I am reading a passage from the book Foundation of Mathematics by Ian Stewart, and I need some help to make sure I understand it properly.
A real number can be expresed by the following unique decimal expansion:
$a_0.a_1a_2\cdot \cdot \cdot a_n \le x \lt a_0.a_1a_2\cdot \cdot \cdot a_n + \frac{1}{10^n}$ [formula 1]
Apparently this approach does not allow for the expansion of $0.9999 \cdot \cdot \cdot\cdot $ i.e. infinite $9$s.
That is because if we try to express it like that we have ($0.999999..9$ means $n$ $9s$):
$0.999999..9 \le 0.9999 \cdot \cdot \cdot\cdot \lt 0.999999..9 + \frac{1}{10^n} \implies 0 \lt 1 - 0.9999 \cdot \cdot \cdot\cdot \le \frac{1}{10^n}$ for all $n \in N$
This last condition is impossible by Archimides' condition: since $1 - 0.9999 \cdot \cdot \cdot\cdot \gt 0$ which means that the $2$ numbers are different then it means there must exist $n$ with $\frac{1}{10^n} \lt 1 - 0.9999 \cdot \cdot \cdot\cdot$. But according to the inequality we have $ 1 - 0.9999 \cdot \cdot \cdot\cdot \le \frac{1}{10^n}$
Is my explanation so far correct? The book just mentions it is impossible but I want to make sure my understanding of why it is impossible is correct.
Now an alternative way that could allow to express $0.9999 \cdot \cdot \cdot\cdot$ would be to change the equality signs i.e.:
$a_0.a_1a_2\cdot \cdot \cdot a_n \lt x \le a_0.a_1a_2\cdot \cdot \cdot a_n + \frac{1}{10^n}$ [formula 2]
This would lead to:
$0 \le 1 - 0.9999 \cdot \cdot \cdot\cdot \lt \frac{1}{10^n}$
Now since $0 \le 1 - 0.9999 \cdot \cdot \cdot\cdot$ we have $2$ cases:
Case $1$: $0 \lt 1 - 0.9999\cdot \cdot \cdot\cdot$ which is the same case mentioned earlier which leads to impossibility
Case $2$: $0 = 1 - 0.9999\cdot \cdot \cdot\cdot$ which means that $1$ and $0.999\cdot \cdot \cdot\cdot$ are the same number.
Hence we are able to express $0.999\cdot \cdot \cdot\cdot$ as a decimal expression of $1$.
Is my understanding so far correct?
Then the text follows saying that the [formula 2] does not allow the expansion of $1.000\cdot\cdot\cdot$ i.e. infinite $0$s
This part I am not sure I get it either.
I suppose the equation becomes (with $n$ digits at left and right side of the inequality) :
$1.0000 \lt 1.000\cdot \cdot \cdot \le 1.0001$
and by doing the same manipulation as earlier we end up with:
$0 \le 1.0001 - 1.000\cdot \cdot \cdot \cdot \cdot\cdot\lt \frac{1}{10^n}$
and by applying the same use case we get that $1.0001$ with the $n$ digits is the same expression as $1.000\cdot \cdot \cdot \cdot$
But somehow I don't think I understand the idea properly. Could someone please help me?
Update:
I realized that this is has to be wrong:
$1.0000 \lt 1.000\cdot \cdot \cdot \le 1.0001$
Because $1.000\cdot \cdot \cdot$ is actually $1$ hence we can't have $1.0000 \lt 1.000\cdot \cdot \cdot$ and so the formula 2 is invalid.
But formula 1 should hold because $1.0000 \le 1.000\cdot \cdot \cdot \cdot$