My discovery about mathematics

Let C=∪_i∈I C_i be a classification of C by an equivalence relation　～.
If C/～={C_i}_i∈I is a set,∪(C/～)=C is a set. If C isn't a set,C/～ isn't a set.
Z is countable and isn't a set. Z/mZ isn't a set.
Let A be a ring. Let M,N be A modules. If Z^(M×N) is a set,
{f|(f∈Z^(M×N))∧∀m∀n(((m,n)≠(0,0))→(f((m,n))=0))}(≅Z(0,0)) is a set.
But this is countable and isn't a set. Z^(M×N) isn't a set. So M⊗_AN isn't a set.
By the axiom of infinity,∃I((1∈I)∧∀x((x∈I)→(x+1∈I))).
N"={x|(x∈I)∧∀y((1∈y)∧∀z((z∈y)→(z+1∈y))→(x∈y))} is a set.
If m-1∉N" for m∈(N"-N),(1∈(N"-{m}))∧∀n((n∈(N"-{m}))→(n+1∈(N"-{m}))).
N"⊂(N"-{m}) This is contradiction. m-1∈N". If m-1∈N,m∈N. So m-1∈(N"-N).
(N"-N)∋m-2,m-3,…(infinitely)
Theorem 4. When M(∋c≠0) is a module, M^(N") isn't a set.
Proof. If M^(N") is a set, B={f|(f∈M^(N"))∧∀n(((f(n)=0)∨(f(n)=c))
∧((n∈N")∧(n-1∈N")∧(f(n)=c)→(f(n-1)=c)))} is a set. Assume f∈B and m∈N"-N.
If f(m)=c, 0≠c=f(m)=f(m-1)=f(m-2)=…(infinitely). f∉M^(N") then.
So f(m)=0. B={(0,0,…),(c,0,0,…),(c,c,0,0,…),(c,c,c,0,0,…)…} is countable and isn't a
set. This is contradiction.♦
Let A(∋1≠0) be a ring. A^(S) may not be a set for a set S. You cannot define M⊗_AN by
using A^(M×N). Generally,you cannot get the surjection A^(M)→M and cannot get the
projective resolution for the A module M. You cannot define Tor and Ext for A modules.
Put N(x)=(x is a natural number.)⇔(x=1)∨(x=2)∨…(infinitely)
,Q(x)=(x is a rational number.),R(x)=(x is a real number.).
Theorem 5. 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∃1 n∃1 r((x=(-1)^s(n+∑_i∈r 1/2ⁱ))∧((s=0)∨(s=1))∧(N(n)∨(n=0))∧
∀y((y∈r)→N(y))∧∀y(N(y)→∃m(N(m)∧(y≤m)∧(m∉r)))).
And ∀x∀y(N(x)∧N(y)→R(x/y)).
r is countable or finite. Countable r doesn't exist as a set. So r is finite when R(x).
R(x)⇒(x=m/2^h)∧(m∈Z)∧N(h).
R(1/3) is false. This is contradiction.♦
You cannot formalize propositions of the natural number theory. Fermat conjecture
∀x∀y∀z∀p(N(x)∧N(y)∧N(z)∧N(p)→(x^p+2+y^p+2≠z^p+2)) is nonsense
in the standard mathematics treating the real numbers.
Theorem 6. You must not use R(x)　in the standard mathematics.
Proof. If you can use R(x),you can use R((0,x,∅))∧(x≠0)⇔N(x).
This is contradiction by theorem 5.♦
Riemann hypothesis ∀x∀y(R(x)∧R(y)∧(ζ(x+y√-1)=0)→(x=1/2)∨(y=0)) is nonsense in
the standard mathematics.
Theorem 7. You must not use Q(x) in the standard mathematics treating the real numbers.
Proof. By Q((0,x,∅))∧(x≠0)⇔N(x).♦
BSD conjecture is nonsense in the standard mathematics treating the real numbers.

                                5

Countable sets which you use intuitively aren't sets. But you need countable sets to
treat the real numbers. So you cannot treat the real numbers and cannot evolve the
analytics and the geometry. Millennium problems except P=NP problem are nonsense.