My discovery about mathematics

Naoto Meguro. Amateur.
MSC 2010 Primary 15A78; Secondary 08C99 .
Key Words and Phrases. The infinite sums,the countable set,the continuum.
The abstract. Existence of N leads existence of some infinite sums leading contradiction.

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Existence of N lesds existence of some infinite sums leading contradiction. So you must not
use the countable set and cannot define R. This is an answer to the second problem
of Hilbert.

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Example 1. Put L=Z^N and F=Z^(N)=⊕_i∈N Z. L/F　is a module whose zero is 0"=F+(0,0,…).
Let's write ((n₁,n₂,…))=F+(n₁,n₂,…).
0"+0"+…(infinitely)=((n₁,0,0,…))+((0,n₂,0,…))+((0,0,n₃,0,…))+…
=((n₁,n₂,n₃,…)). So ((1,1,…))=0"+0"+…=((2,2,…)). This is contradiction.
0"+0"+…(infinitely) doesn't exist in the consistent mathematics.
Example 2. When L=Z_p and F=Z, L/F is a module whose zero is F+0.
(F+a)+(F+b)=F+a+b, (F+a+b)+(F+c)=F+a+b+c,…. Repeating this infinitely,you get
F+0=(F+0)+(F+0)+…(infinitely)=(F+1)+(F+p)+(F+p²)+…=F+(1+p+p²+…(infinitely)) ∈L/F.
(1+p+p²+…)-0∈F=Z. This is contradiction. (F+0)+(F+0)+… doesn't exist.
Example 3. When L=R and F=Q,
F+0=(F+0)+(F+0)+…(infinitely)=(F+3)+(F+0.1)+(F+0.04)+…=F+3.1415…=F+π ∈L/F.
π-0∈F=Q. This is contradiction.
You may set up that ∀x∀y(∑_i∈{x} y=y and ∀x∀y∀z((x∉y)→(∑_i∈y∪{x} z=(∑_i∈y z)+z)) as axioms.
∑_i∈{1} 0"=0", ∑_i∈{1,2} 0"=0"+0",…,∑_{i∈{1,2,3,…}} 0"=0"+0"+…(infinitely).
Existence of N leads existence of 0"+0"+… and the contradiction of example 1.
You must not treat the countable sets. ∪2^N=N. You must not treat 2^N and the continua.

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You must reconstruct the modern mathematics without using countable sets and continua.
You must not treat the continuum Z_p. Fermat conjecture may be still unsolved.
Theory treating R isn't consistent.
The millennium problems except P vs NP problem are provable nonsense then.

Naoto Meguro. Amateur.
MSC 2010. Primary 03C62;Secondary 03C55.
Key Words and Phrases. Peano's axioms,Fermat conjecture,Beal conjecture.
The abstract. Peano's axioms don't prove Fermat conjecture and Beal conjecture.

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I indicate Peano's axioms(PA) don't prove Fermat conjecture(FC) and Beal conjecture(BC).

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Assume that the standard mthematics is consistent. Put N={0,1,2,…}.
Put X={0∈N", ∀x∀y(((x∈N")→(s(x)∈N"))∧((s(x)=s(y))→(x=y))∧(s(x)≠0))∧((x≠0)→∃z(S(z)=x)),
∀x((0∈x)∧∀y((y∈x)→(s(y)∈x))→(N"⊂x)),∀x∀y((x∈N")∧(y∈N")→(x+0=x)∧(x+s(y)=s(x+y))
∧(x0=0)∧(xs(y)=xy+x))}
Let d be a free individual symbol. X doesn't include d.
Theorem 1. X has a model M for which M |= (d∈N")∧(d≠n) for ∀n∈N.
Proof. X∪{d∈N",d≠0,d≠1,d≠2,…,d≠m} 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≠0,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.♦
Let M be that in theorem 1. Let O be the object domain of M. Define M" and x^y by
M" |= x^y=the usual one when y∈N and M" |= x^y=0 for the other cases
and M |= p(c₁,…,c_m) ⇔ M" |= p(c₁,…,c_m). p(x₁,…x_m) is an arbitrary predicate symbol
and c_n∈O. The object domain of M" is O.
M" is a model of PA=X∪{∀x∀y((x∈N")∧(y∈N")→(x⁰=1)∧(x^s(y)=(x^y)x))}.
If N∈O, M |= N"=N. But M |= d∈N" and d∉N for M. So N∉O.
{x| M" |= (x∈N")∧(n^x≠0)}=N∉O for ∀n∈N-{0}. You cannot prove M" |= ∀x((x∈N")→(n^x≠0)) by induction.
Put FC=∀x∀y∀z∀n((x∈N")∧(y∈N")∧(z∈N")∧(n∈N")∧(xyz≠0)→(xⁿ⁺³+yⁿ⁺³≠zⁿ⁺³)).
Theorem 2. PA doesn't prove FC.
Proof. M" |= (d∈N")∧(2^d+3=0=0+0=2^d+3+2^d+3). M" |= ¬FC. ¬(PA |- FC). ♦
Theorem 3. PA doesn't prove BC.
Proof. M" |= (d∈N")∧(2^d+3+3^d+4=0+0=0=5^d+5. M" |= ¬BC. ¬(PA |- BC).♦
In the standard mathematics based on PA,you can set up p^d-0=0-0, p^d/1=0/1,(p^d,p^d,…)=(0,0,…)
p^d=0 as a real number. ζ(d)=2^d/(2^d-1)×3^d/(3^d/(3^d-1)…=0 (d≥1) then.
The standard mathematics based on PA doesn't prove Riemann hypothesis.

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You could not evolve the natural number theory by only PA.