trying to make an axiomatic system for "natural" numbers.
natural numbers are the "numbers" that "count" the "finite" "sets". 
0 1 2 3 4 5 6 7 8 9 ...
"zero" is included for the "empty set"
there is no set with 1/2 , -1, sqrt(2) or i members.


(N,+)
a+b=b+a
(a+b)+c=a+(b+c)
a+0=a
if a+b=a+c then b=c
if a+b=0 then a=b=0


(N-{0},.)
ab=ba
(ab)c=a(bc)
a1=a
if ab=ac then b=c
if ab=1 then a=b=1


a(b+c)=ab+ac


a≤b == there exists x in N such that a+x=b
a|b == there exists x in N-{0} such that ax=b

both are partial orders:
reflective, anti-symmetric, transitive


this is a definition: a0=0
what about this: a+∞=∞ ?


0+1=1
0 ≤ 1


1+1 is not 0
1+1 is not 1
1+1=2
1 ≤ 2


if a≤b and c≤d then ac≤bd

a+x=b and c+y=d, using distributivity
bd = (a+x)(c+y) = ac + (ay+cx+xy)

if √2 ≤ 1 then 2 ≤ 1 contradiction
if 1 ≤ √2 then √2 ≤ 2


no finite model
cancelation law means in + table every row has a 0 in it.
a+b=0 is not possible, unless a=b=0



[1,∞)∪{0} is a model!
yet 1 and √2 are not comparable


model ℕ[√2] = { a+b√2 | a,b∈ℕ }

1 ≤ √2 ?
exist(x) 1+x=√2 ? x=√2-1 ∉ ℕ[√2]


model ℕ(√2) ?  NO
(√2-1)(√2+1)=2-1=1

model ℕ(√3) ?



these axioms are not enough to prove or disprove the ≤ being a total order.




well ordering principle?