My discovery about mathematics

The algebraic number field is countable and isn't a set. Galois theory may be a fake.
Put N(x)=(x is a natural number.)=(x=1)∨(x=2)∨…(infinitely),
R(x)=(x is a real number.).
Theorem 6. You must not use N(x) in the standard mathematics treating the real
numbers.
Proof. Assume that you can use N(x). In the standard mathematics,
R(x)=∃s∃n∃r((x=(-1)^s(n+∑_i∈r 1/2ⁱ))∧((s=0)∨(s=1))∧((n=0)∨N(n))∧∀y((y∈r)→N(y))).
This r is finite or countable. Countable r doesn't exist as a set by theorem 5.
r is finite when R(x). R(x)⇒(x=m/2^h (m∈Z, N(h))).
R(1/3) is false. But R(1/3) is true in the standard mathematics. This is contradiction. ♦
Theorem 7. You must not use R(x) in the standard mathematics treating the real numbers.
Proof. Assume that you can use R(x). You may set up that ∀x(R(x)→([x+1]=[x]+1))
and ∀x(R(x)∧(0≤x)∧(x≤1)∧(x≠1)→([x]=0)). You can use
R(x)∧(x=[x])∧(x≥1) ⇔ N(x).　This is contradiction by theorem 6.♦
Riemann hypothesis ∀x∀y(R(x)∧R(y)∧(ζ(x+√-1y)=0)→(x=1/2)∨(y=0)) is nonsense.
You are to use R(x) in the mathematics treating the real numbers. So you get
the next inconsistency theorem by theorem 7.
Theorem 8. The standard mathematics treating the real numbers isn't consistent.
Millennium problems except P=NP problem are nonsense. They are proved and are denied by
contradiction led by R(x) which you need to express the problems.
1={∅},2=1∪{1},3=2∪{2},… By the axiom of infinity, ∃P((1∈P)∧∀x((x∈P)→(x∪{x}∈P)).
N"={x|(x∈P)∧∀y((1∈y)∧∀z((z∈y)→(z∪{z}∈y))→(x∈y)) is a set.
Let's write y=x-1 when x=y∪{y}. (x-1)-1=x-2,(x-2)-1=x-3,…
Theorem 9. (x∈N")∧(¬N(x))∧N(k)⇒(x-k∈N")∧(¬N(x-k))
Proof. Assume that (x∈N")∧(¬N(x)).
If ¬(x-1∈N"),(1∈N"-{x})∧∀y((y∈N"-{x})→(y∪{y}∈N"-{x})). N"⊂N"-{x}
This is contradiction. So x-1∈N". If N(x-1),N(x). so ¬N(x-1).
(x∈N")∧(¬N(x))⇒(x-1∈N")∧(¬N(x-1))⇒(x-2∈N")∧(¬N(x-2))⇒…(infinitely)♦
Theorem 10. F₂^(N")(≅⊕_i∈N" F₂e_i) isn't a set.
Proof. If F₂^(N") is a set,B={f|(f∈F₂^(N"))∧∀x((x∈N")∧(x-1∈N")∧(f(x)=1')→(f(x-1)=1'))}
is a set. Assume that (f∈B)∧(x∈N")∧(¬N(x)). If f(x)=1', 0'≠1'=f(x)=f(x-1)=f(x-2)=…(infinitely).
f∉F₂^(N") then. So f(x)=0'. f(y)=1'⇒N(y) B={f₀}∪{f₁}∪{f₂}∪…(infinitely)
f₀(x)=0' for x∈N". f_k(x)=1' for x∈{1}∪…∪{k}. f_k(x)=0' for x∈N"-({1}∪…∪{k}).
B is countable and isn't a set. This is contradiction.♦
Let A(∋1) be a commutative ring. Let M,N be A modules. A^(S) may not be a set for a set S by theorem 10.
Generally,you cannot get the surjection A^(M)→M and the projective resolution of M. You cannot define
M⊗_AN by A^(M×N) or Z^(M×N). You must examine theorems got by using ⊗ or Tor( Mordell conjecture etc.).

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The physics treats the real numbers and isn't consistent. Every phenomenon
like free productivity is possible. This is an answer to the 6th problem of
Hilbert. You may have to admit inconsistency of the mathematics too.

                                       1

I indicate that countable sets which you use intuitively aren't sets. The standard mathematics isn't
consistent. Millennium problems are nonsense.

                                       2

Let C={c₁}∪{c₂}∪…(infinitely) (c_i≠c_k for i≠k) be a countable one. Every c_i is a set.
Theorem 1. If C is a set, ∑_i∈C x =∑_i∈C-{c1} x for ∀x.
Proof. When m∉I, ∑_i∈I∪{m} x_i =(∑_i∈I x_i)+x_m. ∑_i∈{m} x_i=x_m.
So ∑_i∈{c1}x_i=x_c1, ∑_{i∈{c1}∪{c2}} x_i=x_c1+x_c2,…,
∑_{i∈{c1}∪…∪{cm}} x_i=x_c1+…+x_cm,…,∑_{i∈{c1}∪{c2}∪…(infinitely)} x_i
=x_c1+x_c2+…(infinitely). ∑_i∈C x=x+x+…(infinitely) Similarly,∑_i∈C-{c1} x=
∑_{i∈{c2}∪{c3}∪…(infinitely)} x=x+x+…(infinitely)=∑_i∈C x. ♦
Let L be a module.
Theorem 2. When x_i,y_i∈L for ∀i∈C, ∑_i∈C x_i +∑_i∈C y_i =∑_i∈C (x_i+y_i)
if C is a set.
Proof. Put B(I)=∑_i∈I x_i + ∑_i∈I y_i. B({c₁})=x_c1+y_c1,
B({c₁}∪{c₂})=(x_c1+y_c1)+(x_c2+y_c2),… B({c₁}∪…∪{c_m})=(x_c1+y_c1)+
…(x_cm+y_cm),…B(C)=(x_c1+y_c1)+(x_c2+y_c2)+…(infinitely)=∑_i∈C (x_i+y_i) ♦
Let 0" be the zero of L. Assume that L∋b≠0". Put a=∑_i∈C 0", d=∑_i∈C b,
d"=∑_i∈C (-b) .
Theorem 3. a≠0" if C is a set.
Proof. Assume that a=0". Put (b_c1,b_c2,…)=(b,0",0",…).
b=a+b=(∑_i∈C-{c1} 0")+b =∑_i∈C b_i
0"=a=∑_i∈C (b+(-b)) =d+d"=((∑_i∈C-{c1} b)+b)+d"=(d+b)+d"
=(∑_i∈C b + ∑_i∈C b_i )+∑_i∈C (-b)=∑_i∈C ((b+b_i)+(-b))=∑_i∈C b_i=b This is contradiction.♦
Put R=∪{F₂^I|I∈2^C-{∅}}={{a_i}_i∈I|∅≠I⊂C, a_i∈F₂} F₂={0',1'}, 0'+0'=1'+1'=0', 0'+1'=1'+0'=1'
Define {a_i}_i∈I～{a"_i}_i∈I"⇔ {i|(i∈I)∧(a_i=1')}={i|(i∈I")∧(a"_i=1')}. ～ is an equivalence relation.
Define {a_i}_i∈I+{a"_i}_i∈I"={d_i}_i∈I∪I" (d_i=a_i+a"_i for i∈I∩I", d_i=a_i for i∈I-I",
d_i=a"_i for i∈I"-I) Let C(u) be the equivalence class of u. If C is a set,R and R"={C(u)|u∈R} are sets.
u,u",v,v"∈R, u～u",v～v" ⇒ u+v～u"+v". You can define C(u)+C(v)=C(u+v). R" is a module then.
Theorem 4. When L=R" and b=C({1'}_i∈{c1})≠0", a=0" if C is a set.
Proof. Set up that L=R". 0"=C({0'}_i∈I) (∅≠I⊂C) I∩I"=∅⇒{a_i}_i∈I∪I"={a_i}_i∈I+{a_i}_i∈I"
So C({a_i}_{i∈{c1}∪{c2}})=C({a_i}_i∈{c1})+C({a_i}_i∈{c2}),…,C({a_i}_{i∈{c1}∪…∪{cm}})=C({a_i}_i∈{c1})+…+C({a_i}_i∈{cm}),…,
C({a_i}_i∈C)=C({a_i}_{i∈{c1}∪{c2}∪…(infinitely)})=C({a_i}_i∈{c1})+C({a_i}_i∈{c2})+…(infinitely)=∑_k∈CC({a_i}_i∈{k}).
0"=C({0'}_i∈C)=∑_k∈CC({0'}_i∈{k})=∑_k∈C0"=a ♦
Theorem 5. C isn't a set.
Proof. By theorem 3 and theorem 4.♦