Proof: Numbers with repeating blocks of digits are rational

In summary: 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)^nuse:\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.
  • #1
e(ho0n3
1,357
0
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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  • #2
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:

[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]
 
  • #3
Nice. I had totally forgotten about the geometric series. My head has overloaded with maths. Thanks again.
 
  • #4
n=0.3333...
10n=3.3333...
=>10n-n=9n=3.0
=>n=3/9=1/3
 
  • #5
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.
 
  • #6
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
 
  • #7
.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.
 
  • #8
have you tried using different bases other than 10?
 

FAQ: Proof: Numbers with repeating blocks of digits are rational

What is a rational number?

A rational number is any number that can be expressed as a ratio of two integers. This means that the number can be written as a fraction, where the numerator and denominator are both whole numbers.

How do you know if a number is rational?

A number is rational if it can be written as a repeating or terminating decimal. This means that the digits after the decimal point either repeat in a pattern or end after a certain number of digits.

What are repeating blocks of digits?

Repeating blocks of digits are sequences of numbers that repeat in a pattern after the decimal point. For example, in the number 0.333333..., the block of digits "3" repeats infinitely.

How can you prove that numbers with repeating blocks of digits are rational?

The proof involves using basic algebra and the concept of geometric series. By setting up the repeating decimal as a geometric series, we can manipulate the equation to show that it is equal to a fraction with two integers as the numerator and denominator, thus proving that it is a rational number.

Are all rational numbers also numbers with repeating blocks of digits?

Yes, all rational numbers can be expressed as numbers with repeating blocks of digits. This is because rational numbers can be written as fractions, and all fractions can be converted into repeating or terminating decimals.

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