Recurring decimals

Question

Does anybody have any idea why when you divide a any number, say, by 11 (excluding multiples of 11 of course) you obtain a recurring decimal? I know that it must either terminate or recur, but why does 1/7 recur while 1/5 doesn't for example?

Answer 1

You get a finite decimal precisely when the divisor has the form $2^m5^n$ for non-negative integers $m$ and $n$. This is because $2$ and $5$ are the only prime factors of $10$, the base in which we write our decimals. If we used base $7$, $\frac17$ would have the finite expansion $0.1$, while $\frac15$ would have a non-terminating expansion, $0.\overline{1254}$, if my quick and dirty arithmetic is correct.

To see why this is, note that if a fraction has the terminating expansion $0.d_1d_2\ldots d_n$ in some base $b$, then it’s equal to

$$\frac{d_1d_2\ldots d_n}{b^n}\;.$$

where $d_1d_2\ldots d_n$ is to be understood as an $n$-place integer base $b$. Call that integer $m$; then the fraction is equal to $\dfrac{m}{b^n}$, so even if it can be further reduced, the only possible prime factors of the denominator are the prime factors of $b$. In the decimal system those are of course $2$ and $5$.

Answer 2

As already pointed out this observation holds if the denominator can be written as $b=2^m5^n$ for $m,n\in\mathbb{N}_0$. This is due to the following:

Let $q=\frac{a}{b},a,b\in\mathbb{N}$ with $a<b$ and $\gcd(a,b)=1$. Then $q=\sum\limits_{i=1}^\infty r_i g^{-i}$ for $g\in \mathbb{N},g\geq 2,r_i\in\{0,\dots,g-1\}$ and for infinitely many $r_i\in\{0,\dots,g_2\}$. Then exists $l\in\mathbb N$ with $r_i=0$ for all $i>l$ if and only if $b\mid g^N$ for any $N\in\mathbb N$.

The proof is rather simple. Let's first assume $$\frac{a}{b}=\sum\limits_{i=1}^lr_ig^{-i} \Leftrightarrow ag^l=b\sum\limits_{i=1}^l r_ig^{l-i}.$$ With $\gcd(a,b)=1$ we get $b\mid g^l$.

For $g^l=bk$ for any $l,k\in\mathbb N$ we have $ag^l=bka$. Because $a<b$ we have $ka<kb=g^l$ and therefore we get $ka=\sum\limits_{i=0}^{l-1}s_ig^i$ with suitable $s_i$. Substitute $r_i=s_{l-i}$ and we get $$ag^l=b\sum\limits_{i=1}^l r_ig^{l-i}.$$

For base $10$ we have $g=10$ and therefore $b\mid 10^N \Leftrightarrow b=2^m5^n$.

Answer 3

Suppose that a decimal terminates after $n$ digits, then it can be expressed as $$\frac r{10^n}$$ Where $r$ is an integer with at most $n$ digits and the denominator of the fraction in lowest terms has only the prime factors $2$ and $5$.

If the decimal is a pure recurring decimal $d$ in which $n$ digits recur then $$10^nd=r+d$$ where $r$ is an integer consisting of the recurring digits of $d$, and $$d=\frac r{10^n-1}$$and we note that $10^n-1$ is not divisible by $2$ or $5$.

If there are a few non-recurring digits, and then the decimal expansion recurs, you can combine the two approaches.

Suppose you have the fraction $\frac pq$. There are only finitely possible remainders $0,1,2 \dots q-1$ on division by $q$, so eventually you either get zero recurring (and the decimal terminates) or you find yourself doing the same cycle of divisions over again, and the decimal recurs.

Answer 4

Here's a proof, working up from easy stuff.

1. If $a/11$ and $b/11$ are repeating decimals with the same "period", then so is $a + b$, because we just add things term by term (and the "carries" that occur in one period will occur in every other period as well), with the caveat than in $a+b$, the "repeating part" might be zero (exactly if $a+b$ is divisible by 11).

2. If $a/11$ is a repeating decimal (with nonzero "repeat"), then so is $ka/11$, where $k = 2, 3, \ldots, 10$. Reason: apply statement 1 to $b = a$ to show that $2a/11$ repeats. Why is the repeating part nonzero? Because if it were zero, that'd mean that for some $n$, $10^n \cdot 2a/11$ is an integer $k$, so $$ a = k \frac{11}{2 \cdot 10^{n}} $$ so that $$ 10^{n+1}a = k \cdot 10^{n+1}\frac{11}{2 \cdot 10^{n}} = 55 \cdot k $$ which means that after $n+1$ decimal digits of $a$, the right hand side is an integer...which means it has to be zero.

A similar argument works for $3a$, etc.

3. $1 / 11$ is a repeating decimal with period 2. (By direct observation.)

4. So $2/11, 3/11, \ldots, 10/11$ are all repeating decimals with period 2.

5. Applying the argument in part 2 to $a = 10$ and $b = 2$ shows that $12/11$ is a repeating decimal with period 2.

6. A similar argument, repeated enough, shows that every 2-digit number (except multiples of 11), is a repeating decimal with period 2.

7. Even more arguments, along the same line, show that every 3-digit integer not divisible by 11 is a repeating decimal with period 2.

And by this time, you get the idea.