Proof: Numbers with repeating blocks of digits are rational

e(ho0n3

Hi everyone,

I need to prove that any number with a repeating block of digits is a rational number. Someone told me I should first find a method of constructing a rational number in the form a/b from a number with repeating blocks of digits (and to do it with very 'easy' numbers first). I'm still stumped though.

For example, given 0.33333..., how do I show that it equals 1/3?

jcsd

You can express 0.333.. as a geometric series:

[tex]\sum^{\infty}_{n=1} \frac{3}{10^n} = 3\sum^{\infty}_{n=1} \left(\frac{1}{10}\right)^n[/tex]

use:

[tex]\sum^{\infty}_{n=1}r^n = \frac{r}{1-r}[/tex]

[tex]3\left(\frac{\frac{1}{10}}{1 - \frac{1}{10}}\right) = \frac{1}{3}[/tex]

e(ho0n3

Nice. I had totally forgotten about the geometric series. My head has overloaded with maths. Thanks again.

Gokul43201

n=0.3333...
10n=3.3333...
=>10n-n=9n=3.0
=>n=3/9=1/3

Gokul43201

Let n == 0.abc...kabc...kabc... repeating blocks of (abc...k), each block having r digits
Then n*10^r == abc...k(point)abc...kabc...kabc... i.e. move the decimal point r places to the right.
Now subtract, n*(10^r - 1)==abc...k digits after the decimal point vanish
So n== abc...k/(10^r - 1)= p/q, a rational number

QED

Plz excuse the freedom I've exercised with notation.

Gokul43201

Alternatively, jcsd's geometric sum method can be generalized for repeating blocks of any size.

PS: Also look at recent post on n/7, n=1,2,...6

CTS

.ABC...Z (with repeating length L)=
[tex]A*\sum_{k=1}^\infty \frac{1}{10^k^L}+B*\sum_{k=1}^\infty \frac{1}{10^k^L*10}+C*\sum_{k=1}^\infty \frac{1}{10^k^L*10^2}+...+[/tex][tex]Z*\sum_{k=1}^\infty \frac{1}{10^k^L*10^L/10}[/tex]

Because A, B,..., Z, are rational and [tex]\sum_{k=1}^\infty \frac{1}{10^k^X}[/tex] is rational for any X, the above sum is also rational.

lvlastermind

have you tried using different bases other than 10?