Editing Duodecimal (section)

=== Fractions ===
Duodecimal [[Fraction (mathematics)|fractions]] for rational numbers with [[Smooth number|3-smooth]] denominators terminate:
* {{sfrac|2}} = 0.6
* {{sfrac|3}} = 0.4
* {{sfrac|4}} = 0.3
* {{sfrac|6}} = 0.2
* {{sfrac|8}} = 0.16
* {{sfrac|9}} = 0.14
* {{sfrac|10}} = 0.1 (this is one twelfth, {{sfrac|{{d2}}}} is one tenth)
* {{sfrac|14}} = 0.09 (this is one sixteenth, {{sfrac|12}} is one fourteenth)

while other rational numbers have [[recurring decimal|recurring]] duodecimal fractions:
* {{sfrac|5}} = 0.{{Overline|2497}}
* {{sfrac|7}} = 0.{{Overline|186{{D2}}35}}
* {{sfrac|{{d2}}}} = 0.1{{Overline|2497}} (one tenth)
* {{sfrac|{{d3}}}} = 0.{{Overline|1}} (one eleventh)
* {{sfrac|11}} = 0.{{Overline|0{{D3}}}} (one thirteenth)
* {{sfrac|12}} = 0.0{{Overline|{{D2}}35186}} (one fourteenth)
* {{sfrac|13}} = 0.0{{Overline|9724}} (one fifteenth)

{| class="wikitable"
! Examples in duodecimal
! Decimal equivalent
|-
| 1 × {{sfrac|5|8}} = 0.76
| 1 × {{sfrac|5|8}} = 0.625
|-
| 100 × {{sfrac|5|8}} = 76
| 144 × {{sfrac|5|8}} = 90
|-
| {{sfrac|576|9}} = 76
| {{sfrac|810|9}} = 90
|-
| {{sfrac|400|9}} = 54
| {{sfrac|576|9}} = 64
|-
| 1{{d2}}.6 + 7.6 = 26
| 22.5 + 7.5 = 30
|}

As explained in [[recurring decimal]]s, whenever an [[irreducible fraction]] is written in [[radix point]] notation in any base, the fraction can be expressed exactly (terminates) if and only if all the [[prime factor]]s of its denominator are also prime factors of the base.

Because <math>2\times5=10</math> in the decimal system, fractions whose denominators are made up solely of multiples of 2 and 5 terminate: {{sfrac|8}}&nbsp;=&nbsp;{{sfrac|(2×2×2)}}, {{sfrac|20}}&nbsp;=&nbsp;{{sfrac|(2×2×5)}}, and {{sfrac|500}}&nbsp;=&nbsp;{{sfrac|(2×2×5×5×5)}} can be expressed exactly as 0.125, 0.05, and 0.002 respectively. {{sfrac|3}} and {{sfrac|7}}, however, recur (0.333... and 0.142857142857...).

Because <math>2\times2\times3=12</math> in the duodecimal system, {{sfrac|8}} is exact; {{sfrac|20}} and {{sfrac|500}} recur because they include 5 as a factor; {{sfrac|3}} is exact, and {{sfrac|7}} recurs, just as it does in decimal.

The number of denominators that give terminating fractions within a given number of digits, {{math|''n''}}, in a base {{math|''b''}} is the number of factors (divisors) of <math>b^n</math>, the {{math|''n''}}th power of the base {{math|''b''}} (although this includes the divisor 1, which does not produce fractions when used as the denominator). The number of factors of ''<math>b^n</math>'' is given using its prime factorization.

For decimal, <math>10^n=2^n\times 5^n</math>. The number of divisors is found by adding one to each exponent of each prime and multiplying the resulting quantities together, so the number of factors of ''<math>10^n</math>'' is <math>(n+1)(n+1)=(n+1)^2</math>.

For example, the number 8 is a factor of 10<sup>3</sup> (1000), so <math display="inline">\frac{1}{8}</math> and other fractions with a denominator of 8 cannot require more than three fractional decimal digits to terminate. <math display="inline">\frac{5}{8}=0.625_{10}.</math>

For duodecimal, <math>10^n=2^{2n}\times 3^n</math>. This has <math>(2n+1)(n+1)</math> divisors. The sample denominator of 8 is a factor of a gross <math display="inline">12^2=144</math> (in decimal), so eighths cannot need more than two duodecimal fractional places to terminate. <math display="inline">\frac{5}{8}=0.76_{12}.</math>

Because both ten and twelve have two unique prime factors, the number of divisors of ''<math>b^n</math>'' for {{math|''b'' {{=}} 10 or 12}} grows quadratically with the exponent {{math|''n''}} (in other words, of the order of <math>n^2</math>).