My discovery about mathematics

                 1

I indicate defect of the formal algebra.

                 2

For u=∑_i∈I c_i and u"=∑_i∈I" c"_i, u+u"=∑_i∈I∪I"d_i (d_i=c_i+c"_i for i∈I∩I",
d_i=c_i for i∈I-I", d_i=c"_i for i∈I"-I) in the formal algebra. And c_m=∑_i∈{m}c_i.
So ∑_i∈I c_i=c_m+∑_i∈I-{m} c_i when m∈I.
And ∑_i∈I c_i + ∑_i∈I c"_i=∑_i∈I (c_i+c"_i).
If x is a countable set, ∑_i∈xy=y+y+…(infinitely). So you may set up that
∀x∀y(∑_i∈Nx=∑_i∈N-{y}x)　(N={1,2,3,…}).
Let 0" be the zero of a module L(∋c≠0"). Put e=∑_i∈N0",
c"=∑_i∈Nc, e"=∑_i∈N(-c).
Theorem 1. e≠0"
Proof. Assume e=0". Put (b₁,b₂,…)=(c,0",0",…).
c=c+e=c+∑_i∈N-{1}0"=∑_i∈Nb_i
0"=e=∑_i∈N(c+(-c))=c"+e"=(c+∑_i∈N-{1}c)+e"
=(c+c")+e"=∑_i∈N((b_i+c)+(-c))=∑_i∈Nb_i=c This is contradiction.♦
But you can make a module for which e=0".
Put R"={∑_i∈Ic_itⁱ| ∅≠I⊂N, c_i∈Z }. t is an indeterminate element.
Define u～u"⇔　（The coefficients in u except the zero correspond with those in u".)
for u,u"∈R". ～ is an equivalence relation.
u～u",v～v"⇒ u+v～u"+v". R"/～　becomes a module.
Let's write u=u" when u～u". You get a module R≅R"/～ by it.
For u=∑_i∈Ic_itⁱ and u"=∑_i∈I"c"_itⁱ, u=u" in R ⇔ (c_i=c"_i for i∈I∩I", c_i=0 for i∈I-I",
c"_i=0 for i∈I"-I.).
The zero of R equals ∑_i∈I0tⁱ (∅≠I⊂N).
Theorem 2. When L=R,e=0".
Proof.Set up that L=R. You may set up that c=∑_i∈{1}1tⁱ≠0".
0tⁱ=∑_s∈{i}0t^s=0" for ∀i∈N. e=∑_i∈N0"=∑_i∈N0tⁱ=0" ♦
The formal algebra isn't consistent by theorem 1 and theorem 2.

                              3

The standard mathematics isn't consistent. You must examine it soon.