Find the rational number whose decimal expansion is $0.3344444444\dots$ 
Find the rational number whose decimal expansion is $0.3344444444...$

This is a homework question. I am not sure if I understand the meaning of the question. As per wiki article I think what I have to do is that I write the  decimal expansion as $0 +  3\frac{1}{10}$ + $ 3\frac{1}{100}$ + $ 4\frac{1}{1000}\dots$
Please tell me  if this is incorrect. 
 A: HINT:
$$0.3344444444\cdots=\dfrac{33}{100}+4\sum_{r=3}^\infty\dfrac1{10^r}$$
A: Multiply by $\dfrac94$ to turn the $4$s into $9$s, giving 
$$0.7524999999\cdots$$ which equals $$0.7525\,.$$
Your number is
$$\frac{7525}{10000}\frac{4}{9}=\frac{301}{900}.$$
A: There is a trick, which goes like: find the repeating pattern and divide it with as many nines. For example 
$$\frac{1090}{9999}=0.10901090....$$
In this case we have
$$0.33444...=0.33+0.0044...=\frac{33}{100}+\frac{1}{100}0.44...=\frac{33}{100}+\frac{1}{100}\frac{4}{9}=\frac{297+4}{900}=\frac{301}{900}.$$
A: You can move the decimal point one position to the left by multiplying by $10$.
$$0.3344444444\cdots\times10=3.344444444\cdots$$
Then subtracting these two numbers, a finite number of decimals remain and
$$0.3344444444\cdots\times9=3.01\,.$$
So the fraction is
$$\frac{3.01}{9}=\frac{301}{900}.$$
A: What you need to do is to find two integers $a,b$ such that $\frac ab=0.3344444\dots$.
You have (at least) two ways of solving the problem:

First option
call the number $0.334444444444\dots$ simply $x$, i.e.
$$x=0.3344444444\dots$$
Then, multiply this equation by $100$ to get
$$100x=33.\overline 4$$
Multiply it again by $10$ to get
$$1000 x = 334.\overline{4}$$
Now, try to subtract the above equations. What do you get?

Second option:
You can write $x=0.33\overline 4$ as $$0.3 + 0.03 + 0.004 + 0.0004 + \dots$$
which we can rewrite as $$x=0.33 + 0.004\cdot (1 + 0.1 + 0.01 + \dots)$$
or further as
$$x=\frac{33}{100} + \frac{4}{1000}\cdot \sum_{i=0}^\infty \left(\frac{1}{10}\right)^{i}$$
Now, remember how you were taught what the sum of a geometric series is?
A: Let $x = 0.33444 \ldots$, then $100x = 33.44444 \ldots $ so that $100x - x = 99x = 33.11 $ and $x=\frac{3311}{9900}$.
A: General method:


*

*Let $x$ denote the input number

*Let $|n|$ denote the number of decimal digits in $n$

*Split $x$ into the following parts:


*

*$\color\red{A}=$ the integer part, i.e., $\lfloor{x}\rfloor$

*$\color\green{B}=$ the fraction part's non-periodic prefix

*$\color\orange{C}=$ the fraction part's periodic postfix



Then:
$$x=\frac{\color\red{A}\cdot(10^{|\color\green{B}|+|\color\orange{C}|}-10^{|\color\green{B}|})+\color\green{B}\cdot(10^{|\color\orange{C}|}-1)+\color\orange{C}}{10^{|\color\green{B}|+|\color\orange{C}|}-10^{|\color\green{B}|}}$$

For example, if $x=0.33\overline{4}$:


*

*$\color\red{A}=0$

*$\color\green{B}=33$

*$\color\orange{C}=4$


Then:
$$x=\frac{\color\red{0}\cdot(10^{|\color\green{33}|+|\color\orange{4}|}-10^{|\color\green{33}|})+\color\green{33}\cdot(10^{|\color\orange{4}|}-1)+\color\orange{4}}{10^{|\color\green{33}|+|\color\orange{4}|}-10^{|\color\green{33}|}}=\frac{301}{900}$$
A: When you learn about representation of rational numbers in decimal form then you study the following results:


*

*If $p/q$ is rational where $p, q$ are coprime to each other and $q > 0$ then the decimal representation of $p/q$ is terminating if and only if $q$ has no prime factors other than $2$ and $5$. If $q$ has a prime factor other than $2$ and $5$ then its decimal representation is non-terminating but repeating after a certain point.

*Every terminating decimal or a non-terminating but repeating decimal is a rational number.


There are simple well defined rules (which can be justified if you are willing to seek justification) to obtain decimal representation of a rational number and to express an appropriate decimal (terminating or non-terminating but repeating) into the form $p/q$ where $p, q$ are integers. For converting a decimal number into the form $p/q$ we have following rules:


*

*If the decimal is terminating then count number of digits after decimal point say this count is $n$. Now for $p$ you have the integer obtained by removing the decimal and $q = 10^{n}$ i.e. $q$ is $1$ followed by $n$ zeroes. Thus $0.123 = 123/1000$.

*If the decimal is non-terminating but repeating then the decimal has initially a non-repeating block of digits (which may be non-existent also) and lets call it $A$ and a repeating block of digits and we call it $B$. Let number of digits in non repeating block be $m$ and the number of digits in repeating block be $n$. The denominator $q$ then consists of $n$ $9$'s followed by $m$ $0$'s. The numerator $p$ is obtained as $p = C - A$ where $C$ is the number formed by removing the decimal (including both non-repeating and repeating blocks) and $A$ is the number formed from non-repeating block only.


Now we come to your current question regarding the decimal $0.33444\ldots$. This is written as $0.33\bar{4}$ so that the block $33$ is non-repeating and $4$ is repeating. The number formed after removing decimal is $334$ and from it we subtract the number formed by non-repeating part i.e. $33$ and thus get numerator $p$ as $334-33 = 301$. The repeating block has only one digit and non-repeating block has two digits hence the denominator $q$ is made of one $9$ followed by two $0$'s so that $q = 900$ and hence the given decimal represents the number $p/q = 301/900$.
A: Your not incorrect in attempting to start finding the answer to this question:
$$0.334444...=0+.3+.03+.004+...$$
$$=\frac{3}{10}+\frac{3}{100}+\frac{4}{1000}+...$$
From which you can continue in this fashion:
$$=\frac{33}{100}+4\sum_{n=3}^{\infty} \frac{1}{10^n}$$
$$=\frac{33}{100}+4\left(\sum_{n=0}^{\infty} (\frac{1}{10})^n-\frac{1}{100}-\frac{1}{10}-1 \right)$$
$$=\frac{33}{100}+4\left(\frac{1}{1-\frac{1}{10}}-\frac{1}{100}-\frac{1}{10}-1 \right)$$
$$=\frac{301}{900}$$
A: One more solution, splitting up $0.33\bar{4}$ differently than everyone else has so far for fun: $$0.33\bar{4}=0.33\bar{3}+0.00\bar{1}=0.\bar{3}+0.00\bar{1}=0.\bar{3}+\frac{0.\bar{1}}{100}=\frac{1}{3}+\frac{\frac{1}{9}}{100}=\frac{1}{3}+\frac{1}{900}=\frac{300}{900}+\frac{1}{900}=\frac{301}{900}$$
