# Apostol's Analysis Problem 1.22

1. Jan 2, 2012

### julypraise

1. The problem statement, all variables and given/known data
Given x>0, suppose a_0 = [x], the largest integer less than or equal to x. Assuming a_0, a_1, . . . , a_n-1 are defined, define a_n as the largest integer such that a_0.a_1...a_n less than or equal to x. Now define r_n = a_0.a_1...a_n, r_0 = a_0. Prove that the sup of the set of all the r_n's is x.

2. Relevant equations

3. The attempt at a solution
I did proof by contradiction but does not work. I've got stuck. Just give me a useful hint, don need full solution. I've analyzed all the meanings carefully, but seems I need a trick to solve this, or maybe not. I even tried to use the notion of limit: I thought about |r_n - x|. It goes 0. But this fact it is even harder to prove.

Okay, while I'm writing this thread, I've got an idea. Please have a look:
Suppose there exists an upper bound p of the set such that p is less than x. Then, since |p-x|>0, for some n, 1/(10^n) < |p-x|. Thus p + 1/10^n < x. Note that a_0.a_1...a_n <= p. Thus a_0.a_1...(a_n + 1) < x; but this is a contradiction.

I've used a huge assumption that "for some n, 1/(10^n) < |p-x|." But I didn't learn this yet in this book. Is there any other way not using it?

2. Jan 4, 2012

### TheFurryGoat

I don't think you haven't copied the problem correctly.
The book i'm looking at also gives a fixed integer $k\geq2,$ and $a_n$ is defined to be the largest integer s.t. $\displaystyle \sum_{i=0}^n{\frac{a_i}{k^i}} \leq x.$ You should be able to prove that $\sup\{r_n:n\geq 0\}=x$ by first proving that $0\leq a_i \leq k-1, \forall i\geq 1$ and that $r_n\leq r_{n+1},\ \forall n.$ After these I think you can use the idea you got to finish the proof, but don't give $k$ any particular value.