Naoto Meguro. Amateur.
MSC 2010. Primary 08A65 ;Secondary 26E99 .
Key Words and Phrases. Countable sums, the real number theory.
The abstract. The real number theory by countable sums isn't consistent.
The usual algebra doesn't treat infinite sums of elements of the module. I try to think them and lead
contradiction of the real number theory etc..
Put N"={0}∪N. Define ∑n∈N" a(n)=a(0)+a(1)+…(infinitely) =…((a(0)+a(1))+a(2))+a(3))+…(infinitely)
and (∑n∈N" a(n))+(∑n∈N" b(n))=∑n∈N" (a(n)+b(n)).
Let M be a Z-module. Put E(M)={∑n∈N" c(n)|c(n)∈M} and 0"=∑n∈N"0 (0∈M).
E(M) is a module whose zero is 0".
Theorem 1. c∈M⇒ 0"=c+0+0+0+…(infinitely).
Proof. E(M)∋∑n∈N" c=…((c+c)+c)+c)+…=…((2c+c)+c)+…=2c+c+c+….
(∑n∈N" -c)+(∑n∈N" c)=(-c+c)+(-c+c)+…=(-c+2c)+(-c+c)+(-c+c)+…=0"=c+0+0+…(infinitely). ♦
Define (∑n∈N" a(n))(∑n∈N" b(n))=∑n∈N"(∑0≤k≤n a(k)b(n-k)).
Let A be a commutative ring. E(A) is a ring then.
Theorem 2. E(A)={0"}
Proof. By theorem 1, 0"=1+0+0+…(infinitely). If x∈E(A), x=(1+0+0+…(infinitely))x=0" x=0". ♦
You may define R={∑n∈N" c(n)|(c(0)∈Z)∧∀n((n∈N)→(c(n)=0)∨(c(n)=1/2n))}. By theorem 2,
{0"}=E(Q)⊃R. R={0"}? 0"=E(k[x1,…,xm])⊃k[[x1…,xm]]={0"}?
{0"}=E(Z)⊃Zp. Zp={0"}? {0"}=E(C)∋∑n∈N" 1/(n+1)s (s∈C). ζ(s)=0" for ∀s∈C?
What I do is only indication of contradiction. The mathematics using R or k[[x1,…,xm]] or Zp is doubtful
until someone corrects the above defect of it. If an irrational number exists in the real world, the laws of nature allow
0.000…=1.000…=2.000…=…. I proposed an experiment about it to European Nuclear Society etc..
MSC 2010. Primary 08A65 ;Secondary 26E99 .
Key Words and Phrases. Countable sums, the real number theory.
The abstract. The real number theory by countable sums isn't consistent.
1
The usual algebra doesn't treat infinite sums of elements of the module. I try to think them and lead
contradiction of the real number theory etc..
2
Put N"={0}∪N. Define ∑n∈N" a(n)=a(0)+a(1)+…(infinitely) =…((a(0)+a(1))+a(2))+a(3))+…(infinitely)
and (∑n∈N" a(n))+(∑n∈N" b(n))=∑n∈N" (a(n)+b(n)).
Let M be a Z-module. Put E(M)={∑n∈N" c(n)|c(n)∈M} and 0"=∑n∈N"0 (0∈M).
E(M) is a module whose zero is 0".
Theorem 1. c∈M⇒ 0"=c+0+0+0+…(infinitely).
Proof. E(M)∋∑n∈N" c=…((c+c)+c)+c)+…=…((2c+c)+c)+…=2c+c+c+….
(∑n∈N" -c)+(∑n∈N" c)=(-c+c)+(-c+c)+…=(-c+2c)+(-c+c)+(-c+c)+…=0"=c+0+0+…(infinitely). ♦
Define (∑n∈N" a(n))(∑n∈N" b(n))=∑n∈N"(∑0≤k≤n a(k)b(n-k)).
Let A be a commutative ring. E(A) is a ring then.
Theorem 2. E(A)={0"}
Proof. By theorem 1, 0"=1+0+0+…(infinitely). If x∈E(A), x=(1+0+0+…(infinitely))x=0" x=0". ♦
You may define R={∑n∈N" c(n)|(c(0)∈Z)∧∀n((n∈N)→(c(n)=0)∨(c(n)=1/2n))}. By theorem 2,
{0"}=E(Q)⊃R. R={0"}? 0"=E(k[x1,…,xm])⊃k[[x1…,xm]]={0"}?
{0"}=E(Z)⊃Zp. Zp={0"}? {0"}=E(C)∋∑n∈N" 1/(n+1)s (s∈C). ζ(s)=0" for ∀s∈C?
3
What I do is only indication of contradiction. The mathematics using R or k[[x1,…,xm]] or Zp is doubtful
until someone corrects the above defect of it. If an irrational number exists in the real world, the laws of nature allow
0.000…=1.000…=2.000…=…. I proposed an experiment about it to European Nuclear Society etc..