My discovery about mathematics

Naoto Meguro. Amateur.
MSC 2010. Primary ;Secondary .
Key Words and Phrases .
The abstract.

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Define ∑_n∈N a(n)=a(1)+a(2)+a(3)+…=…((a(1)+a(2))+a(3))+a(4))+…(infinitely),
( ∑_n∈Na(n))+(∑_n∈Nb(n))=∑_n∈N(a(n)+b(n)),
Let M be a module. Let 0 be the zero of M. Put E(M)={∑_n∈N c(n) | c(n)∈M.}, 0"=∑_n∈N 0.
Theorem 1.c∈M⇒ 0"=c+0".
Proof. c∈M ⇒ E(M)∋∑_n∈N c=c+c+c+…=…((c+c)+c)+c);…(infinitely)=…((2c)+c)+c)+…(infinitely)=2c+c+c+…
(∑_n∈N -c)+(∑_n∈N c)=(-c+c)+(-c+c)+…=(-c+2c)+(-c+c)+(-c+c)+…=0"=c+0". ♦
Put e(1)=(1,0,0,0,…(infinitely)), e(2)=(0,1,0,0,…),…. Let A be a commutative ring.(0,1∈A.)
Put F(A)={∑_n∈N a(n)e(n) |a(n)∈A.}
Theorem 2. F(A)={(0",0",…(infinitely))}.
Proof. F(A)∋(a(1),0,0,…)+(0,a(2),0,0,…)+…=(a(1)+0+0+…,0+a(2)+0+0+…,0+0+a(3)+0+0+…,…)=(0",0",0",…) by theorem 1.♦
⊕_n∈N Ae(n)={∑_n∈N a(n)e(n) | a(n)∈A, ∃p∀n((n∈N)∧(n≥p)→(a(n)=0))}⊂F(A)={(0",0",…)}.
Let M={m(1),m(2),…} be a countable set. Define e(m(i))=e(i). By theorem 2, ⊕_m∈M Ae(m)={(0",0",…)}. When M is an A-module,
⊕_m∈MAe(m)∋∑_m∈Ma(m)e(m) → ∑_m∈Ma(m)m∈M doesn't become a mapping.

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