My discovery about mathematics

Let's make a non-standard model of the complex number theory.
Put C"=C^N, N"=N^N,Z"=Z^N,R"=R^N,{0,1}"={0,1}^N.
Define f((a₁,a₂,…), (b₁,b₂,…)…,(p₁,p₂,…))=(f(a₁,…,p₁), f(a₂,…p₂),…) for f(a,b,,…,p).
(a₁,a₂,…)+(a"₁,a"₂,…)=(a₁+a"₁,a₂+a"₂,…) etc. then.
Let's write a=(a,a,a,…). 2^k=(2,2,…)^(k,k,…)=(2^k,2^k,…)
(a₁,a₂,…)/2^k=(a₁/2^k,a₂/2^k,…)
Define ∑_k∈N (a_{k 1},a_{k 2},…)/2^k=(∑_k∈Na_{k 1}/2^k,∑_k∈N a_{k 2}/2^k,…).
R"={m+∑_k∈N b_k/2^k| m∈Z",b_k∈{0,1}"}.
You get a non-standard model of the complex number theory by the above.
Put {c₁, c₂,…}={c |(c∈C)∧(ζ(c)=0)∧(0≤Re(c)≤1)} and d=(c₁,-2,c₂, -4,…).
ζ(d)=(ζ(c₁),ζ(-2),…)=(0,0,…)=0, Re(d)=(Re(c₁),-2, Re(c₂),-4,…)≠(1/2,1/2,…)=1/2,
d/(-2)=(c₁/(-2),1,c₂/(-2),2,…)∉N". Put RH=∀s((ζ(s)=0)∧(Re(s)≠1/2)→(s/(-2)∈N")).
RH is false in the above model. The complex number theory doesn't prove RH.
The above model doesn't treat the predicate symbol x≤y. Z" isn't a domain. The above isn't final solution.