When I was writing about Dozenal Number System (or Duodecimal Number System), this thought struck me.
In Decimal Number System, just looking at the last digit (least significant digit or unit’s digit) we can say whether the whole number is divisible by 2, 5 and 10. Looking at the last two digits we can say its divisibility by 2, 4, 5, 10, 20, 25, 50 and 100. Looking at the last three digits the divisibility list grows little more and so on. It also means that if there is a fraction of the form k/2, k/5 or k/10 (where ‘k’ is an integer and the fractions are in their minimal form), in real number format, there will be one digit after the fractional point. Similarly, fractions of the form k/4, k/20, k/25, k/50 and k/100 will have two fractional digits. We know that some fractions like 1/3 and 1/7 will never end becoming recurring fractions. It means, to know whether the divisibility by 3 or 7, we need to see the whole number, not just a constant number of digits. What makes some denominators to end after constant number of digits (after the fractional point) and some repeat forever?
To understand the behavior of recurring fractions, let’s dissect the base number. Prime factorization of the base number 10 is 10=2*5. This should explain the divisibility list for the last digit {2, 5, 2*5}. When two digits are considered, the prime factorization comes down to 10*10=2*2*5*5, which gives us the divisibility list for the last two digits {2, 2*2, 5, 2*5, 2*2*5, 5*5, 2*5*5, 2*2*5*5}. Consider any number of digits; they are made up of only 2s and 5s. That’s why if the denominator of a fraction (in minimal form) is made up of only 2s and 5s (and no other prime numbers), only then it ends in constant number of fractional digits. For example, 1/2000 where 2000=2^4 * 5^3 (four 2s and three 5s) would surely end with four fractional digits. And, for example, 1/6 where 6 has a 2 and a 3 will end up as recurring fraction because it has a prime factor other than 2s and 5s.
Recurring fractions are not just limited to Decimal NS, they occur in any n-base Number System (where ‘n’ is any finite integer >= 2). If the denominator of a fraction (in minimal form) has a prime number which is not in the base number, then the fraction ends up as a recurring fraction. For example, in Dozenal NS, the base 12 is made up of 12=2*2*3. A fraction in Dozenal NS, 1/5 for example, would be a recurring fraction. We can see that in any n-base NS, reciprocal of a prime number which is not in the base number, would be a recurring fraction. So, isn’t there any way of having a number system without the fuss of recurring fractions?
I think there is a way!
Regards,
Channa Bankapur
In Decimal Number System, just looking at the last digit (least significant digit or unit’s digit) we can say whether the whole number is divisible by 2, 5 and 10. Looking at the last two digits we can say its divisibility by 2, 4, 5, 10, 20, 25, 50 and 100. Looking at the last three digits the divisibility list grows little more and so on. It also means that if there is a fraction of the form k/2, k/5 or k/10 (where ‘k’ is an integer and the fractions are in their minimal form), in real number format, there will be one digit after the fractional point. Similarly, fractions of the form k/4, k/20, k/25, k/50 and k/100 will have two fractional digits. We know that some fractions like 1/3 and 1/7 will never end becoming recurring fractions. It means, to know whether the divisibility by 3 or 7, we need to see the whole number, not just a constant number of digits. What makes some denominators to end after constant number of digits (after the fractional point) and some repeat forever?
To understand the behavior of recurring fractions, let’s dissect the base number. Prime factorization of the base number 10 is 10=2*5. This should explain the divisibility list for the last digit {2, 5, 2*5}. When two digits are considered, the prime factorization comes down to 10*10=2*2*5*5, which gives us the divisibility list for the last two digits {2, 2*2, 5, 2*5, 2*2*5, 5*5, 2*5*5, 2*2*5*5}. Consider any number of digits; they are made up of only 2s and 5s. That’s why if the denominator of a fraction (in minimal form) is made up of only 2s and 5s (and no other prime numbers), only then it ends in constant number of fractional digits. For example, 1/2000 where 2000=2^4 * 5^3 (four 2s and three 5s) would surely end with four fractional digits. And, for example, 1/6 where 6 has a 2 and a 3 will end up as recurring fraction because it has a prime factor other than 2s and 5s.
Recurring fractions are not just limited to Decimal NS, they occur in any n-base Number System (where ‘n’ is any finite integer >= 2). If the denominator of a fraction (in minimal form) has a prime number which is not in the base number, then the fraction ends up as a recurring fraction. For example, in Dozenal NS, the base 12 is made up of 12=2*2*3. A fraction in Dozenal NS, 1/5 for example, would be a recurring fraction. We can see that in any n-base NS, reciprocal of a prime number which is not in the base number, would be a recurring fraction. So, isn’t there any way of having a number system without the fuss of recurring fractions?
I think there is a way!
Regards,
Channa Bankapur
If you were to divide by any of those numbers in a given base, these are the numbers that'll terminate:
ReplyDeleteBase 10: 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, 100, 125, 128, 160, 200, 250, 256...
Base 12: 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, 144, 162, 192, 216, 243, 256...