Number Theory proof (the number of decimal places in the cycle of the reciprocal of any prime number must be smaller than the prime number itself)

The number of decimal places in a repeating prime number must be smaller than that in the prime number itself.

Generally, decimal places with the reciprocal of prime numbers are complex. Because the reciprocal of prime numbers is also a fraction, it must be a round-robin decimal (except for 2 and 5 ), but what is the maximum number of decimal places in the cycle of prime numbers? Is it possible to be very large and infinitely long? This is the problem to be solved in this article?

Description of the number of cyclic decimal places after the reciprocal of prime numbers:
1/3 = 0. 33..., round robin decimal point = 3, round robin decimal point = 1
1/7 = 0. 142857142857..., round robin decimal point = 142857, round robin decimal point = 6
1/11 = 0. 0909..., round robin decimal point = 09, round robin decimal point = 2
1/13 = 0. 076923076923..., round robin decimal point = 076923, round robin decimal point = 6
1/17 = 0. 05882352941176470588235294117647..., round robin decimal point = 0588235294117647, round robin decimal point = 16
1/19 = 0. 052631578947368421052631578947368421..., round robin decimal point = 052631578947368421, round robin decimal point = 18
1/23 = 0. 04347826086956521739130434782608695652173913..., round robin decimal point = 0434782608695652173913, round robin decimal point = 22
1/23 = 0. 03448275862068965517241379310344827586206896551724137931..., round robin decimal point = 0344827586206896551724137931, round robin decimal point = 28
1/31 = 0. 032258064516129032258064516129..., round robin decimal point = 032258064516129, round robin decimal point = 15
1/37 = 0. 027027..., looping decimal point = 027, repeating decimal point = 3
1/41 = 0. 0243902439..., round robin decimal point = 02439, round robin decimal point = 5
...

1/97 = 0. 010309278350515463917525773195876288659793814432989690721649484536082474226804123711340206185567..., loop decimal places = 96

From the list above, we can see that there are generally many decimal places in the back-to-back cycle of prime numbers, but we have not seen that the number of digits is greater than the prime number.ProgramThe same is true for the number less than 1000. Therefore, I would like to confirm this conclusion.

The following is a proof of concept:

Assume that p is a prime number. The number of decimal places in 1/p is X, and 1/P can be expressed as X/999... 9. The denominator has x 9 in total,

For example:
1/3 = 0. 33..., loop decimal point = 3, loop decimal point = 1, then 1/3 = 3/9
1/7 = 0. 142857142857..., the number of decimal places in the loop = 142857, and the number of decimal places in the loop = 6, then 1/7 = 142857/999999
1/37 = 0. 027027..., loop decimal point = 027, loop decimal point = 3, then 1/37 = 027/999

Certification Process:
Suppose 1/p = 0. x1x2... XN..., x1x2... XN is a circular decimal number. There are n digits in total. Let's do some transformations:

10 ^ N * (1/p)
= 10 ^ N * (0. x1x2... XN...) = x1x2... xn + 0. x1x2... XN ......

=> 10 ^ N * (0. x1x2... XN...)-0. x1x2... XN... = x1x2... XN

=> (10 ^ N-1) * (0. x1x2... XN...) = x1x2... XN

=> 0. x1x2... XN... = x1x2... xN/(10 ^ N-1)

=> 0. x1x2... XN... = x1x2... xN/999... 9 (n-1 9 in total)

Suppose K = 999... 9, there is always a p-9,If we can find an integer x, x smaller than K and 1/p = x/K, then our proposition is proved..

To prove this conclusion, we need to reference the ferma's theorem,

See:
Baidu Encyclopedia: http://baike.baidu.com/view/263807.htm

The ferma's theorem is an important theorem in number theory. Its section contains:If P is a prime number and (A, P) = 1, then a ^ (p-1) 1_1 (mod P)

Suppose P is a prime number. Here we take a = 10, (10, P) = 1.
Conclusion:

10 ^ (p-1) 1_1 (mod P)

10 ^ (p-1)-1 Jun 0 (mod P)

10 ^ (p-1)-1 = 999... 9, P-1, abbreviated as K
K then 0 (mod P), so there is an integer x, making
X * P = K
=> 1/p = x/K
Because 1/p is less than 1, x is certainly less than K. Because K is composed of a number of p1-9, the number of digits of X is smaller than P.

This proposition proves.