Repeating Decimals

Question

I'm just wondering how do we simplify repeating decimals into a fraction in general?

Like, for example,

$$0.5656\dots$$

$$0.12424\dots$$

$$4.23777\dots$$

Thanks!

Answer 1

I’ll illustrate the technique with $1.23456456456\dots$; you should be able to generalize it pretty easily.

Let $x=1.23456456456\dots$. Then $10^3x=1234.56456456\dots$, so

$$\begin{align*} 999x&=1234.564\,564\,564\dots-1.234\,564\,564\dots\\ &=1233.330\,000\,000\dots\\ &=1233.33\,. \end{align*}$$

Multiply by $10^2$ to get rid of the decimals: $99900x=123333$. Now just solve for $x$.

At the first step I simply shifted the decimal point by the length of the repeating block. That ensured that the subtraction would leave me with a terminating decimal.

Answer 2

You can have a look at here here. That link should answer your question.

Here, I'll just give you an example, if you want the step-by-step instruction, you can look at the link above.

$\begin{align}4.12222... &= 4.1 + 0.0222...\\ & = 4.1 + \frac{0.2222...}{10} \\ & = \frac{41}{10} + \frac{\frac{2}{9}}{10} \\ & = \frac{41}{10} + \frac{2}{90}\\ &= \frac{371}{90} \end{align}$

Answer 3

Others already indicated how to show it, but here's the result. $$0.a_1a_2\dots a_n a_1 a_2 \dots = \frac{a_1a_2\dots a_n}{\underbrace{999\cdots9}_{n \text{ nines}}}.$$

Answer 4

There is a method for this, for example, let's take $4.23777\ldots$. The decimal not repeating part is $4.23$ and the repeating part is $0.00777\ldots$. The idea is to seperate the repeating decimal part and then cancel it. Write $x=4.23777\ldots$ and do the following $$x=4.23777\ldots$$ $$100x=423.777\ldots$$ $$1000x=4237.77\ldots$$ $$1000x-100x=4237.77\ldots-423.777\ldots=(4237-423)+(0.777\ldots-0.777\ldots)=3814$$ $$900x=3814$$ $$x=\frac{3814}{900}=\frac{1907}{450}$$ Therefore $4.23777\ldots=\frac{1907}{450}$.

Answer 5

Let's say your number have a form $$ N = I_1I_2\ldots I_n.D_1D_2\ldots D_m \left(P_1 P_2 \ldots P_k\right) $$ where $I_1I_2\ldots I_n$ is an integer part, $D_1D_2\ldots D_m$ is non repeating part of the decimal, and $P_1P_2\ldots P_k$ is repeating part of the decimal. So, it's most general form of such number. For example $$ N = 455.01102(979665) = 455.01102\ 979665\ 979665\ 979665 \ldots $$ so $$ I_1 = 4;\quad I_2 = 5; \quad I_3 = 5 \\ D_1 = 0; \quad D_2 = 1; \quad D_3 = 1; \quad D_4 = 0; \quad D_5 = 2 \\ P_1 = 9; \quad P_2 = 7;\quad P_3 = 9; \quad P_4 = 6; \quad P_5 = 6; \quad P_6 = 5 $$ Then, you can rewrite is as $$ N = I_1I_2\ldots I_n.D_1D_2\ldots D_m + 0.\underbrace{000\ldots 0}_{\text{$m$ times}}\left(P_1 P_2 \ldots P_k\right) = \\ = I_1I_2\ldots I_n.D_1D_2\ldots D_m + 10^{-m} \times 0.\left(P_1 P_2 \ldots P_k\right) $$ So, you really have to consider $P = 0.\left(P_1 P_2 \ldots P_k\right)$ part only.

As it was suggested, multiply the whole thing to $10^k$ and subtract initial number, so $$ 10^kP-P = \underbrace{999\ldots 9}_{\text{$k$ times}}P = P_1P_2 \ldots P_k.(P_1P_2\ldots P_k)-0.(P_1P_2\ldots P_k) = P_1P_2\ldots P_k$$ so you can find now $$ P = \frac {P_1P_2\ldots P_k}{\underbrace{999\ldots 9}_{\text{$k$ times}}} $$ and finally $$ N = I_1I_2\ldots I_n + \frac {D_1D_2\ldots D_m}{10^m} + \frac {P_1P_2\ldots P_k}{10^m\underbrace{999\ldots 9}_{\text{$k$ times}}} $$

Answer 6

$\textbf{Hint:}$ Look at the decimals of $x$ and $100x$, where $x=0.5656...$, for example, to get an idea of a general procedure. Obviously what you are looking for is an expression in fractions for $x$.