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

Represent each of the following rational numbers in fraction form.

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

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.

Solutions

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