### Post by on Aug 16, 2015 20:30:15 GMT -8

Because hexadecimal's only prime factor is two, only dividing by twos gives terminating fractions.

1/2 = 0.8

1/3 = 0.5555555....

1/4 = 0.4

1/5 = 0.333333.....

1/6 = 0.2AAAAAA....

1/7 = 0.249249249249249249249....

1/8 = 0.2

1/9 = 0.1C71C71C71C71C71...

1/A = 0.19999999999...

.. and so on. However, if you have discovered a new real number, and you can't prove if it is rational or not.... write its digits in hexadecimal. It certainly isn't a common fraction like 5/6 or 8/3, so it will recur in any integer base except an absurdly large one. The thing about hexadecimal is its recurring periods tend to be shorter, so you could spot a pattern in the digits and prove the newly discovered number is rational. Any indefinitely repeating sequence indicates the number can't be irrational.

Why would hexadecimal's periods tend to be shorter? Because sixteen is a fourth power - each hexadecimal digit corresponds to two quaternary digits and four binary ones. Let's say you have a repeating period in binary with an even length. Write the same number in quaternary (base 4, or 2^2), and the pattern is cut in half. If it is STILL an even number of digits in quaternary, change it into hexadecimal (square of four) and it will be halved yet again.

The biggest source of long periods is cyclic primes. Any prime above 2 is odd, so the maximum number of digits it can recur is (prime - 1), which is even. For instance, one seventeenth in decimal is 0.0588235294117647058823529411764705882352.... with a sixteen-digit cycle. Of course, not all prime factors not in the factorization of a given base are cyclic, but only square bases are *guaranteed* to have no cyclic primes. One exception is nonary (base nine) where the prime 2 is cyclic. Oh, big deal, a 1-digit period. Square bases prevent any of these lengthy recurrences since their lengths in non-square bases are always even. Hexadecimal does the double whammy because some of the lengths are multiples of four and get "doubly halved."

But hey, what if you still can't spot a recurring period in a mysterious number using hexadecimal? If you're lucky, the period in binary may have a multiple of 3 length (but odd), in which case you could cut it by three and notice it using octal (base eight, 2^3). An octal digit is three bits, so.. yeah, I already told you what that does, lolz

So, is there a useful base which has shorter recurring periods on average and still terminates more fractions?!? Crud, the smallest of such is 2^2 * 3^2 = 4 * 9 = 6^2 = thirty-six. 36(decimal) is a highly composite number, with nine factors (1, 2, 3, 4, 6, 9, 12, 18, 36), so it's a nice base if you want to use all those symbols, and maybe create a multiplication table with 1296 numbers.

1/2 = 0.8

1/3 = 0.5555555....

1/4 = 0.4

1/5 = 0.333333.....

1/6 = 0.2AAAAAA....

1/7 = 0.249249249249249249249....

1/8 = 0.2

1/9 = 0.1C71C71C71C71C71...

1/A = 0.19999999999...

.. and so on. However, if you have discovered a new real number, and you can't prove if it is rational or not.... write its digits in hexadecimal. It certainly isn't a common fraction like 5/6 or 8/3, so it will recur in any integer base except an absurdly large one. The thing about hexadecimal is its recurring periods tend to be shorter, so you could spot a pattern in the digits and prove the newly discovered number is rational. Any indefinitely repeating sequence indicates the number can't be irrational.

Why would hexadecimal's periods tend to be shorter? Because sixteen is a fourth power - each hexadecimal digit corresponds to two quaternary digits and four binary ones. Let's say you have a repeating period in binary with an even length. Write the same number in quaternary (base 4, or 2^2), and the pattern is cut in half. If it is STILL an even number of digits in quaternary, change it into hexadecimal (square of four) and it will be halved yet again.

The biggest source of long periods is cyclic primes. Any prime above 2 is odd, so the maximum number of digits it can recur is (prime - 1), which is even. For instance, one seventeenth in decimal is 0.0588235294117647058823529411764705882352.... with a sixteen-digit cycle. Of course, not all prime factors not in the factorization of a given base are cyclic, but only square bases are *guaranteed* to have no cyclic primes. One exception is nonary (base nine) where the prime 2 is cyclic. Oh, big deal, a 1-digit period. Square bases prevent any of these lengthy recurrences since their lengths in non-square bases are always even. Hexadecimal does the double whammy because some of the lengths are multiples of four and get "doubly halved."

But hey, what if you still can't spot a recurring period in a mysterious number using hexadecimal? If you're lucky, the period in binary may have a multiple of 3 length (but odd), in which case you could cut it by three and notice it using octal (base eight, 2^3). An octal digit is three bits, so.. yeah, I already told you what that does, lolz

So, is there a useful base which has shorter recurring periods on average and still terminates more fractions?!? Crud, the smallest of such is 2^2 * 3^2 = 4 * 9 = 6^2 = thirty-six. 36(decimal) is a highly composite number, with nine factors (1, 2, 3, 4, 6, 9, 12, 18, 36), so it's a nice base if you want to use all those symbols, and maybe create a multiplication table with 1296 numbers.