Problem 56
Question
Express the repeating decimal as a fraction. $$ 0.2 \overline{53} $$
Step-by-Step Solution
Verified Answer
The repeating decimal 0.2\overline{53} expressed as a fraction is \( \frac{251}{990} \).
1Step 1: Define the Decimal as a Variable
Let \( x = 0.2\overline{53} \), which means \( x = 0.2535353\ldots \)
2Step 2: Shift the Decimal
Multiply both sides of the equation \( x = 0.2535353\ldots \) by 1000 to shift the repeating part to the right of the decimal:\[ 1000x = 253.5353\ldots \]
3Step 3: Set Up Another Equation to Align Repeating Parts
Multiply both sides of the equation \( x = 0.2535353\ldots \) by 10 to align the repeating parts:\[ 10x = 2.5353\ldots \]
4Step 4: Subtract the Equations
Subtract the second equation from the first to eliminate the repeating part:\[ 1000x - 10x = 253.5353\ldots - 2.5353\ldots \]Simplify this to:\[ 990x = 251 \]
5Step 5: Solve for x
Divide both sides by 990 to find \( x \):\[ x = \frac{251}{990} \]
6Step 6: Simplify the Fraction
The fraction \( \frac{251}{990} \) is already in simplest form because 251 is a prime number and does not divide 990.
Key Concepts
FractionsDecimal ConversionAlgebraic ManipulationSimplest Form
Fractions
Fractions are an essential part of mathematics, helping us express numbers that aren't whole. A fraction consists of a numerator, the top part, which indicates the number of parts we have. The denominator, the bottom part, shows how many equal parts the whole is divided into.
For example, the fraction \( \frac{251}{990} \) tells us that we have 251 parts out of 990. This represents the repeating decimal we've been dealing with in the exercise, showing us that fractions can help convert complex decimal numbers back into a simpler form.
It's important to recognize that fractions can be more intuitive than decimals because they display a clear ratio. This ratio shows how a number relates to the whole more transparently than a decimal might.
For example, the fraction \( \frac{251}{990} \) tells us that we have 251 parts out of 990. This represents the repeating decimal we've been dealing with in the exercise, showing us that fractions can help convert complex decimal numbers back into a simpler form.
It's important to recognize that fractions can be more intuitive than decimals because they display a clear ratio. This ratio shows how a number relates to the whole more transparently than a decimal might.
Decimal Conversion
When converting a repeating decimal to a fraction, we employ algebraic techniques to make the process seamless. In our example from the exercise, we see the repeating decimal, which is written as \( 0.2\overline{53} \).
This decimal repeats the '53' indefinitely. To make this conversion practical, we first assign this repeating decimal value to a variable, like \( x = 0.2535353\ldots \).
Next, we shift the decimal by multiplying by powers of 10. In this case, multiplying by 1000 helps us shift the repeating segment past the decimal point. This setup allows us to manipulate the decimal easily, setting up situations where subtraction can effectively eliminate the repeating part, bringing us closer to a fractional representation.
This decimal repeats the '53' indefinitely. To make this conversion practical, we first assign this repeating decimal value to a variable, like \( x = 0.2535353\ldots \).
Next, we shift the decimal by multiplying by powers of 10. In this case, multiplying by 1000 helps us shift the repeating segment past the decimal point. This setup allows us to manipulate the decimal easily, setting up situations where subtraction can effectively eliminate the repeating part, bringing us closer to a fractional representation.
Algebraic Manipulation
Algebraic manipulation is a valuable tool in converting decimals to fractions. First, we label our repeating decimal as a variable, \( x \).
From the step-by-step solution, we understand that multiplying by 10 or 1000 positions the repeating decimal further right. This setup creates two equations, with one equation allowing subtraction from the other.
By subtracting these equations, we effectively eliminate the repeating decimal segment. You end up with a simple equation that helps isolate \( x \) on one side.
For instance, the subtraction performed in the step-by-step solution gives us \( 990x = 251 \), which is much easier to solve, thus streamlining the conversion journey.
From the step-by-step solution, we understand that multiplying by 10 or 1000 positions the repeating decimal further right. This setup creates two equations, with one equation allowing subtraction from the other.
By subtracting these equations, we effectively eliminate the repeating decimal segment. You end up with a simple equation that helps isolate \( x \) on one side.
For instance, the subtraction performed in the step-by-step solution gives us \( 990x = 251 \), which is much easier to solve, thus streamlining the conversion journey.
Simplest Form
The simplest form of a fraction keeps the mathematical expression clean and easy to understand. To achieve the simplest form, divide both the numerator and the denominator by their greatest common divisor (GCD).
In our specific case \( \frac{251}{990} \) doesn't change because 251 is a prime number. Prime numbers only have divisors of 1 and themselves, hence they can't simplify further with the denominator.
Writing a fraction in its simplest form makes it more straightforward to comprehend and use in calculations. It's an integral part of polishing the solution and ensuring it's as efficient as possible. Keeping calculations simple is key to better mathematical understanding.
In our specific case \( \frac{251}{990} \) doesn't change because 251 is a prime number. Prime numbers only have divisors of 1 and themselves, hence they can't simplify further with the denominator.
Writing a fraction in its simplest form makes it more straightforward to comprehend and use in calculations. It's an integral part of polishing the solution and ensuring it's as efficient as possible. Keeping calculations simple is key to better mathematical understanding.
Other exercises in this chapter
Problem 55
An arithmetic sequence has first term \(a=5\) and common difference \(d=2 .\) How many terms of this sequence must be added to get 2700\(?\)
View solution Problem 55
Write the sum without using sigma notation. $$\sum_{k=0}^{6} \sqrt{k+4}$$
View solution Problem 56
Depreciation The purchase value of an office computer is \(\$ 12,500\) . Its annual depreciation is \(\$ 1875 .\) Find the value of the computer after 6 years.
View solution Problem 56
Write the sum without using sigma notation. $$\sum_{k=6}^{9} k(k+3)$$
View solution