Chapter 3.3, Problem 76CCR

To determine

To make: the true statement.

Answer to Problem 76CCR

The true statement is $\frac{1}{5}<0.\overline{5}$ .

Explanation of Solution

Given information:

The statement is $\frac{1}{5}•0.\overline{5}$ .

Calculation : first convert $0.\overline{5}$ in fraction form,

  $\begin{array}{l}N=\mathrm{0.555...}\text{ [let }N\text{ represent the number] [eq}\text{. 1]}\\ \text{10}N=10\left(\mathrm{0.5555...}\right)\text{ [multiply each side by 10]}\\ \text{10}N=\mathrm{5.5555...}\text{ [multiply] [eq}\text{. 2]}\\ 10N-N=\mathrm{5.5555...}-\mathrm{0.5555...}\text{ [Subtract eq}\text{.1 from eq}\text{. 2]}\\ \text{9}N=5\text{ [subtract]}\\ N=\frac{5}{9}\text{ [divide]}\end{array}$

to make the true statement, first convert decimal into fraction then equal the denominator,

  $\begin{array}{l}\frac{1}{5}•0.\overline{5}\\ \frac{1}{5}•\frac{5}{9}\text{ [0}\text{.<mover accent="true"><mo>5</mo><mo>¯</mo></mover> =}\frac{5}{9}\text{]}\\ \frac{9}{45}•\frac{25}{45}\text{ [equal the denominator]}\end{array}$

Now, $\frac{25}{45}$ is greater than $\frac{9}{45}$ .

  $\frac{9}{45}<\frac{25}{45}$

Thus, the true statement is $\frac{1}{5}<0.\overline{5}$ .