S([n₁,n₂,…])=[S(n₁),S(n₂),…]. By these, you get a non-standard model M for PA.
When you define [n₁,n₂,…]+[n"₁,n"₂,…]=[n₁+n"₁,n₂+n"₂,…], [n₁,n₂,…][n"₁,n"₂,…]=[n₁n"₁,n₂n"₂,…],
[n₁,n₂,…]^{[n"₁,n"₂,…]}=[n₁^n"₁,n₂^n"₂,…],
[n₁,n₂,…]+[1,1,…]=S([n₁,n₂,…]) and [n₁,n₂,…]+S([n"₁,n"₂,…])=S([n₁,n₂,…]+[n"₁,n"₂,…]).
[n₁,n₂,…][1,1,…]=[n₁,n₂,…] and [n₁,n₂…]S([n"₁,n"₂,…])=[n₁,n₂,…][n"₁,n"₂,…]+[n₁,n₂,…].
[n₁,n₂,…]^[1,1,…] and [n₁,n₂,…]^{S([n"₁,n"₂,…])}=([n₁,n₂,…]^{[n"₁,n"₂,…]})[n₁,n₂,…]. Let's write n=[n,n,n,…] for n∈N.
Theorem 1. N" doesn't include the prime numbers in M.
Proof. Put A={n| a_n≠1} for (a₁,a₂,…). If a=[a₁,a₂,…]≠[1,1,…]=1,A is an infinite set.
You can set up that A=B∪C, B∩C=∅, B and C are infinite sets. Define b_n=a_n when n∈B ,b_n=1 when n∉B,
c_n=a_n when n∈C, c_n=1 when n∉C. Put b=[b₁,b₂,…] and c=[c₁,c₂,…]. a=bc b≠1,c≠1.♦
PA don't prove conjectures treating prime numbers like Goldbach conjecture.
Put P(p)=(p∈N")∧(p≠1)∧∀x∀y((x∈N")∧(y∈N")∧(p=xy)→(x=1)∨(y=1)). By theorem 1,M |= ¬∃pP(p).
M |= ¬GC=¬∀n((n∈N")→∃p∃q(P(p)∧P(q)∧(2(n+1)=p+q)). So ¬(PA |- GC).
¬(PA |- ∀n((n∈N")→∃p∃k((k∈N")∧(p=n+k)∧P(p)∧P(p+2))),
¬(PA |- ∀n((n∈N")→∃p∃k((k∈N")∧P(p)∧(p=n+k)∧∃m((m∈N")∧(p=2^m-1)))).
Put X=PA∪{C"⊃N",C"⊃S, ∀x∀y∀z((x∈C")∧(y∈C")∧(z∈C")→(x+y∈C")∧(x+0=x)∧(x-y∈C")∧(x-x=0)∧(xy∈C")∧(x1=x)∧(x^y∈C")
∧((x+y)+z=x+(y+z))∧((xy)z=x(yz))∧(x+y=y+x)∧(xy=yx)∧(x(y+z)=xy+xz)∧∀s((s∈S)→(x/s∈C")∧(s/s=1))),}.
X has a non-standard model M" in which C"=C^N/~ and S=[C-{0},C-{0},…] and N"=[N,N,…] and
The "subsets of C" "are [M"₁,M"₂,…] (M"_k⊂C".) and [z₁,z₂,…]-[z"₁,z"₂,…] =[z₁-z"₁,z₂-z"₂,…]
and [z₁,z₂,…]/[z"₁,z"₂,…]=[z₁/z"₁,z₂/z"₂,…] , Re([z₁,z₂,…])=[Re(z₁),Re(z₂),…].
In the standard model, C"=C S=C-{0},N"=N.
Put X"=X∪{ ∀s∀A((s∈C")∧(A⊂C")→(∀p(p∉A)→(∏_p∈A(p^s-1)/p^s=1))),
∀s∀A∀q((s∈C")∧(A⊂C")∧(q∈N")→(∀p((p∈A)↔(p=q))→(∏_p∈A(p^s-1)/p^s=(q^s-1)/q^s))),…., ∀p((p∈P)↔P(p))}.
X"-X includes only definitions of ∏_{p∈{a,b,…,z}}(p^s-1)/p^s and P. So M" becomes a model of X".
Like theorem 1, M" |= P=∅=[∅,∅,…]. M" |= 1/ζ(s)=∏_p∈P (p^s-1)/p^s=∏_{p∈[∅,∅,…]} (p^s-1)/p^s)=1. M" |= ζ(s)=1≠0
M" |= RH=∀s((s∈C")∧(ζ(s)=0)∧(Re(s)≠1/2)→(s/(-2)∈N")). ¬(X" |- ¬RH).

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PA seem too useless.