Proof that a repeating decimal has non-repeating digits after decimal iff denuminator has factors of 2 or 5 besides other prime factors

Question

I'm studying a lesson about fractions. It classifies rational numbers into three categories based on their decimal representation.

1. Terminating Decimal: If a reduced fraction'denuminator has only prime factors of 2 or 5 or both (like $\frac{3}{2^3\times 5}$ or $\frac{3}{2^3}$ or $\frac{3}{5}$), then the decimal represention terminates at some point and is not periodic.

2. Simple Repeating Decimal: If a reduced fraction's denuminator has prime factors which are neither 2 nor 5 (like $\frac{3}{7}$ or $\frac{5}{13}$ or $\frac{1}{3}$), then the decimal representation is periodic and the repeating digits appear right after the decimal point.

3. Composite Repeating Decimal: If a reduced fraction's denuminator has prime factors 2 or 5 or both and also other prime factors (like $\frac{3}{2\times 5^2 \times 7}$ or $\frac{3}{2^3 \times 13}$ or $\frac{3}{5 \times 3}$), then the decimal representation has some non-repeating digits after the decimal point and after that the repeating digits appear.

I've searched for hours but did not found anything that classifies repeating decimals into simple and composite. I also found this question which is the same as mine but did not have a satisfying answer and was asked a little differently.

I want a proof of this:

There are factors of 2 or 5 or both on a reduced fraction's denuminator besides other factors $\iff$ fraction's decimal representaion has some non-repeating digits after the decimal point.

Update:-------------------------

I think if I can prove the following, then I can use that to prove the thing I mentioned above.

A reduced fraction's denuminator has only prime factors other than 2 and 5 $\iff$ Fraction's decimal representation has no non-repeating digits after decimal point.

Answer 1

This answer assumes that we may use the followings :

• All rational numbers are either terminating decimal or repeating decimal numerals. (see here)

• A number has a terminating decimal expansion if and only if it is rational and when in lowest terms, its denominator is coprime to all primes other than $2$ and $5$. (see here)

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We want to prove $(1)\iff (2)$ where

$(1)$ A reduced fraction's denominator has factors of $2$ or $5$ or both besides other prime factors.

$(2)$ The decimal representation has some non-repeating digits after the decimal point.

We prove $(1)\iff (2)$ using the following lemma :

(A proof of the lemma is written at the end of this answer)

Lemma : A reduced fraction's denominator is coprime to $10$ if and only if the decimal representation is periodic and the repeating digits appear right after the decimal point.

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Proof that $(1)\implies (2)$ :

Since the denominator is not of the form $2^a5^b$, the decimal representation is periodic. Since the denominator is not coprime to $10$, it follows from the lemma that the repeating digits do not appear right after the decimal point. So, the decimal representation has some non-repeating digits after the decimal point.$\ \square$

Proof that $(2)\implies (1)$ :

It follows from the lemma that the reduced fraction's denominator is not coprime to $10$. Suppose that the denominator is of the form $2^a5^b$. Then, it has a terminating decimal expansion, which contradicts that the decimal representation is periodic. So, the denominator has factors of $2$ or $5$ or both besides other prime factors.$\ \square$

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Finally, let us prove the lemma :

Lemma : A reduced fraction's denominator is coprime to $10$ if and only if the decimal representation is periodic and the repeating digits appear right after the decimal point.

Proof of the lemma :

("if" part)

Let $r'=\overline{r_1r_2\cdots r_t}$ be the repeating digits and $N$ be the integer part. Then, $f$ can be written as $f=N+\dfrac{r'}{10^t-1}=\dfrac{N(10^t-1)+r'}{10^t-1}$ whose denominator is coprime to $10$.

("only if" part)

Since the reduced fraction's denominator is coprime to $10$, the decimal representation is periodic.

We may suppose that $f=\dfrac nd\lt 1$ and that $f=\dfrac nd=\overline{0.b_1b_2\cdots b_s[R][R]\cdots}$ where $[R]=\overline{r_1r_2\cdots r_t}$ represents repeating digits.

Let $r_0$ be the reminder when we divide $$10^{s}n=\overline{b_1b_2\cdots b_s.[R][R]\cdots}$$ by $d$. Since $\gcd(10^sn,d)=1$, we see that $\gcd(r_0,d)=1$ with $0<r_0<d$.

Since $\dfrac{r_0}{d}=\overline{0.[R][R]\cdots}$, we get $\dfrac{r_0}{d}=\dfrac{\overline{r_1r_2\cdots r_t}}{10^t-1}$.

It follows from this that $d$ is a divisor of $r_0(10^t-1)$. Since $\gcd(d,r_0)=1$, we see that $d$ is a divisor of $10^t-1$.

Let $u$ be the smallest positive integer $x$ such that $10^x\equiv 1\pmod d$.

Then, $u$ is a divisor of $t$. (The reason is as follows : There are non-negative integers $y,r$ such that $t=uy+r$ and $0\le r\lt u$, so $1\equiv 10^t\equiv 10^{uy+r}\equiv (10^u)^y\cdot 10^r\equiv 10^r\pmod{d}$. Now, $r\gt 0$ contradicts that $u$ is the smallest positive integer such that $10^x\equiv 1\pmod d$. So, we have $r=0$ which implies that $u$ is a divisor of $t$.)

Since $10^u-1$ is a multiple of $d$, we see that $n(10^u-1)$ is also a multiple of $d$.

Therefore, $\dfrac{n(10^u-1)}{d}$ is an integer.

So, $0\lt \dfrac nd\lt 1$ implies $0\lt \dfrac{n(10^u-1)}{d}\lt 10^u-1$. So, $\dfrac{n(10^u-1)}{d}$ is a positive integer, and the number of the digits is at most $u$.

Letting $\dfrac{n(10^u-1)}{d}=\overline{c_1c_2\cdots c_u}$, we have $$\dfrac{n}{d}=\dfrac{\overline{c_1c_2\cdots c_u}}{10^u-1}=\overline{0.[R'][R']\cdots}$$ where $[R']=\overline{c_1c_2\cdots c_u}$.

This means that the repeating digits appear right after the decimal point.$\ \blacksquare$

Answer 2

Let’s show that a real number between $0$ and $1$ has a repeating decimal if and only if it is a fraction whose denominator is relatively prime with $10$.

Consider positive rational $a/b$ less than $1$ where $b>1$ is coprime to $10$. Let $n=\phi(b)$ Then by Euler’s Theorem, $b|10^n-1$, so we have that $r=a/b\cdot (10^n-1)$ is a positive integer less than $10^n$. We can denote the $n$ right digits (possibly 0) of $r$ as $r=r_1…r_n$, so $a/b=r/(10^n-1)=0.\overline{r_1\dots r_n}$ starts repeating immediately.

On the other hand, consider any fraction $\overline{r_1\dots r_n}$ which starts repeating immediately. Denoting $r=r_1\dots r_n$, we have $\overline{r_1\dots r_n}=\frac{r}{10^n-1}$ where we can note that $10^n-1$ is coprime to 10.

Note since (as you’ve noticed) all fractions eventually repeat, either a fraction immediately repeats and so can be reduced to a fraction with denominator relatively prime with $10$ (case 2), or it cannot in which case its denominator in must not be relatively prime with $10$.