
#1
Jun404, 07:07 PM

P: 1,370

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? 



#2
Jun404, 07:25 PM

Sci Advisor
PF Gold
P: 2,226

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



#3
Jun404, 08:27 PM

P: 1,370

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




#4
Jun404, 09:00 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

Proof: Numbers with repeating blocks of digits are rational
n=0.3333...
10n=3.3333... =>10nn=9n=3.0 =>n=3/9=1/3 



#5
Jun404, 09:07 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

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
Jun404, 09:09 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

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
Jun504, 02:41 PM

P: 20

.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
Jun904, 09:09 PM

P: 101

have you tried using different bases other than 10?



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