My discovery about mathematics

                                   1

It was a very difficult problem to prove the axiom of parallels but it is easy to see that
you cannot prove it by a non-standard model. Riemann hypothesis(RH) is so if the standard
real number theory(SRNT) is formed. If SRNT is formed,there is a non-standard model of the
standard mathematics for which the exponential function has a zero point. There are
non-standard counterexamples of Fermat conjecture(FC) and RH for the model.
While you assume SRNT,you cannot prove FC and RH in the formalism.

                                   2

Assume that the standard mathematics is consistent.
Let X be a countable and consistent axiom system of the standard mathematics.
Let d₁,d₂,… be the free individual symbols. The mathematical axioms in X don't include the d_i's. Put d=d₁.
Set up that (1∈N")∈X, ∀n((n∈N")→(n+1∈N")) ∈X,∀n((n∈N"-{1})→∃m((m∈N")∧(n=m+1))) ∈X,
∀x∀y((x∈N")∧(y∈N")→((x=y)↔(x+1=y+1))) ∈X, ∀x((x∈N")→(1/2^x≥0))∈X,
∀x((1∈x)∧∀y((y∈x)→(y+1∈x))→(N"⊂x)) ∈X, ∀x∀y((x≤y)→(1/2^y≤1/2^x)) ∈X,
∀n((n∈N")→(2ⁿ⁺¹=2ⁿ2)∧(1/2ⁿ⁺¹=(1/2ⁿ)(1/2))∧(2ⁿ=1/(1/2ⁿ))) ∈X, ∀m∀n((m1=m)∧(m(n+1)=mn+m))∈X,
∀s∀n((n∈N")∧(s∈C")∧(Im s≥n)→(|e^s√-1|≤1/2ⁿ))∈X, ∀s((s∈C")→(1/|e^-s|=|e^s|)) ∈X,
∀x∀y((x≤y)∧(y≤x)↔(x=y)) ∈X, ∀x((x∈C")→(0≤|x|)∧((|x|=0)↔(x=0))) ∈X, (N"⊂R")∧(R"⊂C")∈X.
These axioms are formed for the standard model for which N"=N and R"=R and C"=C and e=2.71828….
Theorem 1. X has a model M for which M |= (d∈N")∧(d≥n) for ∀n∈N.
Proof. X∪{d∈N")∪{d≥1}∪…∪{d≥n} has a model M_n for which
M_n |= (d=n)∧(N"=N) and is consistent for ∀n∈N.
So X∪{d∈N"}∪{d≥1}∪{d≥2}∪…(infinitely) is consistent and has a model M.
M is a model of X and M |= (d∈N")∧(d≥n) for ∀n∈N.♦
Theorem 2. M |= 1/2^d=0 for M and d in theorem 1 if SRNT is formed.
Proof. Let D be the object domain of M. You may set up that D is the set of the closed terms.
Define (x≥y ⇔M |= x≥y) and (x=y ⇔M |= x=y) for x∈D and y∈D.
These don't lead contradiction for the axiom system X∪{P| M |= P} is consistent.
M |= d≥n. M |= 1/2^d≤1/2ⁿ. 1/2^d≤1/2ⁿ for ∀n∈N. M |= d∈N". M |= 0≤1/2^d. 0≤1/2^d .
1-1/2ⁿ=1/2+…+1/2ⁿ→1/2+1/2²+…(infinitely)=∑_m∈N 1/2^m=s (n→∞, n∈N). If SRNT is formed,
R∋s=1/2+s/2. s=1. 1/2ⁿ→0 (n→∞, n∈N). 0≤1/2^d≤1/2ⁿ for ∀n∈N. 1/2^d→0 (n→∞, n∈N).1/2^d=0. M |= 1/2^d=0.♦
1/2^d≠0 ⇒ ¬(1/2ⁿ→0 (n→∞, n∈N)) ⇒ (s≠1)∧(R∋s=1/2+s/2) ⇒ R isn't a field.
Put FC=∀x∀y∀z∀n((x∈N")∧(y∈N")∧(z∈N")∧(n∈N")∧(n≥3)→(xⁿ+yⁿ≠zⁿ)).
Theorem 3. ¬(X |- FC) if SRNT is formed.
Proof. For M and d in theorem 1, M |= 1/2^d+1=(1/2^d)(1/2)=0(1/2)=0=1/2^d.
M |= 2^d=1/(1/2^d)=1/(1/2^d+1)=2^d+1. By the definition of multiplication m(n+1)=mn+m,
M |= 2^d=2^d+1=2^d2=2^d(1+1)=2^d1+2^d=2^d+2^d.
M |= (d∈N")∧(2∈N")∧(d≥3). So M |= ¬FC. If X |- FC, M |= FC. This is contradiction.♦
You cannot formalize the known proof of FC if SRNT is formed.
Put RH=∀s(s∈C")∧(0≤Re s≤1)∧(ζ(s)=0)→(Re s=1/2)).
Theorem 4. ¬(X |- RH) if SRNT is formed.
Proof. You may set up that ∀s((s∈C")→(0/Γ(s)=0))∈X and ∀s((s∈C")∧(s≠1)→(0ζ(s)=0))∈X
and ∀s((s∈C")→(ζ(s)=(2π)^sζ(1-s)/2Γ(s)cos(πs/2))) ∈X and ∀s((s∈C")∧(0≤Re s≤1)→((2π)^s0=0)) ∈X
for these are formed for the usual Γ(s) and the usual ζ(s) and the usual (2π)^s and C"=C.
For M and d in theorem 1, M |= ∀s((s∈C")∧(Im s≥d)→(0≤|e^s√-1|≤1/2^d=0)).
So M |= ∀s((s∈C")∧(Im s≥d)→(0≤|1/cos s|=2/|0+e^-s√-1|=2|e^s√-1|≤2(1/2^d)=0)).
M |=∀s((s∈C")∧(0≤Re s≤1)∧(Im πs/2≥d)→(ζ(s)=0)).
M |= ζ(1+d√-1)=0. M |= ¬RH. If X |- RH , M |= RH. This is contradiction.♦

                              3

If the standard mathematics is consistent,you can set up that X is the set of the theorems
of the standard mathematics. If FC is a theorem of the standard mathematics, FC∈X then.
So X |- FC. This is contradiction by theorem 3 if SRNT is formed. SRNT may not be consistent.