Let's make a non-standard model of the complex number theory.
Put C"=CN, N"=NN,Z"=ZN,R"=RN,{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,…). 2k=(2,2,…)(k,k,…)=(2k,2k,…)
(a1,a2,…)/2k=(a1/2k,a2/2k,…)
Define ∑k∈N (ak 1,ak 2,…)/2k=(∑k∈Nak 1/2k,∑k∈N ak 2/2k,…).
R"={m+∑k∈N bk/2k| m∈Z",bk∈{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.
Put C"=CN, N"=NN,Z"=ZN,R"=RN,{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,…). 2k=(2,2,…)(k,k,…)=(2k,2k,…)
(a1,a2,…)/2k=(a1/2k,a2/2k,…)
Define ∑k∈N (ak 1,ak 2,…)/2k=(∑k∈Nak 1/2k,∑k∈N ak 2/2k,…).
R"={m+∑k∈N bk/2k| m∈Z",bk∈{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.