in your explorations you really should try DOING LONG DIVISION BY HAND and don't just use a calculator and notice patterns. if you do the long division, it is really obvious.
It is so obvious if the repeating cycle has a length of 6 then of course there will be 6 different states that exist in the long division algorithm, so of course it will repeat. I can also prove that for some base b, the repeating part of the digital representation formed by 1/n will have a length of at most n-1, because there can only be n states. But now define a function L(n) which is defined as the length of the repeating part of the base b division 1/n. for terminating numbers, this is 0.
in your explorations you really should try DOING LONG DIVISION BY HAND and don't just use a calculator and notice patterns. if you do the long division, it is really obvious.
DarthPickley 7 months ago
It is so obvious if the repeating cycle has a length of 6 then of course there will be 6 different states that exist in the long division algorithm, so of course it will repeat. I can also prove that for some base b, the repeating part of the digital representation formed by 1/n will have a length of at most n-1, because there can only be n states. But now define a function L(n) which is defined as the length of the repeating part of the base b division 1/n. for terminating numbers, this is 0.
DarthPickley 7 months ago