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Chapter 8: Problem 42

Write each repeating decimal first as a geometric series and then as a fraction (a ratio of two integers). $$0 . \overline{6}=0.666 \dots$$

Short Answer

Expert verified
Question: Convert the repeating decimal 0.666... into a fraction. Answer: The repeating decimal 0.666... can be expressed as the fraction $$\frac{2}{3}$$.

Step by step solution

01

Express the repeating decimal as a geometric series

To express the repeating decimal as a geometric series, we can break it down as follows: $$0.666... = 6 \cdot \frac{1}{10} + 6 \cdot \frac{1}{100} + 6 \cdot \frac{1}{1000} + \cdots$$ This is an infinite geometric series with the first term, a, being $$6 \cdot \frac{1}{10}$$ and the common ratio, r, being $$\frac{1}{10}$$.
02

Use the sum formula for an infinite geometric series to represent the series as a fraction

The sum formula for an infinite geometric series is: $$S = \frac{a}{1-r}$$ We can now plug in the values of a and r that we found in Step 1: $$S = \frac{6 \cdot \frac{1}{10}}{1-\frac{1}{10}}$$ Now we simplify the expression: $$S = \frac{\frac{6}{10}}{\frac{9}{10}}$$
03

Simplify the fraction

We can simplify the fraction by multiplying the numerator and denominator by 10, which will eliminate the decimals: $$S = \frac{6}{9}$$ Now, we can simplify this further by dividing both the numerator and denominator by their greatest common divisor, which is 3: $$S = \frac{6 \div 3}{9 \div 3}$$ Which gives us the final answer: $$S = \frac{2}{3}$$ So, the repeating decimal $$0.\overline{6} = 0.666...$$ can be expressed as the fraction $$\frac{2}{3}$$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Infinite Series
An infinite series is a sequence of numbers that continue indefinitely. In the case of a repeating decimal like 0.666..., it can be expressed as an infinite series. This is achieved by breaking it down into a sum of smaller parts:
  • Each term in our series is derived from the repeating digit, in this case 6, multiplied by decreasing powers of 10.
  • The series becomes: \( 6 \times \frac{1}{10} + 6 \times \frac{1}{100} + 6 \times \frac{1}{1000} + \cdots \)
This creates a geometric series where the first term \( a \) is \( 6 \times \frac{1}{10} \) and the common ratio \( r \) is \( \frac{1}{10} \). Geometric series are special because they allow us to find the sum using a specific formula, even when the series goes on forever! This formula is essential when working with repeating decimals.
Repeating Decimal
Repeating decimals are decimals where one or more digits repeat infinitely. For example, 0.666... is a repeating decimal where the digit 6 repeats.
  • To express a repeating decimal as a geometric series, observe the pattern: the repeating section is multiplied by decreasing powers of 10.
  • This turns a seemingly endless number into a manageable mathematical expression.
Repeating decimals often have a fascinating link with fractions. Most infinitely repeating decimals can be represented as a fraction 鈥� a ratio between two integers! Understanding this link helps us convert repeating decimals to fractions using mathematical techniques like geometric series.
Fraction Conversion
Fraction conversion is the process of representing a number, like a repeating decimal, as a fraction. This involves some mathematical steps:
  • First, express the repeating decimal as a geometric series.
  • Use the sum formula for an infinite geometric series: \( S = \frac{a}{1-r} \) to find the equivalent fraction.
  • In our case, \( a = 6 \times \frac{1}{10} \) and \( r = \frac{1}{10} \), so the sum \( S = \frac{6 \times \frac{1}{10}}{1-\frac{1}{10}} \).
  • Simplify to get \( S = \frac{6}{9} \). Further simplify by dividing by the greatest common divisor, which is 3, leading to \( \frac{2}{3} \).
This shows that the repeating decimal 0.666... is equivalent to the fraction \( \frac{2}{3} \), unveiling the underlying relationship between these two forms of numbers.

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Most popular questions from this chapter

A fishery manager knows that her fish population naturally increases at a rate of \(1.5 \%\) per month. At the end of each month, 120 fish are harvested. Let \(F_{n}\) be the fish population after the \(n\) th month, where \(F_{0}=4000\) fish. Assume that this process continues indefinitely. Use infinite series to find the longterm (steady-state) population of the fish.

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the limit of the sequence or state that the sequence diverges. $$a_{n+1}=\frac{1}{2}\left(a_{n}+2 / a_{n}\right) ; a_{0}=2$$

Use Theorem 8.6 to find the limit of the following sequences or state that they diverge. $$\left\\{\frac{3^{n}}{n !}\right\\}$$

The Fibonacci sequence \(\\{1,1,2,3,5,8,13, \ldots\\}\) is generated by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1\). a. It can be shown that the sequence of ratios of successive terms of the sequence \(\left\\{\frac{f_{n+1}}{f_{n}}\right\\}\) has a limit \(\varphi .\) Divide both sides of the recurrence relation by \(f_{n},\) take the limit as \(n \rightarrow \infty,\) and show that \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}}=\frac{1+\sqrt{5}}{2} \approx 1.618\). b. Show that \(\lim _{n \rightarrow \infty} \frac{f_{n-1}}{f_{n+1}}=1-\frac{1}{\varphi} \approx 0.382\). c. Now consider the harmonic series and group terms as follows: $$\sum_{k=1}^{\infty} \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)$$ $$+\left(\frac{1}{9}+\cdots+\frac{1}{13}\right)+\cdots$$ With the Fibonacci sequence in mind, show that $$\sum_{k=1}^{\infty} \frac{1}{k} \geq 1+\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{3}{8}+\frac{5}{13}+\cdots=1+\sum_{k=1}^{\infty} \frac{f_{k-1}}{f_{k+1}}.$$ d. Use part (b) to conclude that the harmonic series diverges. (Source: The College Mathematics Journal, 43, May 2012)

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$\ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}<1+\ln n.$$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$\frac{1}{n+1}>\ln (n+2)-\ln (n+1).$$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\). e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 . f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\}\), estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)
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