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Chapter 10: Problem 25

Express each of the numbers in Exercises \(19-26\) as the ratio of two integers. $$1.24 \overline{123}=1.24123123123 \ldots$$

Short Answer

Expert verified
The number is \( \frac{1240}{999} \).

Step by step solution

01

Define the Repeating Decimal

Let \( x = 1.24\overline{123} \), which means \( x = 1.24123123123\ldots \). This is a repeating decimal where the block "123" repeats indefinitely.
02

Multiply to Eliminate the Repeat

First, identify the length of the repeating block, which is 3 digits. Multiply \( x \) by \( 10^3 = 1000 \) to shift the repeating decimal to the right: \( 1000x = 1241.23123123\ldots \).
03

Subtract to Isolate the Decimal Part

Set up an equation by subtracting \( x \) from \( 1000x \): \[1000x - x = 1241.23123123\ldots - 1.24123123\ldots\]This gives: \[999x = 1241 - 1\]which simplifies to \( 999x = 1240 \).
04

Solve for x

Divide by 999 to find \( x \):\[x = \frac{1240}{999}\]
05

Simplify the Ratio if Possible

Check if 1240 and 999 can be reduced. Calculate the greatest common divisor (GCD), which is 1 for this pair, hence \( \frac{1240}{999} \) is already in its simplest form.

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Key Concepts

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

Ratio of Integers
A repeating decimal like \(1.24\overline{123}\) can be expressed as a ratio of two integers, which means writing it as a fraction \(\frac{a}{b}\), where both \(a\) and \(b\) are whole numbers. This ratio is a way of representing the repeating decimal in a simpler, more manageable form. Here's how we do it:
  • Start by defining the repeating decimal with a variable, such as \(x\). For this example, let \(x = 1.24\overline{123}\).

  • The goal is to eliminate the repeating part by leveraging the nature of base-10 numbers 鈥� specifically, multiplying \(x\) by a power of 10 that matches the length of the repeating sequence.
By employing these steps, we convert our repeating decimal into a straightforward ratio of integers, making the number easier to understand and use. This will also help in calculations and provide a more precise representation than a lengthy decimal string.
Conversion of Decimals
Converting a repeating decimal to a fraction involves a bit of mathematical trickery. The key is manipulating the number so the repeating cycle becomes apparent. Here鈥檚 the method:
  • Identify the decimal part that repeats. In the number \(1.24\overline{123}\), the sequence \(123\) repeats endlessly.

  • Once identified, multiply the whole number by an appropriate power of ten to shift this repeating part to the right of the decimal point. For \(x = 1.24\overline{123}\), we multiply by 1000 (\(10^3\)) because the repeating block has 3 digits: \(1000x = 1241.23123123...\).
Then, subtract the original number \(x\) from our adjusted number \(1000x\). This effectively isolates the repeating decimal, making it manageable and ready for conversion into a clean fraction. It鈥檚 like keeping track of a pattern, allowing us to write repetitive numbers as neat fractions.
Simplifying Fractions
Simplifying fractions makes them easier to work with and understand. Once we have our fraction from the repeating decimal, in this case \(\frac{1240}{999}\), we check whether it can be reduced further.
  • The goal is to find the greatest common divisor (GCD) of the numerator (1240) and the denominator (999).

  • If the GCD is greater than 1, divide both terms by this number to simplify the fraction.
  • However, in this example, the GCD is 1, meaning \(\frac{1240}{999}\) is already in its simplest form since no smaller numbers can equally divide both terms.
Simplifying is crucial because it helps in easier computation and clearer understanding. In mathematics, especially with fractions derived from repeating decimals, ensuring they are reduced to their simplest form is part of maintaining clarity.

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Most popular questions from this chapter

What happens if you add a finite number of terms to a divergent series or delete a finite number of terms from a divergent series? Give reasons for your answer.

Newton's method, applied to a differentiable function \(f(x),\) begins with a starting value \(x_{0}\) and constructs from it a sequence of numbers \(\left\\{x_{n}\right\\}\) that under favorable circumstances converges to a zero of \(f .\) The recursion formula for the sequence is $$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$. a. Show that the recursion formula for \(f(x)=x^{2}-a, a>0\) can be written as \(x_{n+1}=\left(x_{n}+a / x_{n}\right) / 2\) b. Starting with \(x_{0}=1\) and \(a=3,\) calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain.

a. Suppose that \(f(x)\) is differentiable for all \(x\) in [0,1] and that \(f(0)=0 .\) Define sequence \(\left\\{a_{n}\right\\}\) by the rule \(a_{n}=n f(1 / n)\) Show that \(\lim _{n \rightarrow \infty} a_{n}=f^{\prime}(0) .\) Use the result in part (a) to find the limits of the following sequences \(\left\\{a_{n}\right\\}\) \(\mathbf{b} \cdot a_{n}=n \tan ^{-1} \frac{1}{n} \quad\) c. \(a_{n}=n\left(e^{1 / n}-1\right)\) d. \(a_{n}=n \ln \left(1+\frac{2}{n}\right)\)

Show that the error \(\left(L-s_{n}\right)\) obtained by replacing a convergent geometric series with one of its partial sums \(s_{n}\) is \(a r^{n} /(1-r)\)

Prove that a sequence \(\left\\{a_{n}\right\\}\) converges to 0 if and only if the sequence of absolute values \(\left\\{\left|a_{n}\right|\right\\}\) converges to 0 .
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