- #1

marcin w

- 12

- 0

Given a positive real number x, let a0 denote the largest integer <= x. Having chosen a0, we let a1 denote the largest integer such that

a0 + (a1)/10 <= x.

More generally, having chosen a0, a1,...,an-1, we let an denote the largest integer such that

a0 + (a1)/10 + ... + (an)/10^n <= x.

Then let S denote the set of all numbers obtained this way for n = 0, 1, 2, ...

I understand the construction of set S because I have proven before that the greatest integer in a positive real number exists, hence the set is nonempty. I also see that

a0 + (a1)/10 + ... + (an + 1)/10^n is an upper bound, so this property with the nonemptyness of S guarantees that S has a supremum, say b. The punchline is that I want to verify that x = b.

I'm not trying to prove here that .999... = 1. I know that and can prove it using geometric series and algebraic arguments. I want to rigorously prove the above using just the bare essentials (ie. axioms) of the real number system.

I was thinking of arguing by contradiction that x > b or x < b is impossible. My other thought for a rough sketch was that to prove that the infinum of the set for decimal expansion of x - b is 0, but I don't know where to start really. Thanks in advance.