5
$\begingroup$

The repeating decimal .36666... in base 8 can be written in a fraction in base 8. I understand simple patterns such as 1/9 in base 10 is .1111.... so 1/7 in base 8 is .1111. But I'm not too sure how to convert this decimal in this base to the fraction in the same base.

$\endgroup$
  • $\begingroup$ What about converting it to decimal, then base 8? It is $\frac{4}{11}$ and then convert to base 8? $\endgroup$ – Joshua Lochner Jul 24 '16 at 20:20
7
$\begingroup$

\begin{align} 0.3\bar{6}_8 &= \frac{3}{8} + 6\left(\frac{1}{8^2}+\frac{1}{8^3} + \cdots\right)\\ &= \frac{3}{8} + \frac{6}{8^2}\left(1+\frac{1}{8}+ \frac{1}{8^2} +\cdots\right)\\ &= \frac{3}{8} + \frac{6}{8^2}\frac{1}{1-(1/8)} & \text{geometric series}\\ &= \frac{3}{8} + \frac{3}{28}\\ &= \frac{27}{56}\\ &= \frac{33_8}{70_8}. \end{align}

$\endgroup$
  • $\begingroup$ I noticed on second to step you converted the numerator and denominator separately to base 8 from base 10. So this that conversion apparently works in general? $\endgroup$ – Ian Limarta Jul 24 '16 at 20:28
  • 1
    $\begingroup$ @cavern101 I guess you should ask yourself what a "fraction in base 8" means. $\endgroup$ – angryavian Jul 24 '16 at 20:32
  • $\begingroup$ @cavern101: note that you could just as well do the same calculation as this answer does, writing everything in base 8 at all times, so it starts $(\frac{3}{10} + 6(\frac{1}{10^2} + \frac{1}{10^3} + ...)$, and then there would be no conversion to base 8 at the end. A base is just a way of writing down a number, it's still the same number regardless of how you write it. The only difficulty in doing it that way is that mental arithmetic in a base other than 10 is tricky, since a lot of things you remember (including multiplication tables) don't directly apply. $\endgroup$ – Steve Jessop Jul 25 '16 at 11:20
5
$\begingroup$

You could just do all your thinking in base 8. To save writing all the subscripts in the following computations I'll omit the base 8 designation. Legal digits are $0$ through $7$. It's a little mindbending, but only because we're used to base 10.

Let $x=0.3666\ldots$. Then

$$ 10x = 3.666\ldots = 3 + 6/7 = (25 + 6)/7 = 33/7 $$

so $x=33/70$.

I used the facts that multiplying by 10 just shifts the "decimal" point, $3 \times 7 = 25$ and $6/7 = 0.66\ldots$.

$\endgroup$
  • 1
    $\begingroup$ I think it's worth mentioning the justification for 0.666... = 6/7, which is that 0.111... = 1/(10-1) for any base. So it's 1/9 in base 10, and 1/7 in base 8. $\endgroup$ – hobbs Jul 25 '16 at 3:09
  • $\begingroup$ @hobbs, that is, $$0.3666\ldots=\frac{3_8}{10_8}+\frac{6_8}{10_8}\frac{1_8}{7_8}$$ $\endgroup$ – J. M. is a poor mathematician Jul 25 '16 at 6:01
  • $\begingroup$ @hobbs Where $a$ is a digit, $0.aaa... \equiv \frac{a}{10-1}$. $\endgroup$ – wizzwizz4 Jul 25 '16 at 7:29
  • $\begingroup$ @hobbs I didn't elaborate on $6/7 = 0.666\ldots$ since the OP knows about $1/7 = 0.111\ldots$. $\endgroup$ – Ethan Bolker Jul 25 '16 at 12:11
3
$\begingroup$

We use subscript $8$ to denote base $8$ numbers, and all other numbers are base $10$. Take $x = 0.3\overline{6}_8$. Then $7x = 8x - x = 3.\overline{6}_8 - 0.3\overline{6}_8 = 3.3_8 = 27/8$. It follows that $x = 27/56 = 33_8/70_8$.

$\endgroup$
1
$\begingroup$

We compute directly in base $8$ as in base $10$:

Set $y=0.3666\dots$, $x=0.666\dots$. Then \begin{align*}8y&=3+0.666\dots=3+x\\ 8x&=6.666\dots=6+x \end{align*} Thus $$7x=6,\quad\text{so}\enspace x=\frac 67\enspace\text{and}\enspace 8y=3+\frac 67=\frac{3\times 7+6}7=\frac{25+6}7 =\frac{33}7,$$ because in base $8$, $\;3\times 7=25,\;5+6=13$, so that finally $\; y=\dfrac{33}{8\times7}=\color{red}{\dfrac{33}{70}}$.

$\endgroup$
  • $\begingroup$ It's actually ${33}/{70}_8$ $\endgroup$ – Ian Limarta Jul 24 '16 at 20:48
  • $\begingroup$ Oh! yes. I forgot the final division. Thanks for pointing it. $\endgroup$ – Bernard Jul 24 '16 at 20:52
  • $\begingroup$ Your welcome sir! $\endgroup$ – Ian Limarta Jul 24 '16 at 20:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.