Dividing by $b^n-1$ in base $b$ results in repeating decimals. Can this be proven with modular arithmetic?

Question

I'm trying to do a video about mathematical discovery and proof. Idea goes something like this, with appropriate demonstrations along the way:

• notice that $\frac19 = 0.\overline1$, $\frac29 = 0.\overline2$, etc. Any digit $d_0$ in $\frac{d_0}9=0.\overline{d_0}$.
• notice that $\frac{12}{99} = 0.\overline{12}$, $\frac{34}{99} = 0.\overline{34}$. Any two digits $d_0$ and $d_1$ over two nines gives $\frac{d_0d_1}{99}=\overline{d_0d_1}$.
• same for $\frac{123}{999} = 0.\overline{123}$,$\frac{1234}{9999} = 0.\overline{1234}$, $\frac{12345}{99999} = 0.\overline{12345}$, and so on. For a number $n$ digits long, dividiing by that many 9's, we always get $\frac{(d_0d_1d_2 \cdots d_n)}{(999 \cdots9)_n}=(\overline{d_0d_1d_2 \cdots d_n})$.

Of course, the 9's aren't a coincidence; 9 is one less than the base of the number system. So it should be true that for any base, the above fractions can be modified (assuming they're legal fractions in the base). That means I should be able to prove for base $b$ (and here my notation gets a little iffy) $\frac{{d_0}_b}{b-1}=0.\overline{{d_0}}_b$, $\frac{{d_0d_1}_b}{b^2-1}=0.\overline{d_0d_1}_b$, and in general $\frac{{(d_0d_1d_2 \cdots d_n)}_b}{b^n-1}=\overline{(d_0d_1d_2 \cdots d_n)}_b$.

Mechanically you can see what's going on if you do ${12345}\div{99999}$ written out in long division (which I can't do in $\LaTeX$). Each step in the long division yields a rotated form of 12345 (i.e. 23451, 34512, etc.) and the final quotient is indeed $0.\overline{12345}$.

The in-general case is tricky since all the digits are different. You can see that $(d_0d_1d_2 \cdots d_n)_b$ divided by $b^n-1=\overline{(d_0d_1d_2 \cdots d_n)}_b$ if you again do long division, and you can see why the numbers rotate. But the best I can do is handwave over the $\cdots$ in the division.

I know the proofs of divisibility by 9 and by divisibility by $b-1$ in base $b$. One uses mods and one uses induction.

I lean towards induction when I notice $\frac{\overbrace{111 \cdots 1}^{n}}{b^n-1}=\frac{b^{n-1}+b^{n-2}+ \cdots + b + 1}{b^n-1}=\frac{b^{n-1}+b^{n-2}+ \cdots + b + 1}{(b-1)(b^{n-1}+b^{n-2}+ \cdots + b + 1)}=\frac{1}{b-1}=0.\overline{1}$, but once you introduce different digits this really doesn't lead anywhere.

It's pretty clear that I don't need to introduce true decimals into the proof. If I stick to the realm of integers, I can use the scads of theorems I learned about mods in my Abstract Algebra course, but as of now nothing works out. And I've tried turning the infinite decimals into geometric sums, to no avail.

So I'm looking for help on proving $\frac{{(d_0d_1d_2 \cdots d_n)}_b}{b^n-1}=\overline{(d_0d_1d_2 \cdots d_n)}_b$.

Answer 1

Let $a$ be the numerator of the fraction.

The value of a "decimal" fraction with an $n$-digit repeating part equal to $a$ (with leading 0's if needed) is:

$$\frac{a}{b^n} + \frac{a}{b^{2n}} + \frac{a}{b^{3n}} + \frac{a}{b^{4n}} + ... = \sum_{k=1}^\infty{\frac{a}{b^{nk}}}$$

This is a geometric series with initial value $\frac{a}{b^n}$ and ratio $r = \frac{1}{b^n}$. If $b > 1$, then $|r| < 1$, so the sum converges to $\frac{\frac{a}{b^n}}{1-\frac{1}{b^n}} = \frac{a}{b^n - 1}$.

Answer 2

HINT.-$1+\dfrac{1}{b^n}+\dfrac{1}{b^{2n}}+\cdots+\dfrac{1}{b^{kn}}+\cdots=\dfrac{b^n}{b^n-1}$ $$\dfrac{1}{b^n}+\dfrac{1}{b^{2n}}+\cdots=\dfrac{1}{b^n-1}$$ From this you do have multiplying both $LHS$ and $RHS$ by $(d_0d_1d_2 \cdots d_n)$ $$\frac{{(d_0d_1d_2 \cdots d_n)}_b}{b^n-1}=0.\overline{(d_0d_1d_2 \cdots d_n)}_b$$ where you assume that the numerator is less than $b^n-1$ (if each $d_i=b-1$ then the quotient is equal to $1$ and the number $b-1$ is supposed to be a "digit" in base $b$, analogue to $9$ in base $10$)