Expressing $1.24\overline{123}=1.24\;123\;123\;123\;\ldots$ as the ratio of two integers

Question

So I'm supposed to express:

$$1.24\overline{123}=1.24\;123\;123\;123\;\ldots$$ as the ratio of two integers.

So I got $$1+\frac{24}{100}+\frac{123}{10^5}+\cdots$$

I don't know if this is correct, and if it is, I also don't know how I would get the common ratio from this. Would the common ratio just be $\frac{1}{10^5}$? Am I doing this wrong?

Answer 1

Write it out a little longer:

$$\begin{align}1.24\overline{123} &= 1 + \frac{24}{100} + \left(\frac{123}{10^5} + \frac{123}{10^8} + \frac{123}{10^{11}} + \cdots\right)\\& = 1 + \frac{24}{100} + \frac{123}{10^5}\left(10^0 + 10^{-3} + 10^{-6} + \cdots\right)\end{align}$$

can you see the geometric series yet?

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If you don't want to go through geometric series, there is also another way, more standardly taught in high schools. Starting with

$$\begin{align}x &= 1.24\overline{123}\\ 1000x&=1241.23\overline{123}\end{align}$$

you can subtract the two equations and get

$$999x = 1241.23\overline{123} - 1.24\overline{123} = 1239.99 = \frac{123999}{100}$$

and you can calculate $x$ from this easily.

Answer 2

If $a=1.24 123 123 123\cdots$

As $\overline{123}$ contains $3$ digits,

$10^3a=1241.23 123 123\cdots$

$a(10^3-1)=1241.23-1.24=?$

Answer 3

No, it is not correct:$$1+\frac{24}{100}+\frac{123}{10^5}=1.24124.$$

Note that\begin{align}0.\overline{123}&=0.123\,123\,123\ldots\\&=\frac{123}{1000}+\frac{123}{1000^2}+\frac{123}{1000^3}+\cdots\\&=\frac{123}{1000}\left(1+\frac1{1000}+\frac1{1000^2}+\cdots\right)\\&=\frac{123}{1000}\times\frac{1000}{999}\\&=\frac{123}{999}.\end{align}So,$$1.24\overline{123}=1+\frac{24}{100}+\frac{123}{99\,900}=\frac{41\,333}{33\,300}.$$

Answer 4

$$\frac{1000\cdot1.241\overline{231}-1.241\overline{231}}{999}=\frac{1241.231\overline{231}-1.241\overline{231}}{999}=\frac{1239.99}{999}=\frac{123999}{99900}.$$