Decimals, Percents, and Real Numbers
7.3 Repeating Decimals
Repeating Decimals
Previously, we discussed how to convert a fraction to a decimal by creating powers of $10$ in the denominator with the fundamental law of fractions.
$$\frac{7}{8}=\frac{7}{2^3}=\frac{7\cdot 5^3}{2^3\cdot 5^3}=\frac{875}{1000}=0.875$$
We also noted that if the denominator of a fraction contains a prime factor that is not $2$ or $5$, then the decimal representation of the fraction will be a repeating decimal. This is because it would be impossible to make the denominator a power of $10$. To calculate $\frac{1}{7}$, we can use long division.
Once we get back around to $1$, the algorithm will repeat itself. This is one way to see why $\frac{1}{7}=0.\overline{142857}$. The line over the $6$ digits indicates that the digits repeat indefinitely. This repeating part of the decimal expansion is called the repetend.
Example: The Repetend
Decide if the following fractions repeat, and if so, find the repetend.
- $\frac{1}{9}$
- $\frac{1}{13}$
- $\frac{1}{6}$
Show Solution
All three repeat.
- $\frac{1}{9}=0.\overline{1}$, so the repetend is $1$.
- $\frac{1}{13}=0.\overline{076923}$, so the repetend is $076923$.
- $\frac{1}{6}=0.1\overline{6}$, so the repetend is $6$.
Notice that for $\frac{1}{6}$ there is a non-repeating lead-in digit $1$ in the tenths place, but the repeating block is still just the digit $6$.
MyOpenMath: Identify the repetend of a repeating decimal
It is possible to determine the length of the repetend before actually calculating the decimal expansion, but that is a bit more advanced than what we will cover here. Just know that they can be quite long, sometimes even longer than what your calculator can display. It is best to rely on factoring the denominator to determine if a decimal will repeat or terminate instead of trying to calculate the decimal expansion directly.
Writing Repeating Decimals as Fractions
Converting repeating decimals might seem like a daunting task at first, but a little algebra can make it very manageable. Consider the familiar case of $\frac{1}{3}=0.\overline{3}$.
Example: Converting $0.\overline{3}$ to a fraction
Show Solution
Let $x=0.\overline{3}$.
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Multiply both sides by $10$ to shift the decimal one place to the right.
$$10x=3.\overline{3}$$
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Subtract the original equation.
$$10x-x=3.\overline{3}-0.\overline{3}$$
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The repetend disappears, leaving
$$9x=3$$
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Solve for $x$.
$$x=\frac{3}{9}=\frac{1}{3}$$
This method can be generalized. The step that needs adjusting is the one in which we multiply by $10$. In truth, we need to multiply by a power of $10$ that is large enough to move an entire repetend to the left one time. The length of the repetend is called the period of the repetend.
Consider the repeating decimal $0.235235235235\ldots=0.\overline{235}$. The repetend is $235$ and its period is $3$. We can use the same method as before, but we will multiply by $10^3=1000$ instead of $10$.
Example: Converting $0.\overline{235}$ to a fraction
Show Solution
Let $x=0.\overline{235}$.
-
Multiply both sides by $1000$ to shift the decimal three places to the right.
$$1000x=235.\overline{235}$$
-
Subtract the original equation.
$$1000x-x=235.\overline{235}-0.\overline{235}$$
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The repeating part subtracts away.
$$999x=235$$
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Solve for $x$.
$$x=\frac{235}{999}$$
And one more for some fun: consider the repeating decimal $0.\overline{9}$.
Example: Convert to a fraction
Convert $0.\overline{9}$ to a fraction.
Show Solution
Let $x=0.\overline{9}$.
$$10x=9.\overline{9}$$
Subtracting gives
$$10x-x=9.\overline{9}-0.\overline{9}$$
so
$$9x=9$$
and therefore
$$x=1.$$
So $0.\overline{9}=1$.
Example: Convert repeating decimals to fractions
Convert each repeating decimal to a fraction.
- $0.2\overline{1}$
- $0.417\overline{27}$
- $0.\overline{8675309}$
Show Solution
- Let $x=0.2\overline{1}=0.21111\ldots$
$$10x=2.11111\ldots$$ $$100x=21.11111\ldots$$
Subtract to eliminate the repetend:
$$100x-10x=21.11111\ldots-2.11111\ldots=19$$ $$90x=19\Rightarrow x=\frac{19}{90}.$$
- Let $x=0.417\overline{27}=0.417272727\ldots$
The non-repeating part has length $3$ and the repetend has length $2$, so use two shifts:
$$100000x=41727.272727\ldots$$ $$1000x=417.272727\ldots$$
Subtract:
$$99000x=41310$$ $$x=\frac{41310}{99000}=\frac{459}{1100}.$$
- Let $x=0.\overline{8675309}$.
The repetend has length $7$, so multiply by $10^7$:
$$10^7x=8675309.\overline{8675309}$$
Subtract the original equation:
$$10^7x-x=8675309$$ $$9999999x=8675309\Rightarrow x=\frac{8675309}{9999999}.$$
MyOpenMath: Repeating decimal to fraction
Ordering Repeating Decimals
Because we're still working with rational numbers, we still have order. And, we can still find infinitely many rational numbers between two given rational numbers.
Example: Decimal between two repeating decimals
Find a rational number in decimal form between $0.\overline{35}$ and $0.\overline{351}$.
Show Solution
Write a few digits:
$$0.\overline{351}=0.351351351\ldots$$ $$0.\overline{35}=0.353535353\ldots$$
A number between them is
$$0.\overline{352}=0.352352352\ldots$$
since
$$0.351351351\ldots < 0.352352352\ldots < 0.353535353\ldots$$
MyOpenMath: Finding a number between repeating decimals
Example: Ordering repeating decimals
Write the following decimals in order from least to greatest.
- $1.\overline{45}$
- $1.\overline{454}$
- $1.\overline{455}$
- $1.45$
- $1.\overline{445}$
Show Solution
Write enough digits to compare place values directly:
- $1.\overline{445}=1.445445445\ldots$
- $1.45=1.450000000\ldots$
- $1.\overline{454}=1.454454454\ldots$
- $1.\overline{45}=1.454545454\ldots$
- $1.\overline{455}=1.455455455\ldots$
So the order is
$$1.\overline{445}<1.45<1.\overline{454}<1.\overline{45}<1.\overline{455}.$$