Chapter 11: Problem 21
Exer. \(21-24:\) The overbar indicates that the digits underneath repeat indefinitely. Express the repeating decimal as a series, and find the rational number it represents. $$ 0 . \overline{23} $$
Short Answer
Expert verified
The repeating decimal \( 0.\overline{23} \) represents the rational number \( \frac{23}{99} \).
Step by step solution
01
Identifying the Repeating Decimal
The given decimal is \( 0.\overline{23} \), meaning that the digits 23 repeat indefinitely. We need to convert this repeating decimal into a series first.
02
Expressing the Decimal as a Series
The repeating decimal \( 0.\overline{23} \) can be expressed as a series: \( 0.23 + 0.0023 + 0.000023 + \ldots \).
03
Expressing the Series with Fractions
Rewrite the series using fractions: \( \frac{23}{100} + \frac{23}{10000} + \frac{23}{1000000} + \ldots \). Each additional term moves the decimal point two places further.
04
Identifying the Series as a Geometric Series
Observe that the series \( \frac{23}{100} + \frac{23}{10000} + \frac{23}{1000000} + \ldots \) is a geometric series with the first term \( a = \frac{23}{100} \) and common ratio \( r = \frac{1}{100} \).
05
Using the Formula for Sum of Infinite Geometric Series
The sum \( S \) of an infinite geometric series is given by the formula \( S = \frac{a}{1 - r} \). Here, \( a = \frac{23}{100} \) and \( r = \frac{1}{100} \).
06
Calculating the Sum of the Series
Plug these values into the formula: \[ S = \frac{\frac{23}{100}}{1 - \frac{1}{100}} = \frac{\frac{23}{100}}{\frac{99}{100}} = \frac{23}{99} \].
07
Expressing the Repeating Decimal as a Rational Number
The rational number that represents the repeating decimal \( 0.\overline{23} \) is \( \frac{23}{99} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometric Series
In mathematics, a geometric series is a series with a constant ratio between successive terms. This type of series is particularly helpful when dealing with repeating decimals. When you see a number like \(0.\overline{23}\), it means the digits "23" repeat indefinitely. For such repeating decimals, we can express them as an infinite sequence. For instance, the repeating decimal \(0.\overline{23}\) can be expanded into a series: \(0.23 + 0.0023 + 0.000023 + \ldots\).
- Notice how each term is a fraction: \(\frac{23}{100}, \frac{23}{10000}, \frac{23}{1000000}\), and so on.
- Each term is obtained by multiplying the previous term by \(\frac{1}{100}\).
- This constant multiplier is known as the common ratio \(r\).
Rational Number Representation
Once you've identified a repeating decimal and expressed it as a geometric series, the next step is to find its representation as a rational number. A rational number is any number that can be expressed as the quotient or fraction of two integers. This is key to transforming a non-terminal decimal into a clean fraction form.For the repeating decimal \(0.\overline{23}\), our task is to express it as a quotient of integers. From the series expression \(\frac{23}{100} + \frac{23}{10000} + \frac{23}{1000000} + \ldots\), we know each term individually contributes to the total sum we are seeking. When structured into a mathematical formula, a series of this kind, known as a geometric series, can be summed up using a special formula, leading to its rational number form. In this case, through various simplifications, the repeating decimal \(0.\overline{23}\) equates to the fraction \(\frac{23}{99}\). Therefore, anytime you find a repeating decimal, there is a rational number lurking beneath its surface, waiting to reveal a pure fraction.
Sum of Infinite Series
Finding the sum of an infinite series might seem daunting, but with a geometric series, it's straightforward thanks to a handy formula. When the common ratio \(|r|\) is less than 1, the infinite series does converge to a specific sum.For a geometric series with initial term \(a\) and common ratio \(r\), the sum \(S\) is given by:\[ S = \frac{a}{1 - r} \]Let's apply this to our example. We identified the repeating decimal \(0.\overline{23}\) as a series \( \frac{23}{100} + \frac{23}{10000} + \frac{23}{1000000} + \ldots \). Here, \(a = \frac{23}{100}\) and \(r = \frac{1}{100}\). Plugging these into the formula, we get:\[ S = \frac{\frac{23}{100}}{1 - \frac{1}{100}} = \frac{\frac{23}{100}}{\frac{99}{100}} = \frac{23}{99}\]This calculation confirms that the sum of these infinitely repeating terms is indeed \(\frac{23}{99}\). Thus, by using the formula for the sum of an infinite geometric series, infinite decimals are elegantly converted into simple fractions.