- 3,385
- 760
So what is the structure of Ras an abelian group under addition?
The factor group of the free abelian group generated by all real numbers by the subgroup generated by all relations ##(p,q) = \{p \text{ and } q \text{ are represented by the same Dedekind cut}\}##.lavinia said:So what is the structure of Ras an abelian group under addition?
I do not see why this is true.fresh_42 said:The factor group of the free abelian group generated by all real numbers by the subgroup generated by all relations ##(p,q) = \{p \text{ and } q \text{ are represented by the same Dedekind cut}\}##.
Edit: Maybe not a satisfactory algebraic explanation but we have to define the relations which makes to real numbers equal.
... and a prof of mine once said: A mathematician's talent is transmitted to his son-in-lawmathwonk said:mike is emil's son. (artin)
It seems to me that Dedekind cuts assume that you already know what addition is.fresh_42 said:Let ##C = ℝ## be the continuum and basis of the free abelian group ##F##. Further let ##ι : C → F## be the embedding (of the alphabet) and ##id: C → ℝ_+## the identity. Then there is a unique group homomorphism ##π: F → ℝ_+## which extends ##id##, i.e. ##id = π \, ι## because ##F## is free. ##π## is onto. The question is what kernel ##π## has or what makes ##4 + 5 = 9## since ##4 + 5## and ##9## are different elements in ##F##. The Dedekind cuts were what occurs to me. Maybe there is a more algebraic way to define the equivalence classes.
I thought one needs the order in ℝ and the embedding of ℚ. But you are right, this breaks the definition requirements only down on ℚ.lavinia said:It seems to me that Dedekind cuts assume that you already know what addition is.
micromass said:As for the addition structure on ##\mathbb{R}##. It might help to look at ##\mathbb{R}## as a ##\mathbb{Q}##-vector space of dimension ##2^{\aleph_0}##. So we could see it as something like ##\bigoplus_{2^{\aleph_0}~\text{factors}} \mathbb{Q}##