Problem 47
Question
Express each repeating decimal as a fraction in lowest terms. $$ 0 . \overline{47}=\frac{47}{100}+\frac{47}{10,000}+\frac{47}{1,000,000}+\dots $$
Step-by-Step Solution
Verified Answer
The repeating decimal \(0.\overline{47}\) can be expressed as the fraction \(\frac{47}{99}\) in lowest terms.
1Step 1: Identify the geometric series
First, recognize the representation as a geometric series where the first term \(a = \frac{47}{100}\) and the common ratio \(r = \frac{1}{100}\). This series can be written as \(0.\overline{47} = a + ar + ar^2 + \dots \)
2Step 2: Find the sum of the geometric series
The sum \(S\) of an infinite geometric series can be found using the formula \(S = \frac{a}{1 - r}\). Substituting the values of \(a\) and \(r\) from step 1 into the formula, it gives \(S = \frac{\frac{47}{100}}{1 - \frac{1}{100}}\)
3Step 3: Simplify the fraction
Simplify the fraction by subtracting the denominator and simplifying the numerator and denominator separately. = \(\frac{47/100}{99/100} = \frac{47}{99}\)
4Step 4: Check for lowest terms
Check if the fraction is in its lowest terms. Here, since 47 and 99 don't have a common factor apart from 1, \(\frac{47}{99}\) is already in its lowest terms.
Key Concepts
Understanding Geometric SeriesSumming an Infinite Geometric SeriesSimplifying Fractions
Understanding Geometric Series
A geometric series is essentially a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In a geometric series, we denote the first term by 'a' and the common ratio by 'r'.
Imagine a scenario where you're trying to capture recurring decimal numbers like the one in our exercise, where the digits repeat infinitely. To do this, we express the repeating part of the decimal as the sum of an infinite geometric series. For instance, the repeating decimal 0.\(\overline{47}\) can be broken down into \(\frac{47}{100}\), \(\frac{47}{100}\) times \(\frac{1}{100}\), and so on, adhering to a pattern that represents a geometric series.
This concept dissolves what might initially seem like a complicated or infinite problem into a finite and solvable equation. By recognizing this pattern, we leverage the properties of geometric series to make sense of, and eventually solve, expressions representing repeating decimals.
Imagine a scenario where you're trying to capture recurring decimal numbers like the one in our exercise, where the digits repeat infinitely. To do this, we express the repeating part of the decimal as the sum of an infinite geometric series. For instance, the repeating decimal 0.\(\overline{47}\) can be broken down into \(\frac{47}{100}\), \(\frac{47}{100}\) times \(\frac{1}{100}\), and so on, adhering to a pattern that represents a geometric series.
This concept dissolves what might initially seem like a complicated or infinite problem into a finite and solvable equation. By recognizing this pattern, we leverage the properties of geometric series to make sense of, and eventually solve, expressions representing repeating decimals.
Summing an Infinite Geometric Series
Calculating the sum of an infinite geometric series might sound daunting. But if the common ratio, 'r', is between -1 and 1, the series converges to a finite sum. Thankfully, there’s a formula to find the sum of such series: \(S = \frac{a}{1 - r}\).
Here, 'S' represents the sum of the geometric series, 'a' is the first term, and 'r' is the common ratio. The idea is that for an infinite series, adding more and more terms will eventually get us closer and closer to the sum, but it will not change significantly once we reach a certain point.
Let's see this in action with our previous repeating decimal example, 0.\(\overline{47}\). Since we have identified 'a' as \(\frac{47}{100}\) and 'r' as \(\frac{1}{100}\), the formula tells us that the sum is indeed a finite number and particularly, it simplifies to \(\frac{47}{99}\). It's this simple yet powerful formula that allows us to convert what looks like an endlessly repeating decimal into a precise fraction.
Here, 'S' represents the sum of the geometric series, 'a' is the first term, and 'r' is the common ratio. The idea is that for an infinite series, adding more and more terms will eventually get us closer and closer to the sum, but it will not change significantly once we reach a certain point.
Let's see this in action with our previous repeating decimal example, 0.\(\overline{47}\). Since we have identified 'a' as \(\frac{47}{100}\) and 'r' as \(\frac{1}{100}\), the formula tells us that the sum is indeed a finite number and particularly, it simplifies to \(\frac{47}{99}\). It's this simple yet powerful formula that allows us to convert what looks like an endlessly repeating decimal into a precise fraction.
Simplifying Fractions
Once you've utilized the formula for the sum of an infinite geometric series, you're often left with a fraction. The task then becomes to simplify this fraction to its lowest terms. Doing so requires finding the greatest common divisor (GCD) of the numerator and the denominator and then dividing both by the GCD.
If there is no common factor greater than 1, then the fraction is already in its simplest form. For example, in the fraction \(\frac{47}{99}\), neither 47 nor 99 share any common factors aside from 1; hence, this fraction is already in its most streamlined expression.
Simplifying fractions is not just about making them 'look nice'. It's a crucial step for clarity and sometimes for the subsequent mathematical operations. If the numbers are smaller and without shared factors, computations and comparisons become more manageable, and patterns or ratios might become more apparent.
If there is no common factor greater than 1, then the fraction is already in its simplest form. For example, in the fraction \(\frac{47}{99}\), neither 47 nor 99 share any common factors aside from 1; hence, this fraction is already in its most streamlined expression.
Simplifying fractions is not just about making them 'look nice'. It's a crucial step for clarity and sometimes for the subsequent mathematical operations. If the numbers are smaller and without shared factors, computations and comparisons become more manageable, and patterns or ratios might become more apparent.
Other exercises in this chapter
Problem 47
For Exercises \(45-50,\) write out the first three terms and the last term. Then use the formula for the sum of the first n terms of an arithmetic sequence to f
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In Exercises 39-48, find the term indicated in each expansion. $$\left(x^{2}+y\right)^{22} ;$$ the term containing \(y^{14}\)
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Express each sum using summation notation. Use I as the lower limit of summation and i for the index of summation. $$1+2+3+\dots+30$$
View solution Problem 48
Exercises \(46-48\) will help you prepare for the material covered in ehe next section. Each exercise involves observing a pattern in the expanded form of the b
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