Question

a. Consider the number 0.555555...., which can be viewed as the series \(5 \sum_{k=1}^{\infty} 10^{-k} .\) Evaluate the geometric series to obtain a rational value of \(0.555555 \ldots\) b. Consider the number \(0.54545454 \ldots,\) which can be represented by the series \(54 \sum_{k=1}^{\infty} 10^{-2 k} .\) Evaluate the geometric series to obtain a rational value of the number. c. Now generalize parts (a) and (b). Suppose you are given a number with a decimal expansion that repeats in cycles of length \(p,\) say, \(n_{1}, n_{2} \ldots \ldots, n_{p},\) where \(n_{1}, \ldots, n_{p}\) are integers between 0 and \(9 .\) Explain how to use geometric series to obtain a rational form of the number. d. Try the method of part (c) on the number \(0.123456789123456789 \ldots\) e. Prove that \(0 . \overline{9}=1\)

Short answer

Question: Use the method of sum of infinite geometric series to evaluate the following repeating decimals. a) \(0.555555\ldots\) b) \(0.54545454\ldots\) c) Generalize the method for repeating decimals. d) Apply the method to \(0.123456789123456789\ldots\) e) Prove \(0.\overline{9} = 1\)

Step-by-step solution

Part a: Evaluate \(0.555555\ldots\) as a geometric series

To find the sum of this geometric series, we will use the formula for the sum of an infinite geometric series: \(S = \frac{a}{1 - r}\), where \(S\) is the sum of the series, \(a\) is the first term, and \(r\) is the common ratio. In this case, we have \(a = 5 \cdot 10^{-1}\) and the common ratio \(r=10^{-1}\). Plugging the values into the formula, we get: \(S = \frac{5 \cdot 10^{-1}}{1 - 10^{-1}} = \frac{5}{9}\) So, the rational form of \(0.555555\ldots\) is \(\frac{5}{9}\).

Part b: Evaluate \(0.54545454\ldots\) as a geometric series

Again, we will use the formula for the sum of an infinite geometric series. This time, we have \(a = 54 \cdot 10^{-2}\) and the common ratio \(r=10^{-2}\). Plugging these values into the formula: \(S = \frac{54 \cdot 10^{-2}}{1 - 10^{-2}} = \frac{54}{99}\) The rational form of \(0.54545454\ldots\) is \(\frac{54}{99}\). Simplifying this fraction, we get \(\frac{6}{11}\).

Part c: Generalize the method for repeating decimals

To obtain a rational form for a repeating decimal with a cycle of length \(p\), we can represent it as a geometric series of the form: \(S = \frac{n \sum_{k=1}^{\infty}10^{-kp}}{9\ldots9}\), where \(n\) is the \(p\)-digit number that repeats and the denominator of the fraction is a number consisting of \(p\) nines. Using the formula for the sum of an infinite geometric series, we get the rational form of the number: \(S = \frac{\frac{n\cdot10^{-p}}{1-10^{-p}}}{9\ldots9} = \frac{n}{9\ldots9(1-10^{-p})}\)

Part d: Apply the method to \(0.123456789123456789\ldots\)

The repeating decimal has a cycle of length \(p=9\) and starts with \(n=123456789\). So, applying the generalized method, we get the rational form of the number: \(S = \frac{123456789}{999999999(1-10^{-9})} = \frac{123456789}{999999999 - 1} = \frac{123456789}{999999998}\)

Part e: Prove \(0.\overline{9} = 1\)

To prove this statement, we will use the same method as in the previous parts. We have a repeating decimal with a cycle of length \(p=1\) and starts with \(n=9\). So, applying the generalized method, we get the rational form of the number: \(S = \frac{9}{9(1-10^{-1})} = \frac{9}{9(1-0.1)} = \frac{9}{9(0.9)} = \frac{9}{8.1}\) Since \(\frac{9}{8.1} = 1\), we have proven that \(0.\overline{9} = 1\).

Key concepts

Repeating Decimals

A repeating decimal, also known as a recurring decimal, is a decimal number in which a sequence of digits keeps on repeating indefinitely. For example, in the decimal 0.555..., the digit '5' repeats endlessly. Such numbers can be quite intriguing because they can also be represented exactly as fractions—the core idea of converting them to a rational number.

To convert a repeating decimal to a fraction, we use a geometric series. By identifying the first term (\(a\)) and the common ratio (\(r\)) of the series these decimals form, we can determine their sum using the formula for an infinite geometric series: \[ S = \frac{a}{1 - r} \].

This formula lets us capture the essence of endlessly repeating digits into a neat rational form, unfolding the continuous repeating nature into a clear, finite number.

Rational Numbers

Rational numbers are numbers that can be expressed as the quotient or fraction \( \frac{p}{q} \), where \(p\) and \(q\) are integers and \(q eq 0\). These numbers are able to take forms both as terminating decimals and repeating decimals.

For instance, the repeating decimal \(0.555...\) is represented by the fraction \(\frac{5}{9}\), while \(0.545454...\) gives us \(\frac{6}{11}\).

When you see a repeating decimal, you can convert it to a rational number using the insights provided by geometric series, making things like converting endless decimals into finite forms possible and logical. This conversion shows the symmetry and balance present between the infinite nature of repeating decimals and the completeness of rational numbers.

Infinite Series

An infinite series is essentially the sum of infinite terms, like you would find in a repeating decimal when visualized as a sequence. In mathematics, we often encounter series that do not stop, such as those in infinite geometric progressions.

When converting a repeating decimal to a fraction, we are summing an infinite series where each term represents a place value of the repeating sequence. By understanding this infinite geometry, we employ the formula for the sum of an infinite geometric series, \[ S = \frac{a}{1 - r} \], turning what appears endless into a finite solution.

This concept allows for repetitive but predictable cycles like those in repeating decimals to be expressed simply and precisely, leveraging our understanding of geometric series and making sense of numbers that repeat forever by expressing them in a clear, fractional form.