Is 1 divided by any multiple of 3 always non-terminating?

Question

I happened to run across a phenomena of numbers that are the result of 1/(any multiple of 3). In each case I tested (from 3 to 81) the result was non-terminating. I don't have a clue how to prove such a thing and my google-fu was weak so I couldn't find a concise answer for this.

The question: Is 1 divided by any multiple of 3 always a non-terminating number?

Bonus: If not, then how about 1 divided by a power of 3?

Thanks much!

Answer 1

1 divided by any integer is always rational: a rational number is precisely any integer divided by any other integer. It seems that you are calling it irrational because its decimal expansion goes on forever, but many rational numbers have this property too. What sets rational numbers apart from irrational numbers it that the decimal expansion of a rational number eventually starts repeating: for instance, the decimal expansion of 1/3 is just the digit 3 repeating, while the decimal expansion of 1/27 is just the digits 037 repeating.

Now, to answer the question that you seem to have meant to ask: yes, for any positive integer $n$, the decimal expansion of $$ \frac1{3n} $$ goes on forever. More generally, we can say the following: the decimal expansion of $\frac1{k}$ goes on forever if and only if $k$ has prime factors other than $2,5$. To see this, suppose that the decimal expansion of $\frac1k$ ends. Let's say its length is $n$. Then the number $$ \frac{10^n}{k} $$ is an integer, since multiplying by $10^n$ shifts the decimal point $n$ places to the right. Hence, $k$ must evenly divide $10^n$. But then its only prime factors can be $2$ and $5$. Conversely, if $k = 2^a5^b$, then $$ \frac{10^{\max(a,b)}}{k} $$ will be an integer, so the decimal expansion of $k$ must be finite.

Answer 2

Suppose that $\frac{1}{3k}$ did actually have a terminating decimal expansion: $0.a_1a_2a_3\dots a_n$ (where here $a_1a_2a_3\dots a_n$ is the concatenation of the digits $a_1,a_2,\dots$).

Note then that $0.a_1a_2a_3\dots a_n = \dfrac{a_1a_2a_3\dots a_n}{10^n}$

Supposing that $\frac{1}{3k}=\frac{a_1a_2a_3\dots a_n}{10^n}$, by cross multiplication we have:

$3k\cdot (a_1a_2a_3\cdots a_n)= 10^n$, but since $3$ is prime that implies that $3$ should divide evenly into $10^n=2^n5^n$ which is clearly false. Thus by contradiction $\frac{1}{3k}$ never terminates.