If a < b then a^2 < b^2. This is not true if you let a = 0 and b = h. The dual numbers do not have an ordering. You should provide sources and proofs next time because it seems like you are just making things up.
That isn't what an ordered ring is. Your property of a < b → a² < b² doesn't even hold true in the integers. For instance, let a = -2 and b = -1.
The correct property is that if a ≤ b, a + c ≤ b + c, and if a ≥ 0 and b ≥ 0, then ab ≥ 0. It is fairly easy to see that these properties hold for dual numbers.
In my argument a and b are positive and this is true for all ordered rings but not for the dual numbers as you've defined them. Specify the ordering and you will realize h can not be larger nor smaller than 0 because both cases lead to a contradiction.
MikeBattaglia|1 year ago
The correct property is that if a ≤ b, a + c ≤ b + c, and if a ≥ 0 and b ≥ 0, then ab ≥ 0. It is fairly easy to see that these properties hold for dual numbers.
clooper|1 year ago
In any case, I'm dropping out of this thread.