A new number system

Every number system with base b b uses b b digits from 0 0 to b1 b-1 . But does it always have to be like that? Or can we have a number system that doesn't use 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:


To find out the place value of some position i i , we have to calculate the number of possible numbers that use fewer digits than i i .
There are 2 2 possibilities for the first position, 3 3 for the second, then 4 4 and so on, up to i i possible digits t the i1 i-1 th postion. The total number of possibilities and also the place value at position i i is therefore 234(i1)i=i! 2 \cdot 3 \cdot 4 \cdots (i-1) \cdot i = i! .

This means, for example, that writing 14 14 as 210 210 stands for 14=23!+12!+01! 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 {}_F behind each number in this Factorial system.


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.



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] [0,1] into ten parts. Now this digit can only be 0 0 or 1 1 , so ist must divide the interval into two parts, [0,12] [0,\frac 1 2] and [12,1] [\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:


We see that all fractions can be written as a24=a4! \frac a{24} = \frac a {4!} . This also proves that every rational number ab \frac a b can be written as a terminating decimal in this number system because it is possible to expand the fraction to ab!b! \frac {ab!}{b!} and every fraction cb! \frac c {b!} can be found in the numbers that use b1 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?


To find out which numbers periods correspond to, let's start with an easy example x=0.1111=0.1 x = 0.1111\ldots = 0.\overline 1 .

0.1=k=21k!=(k=01k!)11!10!=(k=01k!)2 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 e . So,

0.1=e2 0.\overline 1 = e - 2

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


0.10=k=11(2k)!=(k=01(2k)!)11!=(k=01(2k)!)1=cosh11 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

> 0.01=k=11(2k+1)!=(k=01(2k+1)!)11!=(k=01(2k+1)!)1=sinh11 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


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



Note by Henry U
2 years, 8 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

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 10n=n1010_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 years, 8 months ago

Log in to reply

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

Henry U - 2 years, 8 months ago

Log in to reply

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

Blan Morrison - 2 years, 8 months ago

Log in to reply

That's the same problem as in Hexadecimal, but in base F 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!3.721041 36! \approx 3.72\cdot10^{41} which is more than a 36-digit base 10 number (n! n! passes 10n 10^n at n=25 n = 25 ). For more, we can use lowercase letters too (62!31085 62! \approx 3\cdot10^{85} ) or indroduce greek letters ( 93!10144 93! \approx 10^{144} if we ignore letters that look too similar).

Henry U - 2 years, 8 months ago

Log in to reply

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 290,0002^{90,000} possible digits, or about 5×1027,0925\times 10^{27,092}. Given that I can't even paste the digits of 99,999!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×10ba\times 10^b, in base FF, it would be a×b!a\times b!.

Blan Morrison - 2 years, 7 months ago

Log in to reply

@Blan Morrison Also, it can be calculated that the quantity representing the digits would have about 1.25×1027,0921.25\times 10^{27,092} trailing zeroes.

Blan Morrison - 2 years, 7 months ago

Log in to reply

@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 years, 7 months ago

Log in to reply

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.

A Former Brilliant Member - 2 years, 7 months ago

Log in to reply

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 years, 7 months ago

Log in to reply

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.

A Former Brilliant Member - 2 years, 7 months ago

Log in to reply

@A Former Brilliant Member I would disagree with the arithmetic argument; I feel that arithmetic is possible, just highly impractical.

Blan Morrison - 2 years, 7 months ago

Log in to reply


Isaac YIU Math Studio - 1 year, 10 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...