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

Question

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.

Answer 1

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:

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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?

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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?

Answer 2

HINT:

$$0.3344444444\cdots=\dfrac{33}{100}+4\sum_{r=3}^\infty\dfrac1{10^r}$$

Answer 3

Let $x = 0.33444 \ldots$, then $100x = 33.44444 \ldots $ so that $100x - x = 99x = 33.11 $ and $x=\frac{3311}{9900}$.

Answer 4

Multiply by $\dfrac94$ to turn the $4$s into $9$s, giving $$0.7524999999\cdots$$ which equals $$0.7525\,.$$

Your number is

$$\frac{7525}{10000}\frac{4}{9}=\frac{301}{900}.$$

Answer 5

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 $$0.33444...=0.33+0.0044...=\frac{33}{100}+\frac{1}{100}0.44...=\frac{33}{100}+\frac{1}{100}\frac{4}{9}=\frac{297+4}{900}=\frac{301}{900}.$$

Answer 6

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:

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

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

• $\color\red{A}=0$
• $\color\green{B}=33$
• $\color\orange{C}=4$

Then:

$$x=\frac{\color\red{0}\cdot(10^{|\color\green{33}|+|\color\orange{4}|}-10^{|\color\green{33}|})+\color\green{33}\cdot(10^{|\color\orange{4}|}-1)+\color\orange{4}}{10^{|\color\green{33}|+|\color\orange{4}|}-10^{|\color\green{33}|}}=\frac{301}{900}$$

Answer 7

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

$$\frac{3.01}{9}=\frac{301}{900}.$$

Answer 8

When you learn about representation of rational numbers in decimal form then you study the following results:

1. 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.
2. 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:

1. 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 $0.123 = 123/1000$.
2. 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 $p/q = 301/900$.

Answer 9

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:

$$=\frac{33}{100}+4\sum_{n=3}^{\infty} \frac{1}{10^n}$$

$$=\frac{33}{100}+4\left(\sum_{n=0}^{\infty} (\frac{1}{10})^n-\frac{1}{100}-\frac{1}{10}-1 \right)$$

$$=\frac{33}{100}+4\left(\frac{1}{1-\frac{1}{10}}-\frac{1}{100}-\frac{1}{10}-1 \right)$$

$$=\frac{301}{900}$$

Answer 10

One more solution, splitting up $0.33\bar{4}$ differently than everyone else has so far for fun: $$0.33\bar{4}=0.33\bar{3}+0.00\bar{1}=0.\bar{3}+0.00\bar{1}=0.\bar{3}+\frac{0.\bar{1}}{100}=\frac{1}{3}+\frac{\frac{1}{9}}{100}=\frac{1}{3}+\frac{1}{900}=\frac{300}{900}+\frac{1}{900}=\frac{301}{900}$$