Tiny Nines

Stage: 4 Challenge Level: Challenge Level:1

Good solutions to this were sent in by Chen of The Chinese High School, Singapore, Anders and Sammy of Bentley Park College, Rebecca of Henry Box School and Andrei of School 205, Bucharest, Romania. Rolf sent in a very nicegeneralisation. Well done all of you!

"I observed the pattern, then I proved it. Writing the decimal expansions we get recurring decimals and I use the brackets to denote that the set of digits inside the brackets is repeated over and over indefinitely.

$1/9 = 0.11111\dots = 0.(1)$

$1/99 = 0.010101\dots = 0.(01)$

$1/999 = 0.001001001\dots = 0.(001)$

$1/9999 = 0.000100010001\dots = 0.(0001)$ and so on.

The general pattern is:

$$\frac{1}{(\underbrace{9\dots9}_{n})} = 0.(\underbrace{00\dots0}_{n-1}1)$$

Now, I have to prove it. I multiply the number by $10^n$ $$ a = 0.(\underbrace{00\dots0}_{n-1}1)$$ $$10^{n}a = 1.(\underbrace{00\dots0}_{n-1}1)$$ Now, I subtract the first expression from the second one: $$(10^{n} - 1)a = 1$$ $$a = \frac{1}{10^{n} - 1} = \frac{1}{\underbrace{99\dots9}_{n}}$$ So, I have proved that I am right.''

Here is Rolf's generalization.

"A generalization can be stated as follows: Given a number $(a_1)(a_2)(a_3)(a_4)\dots(a_n)$ where each $(a_i)$ corresponds to the $i^{th}$ digit of the number, we can create an infinitely repeating decimal of the form: $0.a_{1}a_{2}a_{3}a_{4}\dots{a_{n}}a_{1}a_{2}a_{3}a_{4}\dots{a_{n}}\dots$ by writing it as a fraction with repeated nines as the denominator, that is

$0.a_{1}a_{2}a_{3}a_{4}\dots{a_{n}}$ (a sequence of $n \; 9$'s).

A proof of this is given for the sequence $0.234523452345\ldots$

Let $k= 0.234523452345 \ldots$

Then $10000k = 2345.23452345 \ldots$ and $9999k= 2345$.

Hence $k = 2345/9999$ as desired.

This proof can clearly be modified to prove the generalization for any sequence of $(a_{i})$s.''