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What I know is that the existance of multiplicative inverses of a nonzero real number $a$ is $\frac{1}{a}$ and that holds when $a$ is a repeating decimal. So, how to divide an integer by a repeating decimal?

I tried using long division but I just can't get rid of the decimal by multiplying the numerator and the denominator by powers of 10.

Muhammad
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  • A number with a periodic decimal expansion is rational. Did you try writing the number in the form $p/q$ where $p,q\in\Bbb Z$? – Shubham Johri May 20 '19 at 09:01
  • @ShubhamJohri can you show that each repeating decimal is a rational number? – Muhammad May 20 '19 at 09:03
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    If a repeating decimal $d$ has $r$ repeating digits, consider $10^rd-d=(10^r-1)d$ - the tail digits all cancel and you get a terminating decimal, which is a rational number $t$. Then $d$ is the rational number $t/(10^r-1)$ – Mark Bennet May 20 '19 at 09:05

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Let $a$ have a decimal expansion that ultimately becomes periodic, i.e.$$a=\lfloor a\rfloor\color{red}.d_1d_2\cdot\cdot\cdot d_n\overline{q_1q_2\cdot\cdot\cdot q_p}$$where $d_i,q_j$ are digits after the decimal point and $p$ is the period of the decimal representation. Thus,$$10^n\cdot a=\lfloor a\rfloor d_1d_2\cdot\cdot\cdot d_n\color{red}.\overline{q_1q_2\cdot\cdot\cdot q_p}\\10^{n+p}\cdot a=\lfloor a\rfloor d_1d_2\cdot\cdot\cdot d_nq_1q_2\cdot\cdot\cdot q_p\color{red}.\overline{q_1q_2\cdot\cdot\cdot q_p}$$Subtracting the two numbers will get rid of the periodic part, as under:$$(10^{p}-1)10^n\cdot a=\lfloor a\rfloor d_1d_2\cdot\cdot\cdot d_nq_1q_2\cdot\cdot\cdot q_p-\lfloor a\rfloor d_1d_2\cdot\cdot\cdot d_n\\\implies a=\frac{\lfloor a\rfloor d_1d_2\cdot\cdot\cdot d_nq_1q_2\cdot\cdot\cdot q_p-\lfloor a\rfloor d_1d_2\cdot\cdot\cdot d_n}{(10^{p}-1)10^n}$$which is of the form $\frac pq;p,q\in\Bbb Z,q\ne0$.

Consider, for example, $a=3.4\overline{56}\implies 10a=34.\overline{56},1000a=3456.\overline{56}$ and $1000a-10a=990a=3456-34=3422$. Thus, $a=3422/990=1711/495$ and $a^{-1}=495/1711$. You can verify this by finding the decimal representation of $1711/495$.

Shubham Johri
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Suppose we have a number $x$ with a repeating decimal expansion. For example:

$$x=2.7846153846153846153\ldots=2.7\dot84615\dot3$$

Multiply $x$ by $10^d$, where $d$ is the period of the repeat. In our example, $d=6$, so we get $$10^6x=2784615.3\dot84615\dot3$$

Subtract the first equation from the second to get: $$(10^6-1)x=2784615.3-2.7=2784612.6$$ Divide by $999999$: $$x=\frac{2784612.6}{999999}=\frac{27846126}{9999990}=\frac{181}{65}$$

This procedure always ends up with a rational number.

TonyK
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  • Thanks, I have another question, what if the decimal extends for infinite but not regurarly, how can I divide an integer by it? – Muhammad May 20 '19 at 09:40
  • @Mason: your question makes no sense to me. Perhaps you can give an example of what you are looking for? If you need a non-repeating decimal for this, I can suggest $\sqrt 2$ or $\pi$. – TonyK May 20 '19 at 09:49
  • $\sqrt{2}$ is a nonzero real number. I know that there exists a multiplicative inverse for $\sqrt{2}$ since it is a nonzero real number, which is $\frac{1}{\sqrt{2}}$. How do I compute this? – Muhammad May 20 '19 at 10:10
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    So you want to know how to compute the decimal expansion of $1/x$ from the decimal expansion of $x$, for a generic non-zero real number $x$? – TonyK May 20 '19 at 10:13
  • yeah, does this need a new question? – Muhammad May 20 '19 at 10:19
  • @Mason: yes, I think so. – TonyK May 20 '19 at 11:12
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A repeating decimal is a rational number... So if $r$ is a repeating decimal and $k$ is an integer, considering that $r = \frac nm$ you can just note that $$ \frac{k}{r} = \frac{k}{n/m} = \frac{k m}{n}. $$

PierreCarre
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  • Can you show that a repeating decimal is a rational number? – Muhammad May 20 '19 at 09:01
  • @Mason Did you try google on "repeating decimal"? – coffeemath May 20 '19 at 09:03
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    @Mason You can use geometric series to show that. For instance $$0.999(9) = 9 \sum_{i=1}^{\infty} (1/10)^n = 9 \times \frac{1/10}{1-1/10} = 1$$ whenever you have a repeating decimal you can reason in a similar way just by writing the number as the multiplication of the repeating block by an infinite sum of powers of $10$. – PierreCarre May 20 '19 at 09:05