# Converting Decimal Representations of Rational Numbers to Fraction Representations

Alignments to Content Standards: 8.NS.A.1

Represent each of the following rational numbers in fraction form.

1. $0.33\overline{3}$
2. $0.3\overline{17}$
3. $2.1\overline{6}$

## IM Commentary

Standard 8.NS.1 requires students to "convert a decimal expansion which repeats eventually into a rational number." Despite this choice of wording, the numbers in this task are rational numbers regardless of choice of representation. For example, $0.33\overline{3}$ and $\frac13$ are two different ways of representing the same number.

So what is a rational number? Sometimes people define a rational number to be a ratio of integers, but to be consistent with the CCSSM, we would need to say a rational number is any number that is the value of a ratio of two integers. Sometimes people define a rational number based on how it can be represented; here is a typical definition: A rational number is any number that can be represented as $\frac{a}{b}$ where $a$ and $b$ are integers and $b\neq 0$. It is interesting to compare this with the definition of a rational number given in the Glossary of the CCSSM (as well as the more nuanced meaning developed in the standards themselves starting in grade 3 and beyond).

A more constructive definition for a rational number that does not depend on the way we represent it is:

A number is rational if it is a quotient $a\div b$ of two integers $a$ and $b$ where $b\neq 0$.
or, equivalently,
A rational number is a number that satisfies an equation of the form $a=bx$, where $a$ and $b$ are integers and $b\neq 0$.

So $0.33\overline{3}$ is a rational number because it is the result we get when we divide 1 by 3, or equivalently, because it is a solution to $1=3x$. However, numbers like $\pi$ and $\sqrt{2}$ are not rational because neither of them satisfies an equation of the form $a=bx$ where $a$ and $b$ are integers. This is actually tricky to show and is an exercise left to high school or college.

## Solution

The solution for all the parts of this take advantage of the repeating structure of the decimal expansions. Namely, by multiplying by a suitable power of 10 (namely, $10^r$ where $r$ is the length of the repeating segment in the decimal expansion) and subtracting the original number, we can get a multiple of $x$ with a finite decimal expansion.

1. Let $x= 0.33\overline{3}$ Then $$10x = 3.3\overline{3} = 3 + 0.33\overline{3} = 3 + x$$ Subtracting $x$ from both sides gives $9x=3$, so $$0.33\overline{3}=x = \frac39 = \frac13.$$
2. Let $x=0.31717\ldots$

Then \begin{alignat*}{7} 100x&\,\,\, =3&1.&71&7171\ldots\\ x&\,\,\, =&0.&31&7171\ldots \end{alignat*} Now subtracting the two equations gives $99x=31.4$, so $$0.3\overline{17}=x=\frac{31.4}{99}=\frac{314}{990}.$$

3. Let $x=2.166\ldots$. Then \begin{alignat*}{7} 10x&\,\,\, =2&1.&66&66\ldots\\ x&\,\,\, =&2.&16&66\ldots \end{alignat*} Now subtracting the two equations gives $9x=19.5$, so $$2.1\overline{6}=x=\frac{19.5}{9}=\frac{195}{90}.$$