8.NS Converting Decimal Representations of Rational Numbers to Fraction Representations


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

Task

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 $$ 2.1\overline{6}=x=\frac{19.5}{9}=\frac{195}{90}. $$