Proof: Numbers with repeating blocks of digits are rational

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Discussion Overview

The discussion revolves around proving that numbers with repeating blocks of digits are rational numbers. Participants explore various methods to construct rational representations from such numbers, focusing on both specific examples and general approaches.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests starting with simple examples, such as 0.33333..., to demonstrate its equivalence to 1/3.
  • Another participant introduces the geometric series method to express 0.333... as a series, leading to the conclusion that it equals 1/3.
  • A different approach is presented using algebraic manipulation, where n is defined as 0.3333..., and the decimal is shifted to derive the rational form.
  • One participant generalizes the geometric series method for repeating blocks of any size.
  • A further contribution discusses a summation approach for numbers represented with repeating blocks of digits, asserting that the sum remains rational due to the rationality of its components.
  • Another participant inquires about the possibility of using different bases other than 10 for the representation of repeating decimals.

Areas of Agreement / Disagreement

Participants generally agree on the rationality of numbers with repeating blocks of digits, but multiple methods and approaches are presented without a consensus on a single preferred method.

Contextual Notes

Some methods rely on specific assumptions about the representation of numbers, such as the base used for the decimal system, and there are unresolved details regarding the generalization of these methods.

e(ho0n3
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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?
 
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e(ho0n3 said:
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?

You can express 0.333.. as a geometric series:

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

use:

\sum^{\infty}_{n=1}r^n = \frac{r}{1-r}

3\left(\frac{\frac{1}{10}}{1 - \frac{1}{10}}\right) = \frac{1}{3}
 
Nice. I had totally forgotten about the geometric series. My head has overloaded with maths. Thanks again.
 
n=0.3333...
10n=3.3333...
=>10n-n=9n=3.0
=>n=3/9=1/3
 
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.
 
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
 
.ABC...Z (with repeating length L)=
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}+...+Z*\sum_{k=1}^\infty \frac{1}{10^k^L*10^L/10}

Because A, B,..., Z, are rational and \sum_{k=1}^\infty \frac{1}{10^k^X} is rational for any X, the above sum is also rational.
 
have you tried using different bases other than 10?
 

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