Proving if there are two different decimal representations of one number, every "decimal" digit will be either a 0 or 9

Question

Proposition: Let $m.d_1d_2d_3... and n.e_1e_2e_3...$ be different decimal expansions of the same non-negative number. If $m<n$ then $n = m + 1$ and every $e_k$ is 0 and every $d_k$ is 9.

Here is my proof so far:

Since $m.d_1d_2d_3...$ and $n.e_1e_2e_3...$ are the same number, $m.d_1d_2d_3... = m+ \sum_{k=1}^{\infty} \frac{d_k}{10^k} = n.e_1e_2e_3... = n + \sum_{k=1}^{\infty} \frac{e_k}{10^k}$

$ m - n$ = $\sum_{k=1}^{\infty} \frac{e_k}{10^k} - \sum_{k=1}^{\infty} \frac{d_k}{10^k}$

$ m -(m+ 1) = 1 = \sum_{k=1}^{\infty} (\frac{e_k}{10^k} - \frac{d_k}{10^k})$

$ 1 = \sum_{k=1}^{\infty} \frac{e_k-d_k}{10^k}$

= $\lim_{i\to\infty} (\sum_{k=1}^{i} \frac{e_k-d_k}{10^k})$

= $\lim_{i\to\infty} (\sum_{k=1}^{i} (\frac{1}{10})^k * (e_k-d_k)$

= $\frac{1}{9} * \lim_{i\to\infty} \sum_{k=1}^{i} (e_k-d_k)$

So obviously $(e_k-d_k)$ will be 9, and it makes sense because of the digits; but I am having trouble proving it.

I have tried proving it by way of contradiction but sadly to no prevail.

Any advice on how to move forward? Thank you !

Answer 1

If $n>m$ then $n\geq m+1$. Now, note that $m.d_1 d_2\dots \leq m+1$ and $n.e_1e_2\dots\geq n\geq m+1$. So, $$n.e_1e_2\dots\geq n\geq m+1\geq m.d_1 d_2\dots$$ Since we have the equality $n.e_1e_2\dots=m.d_1 d_2\dots$ (since they are just different representations of the same number), we conclude that in fact

$$n.e_1e_2\dots= n=m+1=m.d_1 d_2\dots$$

From here, note that all the $e_i$'s have to be zero (otherwise $n.e_1e_2\dots> n$) and all the $d_i'$s have to be 9 (otherwise $m+1>m.d_1 d_2\dots$).

Answer 2

Let $e_k-d_k=a_k$

Then $1=\sum a_k 10^{-k}=\sum 9\cdot 10^{-k}$. The first of these is by construction and the second is a known sum of a geometric progression (I have left out the limits of summation, which you can supply).

Then $0=\sum (9-a_k)10^{-k}$ and this should lead you to conclude that $a_k=9$ (choose the least $k$ for which this is no true, and show a contradiction)

Answer 3

$$n-m=\sum_{k=1}^{\infty} \frac{e_k}{10^k} - \sum_{k=1}^{\infty} \frac{d_k}{10^k}$$ How $n>m$ and $n,m\in\mathbb{N}$ and how $0\leq d_k,e_k\leq9$,$$0\leq\sum_{k=1}^{\infty} \frac{e_k}{10^k} - \sum_{k=1}^{\infty} \frac{d_k}{10^k}\leq1$$ then $$n-m=1\implies n=m+1$$ Now give that

$$1=\sum_{k=1}^{\infty} \frac{e_k}{10^k} - \sum_{k=1}^{\infty} \frac{d_k}{10^k}$$

$$1+\sum_{k=1}^{\infty} \frac{d_k}{10^k}=\sum_{k=1}^{\infty} \frac{e_k}{10^k} $$ How $0\leq\sum_{k=1}^{\infty} \frac{d_k}{10^k}\leq1$ and $0\leq\sum_{k=1}^{\infty} \frac{e_k}{10^k} \leq1$ then $$\sum_{k=1}^{\infty} \frac{d_k}{10^k}=0$$ and

$$\sum_{k=1}^{\infty} \frac{e_k}{10^k} =1$$ If there is a $d_k\neq0$ then $$\sum_{k=1}^{\infty} \frac{d_k}{10^k}>0$$ If there is a $e_k\neq9$ then

$$\sum_{k=1}^{\infty} \frac{e_k}{10^k} <1$$