# A Rational Number With a Finite Number of Digits

A rational number, by its definition, is a ratio of two integers. Yet, some of them are not expressible using a finite number of digits in decimals, or more formally known as base-ten positional notation. For example, the number one third can be written exactly as $\frac{1}{3}$ as a fraction, yet we cannot express the same number using a finite number of digits in base 10. The best we can do is to imply the indefinite repeating of the digit 3 using elipsis as in: $0.33...$ or more formally with a vinculum such as an overline to indicate the sequence of digits to repeat as in: $0.\overline{3}$. In the case of one seventh, a sequence of digits $142857$ gets repeated as in $0.142857142857...$ or $0.\overline{142857}$. When I was a kid and first learned of division, I was very fascinated by this phenomnea. Today, I present an answer to one of the questions I once had as a kid: when can a rational number be expressed using a finite number of digits?

## Theorem

A rational number $r$ can be represented in the positional notation of an integral base $b\ge 2$ using a finite number of digits if and only if every factor of the denominator of $r$ in its irreducible fraction is also a factor of $b$.

### Understanding the theorem

Let’s break down this statement. If $r$ is a rational number, it is a ratio of two integers by the definition. Thus, there exist two integers $n\in \mathbb{Z}$ and $d\in \mathbb{Z}$ such that $r=\frac{n}{d}$. Furthermore, without loss of generality, we can assume $\frac{n}{d}$ to be the irreducible fraction (also known as the lowest form) of $r$ meaning that $n$ and $d$ do not share a common factor. (If $r$ and $d$ did share common factors, we can keep dividing them by the common factors to arrive at the irreducible fraction.)

Now let $b=\prod _{i=1}^{k}{p}_{bi}^{{q}_{bi}}={p}_{b1}^{{q}_{b1}}{p}_{b2}^{{q}_{b2}}...{p}_{bk}^{{q}_{bk}}$ where ${p}_{bi}\in \mathbb{Z}$ is a prime number and ${q}_{bi}\in \mathbb{Z}$ for every $i$ be the prime factorization of $b$, and $d=\prod _{j=1}^{l}{p}_{dj}^{{q}_{dj}}={p}_{d1}^{{q}_{d1}}{p}_{d2}^{{q}_{d2}}...{p}_{dl}^{{q}_{dl}}$ where ${p}_{dj}\in \mathbb{Z}$ is a prime number and ${q}_{dj}\in \mathbb{Z}$ for every $j$ be the prime factorization of $d$. Then the theorem says that $r$ is expressible using a finite number of digits in the positional notation of base $b$ if and only if every prime number ${p}_{d1},{p}_{d2},...,{p}_{dl}$ in the prime factorization of $d$ also appears in the prime factorization of $b$ (i.e. if $t={p}_{di}$ for some $i$, then $t={p}_{bj}$ for some $j$).

For example, one nineth is not expressible using a finite number of digits in base 10 because its irreducible fraction $\frac{1}{9}$ contains 3 (since $9={3}^{2}$) as a prime factor in its denominator, which is not a factor of $10=2\cdot 5$. However, $\frac{1}{9}$ is expressible using a finite number of digits in base 3 notation since the denominator only contains the prime factor of 3, which is also a factor of the base. Namely, $\frac{1}{9}$ is 0.01 in base 3. Similarly, $\frac{1}{20}$ is expressible using a finite number of digits in base 10 because its irreducible fraction $\frac{1}{20}$ only contains prime numbers which are also factors of 10 in its denominator: $20={2}^{2}\cdot 5$ Note that reducing the fraction to its irreducible form is important. As an example, $\frac{3}{15}$ is expressible using a finite number of digits in base 10 even though its denominator contains a prime factor of 3, which is not a factor of 10 since its irreducible form $\frac{1}{5}$ only contains the prime factor of 5, which is a factor of 10.

## Proof

### If a rational number can be represented using a finite number of digits in a base, then a prime factor of the denominator of its irreducible fraction is a factor of the base

Let $r$ be a rational number which can be expressed using a finite number of digits in the positional notation of an integral base $b\ge 2$. Then let ${d}_{k}...{d}_{2}{d}_{1}{d}_{0}.{d}_{-1}{d}_{-2}...{d}_{-l}$ be such a sequence of digits presenting $r$ in base $b$ where $k\in \mathbb{N}$ is the place of the highest digit and $l\in \mathbb{N}$ is the place of the lowest digit after the radix point, and ${d}_{i}\in \left\{{d}_{i}\in \mathbb{Z}|0\le {d}_{i} for all $i$ between $-l$ and $k$. Namely, $r=\sum _{i=-l}^{k}{d}_{i}{b}^{i}={d}_{k}{b}^{k}+...+{d}_{1}{b}^{1}+{d}_{0}{b}^{0}+{d}_{-1}{b}^{-1}+...+{d}_{-l}{b}^{-l}$ Let $n=r\cdot {b}^{l}$. Using the definition of $r$, $n=r\cdot {b}^{l}=\left(\sum _{i=-l}^{k}{d}_{i}{b}^{i}\right)\cdot {b}^{l}=\sum _{i=-l}^{k}{d}_{i}{b}^{i+l}$ Substituting $i$ with $j=i+l$ yields: $n=\sum _{j=0}^{k+l}{d}_{j-l}{b}^{j}$ Because ${d}_{j-l}$ and ${b}^{j}$ are both integers, $n$ is also an integer since the sum of products of integers is also an integer. Now consider the fraction $\frac{n}{{b}^{l}}=\frac{r\cdot {b}^{l}}{{b}^{l}}=r$. The denominator only contains the prime factors of b since it’s a power of b. Because reducing this fraction by canceling common factors in the numerator and the denominator does not introduce a new prime factor in the denominator, the irreducible fraction of $r$ must also not have a prime factor which is not a factor of b in its denominator. That is, every prime factor of the denominator of the irreducible fraction of $r$ is also a factor of b as desired.

### If every prime factor of the denominator of a rational number in its irreducible fraction is also a factor of a base, then the number can be represented using a finite number of digits in the base.

Let $r$ be a rational number and $r=\frac{n}{d}$ with $n,d\in \mathbb{Z}$ be its irreducible fraction such that every factor of $d$ is a factor of a base $b\ge 2$. Without loss of generality, assume $d$ is non-negative (since the negative sign can always be moved from the denominator of a fraction to its numerator).

Case 1: If $n$ is 0 or $d$ is 1, $r$ is an integer, and there is a representation of $r$ using a finite number of digit in base $b$.

Case 2: If $n$ is not 0 and $d>1$, let $d=\prod _{i=1}^{k}{p}_{i}^{{q}_{i}}$ where $k\in \mathbb{N}$, ${p}_{i}\in \mathbb{N}$ is a prime number and ${q}_{i}\in \mathbb{N}$ for every $i$ be the prime factorization of $d$ using the fundamental theorem of arithmetic. From the supposition, every prime factor (${p}_{i}$ for every $i=1...k$) of $d$ is also a factor of the base $b$. Let ${p}_{k+1},...{p}_{l}$ be prime factors of b which is not a factor of $d$ if there are any, and $b=\prod _{j=1}^{l}{p}_{j}^{{r}_{j}}$ where $l\in \mathbb{N}$ such that $k\le l$ and ${r}_{i}\in \mathbb{N}$ be the prime factorization of $b$.

Now let $m$ be the maximum number amongst ${q}_{i}$, and consider the quantity $w=\frac{{b}^{m}}{d}$. $w=\frac{{b}^{m}}{d}=\frac{{\left(\prod _{j=1}^{l}{p}_{j}^{{r}_{j}}\right)}^{m}}{\prod _{i=1}^{k}{p}_{i}^{{q}_{i}}}=\frac{\prod _{j=1}^{l}{p}_{j}^{{r}_{j}m}}{\prod _{i=1}^{k}{p}_{i}^{{q}_{i}}}=\prod _{j=1}^{l}{p}_{j}^{{r}_{j}m}\prod _{i=1}^{k}{p}_{i}^{-{q}_{i}}$ $=\left\{\prod _{i=1}^{k}{p}_{i}^{{r}_{i}m}\prod _{j=k+1}^{l}{p}_{j}^{{r}_{j}m}\right\}\prod _{i=1}^{k}{p}_{i}^{-{q}_{i}}=\prod _{i=1}^{k}{p}_{i}^{{r}_{i}m-{q}_{i}}\prod _{j=k+1}^{l}{p}_{j}^{{r}_{j}m}$ Because $m\ge {q}_{i}$ and ${r}_{i}\ge 1$ by the definitions of $m$ and ${r}_{i}$, ${r}_{i}m\ge {q}_{i}$ and ${r}_{i}m-{q}_{i}\ge 0$. Thus, ${p}_{i}^{{r}_{i}m-{q}_{i}}$ is a non-negative integer for every $i=1,...,k$ since it’s a power of a prime number to a non-negative integer. Since ${p}_{j}^{{r}_{j}}$ is also a non-negative integer for all $j=k+1,...,l$ and $m$ is a positive integer by their definitions, the whole product $w$ is also a non-negative integer. Then the quantity $nw$ is also an integer since the product of two integers is also an integer.

Let ${n}_{s}{n}_{s-1}...{n}_{0}$ be the representation of $nw$ in the base $b$. Then consider the representation of $\frac{nw}{{b}^{m}}=\frac{n{b}^{m}}{d{b}^{m}}=\frac{n}{d}$ in base $b$: ${n}_{s}{n}_{s-1}...{n}_{m+1}.{n}_{m}...{n}_{0}$ with the radix point between ${n}_{m}$ and ${n}_{m+1}$. This is the representation of $\frac{n}{d}$ in base $b$ using a finite number of digits as desired.

Q.E.D.

## Intuitive Reasoning

The second half of the proof gives us an intuition as to why this theorem holds. The question of whether we can write down $\frac{n}{d}$ using a finite number of digits in base $b$ really comes down to whether we can write down $\frac{1}{d}$ using a finite number of digits in base $b$ because $\frac{n}{d}$ is just $\frac{1}{d}$ repeated $n$ times. If $\frac{1}{d}$ is expressible using a finite number of digits in base $b$, there ought to be some natural (positive) number $c$ such that c-th digit after the radix point is the last digit. That is, since it’s finite, there must be the last digit. The quantity $\frac{1}{{b}^{c}}$ represents 0 followed by $c$ 0s and then 1. (e.g. if $c=3$ then $0.001$.) The question boils down to whether there is such a $c$.

Suppose there was such a $c\in \mathbb{N}$. Then $\frac{1}{d}$ ought to be some multiple of $\frac{1}{{b}^{c}}$. To give an example, consider a number $0.345$ with $c=3$ in base 10. We can think of this number as $0.001$ added $345$ times or simply $0.001\cdot 345$. Let $t\in \mathbb{Z}$ be such an integer multiple: $\frac{1}{d}=t\cdot \frac{1}{{b}^{c}}$. e.g. $t=345$ with $\frac{1}{{b}^{c}}=\frac{1}{{10}^{3}}=\frac{1}{1000}$ for expressing $0.345$ in base 10. Multiplying both sides of this equation by $d\cdot {b}^{c}$ yields: ${b}^{c}=dt$.

This is a crucial equation. Intuitively, this equation says that ${b}^{c}$ must be some integral multiple of the denominator $d$. e.g. in the case of $0.345$ in base 10, ${b}^{c}={10}^{3}=1000$ must be some integral multiple of the denominator of its irreducible fraction. To see why, let’s turn $0.345$ into a proper fraction with integral numerators and denominators by multiplying the base number 10: $0.345=\frac{3.45}{10}=\frac{34.5}{100}=\frac{345}{1000}$ Now apply the prime factorization and cancel common factors to get the irreducible fraction: $0.345=\frac{345}{1000}=\frac{3\cdot 5\cdot 23}{{2}^{3}\cdot {5}^{3}}=\frac{3\cdot 23}{{2}^{3}\cdot {5}^{2}}=\frac{69}{200}$ Sure enough, $1000=5\cdot 200$. Notice that $t=5$ in this case is simply a common factor we originally had. In fact, $t$ can be thought of as common factors needed in numerators and denominators to make the denominator become a power of 10.

Now, going back to ${b}^{c}=dt$, we see that ${b}^{c}$ must have all prime factors of $d$ and $t$ because ${b}^{c}$ and $dt$ are identical. But this would implies that every prime factor of $d$ must be a prime factor of $b$ since raising $b$ to the power of $c$ does not introduce a new prime factor, and multiplying $d$ with any integer, in particular by $t$, does not eliminate a prime factor in $d$.