Can you prove this theory? (decimal period of rational numbers)?

Question

Can you prove this theory? (decimal period of rational numbers)?

Answer 1

355/113 will recur with a era of length below 113. The definition wins the following and says that 355/113 is rational. the area is in assuming that 355/113, which superficially imitates pi partly of its decimal representation, would not terminate. yet another situation is in the actual incontrovertible truth that the rationals and irrationals are disjoint gadgets with the help of their definitions. i'm curious as to why you theory that 355/113 does no longer recur?

Answer 2

Maybe I've understood what you mean.



As to the proof for the periodic numbers formula (how to get a ratio from a periodic number), its reason lies in the formula for the infinite sum of (1/10)^n as n varies from 1 to infinity: try and think about it!

For the rest of your theory, it seems reasonable to me.

Answer 3

No I can not, however i will blame it on the fact i dont know what your talking about!

Answer 4

Let's consider base N and a number k (the divisor)

and take the powers of N: 1, N, N^2, N^3, ...



There's an infinity of such numbers, however their remainders when divided by k can only be smaller than k, therefore some must be repeating.



Let N^a and N^b (a
Then k | N^a*(N^(b-a)-1). Assuming that k is not divisible by the prime factors of N, it follows that k and N^a are mutual primes, so k | N^(b-a)-1.



Let c=b-a; c>0 because a
But N^c-1 is nothing else than the digit (N-1) repeated c times.

If you choose the minimal such number c, then (N^c-1)/k gives exactly the period of 1/k in base N.





Example: N=10, k=7



powers of N: 1, 10, 10^2=100, 10^3=1000, 10^4=10000, 10^5=100000, 10^6=1000000, ...

remainders when divided by 7: 1, 3, 2, 6, 4, 5, 1, ...

We notice a repeating remainder: 1, for 1=10^0 and 10^6. Then 7 | 10^6-1=999999

999999/7=142857

1/7=0.(142857)