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 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 $i$, 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:

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

• $e$: Euler's number $\approx 2.718$

Functions

• $\cosh(x), \sinh(x)$: The hyperbolic sine and cosine