A new number system

Every number system with base \( b \) uses \( b \) digits from \( 0 \) to \( b-1 \). But does it always have to be like that? Or can we have a number system that don't uses the same number of digits for each position?


Let's construct a number system that uses 2 digits in the first place (next to the decimal point), 3 digits next to that, 4 digits next to that and so on. Counting would look like this:

0123456789101112131415161718192021222324252627...
01101120211001011101111201212002012102112202213003013103113203211000100110101011...

To find out the place value of some position \( i \), we have to calculate the number of possible numbers that use fewer digits than \( i \).
There are \( 2 \) possibilities for the first position, \( 3 \) for the second, then \( 4 \) and so on, up to \( i \) possible digits t the \( i-1 \)th postion. The total number of possibilities and also the place value at position \( i \) is therefore \( 2 \cdot 3 \cdot 4 \cdots (i-1) \cdot i = i! \).

This means, for example, that writing \( 14 \) as \( 210 \) stands for \( 14 = 2 \cdot 3! + 1 \cdot 2! + 0 \cdot 1! \).

To distinguish base 10 and the new number system, let's write a litte \( {}_F \) behind each number in this Factorial system.


Adding

To add two numbers in this system, we simply write one below the other and add position by position. We have to be really careful about carries because the capacitiy of a position varies.

Multiplying

...

Calculating with these numbers is actually not that easy. What's more interesting, is how you can express real numbers between 0 and 1.


Real numbers

To express real numbers, all we have to do is to use our idea from the start again. We said that the first place should have 2 possible digits, the second 3 and so on. In base 10, the first digit after the decimal point divides the interval \( [0,1] \) into ten parts. Now this digit can only be \( 0 \) or \( 1 \), so ist must divide the interval into two parts, \( [0,\frac 1 2] \) and \( [\frac 1 2, 1] \). The next digit can have 3 values, so it divides both intervals in 3, and so on for all digits.

Again, we can make a table, but we have to restrict ouselves to only using 3 digits after the decimal point:

\(0_F\)\(0.001_F\)\(0.002_F\)\(0.003_F\)\(0.01_F\)\(0.011_F\)\(0.012_F\)\(0.013_F\)\(0.02_F\)\(0.021_F\)\(0.022_F\)\(0.023_F\)\(0.1_F\)\(0.101_F\)\(0.102_F\)\(0.103_F\)\(0.11_F\)\(0.111_F\)\(0.112_F\)\(0.113_F\)\(0.12_F\)\(0.121_F\)\(0.122_F\)\(0.123_F\)
\(0\)\(\frac1{24}\)\(\frac1{12}\)\(\frac18\)\(\frac16\)\(\frac5{24}\)\(\frac14\)\(\frac7{24}\)\(\frac13\)\(\frac38\)\(\frac5{12}\)\(\frac{11}{24}\)\(\frac12\)\(\frac{13}{24}\)\(\frac7{12}\)\(\frac58\)\(\frac23\)\(\frac{17}{24}\)\(\frac34\)\(\frac{19}{24}\)\(\frac56\)\(\frac78\)\(\frac{11}{12}\)\(\frac{23}{24}\)
\(0\)\(\frac1{24}\)\(\frac2{24}\)\(\frac3{24}\)\(\frac4{24}\)\(\frac5{24}\)\(\frac6{24}\)\(\frac7{24}\)\(\frac8{24}\)\(\frac9{24}\)\(\frac{10}{24}\)\(\frac{11}{24}\)\(\frac{12}{24}\)\(\frac{13}{24}\)\(\frac{14}{24}\)\(\frac{15}{24}\)\(\frac{16}{24}\)\(\frac{17}{24}\)\(\frac{18}{24}\)\(\frac{19}{24}\)\(\frac{20}{24}\)\(\frac{21}{24}\)\(\frac{22}{24}\)\(\frac{23}{24}\)

We see that all fractions can be written as \( \frac a{24} = \frac a {4!} \). This also proves that every rational number \( \frac a b \) can be written as a terminating decimal in this number system because it is possible to expand the fraction to \( \frac {ab!}{b!} \) and every fraction \( \frac c {b!} \) can be found in the numbers that use \( b-1 \) digits. This is already an improvement. Eery rational number is terminating, and also every terminating number is rational. But what about infinitely long periods?


Periods

To find out which numbers periods correspond to, let's start with an easy example \( x = 0.1111\ldots = 0.\overline 1 \).

\( 0.\overline 1 = \displaystyle \sum_{k=2}^\infty \frac 1 {k!} = \left( \displaystyle \sum_{k=0}^\infty \frac 1 {k!} \right) - \frac 1 {1!} - \frac 1 {0!} = \left( \displaystyle \sum_{k=0}^\infty \frac 1 {k!} \right) - 2 \)

This sum might look familiar to you. It's actually equal to \( e \). So,

\( 0.\overline 1 = e - 2 \)

This means that, in this number system, we can write irrational numbers as periodic decimals.

Similarly,

\( 0.\overline {10} = \displaystyle \sum_{k=1}^\infty \frac 1 {(2k)!} = \left( \displaystyle \sum_{k=0}^\infty \frac 1 {(2k)!} \right) - \frac 1 {1!} = \left( \displaystyle \sum_{k=0}^\infty \frac 1 {(2k)!} \right) - 1 = \cosh 1 - 1 \) \( 0.\overline {01} = \displaystyle \sum_{k=1}^\infty \frac 1 {(2k+1)!} = \left( \displaystyle \sum_{k=0}^\infty \frac 1 {(2k+1)!} \right) - \frac 1 {1!} = \left( \displaystyle \sum_{k=0}^\infty \frac 1 {(2k+1)!} \right) - 1 = \sinh 1 - 1 \)


...


Variables, constants and funtions I used

Variables

  • \( b \): the base of a number system
  • \( i \): the position of a digit
  • \( k \): an index

Constants

Functions

Note by Henry U
2 weeks, 5 days ago

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Ha! I find it very interesting that you decided to gloss over operations. The only place I can see this being used is in a dystopian society where the ruler doesn't want people to think too complex thoughts.

Also, I find it cool that the base factorial system (cool name, by the way) defies the law that \(10_n=n_{10}\). I was applying this base to this problem, and it turns out the answer is different than most bases.

There is one flaw with this base system, however. The number 3628799 can be represented as 987654321; how do you represent 3628800 in base factorial?

Blan Morrison - 2 weeks, 5 days ago

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That's the same problem as in Hexadecimal, but in base \( F \), in theory, we would need infinitely many digits. Of course that's not possible, but we could use the latin alphabet for 26 additional letters. With that we can get up to \( 36! \approx 3.72\cdot10^{41} \) which is more than a 36-digit base 10 number (\( n! \) passes \( 10^n \) at \( n = 25 \)). For more, we can use lowercase letters too (\( 62! \approx 3\cdot10^{85} \)) or indroduce greek letters ( \( 93! \approx 10^{144} \) if we ignore letters that look too similar).

Henry U - 2 weeks, 4 days ago

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If we think of all symbols that can be generated on a 300x300 pixel grid (which is the standard size of a letter image on Google), then that would be \(2^{90,000}\) possible digits, or about \(5\times 10^{27,092}\). Given that I can't even paste the digits of \(99,999!\) into a word processor, I feel that might be enough digits.

As I was writing this, I realized scientific notation is really easy! Instead of writing \(a\times 10^b\), in base \(F\), it would be \(a\times b!\).

Blan Morrison - 2 weeks, 3 days ago

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@Blan Morrison That's a definitely enough letters, even if we can't use all of them.

The scientific notation is really interesting. First, it's easier to see the order of magnitude and, since it represents numbers as a product, it might make multiplication possible. I will think about that.

Henry U - 2 weeks, 2 days ago

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@Blan Morrison Also, it can be calculated that the quantity representing the digits would have about \(1.25\times 10^{27,092}\) trailing zeroes.

Blan Morrison - 2 weeks, 3 days ago

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I will try to give a rule for multiplication, but I think it might be very complicated. When I first thought of this system (actually a few months ago) I immediately threw away the idea of arithmetic and focused on real numbers because I couldn't find simple rules.

About your problem, the strategie is still to minimize the terms, but in base \( b \) this would simply give \( b^d \). In base \( F \), you would end up with \( b! \). Adding \( 10 \) of these gives \( 10d! \) and for this to also have \( d \) digits, \( 10d! < (d+1)! = (d+1) d! \Leftrightarrow 10 < d+1 \Leftrightarrow d > 9 \) must be true. So the minimum number of digits is \( d \), but \( d+1 \) is also possible.

Henry U - 2 weeks, 5 days ago

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Right. The minimum would indeed be \(d\). However, 10, in base factorial is simply 2, and 10+11=21, which still has \(d\) digits.

Blan Morrison - 2 weeks, 5 days ago

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An interesting note, though I can see how this number system is highly impractical. I think the current literature has moved away from this and ventured into the quantum realm; if you dig into it a little bit, you will find the same thing.

Gennady Notowidigdo - 2 weeks, 4 days ago

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I know, the trouble comes with the first idea of infinitely many characters and arithmetic is actually almost impossible, but it might be better than base 10 for expressing decimals.

Henry U - 2 weeks, 3 days ago

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Remove the "almost" part and we will be in agreement. Regardless, it goes back to size of representations (length of strings) vs. size of representatives (no. of characters available); it will always be a tradeoff between one or the other, so I think that base 10 may be the best of the worst (though I now think that hexadecimal system is much better from a computational point of view).

Because of this, I think the idea of quantum representation now has a big push and we will see in the coming generations as to what this technology has to offer us.

Gennady Notowidigdo - 2 weeks, 3 days ago

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@Gennady Notowidigdo I would disagree with the arithmetic argument; I feel that arithmetic is possible, just highly impractical.

Blan Morrison - 2 weeks, 3 days ago

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