I regret that I do not remember the details; reconstructing them now would basically be working from scratch. Obviously the well-definedness of multiplication can be proven in the decimal construction; the only question is how ugly the proof is. The Cauchy construction has the burden of first constructing the rationals; the decimal construction avoids that step, but again, I'm not sure if it turns out to be worth it. On Fri, Feb 12, 2010 at 10:40 AM, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Feb 12, 2010 at 10:26 AM, Allan Wechsler <acwacw@gmail.com> wrote:
I have a vivid memory of performing this construction as an exercise, I think when I was a senior in high school. I was motivated by annoyance at the Cauchy-sequence construction, and was certain that the formal-decimal-expansion approach would prove much more straightforward. I don't remember whether it was, in fact.
In some sense, the difference is that the Cauchy-sequence construction defines an equivalence relation on Cauchy sequences, while the "infinite decimals" construction chooses a canonical representative for each equivalence class (except for rationals with certain denominators, where it ends up choosing two canonical representatives instead.
In defining multiplication on infinite decimals, it's easy to define a sequence that represents the product, but you then have to find an equivalent canonical sequence. If this isn't significantly easier than first showing that the sequence is Cauchy and then showing that *every* Cauchy sequence is equivalent to a canonical sequence, then completing the infinite-decimals construction pretty much includes proving that the two constructions are equivalent.
Or is there a cleverer way to define multiplication of infinite decimals that I'm missing? You can take the products of the finite decimal approximations, but I don't see an easy way to show that each digit is ultimately constant as you take this sequence.
Andy
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