Chapter 10.1, Problem 45E

To determine

To express: the given number in the ratio of two integers.

Answer to Problem 45E

  $1.\overline{414}=\frac{157}{111}$

Explanation of Solution

Given:

The number is,

  $1.\overline{414}=\mathrm{1.414414414.....}$

Formula used:

A geometric progression with initial term $a$ and common ratio $r$ , Then the sum of this geometric progression with $r<1$ is,

  $S=\frac{a}{1-r}$

Calculation:

As, the given number is $1.\overline{414}=\mathrm{1.414414414.....}$

So, it can be written as

  $\begin{array}{l}1.\overline{414}=\mathrm{1.414414414.....}\\ =1+1.414+1.000414+1.000000414+\mathrm{....}\\ \Rightarrow 1.\overline{414}=1+\frac{414}{{10}^3}+\frac{414}{{10}^6}+\frac{414}{{10}^9}+\mathrm{...}\end{array}$

Now, consider the series

  $0.\overline{414}=\frac{414}{{10}^3}+\frac{414}{{10}^6}+\frac{414}{{10}^9}+\mathrm{...}$

Here, it is a geometric progression with initial term $a=0.414$ and common ratio $r=0.001$

The sum of this geometric progression with $r<1$ is,

  $\begin{array}{l}S=\frac{a}{1-r}\\ =\frac{0.414}{\left(1-0.001\right)}\\ =\frac{414}{999}\\ =\frac{46}{11}\end{array}$

So,

  $\begin{array}{l}1.\overline{414}=1+0.\overline{414}\\ =1+\frac{46}{111}\\ =\frac{157}{111}\end{array}$

Hence, $1.\overline{414}=\frac{157}{111}$