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Chapter 6: Problem 243

Interpret the recurring decimal \(0.037037037 \cdots\) (for ever) as an infinite geometric series, and hence find its value as a fraction.

Short Answer

Expert verified
The recurring decimal \(0.037037037\cdots\) equals \(\frac{37}{999}\) as a fraction.

Step by step solution

01

Express as a Geometric Series

The recurring decimal \(0.037037037\cdots\) can be expressed as the infinite series \(0.037 + 0.000037 + 0.000000037 + \ldots\). Identify the first term \(a = 0.037\) and the common ratio \(r = 0.001\).
02

Use the Sum Formula for an Infinite Geometric Series

The sum \(S\) of an infinite geometric series with the first term \(a\) and common ratio \(r\) (where \(-1 < r < 1\)) is given by \( S = \frac{a}{1 - r} \). Substitute \(a = 0.037\) and \(r = 0.001\) into the formula.
03

Calculate the Sum

Calculate the sum using the formula: \( S = \frac{0.037}{1 - 0.001} = \frac{0.037}{0.999}\).
04

Simplify the Fraction

Simplify \( \frac{0.037}{0.999} \) by multiplying both the numerator and the denominator by 1000 to eliminate decimals: \( \frac{37}{999}\).
05

Reduce the Fraction

The fraction \(\frac{37}{999}\) is already in its simplest form because 37 is a prime number and does not divide 999 (since the sum of 999's digits (27) is not divisible by 37). Hence, the fraction is \(\frac{37}{999}\).

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Key Concepts

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

Recurring Decimals
Recurring decimals are numbers that have one or more repeating digits after the decimal point. These can go on indefinitely. A classic example is 0.333... where the digit '3' repeats forever. For the decimal 0.037037037..., the sequence "037" repeats over and over.
  • Recurring decimals are common in situations where a fraction or a number has non-terminating, repeating digits after the decimal point.
  • This happens typically because the denominator has factors of prime numbers other than 2 or 5, which leads to a repeating cycle in the decimal representation.
Understanding how to handle recurring decimals not only helps in converting them to fractions but also in recognizing patterns that can simplify computations. In an infinite geometric series context, these repeating decimals translate neatly into series with a specific common ratio, helping find their fractional equivalent.
Fraction Simplification
Fraction simplification involves reducing fractions to their simplest form. This means expressing the fraction with the smallest possible numerator and denominator. In mathematics, a few steps are typically followed for this:
  • Find the greatest common divisor (GCD) of the numerator and the denominator.
  • Divide both the numerator and the denominator by this GCD.
  • If the GCD is 1, then the fraction is already in its simplest form.
In the problem of converting 0.037037037... to its fractional form, once the fraction was found to be \(\frac{37}{999}\), simplification was straightforward as 37 is a prime number, and does not share any common factors with 999. Thus, \(\frac{37}{999}\) is the simplest form, reflecting a clear, concise representation of the recurring decimal as a fraction.
Convergent Series
A convergent series is essentially a series whose terms tend to zero, and even when added up infinitely, they reach a finite limit. The infinite geometric series used to represent recurring decimals is a common example.
  • An infinite geometric series has a first term \(a\) and a common ratio \(r\), where \(-1 < r < 1\) to ensure convergence.
  • The sum \(S\) of the series is calculated using the formula \( S = \frac{a}{1 - r} \).
  • This sum represents the finite limit the series approaches as more and more terms are added.
In our problem, converting the recurring decimal \(0.037037037\ldots\) into an infinite series allows us to efficiently find its equivalent fraction. By applying this convergence principle, an otherwise endless decimal becomes a precise, manageable number, \(\frac{37}{999}\), demonstrating the practicality of mathematical series in real-world problems.

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Most popular questions from this chapter

(Calkin-Wilf tree) The binary tree in the plane has a distinguished vertex as 'root', and is constructed inductively. The root is joined to two new vertices; and each new vertex is then joined to two further new vertices \(-\) with the construction process continuing for ever (Figure 11 ). Label the vertices of the binary tree with positive fractions as follows: \- the root is given the label \(\frac{1}{1}\) \- whenever we know the label \(\frac{i}{j}\) of a 'parent' vertex, we label its 'left descendant' as \(\frac{i}{i+j},\) and its 'right descendant' \(\frac{i+j}{j}\). (a) Prove that every positive rational \(\frac{r}{s}\) occurs once and only once as a label, and that it occurs in its lowest terms. (b) Prove that the labels are left-right symmetric in the sense that labels in corresponding left and right positions are reciprocals of each other. \(\triangle\)

When John von Neumann \((1903-1957)\) was seriously ill in hospital, a visitor tried (rather insensitively) to distract him with the following elementary mathematics problem. Have you heard the one about the two trains and the fly? Two trains are on a collision course on the same track, each travelling at \(30 \mathrm{~km} / \mathrm{h}\). A super-fly starts on Train \(A\) when the trains are 120 \(\mathrm{km}\) apart, and flies at a constant speed of \(50 \mathrm{~km} / \mathrm{h}-\) from Train \(A\) to Train \(B\), then back to Train \(A\), and so on. Eventually the two trains collide and the fly is squashed. How far did the fly travel before this sad outcome?

Problem 259 A tree is a connected graph, or network, consisting of vertices and edges, but with no cycles (or circuits). Prove that a tree with \(n\) vertices has exactly \(n-1\) edges. The next problem concerns spherical polyhedra. A spherical polyhedron is a polyhedral surface in 3-dimensions, which can be inflated to form a sphere (where we assume that the faces and edges can stretch as required). For example, a cube is a spherical polyhedron; but the surface of a picture frame is not. A spherical polyhedron has \- faces (flat 2-dimensional polygons, which can be stretched to take the form of a disc), \- edges (1-dimensional line segments, where exactly two faces meet), and \- vertices (0-dimensional points, where several faces and edges meet, in such a way that they form a single cycle around the vertex). Each face must clearly have at least 3 edges; and there must be at least three edges and three faces meeting at each vertex. If a spherical polyhedron has \(V\) vertices, \(E\) edges, and \(F\) faces, then the numbers \(V, E, F\) satisfy Euler's formula $$ V-E+F=2 $$ For example, a cube has \(V=8\) vertices, \(E=12\) edges, and \(F=6\) faces, and \(8-12+6=2\)

(Gray codes) There are \(2^{n}\) sequences of length \(n\) consisting of 0 s and 1 s. Prove that, for each \(n \geqslant 2,\) these sequences can be arranged in a cyclic list such that any two neighbouring sequences (including the last and the first) differ in exactly one coordinate position.

Prove by induction that \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}} \geqslant \sqrt{n},\) for all \(n \geqslant 1\)
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