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Chapter 2: Problem 66

(a) Find the fraction equivalent to each of these recurring decimals: (i) \(0.037037037 \cdots\) (ii) \(0.370370370 \cdots\) (iii) \(0.703703703 \cdots\) (b) Let \(a, b, c\) be digits \((0 \leqslant a, b, c \leqslant 9)\). (i) Write the recurring decimal "0.aaaaa \(\cdots\) " as a fraction. (ii) Write the recurring decimal "0.ababababab \(\cdots\). " as a fraction. (iii) Write the recurring decimal "0.abcabcabcabcabc \(\cdots\) " as a fraction.

Short Answer

Expert verified
(a): (i) \( \frac{37}{999} \), (ii) \( \frac{370}{999} \), (iii) \( \frac{703}{999} \). (b): (i) \( \frac{a}{9} \), (ii) \( \frac{ab}{99} \), (iii) \( \frac{abc}{999} \).

Step by step solution

01

Express Recurring Decimal as a Variable

For (a) part (i), let \( x = 0.037037037\cdots \). The digits repeat every 6 steps.
02

Remove the Recurrence

Multiply both sides by 1000 (move the decimal 3 places for one cycle) to get \( 1000x = 37.037037\cdots \). Then multiply by 1 (one full cycle) to get \( 1000000x = 37037.037037\cdots \).
03

Solve for x

Subtract the first equation from the second: \( 1000000x - 1000x = 37037.037037\cdots - 37.037037\cdots \).This simplifies to \( 999000x = 37000 \). Thus, \( x = \frac{37000}{999000} = \frac{37}{999} \).
04

Repeat for Part (a) (ii)

Let \( y = 0.370370370\cdots \). Similarly, set up equations:\( 1000y = 370.370370\cdots \) and \( 1000000y = 370370.370370\cdots \).Subtract to get:\( 999000y = 370000 \), so \( y = \frac{370000}{999000} = \frac{370}{999} \).
05

Repeat for Part (a) (iii)

Let \( z = 0.703703703\cdots \). Set up equations:\( 1000z = 703.703703\cdots \) and \( 1000000z = 703703.703703\cdots \).Subtract to get:\( 999000z = 703000 \), so \( z = \frac{703000}{999000} = \frac{703}{999} \).
06

Solve Part (b) (i)

Let \( a_n = 0.aaaaa\cdots \). Then, \( 10a_n = a.aaaaa\cdots \).Subtracting gives \( 10a_n - a_n = a \), so \( 9a_n = a \) and \( a_n = \frac{a}{9} \).
07

Solve Part (b) (ii)

Let \( b_n = 0.ababab\cdots \). Here, \( 100b_n = abab.abab\cdots \).Subtracting gives \( 100b_n - b_n = ab \), so \( 99b_n = ab \).Thus, \( b_n = \frac{ab}{99} \).
08

Solve Part (b) (iii)

Let \( c_n = 0.abcabcabc\cdots \). We need three digits, so \( 1000c_n = abcabc.abc\cdots \).Subtracting, \( 1000c_n - c_n = abc \), so \( 999c_n = abc \).Therefore, \( c_n = \frac{abc}{999} \).

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Key Concepts

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

Fractions
Fractions are an essential element of elementary mathematics, representing parts of a whole. A fraction consists of a numerator and a denominator. The numerator is the top number depicting how many parts we have, and the denominator is the bottom number indicating into how many parts the whole is divided. For instance, in the fraction \(\frac{3}{4}\), 3 is the numerator and 4 is the denominator, representing three out of four equal parts of a whole entity.
  • Fractions can be represented in different forms, such as equivalent fractions where \(\frac{1}{2}\) is equivalent to \(\frac{2}{4}\).
  • They simplify complex numbers, into easy to understand portions of wholes, which is crucial for problem-solving.
Understanding fractions enables students to explore new mathematical concepts such as decimals, percentages, and ratios with ease.
Converting Decimals to Fractions
Converting repeating decimals to fractions is a common question in mathematics, often posing a good challenge for students. Recurring decimals are decimals in which a specific set of numbers repeat indefinitely. For example, in \(0.037037037\cdots\), the digits 037 repeat.
To convert these recurring decimals into fractions, there is a systematic approach:
  • First, let the decimal be equal to a variable, such as \(x\).
  • Multiply \(x\) by an appropriate power of 10 to shift the repeating part to the left of the decimal point.
  • Set up an equation to represent the original decimal equation subtracted from the multiplied equation, which facilitates elimination of the repeating part.
  • Solve this equation to find \(x\) in fractional form.
This method efficiently handles recurring decimals, converting them into neat fractions that are simpler to interpret than their decimal counterparts.
Mathematical Problem Solving
Mathematical problem-solving involves a series of logical steps to find a solution. It is akin to a puzzle where each step builds upon the last to reach the final outcome. In the context of recurring decimals, problem-solving starts with analyzing the problem and identifying the repeating sequence. Then, algebra is applied by using multiplication and subtraction to eliminate the repeating decimal and solve for the original number in fractional form.
Problem-solving not only helps in converting decimals but also nurtures critical thinking and builds confidence in approaching other complex mathematical tasks:
  • Recognizing patterns is crucial for effectively solving mathematical problems.
  • Algebraic manipulations form a vital part of many problem-solving strategies.
  • Visualization and systematic approaches lead to clearer understanding and better retention.
Utilizing these methods ensures one鈥檚 capability to tackle real-world math problems efficiently.
Elementary Mathematics
Elementary mathematics serves as the foundational structure upon which more complex mathematical theories and problems are built. It covers basic arithmetic operations, understanding of place value, fractions, decimals, percentages, and the fundamental concept of number sense. Recurring decimals play a significant role in elementary mathematics, introducing students to more advanced operations, such as converting numbers between different formats like decimals to fractions.
  • By understanding the concept of recurring decimals and conversions, students build a strong skill set that aids in higher-level math courses.
  • Elementary math stresses the importance of logical thinking and problem-solving skills.
  • It encourages students to approach math in an intuitive and engaging manner, making complex ideas accessible and relatable.
Mastery of elementary mathematics provides a seamless transition to intermediate concepts, ensuring long-term academic success.

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Most popular questions from this chapter

(a) Evaluate the first twelve terms of the sequence of triangular numbers: $$1,1+2,1+2+3,1+2+3+4, \ldots, 1+2+3+\cdots+10+11+12$$ (b) Find and prove a formula for the \(n^{\text {th }}\) triangular number $$T_{n}=1+2+3+\cdots+n$$ (c) Which triangular numbers are also (i) powers of \(2 ?\) (ii) prime? (iii) squares? (iv) cubes?

The decimals listed here all continue forever, recurring in the expected way. Calculate: (a) \(0.55555 \cdots+0.66666 \cdots=\) (b) \(0.99999 \cdots+0.11111 \cdots=\) (c) \(1.11111 \cdots-0.22222 \cdots=\) (d) \(0.33333 \cdots \times 0.66666 \cdots=\) (e) \(1.22222 \cdots \times 0.818181 \cdots=\)

Prove than an integer written in base 11 is divisible by ten precisely when its digit-sum is divisible by ten.

Without using a calculator: (a) Work out (i) \(12345679 \times 9\) (ii) \(7 \times 9 \times 11 \times 13\). (b) Divide (i) 123123123 by 123 (ii) 111111111 by 111 (iii) 111111111 by 37 .

(a)(i) Generate the first twelve terms of the Fibonacci sequence: $$F_{0}, F_{1}, \ldots, F_{11}$$ (ii) Use this to generate the first eleven terms of the sequence of "differences" between successive Fibonacci numbers. Then generate the first ten terms of the sequence of "differences between successive differences". (iii) Find an expression for the \(m^{\text {th }}\) term of the \(k^{\text {th }}\) sequence of differences. (b)(i) Generate the first twelve terms of the sequence of powers of 2 : $$2^{0}, 2^{1}, 2^{2}, \ldots, 2^{11}$$ (ii) Use this to generate the first eleven terms of the sequence of "differences" between successive powers of \(2 .\) Then generate the first ten terms of the sequence of "differences between successive differences". (iii) Find an expression for the \(m^{\text {th }}\) term of the \(k^{\text {th }}\) sequence of differences. (b)(i) Generate the first twelve terms of the sequence of powers of 2 : $$2^{0}, 2^{1}, 2^{2}, \ldots, 2^{11}$$ (ii) Use this to generate the first eleven terms of the sequence of "differences" between successive powers of 2 . Then generate the first ten terms of the sequence of "differences between successive differences". (iii) Find an expression for the \(m^{\text {th }}\) term of the \(k^{\text {th }}\) sequence of differences.
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