My discovery about mathematics

                                1

I show a counter-example of Fermat conjecture by a non-standard model.
You cannot prove Fermat conjecture in the formalism. The known proofs
use the logic which you cannot formalize. If you admit such logic,the
standard mathematics isn't consistent.

                                2

Let X be a countable and consistent axiom system of the standard mathematics.
You may set up that ∀x∀y((x∈N")∧(y∈N")→(x≤y)∨(x≥y))∈X and
∀a((a∈{0,1}^N")→∃x((x∈N")∧∀y((y∈N")∧(x≤y)→(a(y)=0))→(∑_m∈N" a(m)2^m ∈N"))) ∈X
and (1∈N")∈X, (2∈N")∈X , … and ∀x((x∈N")→(2^x∈N")) ∈X
for these are formed for the standard model for which N"=N={1,2,3,…}.
Let d be a free individual symbol. The mathematical axioms in X don't include d.
Theorem 1. X has a model M for which M |= d-m∈N" for ∀m∈N∪{0}.
Proof. X∪{d∈N"}∪{d-1∈N"}∪…∪{d-m∈N"} has a model M_m for which
M_m |= (d=m+1)∧(N"=N) and is consistent for ∀m∈N.
So X∪{d∈N"}∪{d-1∈N"}∪{d-2∈N"}∪…(infinitely) is consistent and has a model M.
M is a model of X. M |= d-m∈N" for ∀m∈N.♦
Put FC(n)=∀x∀y∀z((x∈N")∧(y∈N")∧(z∈N")→(xⁿ+yⁿ≠zⁿ)) (n∈N).
Theorem 2. ¬(X |- FC(n)) for∀n∈N.
Proof. There is a b(∈{0,1}^N) for which |1/2^1/n-∑_m∈Nb(m)/2^m|≤1/2^k for ∀k∈N.
You may set up that (c∈{0,1}^N")∧((∑_m∈N" c(m)/2^m)ⁿ=1/2)∈X and (c(1)=b(1))∈X,(c(2)=b(2))∈X,…
and ∀x((x∈N")→(1/2^x≥0)∧(1/2^x≠0))∈X for these are formed for the standard model for which N"=N.
Let D be the object domain of M in theorem 1. Define x≥y ⇔ M |= x≥y for x∈D and y∈D.
x=y ⇔ (x≥y)∧(y≥x) ⇔ M |= x=y for x∈D and y∈D. Put L={m|(m∈D)∧(M |= m∈N")}.
Put r=∑_m∈L c(m)/2^m and s=∑_m∈N b(m)/2^m. By formal calculation of real numbers,you get
sⁿ=∑_m∈N b"(m)/2^m (b"(m)=0 or =1). |s|≤1 1/2^1/n≤1
|sⁿ-1/2|=|(s-1/2^1/n)(s^n-1+s^n-2/2^1/n+…+1/2^(n-1)/n|≤n/2^n+m≤1/2^m for ∀m∈N.
When b"(1)=1,∃m((m∈N-{1})∧(b"(m)=1))⇒(sⁿ-1/2≥1/2^m). So sⁿ=1/2. When b"(1)=0,
∃m((m∈N-{1})∧(b"(m)=0))⇒(1/2-sⁿ≥1/2^m). So sⁿ=∑_m∈N1/2^m+1. sⁿ-sⁿ/2=1/2² sⁿ=1/2 in this case too.
You may assume ∀x(∀m((m∈x)→(c(m)/2^m≥0))→(∑_m∈x c(m)/2^m ≥0)) and
∀x∀y((x∩y=∅)→∑_m∈x∪y 1/2^m=∑_m∈x 1/2^m +∑_m∈y 1/2^m
and ∑_m∈{k} 1/2^m=1/2^k as axioms about ∑.
k∈L ⇒ M |= k∈N" ⇒ M |= c(k)/2^k≥0 ⇒ c(k)/2^k≥0 L-{m}-N⊂L So q=∑_k∈L-{m}-N c(k)/2^k ≥0.
And m∈L-N⇒ 1/2^1/n=r=q+c(m)/2^m+s≥c(m)/2^m+s≥s=1/2^1/n ⇒ c(m)=0.
M |= c(m)=1 ⇒ m∈N ⇒ M |= d-m∈N" ⇒ d-m∈L.
M |= 2^dr=0+02+02²+… …+c(m)2^d-m+…+c(1)2^d-1+02^d+02^d+1+…=∑_m"∈N" e(m")2^m" ∈N".
(e(m")=c(d-m") when (m"∈L)∧(d-m"∈L) and e(m")=0 when (m"∈L)∧(d-m"∉L).
So m"≥d ⇒e(m")=0. ∑_m∈L e(m)2^m=2^d∑_m∈L c(m)/2^m +∑_m∈L 02^d+m-1 -∑_m∈L 0/2^m)=2^dr+0-0.
You may set up that ∀x((x∈N")→(g(x)∈{0,1}^N")
∧∀m"∀y(((m",y)∈g(x))↔((m"∈N")∧(x-m"∈N")∧(y=c(x-m"))∨((m"∈N")∧(x-m"∉N")∧(y=0)))) ∈X.
M |= d∈N" M |= g(d)=e∈{0,1}^N" So M |= 2^dr∈N".) M |= rⁿ=1/2. M |= 2^d∈N"
M |= (2^dr)ⁿ+(2^dr)ⁿ=(2^d)ⁿ. M |= ¬FC(n). If X |- FC(n), M |= FC(n). This is contradiction.♦
You cannot prove Fermat conjecture in the formalism. The axiom systems which prove
FC(n) aren't consistent.
Theorem 3. The standard mathematics isn't consistent.
Proof. You may set up that FC(4)∈X for FC(4) is formed in the standard mathematics
for which N"=N. X |- FC(4) X isn't consistent then by theorem 2.♦

                                 3

The known proofs of FC(n) use the unique factorization theorem or the tensor product
which you cannot formalize. (2^d(∈N") isn't a product of finite prime numbers.)
The strict formalism doesn't use them and can ignore theorem 3 but cannot prove Fermat conjecture.