Need help for a proof, related to decimal representation of a rational number

Apr 2011
108
2
Somwhere in cyberspace.
The problem appears in Tom Apostol's Calculus, Volume 1, pages 31,32. This is introductory material, related to foundations of real number system.
Quote:
Since we shall do very little with decimals in this book, we shall not develop their properties in any further detail except to mention how decimal expansions may be defined analytically with the help of the Least-Upper-Bound axiom.
If x is a given positive real number, let \(\displaystyle a_{0}\) denote the largest integer \(\displaystyle \le x\) Having chosen \(\displaystyle a_{0}\), we let \(\displaystyle a_{1}\) denote the largest integer such that
\(\displaystyle a_{0}+\frac{a_{1}}{10}\le x\)
More generally, having chosen \(\displaystyle a_{0},a_{1},\ldots,a_{n-1}\) we let \(\displaystyle a_{n}\) denote the largest integer such that
\(\displaystyle a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots+\frac{a_{n}}{10^{n}}\le x\)
Let S denote the set of all numbers
\(\displaystyle a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots+\frac{a_{n}}{10^{n}}\)
obtained this way for n=0,1,2... Then S is nonempty and bounded above, and it is easy to verify that x is actually the least upper bound of S.
No proof is given and as often happens to me, contrary to the claim, I could not do the thing easily.

So I need a proof that sup(S)=x.
For this purpose I should use Least-Upper-Bound axiom which roughly states that any set of real numbers that has upper bound has supremum too (which is real number).

I did try to prove it, resulting somewhat ugly and cumbersome proof, which most surely has weaknesses. I'm giving it here in hope someone could
comment on it. My attempt at the proof uses two intermediate facts.

a/ If a,x,y are real numbers which satisfy equalities \(\displaystyle a\le x \le a + \frac{y}{n}\) for any \(\displaystyle n\ge 1\) then \(\displaystyle a=x\). It is proven in the textbook as consequence of the Least-Upper-Bound axiom
b/ \(\displaystyle \frac{1}{n}>\frac{1}{10^{n}}\) for any \(\displaystyle n \ge 1\).

The proof is by induction on n. Case n=1 is obviously true, so assuming
\(\displaystyle \frac{1}{n}>\frac{1}{10^{n}}\) is true we need to prove \(\displaystyle \frac{1}{n+1}>\frac{1}{10^{n+1}}\)
\(\displaystyle \frac{1}{n+1}>\frac{1}{10^{n+1}} \iff \frac{1}{n+1}> \frac{1}{10^{n}} \frac{1}{10}\)
Since \(\displaystyle \frac{1}{n}>\frac{1}{10^{n}}\)
if we manage to prove that
\(\displaystyle \frac{1}{n+1}> \frac{1}{n} \frac{1}{10}\) we shall have
\(\displaystyle \frac{1}{n+1}> \frac{1}{n} \frac{1}{10} > \frac{1}{10^{n}} \frac{1}{10}\).
This is easy
\(\displaystyle \frac{1}{n+1} > \frac{1}{n} \frac{1}{10} \iff 10n > n+1 \iff 9n>1\) which is true for all \(\displaystyle n\ge 1\).

Now to the proof of the main statement.

By construction of x,

\(\displaystyle a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots+\frac{a_{n}}{10^{n}} \le x < a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots+\frac{a_{n}}{10^{n}} + \frac{1}{10^{n}}\) for any \(\displaystyle n \ge 0\) and x is an upper bound of S.
By Least-Upper-Bound axiom, S has supremum and it's a real number. Denote \(\displaystyle y=sup(S)\). Then
\(\displaystyle a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots+\frac{a_{n}}{10^{n}} \le y\) for any \(\displaystyle n \ge 0\).
Combining the two last inequalities with b/ we can conclude that
\(\displaystyle x < a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots+\frac{a_{n}}{10^{n}} + \frac{1}{10^{n}} \le y + \frac{1}{10^{n}} \le y + \frac{1}{n}\) for any \(\displaystyle n \ge 0\).
And because y is least upper bound we have
\(\displaystyle y \le x < y + \frac{1}{n}\) for any positive integer n. Applying a/ gives us the equality x=y.

I'd appreciate any comments as well as one-liner proof by contradiction for example.
Live long and prosper ;-).
 
Dec 2012
1,145
502
Athens, OH, USA
Hi,
Your proof looks fine to me. I think maybe Tom was thinking along these lines:
Let \(\displaystyle s_n=a_0+{a_1\over10}+\cdots+{a_n\over10^n}\) and \(\displaystyle S=\{s_n:n\geq0\}\). Then sup S=x:

As you point out, for any \(\displaystyle n\geq0,s_n\leq x<s_n+{1\over10^n\) and so x is an upper bound of S. Let \(\displaystyle \epsilon>0\). Choose n with \(\displaystyle {1\over10^n}<\epsilon\) -- n can be such that \(\displaystyle \text{log}_{10}(\epsilon^{-1})<n\). Then \(\displaystyle s_n\in S\) and \(\displaystyle x-\epsilon<s_n\leq x\). Done.
 
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Apr 2011
108
2
Somwhere in cyberspace.
Thank You very much for the proof and sorry for the delayed feedback. This seems quite much like one-liner I asked for!