Question

How do you prove that for all values of `n/p,n!=kp,kinRR`, where p is any prime number that isn't 2 or 5, gives a recurring decimal?

As in `(n/3,n/7,n/9,n/11,"etc.")` giving recurring decimals, but the only non-recurring decimals being given by fractions which are only multiples of `1/2` and `1/5`.

Answer 1

`"See explanation"`

Explanation:

`"When dividing numerically, we can only have at most p"`
`"different remainders. If we encounter a remainder that"`
`"we had before, we get in a cycle."`

`n/p = a_1 a_2 ... a_q . a_{q+1} a_{q+2} ...`

`"Now call "r = n - [a_1 a_2 ... a_q] * p","`
`"then "0<= r < p.`

`r/p = 0.a_{q+1} a_{q+2} ...`

`r_2 = 10 r - p a_{q+1}`

`"Then we have"`
`0 <= r_2 < p`

`"And when dividing further, we repeat with "r_3" between"`
`0" and "p-1". And then "r_4", and so on..."`

`"Whenever we encounter a "r_i" that we have encountered"`
`"before we begin to cycle."`
`"As there are only "p" different "r_i" possible, this will certainly"`
`"happen."`

`"2 and 5 are not special, they give recurring 0 which we also"`
`"can consider as a recurring decimal. And we do not have to"`
`"restrict ourselves to prime numbers."`