recurring numbers

 Catbabel
 Posts: 771
 Joined: Sat Nov 16, 2019 8:18 am
recurring numbers
I had to work out the fraction that 39 represented of 99, and was pleased to find it was 0.39393939....and any fraction x of 99 (up to 99, obviously, and above that the pattern shifts) is the recurrent value 0.x. Is there a name for this property?
Re: recurring numbers
That’s really neat, I never knew that. Haven’t got an answer for you though.Allo V Psycho wrote: ↑Mon Dec 04, 2023 9:51 amI had to work out the fraction that 39 represented of 99, and was pleased to find it was 0.39393939....and any fraction x of 99 (up to 99, obviously, and above that the pattern shifts) is the recurrent value 0.x. Is there a name for this property?
where once I used to scintillate
now I sin till ten past three
now I sin till ten past three
Re: recurring numbers
There's a mathematical explanation for this, and it runs roughly as follows:
The repeating decimal 0.ababababababab.... can be written as
ab * (0.01 + 0.0001 + 0.000001 + other terms with number of zeros increasing by 2), which can be summarised as
ab * sum of 0.01^n as n goes from 1 to infinity.
This is a geometric series, with common ratio 0.01 and can be summed using the formula a/(1r) where a is the first term and r is the common ratio.
1  0.01 is 0.99, and the first term is 0.ab, so it simplifies to:
0.ab/0.99 = ab/99.
A similar technique can be used for other repeating decimals  0.abcabcabcabcabcabc... for instance is abc/999 (the common ratio is 1/1000 in the geometric series for this one)  and can be extended to a method to calculate the fraction for any decimal which eventually repeats.
The repeating decimal 0.ababababababab.... can be written as
ab * (0.01 + 0.0001 + 0.000001 + other terms with number of zeros increasing by 2), which can be summarised as
ab * sum of 0.01^n as n goes from 1 to infinity.
This is a geometric series, with common ratio 0.01 and can be summed using the formula a/(1r) where a is the first term and r is the common ratio.
1  0.01 is 0.99, and the first term is 0.ab, so it simplifies to:
0.ab/0.99 = ab/99.
A similar technique can be used for other repeating decimals  0.abcabcabcabcabcabc... for instance is abc/999 (the common ratio is 1/1000 in the geometric series for this one)  and can be extended to a method to calculate the fraction for any decimal which eventually repeats.
Last edited by geejaytee on Mon Dec 04, 2023 11:56 am, edited 1 time in total.
Re: recurring numbers
Quote rather than edit, ha.
Re: recurring numbers
It's just called a repeating or recurring decimal.
https://en.m.wikipedia.org/wiki/Repeating_decimal
I would be interested to check whether it's the same for any number n , nn, nnn etc expressed in Base n+1.
I think it seems likely
https://en.m.wikipedia.org/wiki/Repeating_decimal
I would be interested to check whether it's the same for any number n , nn, nnn etc expressed in Base n+1.
I think it seems likely
My avatar was a scientific result that was later found to be 'mistaken'  I rarely claim to be 100% correct
ETA 5/8/20: I've been advised that the result was correct, it was the initial interpretation that needed to be withdrawn
Meta? I'd say so!
ETA 5/8/20: I've been advised that the result was correct, it was the initial interpretation that needed to be withdrawn
Meta? I'd say so!
Re: recurring numbers
I think GJT's explanation above applies to bases other than ten. However, I can't make it work with hexadecimal floats in Python, but I think that's because <float>.hex() doesn't actually convert the float to base 16, but instead shows the hexadecimal for the IEEE representation of the decimal float.Gfamily wrote: ↑Mon Dec 04, 2023 12:32 pmIt's just called a repeating or recurring decimal.
https://en.m.wikipedia.org/wiki/Repeating_decimal
I would be interested to check whether it's the same for any number n , nn, nnn etc expressed in Base n+1.
I think it seems likely
e.g. replace:
with:1  0.01 is 0.99, and the first term is 0.ab, so it simplifies to:
0.ab/0.99 = ab/99.
1  0.01 is 0.ff, and the first term is 0.ab, so it simplifies to:
0.ab/0.ff = ab/ff.
Re: recurring numbers
999999 has factors including 7, 13 and 37. But they are factors of no smaller integer made up of repeated 9s, as with 3 and 11. (3, 7, 11, 13 and 37 are the only prime factors of 999999). That is why the recurring decimals for n/7, n/13 and n/37, and compounds of those such as n/91, have a repeated group of 6 figures.
It is easy to understand that any prime p (that is not a factor of 10) must have a recurring decimal of at most p1 digits. This is apparent because when you are doing a long division to it work out, it must cycle back to where you started in at most p1 steps. That's because the remainder is always a number somewhere from 1 to p1. So that presents an interesting way of discovering that any prime p (that is not a factor of 10) must divide exactly into at least one integer made up of repeated 9s of at most p1 digits. Also true for repeated 1s and repeated 3s, for we can take out the factors of 3.
It is easy to understand that any prime p (that is not a factor of 10) must have a recurring decimal of at most p1 digits. This is apparent because when you are doing a long division to it work out, it must cycle back to where you started in at most p1 steps. That's because the remainder is always a number somewhere from 1 to p1. So that presents an interesting way of discovering that any prime p (that is not a factor of 10) must divide exactly into at least one integer made up of repeated 9s of at most p1 digits. Also true for repeated 1s and repeated 3s, for we can take out the factors of 3.